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O}
(1.3) (1.4) (1.5)
= {z : I(x) > OJ.
Notation 1.1.6 Throughout this book, we will use B(x, r), r > 0, to den.ote the open ball about x with radius r. We will also use iii(x, r) r ~ O,to denote the closed ball about x with radius r. Proof: We use the characterization that the function I is lower-semicontinuous if and only if {x : lex) > o} is open for every a e III By redefining I to equal 1 on {x: I(x) > 1}, we may and shall assume that / takes only finite values. E~hausting {x: I(x) > O} by compact subsets, we can find a sequence of points Plo P2, ... and positive reals U1, U2, ••• such that 00
{x: f(x) > O} =
U lR(Pi,Ui). i=l
For each i = 1,2, ... , define the function
Noting that
1,6[-1,1)
= ~ l(Pi) 1,6[-1,1) (Ix -
Pil 2 fun·
is everywhere less than 1, we see that each tPi satisfies
0'5: tPi :'5: I· For each i = 1,2, ... , set ki
{I I:
= sup ':~i a is a multi-index with lal '5:
Then the series 00
i} .
2-;
L 1 +k. tPi i=l
•
converges unifonnly, as do all its term-by-term derivatives of any order. The function defined by the series has the desired properties. • Proof of the Coo Varieties Theorem 1.1.4: Let V be a closed subset of lllN. We apply Lemma 1.1.5 with I = Xcv, the characteristic function of the complement of V. The resulting function
o. Choose a continuous function 9 with compact support such that 11/- gll£1 < £2. Set T.
=
{z E
]RN :
I
lim sup .eN r-+O+
-lim inf .eN T-+O+
(lB~z, r »JB(z,r) r I(t) dt
(lB~Z, r » lB(z,r) f I(t) dtl > E}.
We wish to estimate the measure of T •. Now .eN(T.)
~
.eN
+
.eN ({ x E
({x
liminf r-+O+
e]RN :
lim sup r-+O+
.eN(~ Z, r »JB(z,r) r I/(t) -
]RN : llim sup .eN (lB~ r-+O+
z, r
»
.eN(lB~ :1:, r » JB(z,r) r g(t)dtl >
r
JB(z,r) E/
({x e]RN : liminf .eN(~ r » r
+
.eN
-
AI +A2 +A3.
r-+O+
:1:,
g(t)ldt > £/3})
get) dt
3})
JB(z,r)
Ig(t) -/(t)ldt > E/3})
It is plain that A2 = 0 since 9 is continuous. We estimate Al and A3 in
89
3.5. COVERING LEMMAS just the same way, so we shall. concentrate on the former. Now
Al
({z e ~ : ~p £N(~z, r» k."r) e~ > 3})
~
£N
=
£N({z
~
C E/3 111 - gllL1
<
-·E
=
C·E.
C
:M(f -g)
I/(t) - g(t)1 dt
> E/3})
E/
2
E/3
It foll..ows from this last estimate that Al = 0 (if instead Al = ,\ > 0 then take E = '\/(2C) to derive a contradiction). By similar reasoning, As = o. Thus, for E sufficiently small, £N (T.) = o. This gives the result that the desired limit exists. That the limit agrees with the original function (representative) I almost everywhere follows from a similar argument that we leave to the reader. • Corollary 3.5.6 If A C lItN is Lebesgue measurable, then, for almost every
z·e lItN, it holds that . X () .A
Proof: Set
I
x
r
= r.!.If:+
£N(Anlll(z,r» £N(lII(z,r»·
= XA. l'hen (
f(t) dt = &.N(A n lII(x,
JlI(z,r)
r»
and the corollary follows from Proposition 3.5.4..
•
Definition 3.5.7 A function I : lItN -+ lIt is said to be approximately continuous if, for almost every Xo e ]iN and lor each f > 0, the set {x: I/(x) - l(xo)1 > f} has density 0 at Xo, that is,
0- lim .cN({x:l/(x)-/(xo)l>f}nlll(xo,r» - r-+O+ £N(lII(xo,r» . Corollary 3.5.8 If a function it is approximately continuous.
I : ]iN -+
]i
is Lebesgue measurable, then
Proof: Suppose that I is Lebesgue measurable. Let ql, ~, ... be an enumeration of the rational numbers. For each positive integer i, let Ei be the set of points x ~ {z : I(z) < qi} for which
o
lim
<
r-+:!P
.cN({z:/(z)
CHAPTER 3. MEASURES
90 and let Ei be the set of points x ¢ {z : qi O
·
< 1lDlSUP r-+O+
< I(z)} for which
C.N({z:q.(z)}nB(x,r» i'N(B(}} . 'x,r
By CoroUary 3.5.6 and the Lebesgue measurability of I, we know that =0 and c.N (Ei) = O. Thus we see that
c,N (&)
00
E
= U(Ei UEi} i=1
is also a set of Lebesgue measure zero. Consider any point Xo ¢ E and any qi and q; such that
I(xo} -
f
f
> O. There exist rational numbers
< qi < I(xo} < q; < I(xo} + f.
We have {x : I/(x} - l(xo}1 > f} C {z : I(z} the definition of Ei and Ej we have
and 0=" lim r-+O+
< qi} U {z : q; < J(z)}. By
c. N ({z : qj < I(z)} n B(xo, r». C,N(B(xo,r»
It follows that
0= lim C,N({x:I/(x}-/(xo}l>f}nB(xo,r»." r-+O+ C,N(B(xo,r» Since Xo ¢ E and f > 0 were arbitrary, we conclude that J is approximately continuous. • Remark 3.5.9 The converses of Corollaries 3.5.6 and 3.5.8 are also true, but there is more technical difficulty involved when one deals with a set A or a function J that may not be measurable. The interested reader should see Federer [41, §2.9.12 and §2.9.13. The Besicovitch Covering Theorem We next turn to Besicovitch's covering theorem because its proof is similar to that of Wiener's theorem. It is a purely geometric lemma, as it speaks to how balls may be packed in space.
3.5. COVERING LEMMAS
91
Theorem 3.5.10 (Besicovitch) Let N be a positive integer. There is a constant K = K(N) wi~.~ following property. Let 8 = {Bj}~l be any finite collection 0/ open balls in RN with the property that no ball contains· the center 0/ any other. 77len we may write
so that each 8j, j
= 1, ... ,K, is a collection o/pairwise disjoint balls.
In general it is not known what the best possible K is for any given dimension N. Interesting progress on this problem has been made in Sullivan [1]. Certainly our proof will give little indication of the best K. We shall see that the heart of this theorem is the following lemma about balls. We shall give two different proofs of this lemma. One, contrary to our avowed philosophy in the present subsection, will in fact depend on measure-or at least on the notion of volume. The second proof will rely instead on trigonometry. . Lemma 3.5.11 There is a constant K = K(N), depending only on the dimension 0/ our space aN, with the following property: Let Bo = B(xo, TO) be a ball 0/ fixed radius. Let Bl ~ B(Xl,·Tl),B2 = B(X2,T2), .•. ,Bp = B(xp, Tp) be balls such that
(i) Each B j has non-empty inters·ection with B o, j (ii) The radii Tj
~
= 1, ... ,Pi
TO lor all j = 1, ... ,Pi
(iii) No ball Bj contains the center 0/ any other B" lor k
¥-
j, j,k
=
O, .•• ,p.
Thenp
~
K.
Here is what the lemma says in simple terms: fix the ball Bo. Then at most K pairwise disjoint balls of (at least) the same size can touch Bo. Note here that being 'pairwise disjoint' and 'intersecting but not containing the center of any other ball' are essentially equivalent: if the balls intersect, but none contains the center of another, then shrinking each ball by a factor of one half makes the balls pairwise disjoint; if the balls are already pairwise disjoint, have equal radii, and are close together, then doubling their size arranges for them to intersect without any ball containing the center of another. First Proof of the Lemma: The purpose of providing this particular proof, even though it relies on the concept of volume, is that it is quick and intuitive. The second proof is less intuitive, but it introduces the important idea of 'directionally limited.' Since any ball Bit j = 1,2, ... ,p, for which Tj > TO can be replaced by the ball of radius TO internally tangent to B j at the point of intersection
CHAPTER 3. MEASURES
:u.
Figure 3.4: Replacing a Larger BaIl of the segment connecting the center of Bj and Bo while maintaining the validity of all three conditions (i)-(iii) (see Figure 3.4), it will suffice to prove the lemma with all balls having radius roo We now assume that all balls have the same radius. With the balls as given, ·the smaller balls Ja(Xl' ro/2), Ja(X2' ro/2), ... , JR(xp, ro/2) are pairwise diSjoint and are all contained in JR(xo,3ro). We calculate that p
eN(u'J=lBj) (ro/2)NYN eN (JR(xo, 3ro» (ro/2)NYN
As a result of this calculation, we see that K(N) exists and does not exceed 6N • • Second Proof of the Lemma: For this argument, see Krantz and Parsons [1]. In fact we shall prove the following more technical statement
Let the universe be the two dimensional plane, ]R2, and let ~ = {reill : 0 $ r < 00,0::; (}::; 1r/6}. Set S = {z E ~: Izl ~ 3}. H a,b E S and if each of the balls lR(a, r), lR(b, s) intersects 111(0, 1), then
la - bl < max(r, 8).
(3.60)
93
3.5. COVERING LEMMAS
A moment's thought reveals that this yields the desired sparseness condition in dimension two. The N-dimensional result is obtained by slicing with two dimensional planes. To prove (3.60), we first note the inequalities (0
_1)2 - (2 - va)a2 ~ 0
if 0
~
({3 - 1)2 - (02 - ..r3a{3 + {32) ~ 0
3;
(3.61) (3.62)
The first of these is proved by noting that the derivative of the left side of (3.61), in the variable 0, is positive when 0 ~ 3; and the inequality is satisfied when 0 = 3. So the result follows from the fundamental theorem of calculus. Similarly, the derivative of the left side of (3.62), in the variable {3, is positive when {3 ~ 0 ~ 3, and the case {3 a 3 is just inequality (3.61), which has already been established. With these inequalities in hand, we introduce polar coordinates in the plane, writing a = oei8 and b = {3e i
= =
(3.63) The law of cosines tells us that
la - bl 2 = 0:2 -
2a{3cos(¢> - 9)
+ (P.
(3.64)
Since cos(¢> - 9) ~ cos7r/6 = .../3/2, it follows that the right side of (3.64) does not exceed a2 - v'3ab + b2 • Tlie inequality (3.63) now follows from • (3.62). H. Federer's concept of a directionally limited metric space-see Federer [4]-formalizes the geometry that goes into the proof of our last lemma. More precisely, it generalizes to abstract contexts the notion that a cone in a given direction can contain only a certain number of points with distance T/ > 0 from the vertex and distance T/ from each other. The interested reader is advised to study that primary source. Now we can present the proof of Besicovitch's covering lemma: Proof of Theorem 3.5.10: Begin as in the proof of the Wiener covering lemma. Select BI to be a ball of maximum radius. Then select B~ to be a ball of maximum radius that is disjoint from Bl. Continue until this selection procedure is no longer possible (remember that there are only finitely many balls in total). Set B1 {BJ}. Now work with the remaining balls. Let B? be the ball with greatest ra.dius. Then select B~ to be the remaining ball with greatest radius, disjoint
=
94
CHAPTER 3. MEASURES
from
B2
B1.
Continue in this fashion until no further selection is possible. Set
= {Bj}.
Working with the remaining balls, we now produce the family Ba, and so forth. Clearly, since in total there are only finitely many balls, this procedure must stop. We will have produced finitely many-say p-nonempty families of pairwise disjoint balls, Bl"'" 8". It remains to say how large p can be. Suppose that p > K(N) + I, where K(N) is as in the lemma. Let Bf be the first ball in the family B". That ball must have intersected a ball in each of the preceding families; by our selection procedure, each of those balls must have been at least as large in radius as Bf. Thus Bf is an open ball with at least K(N) + 1 "neighbors" as in the lemma. But the lemma says that a ball can only have K(N) neighbors. That is a contradiction. • We conclude that p $ K(N) + 1. That proves the theorem. Using the Besicovitcb covering theorem, we now can extend the earlier results on the maximal function and differentiation to Radon measures other than the Lebesgue measure. Definition 3.5.12 Let p be a Radon measure on ]RN and let I be a locally p-integrable function on ]RN. We define the" Inaximal function with respect to p, denoted M,./, by setting M,./(x) = 0 if there exists 0 < r < 00 with p(lII(x, = 0, and
r»
M,.I(x)
1
=O
B(."r)
lI(t)1 dpt
otherwise." Lemma 3.5.13 There is a constant C > O-depending on the dimensilm N and the Radon measure 1', but not on the particular function I -such
that the operator M satisfies p{x
N
E]R
:
M,.I(x) > A} $
C -:xIl/IlL!(RN)
lor any scalar A > O. The constant C is independent 01 A and 01 I. Proof: The same proof may be used for Lemma 3.5.13 as was used for Lemma 3.5.3, except that the Besicovitch covering theorem replaces the Wiener covering lemma and a little more care must be exercised in constructing the initial cover of K so as to arrange that no ball contain the center of any other ball. • As a corollary, we get the following more general differentiation theorem: Proposition 3.5.14 Let p be a Radon measure on]RN and let I be a locally IJ-integrable function on ]RN. Then for p-almost every x E ]RN it holds that lim r-+O+
1 p(B(x,
r»
( JIl(."r)
I(t) dpt
3.5. COVERJNG LEMMAS
95
exists and equals /(x). Lebesgue's Covering Theorem The covering theorems presented thus far can be adapted to a variety of situations (besides Euclidean space); notable among these is the setting of spaces of homogeneous type (see Christ [1], Coifman and Weiss [1], [2], Krantz [5]). The next theorem is an integral part of dimension theory (for which see Hurewicz and Wallman [I]), which is usually studied on a separable metric space. The theorem has important geometric content, and is of wide utility. Note that if U = {U.. } ..EA is a covering of a set S by open sets then a refinement V = {V.s}.sEB of U is an open covering of S with the property that each V.s lies in some U... Theorem 3.5.15 Let S c JRN be any set. LetU = {U.. } .. EA be a covering 0/ S by open sets. Then there is a refinement V = {V.s}.sEB 0/ U that has valence at most N + 1. That is, each point of S is an element 0/ at most N + 1 0/ the V.s. The proof will proceed in several steps. We will introduce a few simple concepts from dimension theory. First, the empty set is declared to have dimension "-1. It is the only set with dimension -1. A point P in a set S is said to have dimension $ N if P has a neighborhood basis {Uj } with the property that aUj n S has dimension $ (N -1). A set is said to have dimension N if it has dimension $ N but not $ (N - 1). A moment's thought reveals that.th~ set of rational numbers, lying in lR, has dimension O. So does the set of irrational numbers. For the first of these, simply note that if P is rational then a ball around P with irrational radius will have boundary possessing intersection with Q that is empty (hence has dimension -1). It will be useful for us to note that if 1 $ m $ N then the set T m of points (Xli X2, ••• , XN) in l\lN with Xli X2, • •• , Xm rational has dimension O. Thus we see that
where S is the set of points in JRN with all irrational coordinates. And each of these sets has dimension zero. Thus we have decomposed JRN into (N + 1) sets of dimension O. While it is certainly true that a discrete set is of dimension 0, the preceding examples show that the converse is not true. Nevertheless, sets of dimension 0 share certain properties with discrete sets. The one of greatest interest for us is contained in the following lemmas. Before beginning, we note that a set C C JRN is said to separate two disjoint subsets AI, A2 of RN if JRN \ C = UI U U2 with UloU2 being relatively open in JRN \ C,
CHAPTER 3. MEASURES
96
disjoint, and Aj C Uj for j = 1,2. As an exercise, the reader may wish to verify that if a set S C lllN is zero dimensional then any two disjoint relatively closed sets in S may be separated by some set 0 c S (or see stateIqent (E), page 15, in Hurewicz and Wallmah [1]). Lemma 3.5.16 1/ E1> ~ are closed, disjoint subsets 0/ RN and if B is a zero dimensional subset 0/ aN, then there is a closed set C in aN such that (i) C separates El and ~ and (ii) B n C = 0. Proof: There certainly exist open sets WI, W2 such that Ej C Wj, j = 1,2, and WIn W 2 = 0. [This is just the property of normality for JRN.J Now WIn B and W 2 n B are disjoint and closed in B. Since B has dimension 0, these two sets can be separated in B (by the remark preceding the statement of this lemma). We conclude that there are disjoint sets U{, U~ satisfying B=U;UU~
with, W j nBc Ui, j =.1,2, and U{, U~ both closed and open in B: Now (U; nW 2 )u (U~
n WI)
= 0,
(3.65)
(Ui n U'2) u (U~ n U'l) = 0,
(3.66)
and therefore (U;
nE2 ) u (U~ n Ed = 0.
(3.67)
But WlJ W2 are open, so that (3.65) implies that (U'l n W 2 ) U (U'e- n = 0. As a result, by the first sentence of the proof, we conclude that (U'l n~) u (U'2 n E l ) = 0. Equations (3.65}-(3.67) allow us now to conclude, since El n ~ = 0, that neither of the disjoint sets El U U{ and E2 U U~ contains a cluster point of the other. Since every subspace of lllN is normal, we find that there is an open set 0 such that
Wd
and on(~ U U~)
In conclusion, the boundary C disjoint from U{ U U~ = B.
=0
= 0.
\ 0 separates El and
~
and is •
Lemma 3.5.17 Let S be a zero dimensional subset o/JRN . Let U = {Ul, U2} be open subsets o/lllN that cover S. Then there is a refinement V = {Vt, V2} ofU that still collers S and such that Vt n V2 = 0.
3.5. COVERING LEMMAS
97
Proof: It is convenient to replace the universe RN with the set X = U1 UU2 and we do so without further comment. Thus D1 = X \ U2 and D2 = X \ U1 are closed, disjoint sets. Since S has dimension 0, we may apply the preceding lemma to get a closed set C that is disjoint from S and separates D1 from D 2 • Thus there exist sets Wh W2 which are open, disjoint, and satisfy Wj ::> Djo j = 1,2, and X\C = W1 UW2 • As a consequence, Wj C Uj, j = 1,2. Also SnC = 0 and W1 UW2 ::> S. This is the desired conclusion. • LeDlDla 3.5.18 Let S be a zero dimensional "Subset of aN. Let U = {Uj}~=r1" be open sets that colier S. Then there exists a refinement V = {V; }~~1 of U that still covers S and such that V; n V" = 0 when j :f: k. Proof: We proceed by induction on p. Note that if p = 1 then the statement is trivial. Now assume that the result has been established for p = k - 1; we use that hypothesis to prove the result for p = k. Thus we are given a covering U =" {Uj of a set sCaN. Let Uk_1 == U"-l U U". Then the collection U' given by
H=l
is still an open covering of S; of course it has (k-l) elements. The inductive" hypothesis applies to the covering U' of S, and we obtain a refinement
of U' that still covers S and is such that the elements of V' are pairwise disjoint. Observe that the set T == snvk_1 has dimension not exceeding 0 (since S does). Notice also that U"-l and U" cover T. By the last lemma, there exist open sets V"-l> V" that refine the cover U,,-l> U" of T and which are disjoint. But then form an open covering of S with all the desired properties.
•
The proof of the Lebesgue covering theorem is now easy: Proof of Theorem 3.5.15: Begin by writing S = Sl US2 U···USN+1,
where each Sj has dimension at most O. Since our open covering certainly covers each Sj, we may apply the last lemma to obtain (N + 1) refinements
CHAPTER 3. MEASURES
98 VI, ... , VN+l of U such that
Vi = {Vii, ... , V;(i)};
(3.68)
Vi covers Si;
(3.69)
V; n Vt' = 0
if 8
I t.
(3.70)
Vl
Now let V be the covering consisting of all the open sets for all possible values of i and j. This covering certainly covers S, for it covers each of the Si. We claim that V has valence at most N + 1. For note that any selection of N + 2 elements of V must contain two elements from a single Vi which are therefore disjoint. So it is impossible for N + 2 elements of V to have a point in common. • It turns out that the property of any open covering having a refinement of valence at most N + 1 characterizes sets of dimension N. Since this is not a study of dimension theory, it would take us far afield to provide all the details of this assertion. We refer the interested reader to Hurewicz and Wallman [1] .
. The Vitali Covering Theorem The Vitali covering theorem is probably the most broadly used covering theorem in all of measure theory. It has the dual advantage of being both intuitively appealing and rather profound. Indeed it says something very important about the structure of sets in Euclidean space. We derive our treatment of Vitali's ideas from Evans and Gariepy [l]i the interested reader may consult that source for further details. We begin with a modification of the Wiener covering theorem: Proposition 3.5.19 Let U be a collection of closed balls in RN such that
sup{diamB: B E U} = M Then there is a countable, disjoint subfamily V
UBe BeU
< 00.
eU
such that
USB. BeV
[Here, as usual the notation 5B denotes the ball with the same center as B but 5 times the radius.] The proof of this proposition is left to the reader. Now the principal theorem of Vitali type that we wish to consider is this:
3.5. COVERING LEMMAS
99
Theorem 3.5.20 (Vitali) Let U c aN be an open set. Let 6 > O. Then there e:£ists a countable collection W = {Bj}~l 0/ disjoint closed balls, each lying in U, such that . ". .
(i) diamBj (li)
~
6/or each j;
CN[U\U~l~;] =0.
The gist of the theorem is that one may "fill up" an arbitrary open set U with small balls, to the extent that the portion not covered has measure zero. Of course it is too much to hope that the entire open set can be exhausted by balls. For example, if U is the interior of a unit square in the plane, then it is easy to see that there is no way to fill U completely with countably many discs of radius not exceeding 1/4. In fact, one can see in this example that the set of points not covered will be uncountable and nowhere dense. Proof: Assume for convenience that U has finite Lebesgue measure. Some remarks about the general case will be given at the end. Fix a number A such that 1 - 1/5N < A < 1, where N is the dimension. . We will construct the required baIls by an iterative procedure. Each step of our process will exhaust a proportion (1 - 9) of that portion of U that is not already covered by balls. For the first step, set U1 = U and let
(h = {B : B is a closed ball lying in Ul and having diameter less than 6}. By the preceding proposition, there iLa countable, disjoint family 'HI C 91 'with the property that Ul C 5B.
U
BE1l1
As a result, we may calculate that CN(Ul)
~
L
£N(5B)
BE1l1
=
5N
L
£N(B)
BE1l1
=
5N
£NCU
B).
BE1l1
We conclude that CN(_U B) BE1l1
~ 5~£N(Ul).
100
CHAPTER 3. MEASURES
Since each element of 1i1 is a subset of Vb we see that
Of course the collection of balls 1i 1 is countable; thus there is a subcollection B I .B 2 , •••• BK, satisfying f:.N (VI \
0
Bi )
~ Am(V.).
)=1
For the second step. set K,
V2
= VI \
U Bi i=1
and let
{h
= {B : B is a closed ball lying in V2 and having diameter less than 6}.
Repeating the arguments from Step 1, we find a collection
BK,+l,
BK, +2,
... , B K, of disjoint elements of ge such that
=
f:.N (V2 \
U i) B
j=M,+1
~ ~
Ae N (V2 ) >.2 f:.N (U).
We continue this process countably many times. The result is a pairwise disjoint collection of closed balls B j, all lying in U, all having diameter less than fJ, such that f:.N (V
\Q Bj) ~
>.p eN (V),
for p = 1.2, .... Since 0 < A < 1, our conclusion follows provided that f:.N (U) < 00. In case f:.N (U) = 00. write V = (UlUl) U T, where each Ut is open with finite measure and T has measure zero. •
3.6
Functions of Bounded Variation
Functions of bounded variation from R to IR are well understood (see Federer (4], §§2.5.16-2.5.18). The situation for functions from aN to R is
101
3.6. FUNCTIONS OF BOUNDED VARIATION
less settled and there would seem to be opportunities for future clarification. Our main references for this section are Evans and Gariepy [1] and Ziemer [1]. The following example motivates the definition of functions of bounded variation.
Example 3.6.1 Suppose I E C'2(n), where n is an open subset of IRN . For any 9 = (gl.!h, ... ,gN) E c:(n;IRN ), with Igi ~ 1 for all x E n, we can use integration by parts to see that
L
Idiv(g)dx
=-
L
(grad/)· gdx
~
LI
grad II dx.
=
Now, set n' n n {x : grad/(x) ¢ O}. If we let K be an arbitrary compact subset of n', then we can choose a non-negative Coo function ¢J with 0 ~ t/J ~ 1, suppl en', and with tf>(x) 1 for x E K. Setting
=
9
= -¢Jgrad f
Igrad 11- 1 ,
we have 9 E cJ(n; KN), with Ig(x)1 ~ 1 for all x En, and in this case, we have
L
I div(g) dx
=
L
tf> Igrad II dx
~
L
Igrad II dx.
Since Ken' was an arbitrary compact set, we conclude that
L
Igrad II dx SU P
=
{L /
19(x)l~IVXEn}.
N diV(9)dx: 9 EC:cn;IR ),
(3.71)
The reader should note that with more effort one can see that (3.71) also holds for / E c 1 (n). Definition 3.6.2 Let n C IRN be open. A function be 01 bounded variation on i/
n
I
E Ll (n) is said to
LID/I:= sup
{fo
Idiv(g)dx: 9 E c:
The space 0/ functions and will be normed by
),
Ig(x)l
0/ bounded variation on
II/liB\'
= II/IILI +
L
~ 1 "Ix En}
<
00.
(3.72)
n will be denoted BV(n)
ID/I·
(3.73)
Basic, but important, properties of BV functions follow easily from the definition.
102
CHAPTER 3. MEASURES
Proposition 3.6.3
III e BV(U) and 9 e
: IL
ldiv(g) dzl
C~(U;lRN), then
~~plgl
k
(3.74)
IDFI .
•
Proof: Immediate from the definition. Proposition 3.6.4 III e BV(U) and u both bounded, then ul e BV(U) with
e
a1 (n)
with lui and Igrad ul
f ID(u/)ISsuplul f IDII+suplgradul f I/ldz. 10 0 10 0 10
(3.75)
Proof: Use the identity udiv(g)
= div(ug) -
(grad u· g)
•
and (3.72).
Theorem 3.6.5 (Semicontinuity) Suppose U C RN is open. 11ft, 12, ... is a sequence in BV(U) that converges in Ltoc(U) to a function I, then
f IDIsI. 10f ID/I S li~inf .o~oo 10
(3.76)
Proof: Let 9 E a~ (Uj RN) with I!!I S 1 be arbitrary. Then for any i we have .lim f Ii div(g)dz S l~infIDIsI. 10f /div(g)dz = 0-+00 10 o~oo The result follows by taking the supremum over all such g. • Corollary 3.6.6 (Completeness) The space 01 functions 01 bounded variation with the nonn given in (3.73) is a Banach space. Proof: The reader may verify that (3.73) does in fact define a norm. We will verify completeness. Suppose 11,12, ... is a Cauchy sequence in the norm II·IIBV. A lortiori, 11,12, ... is a Cauchy sequence in the Ll norm, and thus converges in the Ll norm to a function I e LI(n). We know that Is -Ii converges to Is - / in Ll(U) as i -+ 00, so by the Semicontinuity Theorem 3.6.5 we have lim sup
i~oo
f
10
ID(1s - 1)1 S S
lim sup (liptinf ( ID(fi i~oo 3~OO 10 limsup
i.i~oo
f
10
ID(fi -
1;)1 = 0,
1;)1)
3.6. FUNCTIONS OF BOUNDED VARIATION
103
•
from which the result follows.
The next theorem shows that BV functions can also be characterized as those functions whose partial derivatives are measures. TheorelD 3.6.7 Let 0 c JlN be open. 1// E BV(O) and 1 ~ i ~ N, then there e:dsts a signed Borel measure I'i 0/ finite total variation on 0 such that (3.77)
lor each tP E C~(O). Conversely, i/ there are signed Borel meosure8 1'10 1'2, ... ,I'N such that (3.77) holds/or each 1 ~ i ~ Nand tP E ~(O), then / E "BV(O). Remark 3.6.8" Note that the left-hand side of (3.77) represents the action on the test function tP of the partial derivative of / in the sense of distributions. Proof: Suppose that / E BV(O). The left-hand side of (3.77) defines a linear operator Li on C~(O). By (3.74), we see that L is bounded in the supremum norm. The lliesz representation theorem (see Rudin [2], Theorem 6.19) then insures the existence of a signed measure such that (3.78)
tP E C~(O). (Note that since Li is bounded in the supremum norm, it ~en~~~~J " To see the converse, simply observe that for (0; "liN ) with Igl ~ 1,
for
gee:
we have
1
/div(g)dx
{}
N
=- L i=1
L -
gidpi $
{}
N
L Il'il(O). i=1
•
Definition 3.6.9 II IE BV(O), then DI is used to denote the derivative in the sense 01 distributions. Corollary 3.6.10 II/ E BV(O), then there exists a Radon measure I' on
o and a I'-measurable function u, with lu(x)1 = l/or I' almost every x E 0, such that D/=up.
An important point to make in reference to Theorem 3.6.7 is that the measures representing the partial derivatives of a function of bounded variation typically have a non-trivial singular part with respect to Lebesgue measure. Whenever the singular part with respect ~ Lebesgue measure is non-zero, then the function in question cannot be in any Sobolev space (Sobolev spaces will be treated in some detail in the next chapter). The next example illustrates this phenomenon.
104
CHAPTER 3. MEASURES
Example 3.6.11 Set
J = Xi(O.I)' the characteristic function of the closed unit ball in JltN. By the divergence theorem2 , for 9 e C: (aN , R,N) we have
{ li(o.l)
div(g)dx =
(
g. vd'H. N - 1 ,
ls(o.l)
where v is the outward unit normal to the unit sphere. Comparing this equation with (3.77), we ~ that DJ = _v1l N - 1 LS(O, 1), the measure 1l N - 1 LS(O, 1) is singular with respect to Lebesgue measure. The closure of the space of Coo functions in the BV norm is the Sobolev space ·Hl.l of functions whose partial derivatives in the sense of distributions are represented by Ll functions. Thus in general it is impossible to approximate a BV function in the BV norm by a smooth function: On the other hand, the next theorem gives us some positive information on smooth approximations to BV functions. Theorem 3.6.12 Let J e BV({}). Then there exists a sequence of functions It, 12, ... in Coo({}) such that
(ii)
In IDhl-t In IDfl,
(iii) Dh -" Df Proof: Let '1/;" be a family of mollifiers as in Section 1.3. While it is not essential, it is convenient to suppose that '1/;" satisfies the symmetry (evenness) condition (3.79) ¢,(-x) = 'I/;,,(x). In any case, one can always arrange for (3.79) to hold by replacing 'I/;,,(x) by H¢,,(x) + '1/;,,( -x». Technical difficulties are minimized if we assume that {} = JltN, so we shall first make that assumption. 2We use the tenns Gauss-Green theorem and divergence theorem interchangeably. Both are multidimensional versions of the Fundamental Theorem of Calculus (see Section 5.5)
3.6. FUNCTIONS OF BOUNDED VARIATION
105
Fix I E BV(IllN). By Theorem 1.3.31, we know that 1* tPa -+ I in Ll(IllN), proving (i). Suppose 9 E C~(RN;aN). Then using Fubini's Theorem and the theorem on differentiating under the integral sign, compute
we
I (J * tPa)(z) div(g)(z) d:i: JRN = kN (kNI(Y)tPa(Z-Y)dY)
div(g)(z)dx
=
kNI(Y) (kN tPa(Z-Y)diV(9)(Z)dx) dy
=
kNI(Y) (kNtPa(-Z)diV(9)(Y-Z)dx) dy
=
kNI(Y)diV (kN 1Pa(-z)g(y-z)dx) dy
=
I I(Y) div(g* 1Pa)(y)dy; 1RN
conclusion (iii) now follows, because Theorem 1.3.31 tells us that g*tPa converges to 9 in the supremum norm as (J' -+ O. Also note that I(g * tPa) (z) I $ 1 . holds for all z, showing that
But of course Theorem 3.6.5 implies that
so (ii) also follows. To handle the general case, that I E BV(O) when 0 =F lllN, we must show how to construct a function U E COO(O) with
where e > 0 is given. To this end, choose an open 0 0 CC 0 with
I
10\00
ID/I $
-3t
and construct a sequence of open subsets of 0 with 0 0 CC 0 1 CC ... C 0 and U~OOi = O. Next, introduce a partition of unity .pi, i = 0, 1, ... , such that
1/>0
= 1 on 00,
supp (1/>0) C 010
106
CHAPTER 3. MEASURES
and in general, supp (tPi) c niH \Oi-1o i ~ 1. Of course, we have L:o tPi == Ion n. For each i = 0, 1, ... , we choose tTi > 0 so that with
=
Ui (ftPi) * ""tTl'· the fOllowing conditions hold: supp (Ui)
L
L
= (J gradtPi) * tPtTl'
supp (Uo) c nlo niH \ Oi-l, i ~ 1,
(3.80) (3.81)
ItPil clz < 2-(iH) ~,
(3.82)
< 2-(iH) ~.
(3.83)
c
lUi -
Vi
IVi - fgradtPil clz 00
u= LUi, i=O
and notice that U E COO(n) because each point of n is only in the support of finitely many Ui. The estimate of the Ll norm of the difference between f and U is simply
{ Iu - II clz ~
10
f: 10( IUtPi) * .,pa, - frPildz
The more interesting estimate bounds the quantity 10 IDul." Using the fact that 00
o = grad(E:o tPi) =
L grad 4>i, i=O
we compute Du
= =
00
00
i=O
i=O
:Eu grad tPi) * ""a, + L(4)i D f) * tPa, 00
00
i=O
i=O
:E[U grad tPi) * .,pa, - f grad4>d + L(4)i D f) * .,pa,.
Arguing with Fubini's Theorem as in the case n {1(tPiDf)*""a,IS{
10
_
=]RN, we see that
101+1 \0'_1
IDflj
since each x E n belongs to at most three of the sets able to estimate
ni+l \ Oi-1o we are
LIDul ~
•
107
3.7. DOMAINS WITH FINITE PERIMETER
The final result we will present in this section concerns compactness in the space of functions of bounded variation. -The compactness result is a consequence of the smooth approximation theorem just proved and Rellich's theorem which we will state without proof. A general treatment of Rellich's Theorem can be found in Ziemer [IJ, Section 2.5. Decreasing the regularity of 80 increases the difficulty of proving the result. The relevant version of the theorem with nearly the weakest possible regularity' assumption for 80 is the following: Theorem 3.0.13 (Rellich) Suppose n c lllN is a bounded open set the boundar7J of which can locally be represented as the graph of a Lipschitz function. If h, /2, ... is a sequence of functions in Ll (0) whose first partial derivatives in the sense of distributions are also in Ll(O), and if there is M < 00 such that
f
leI"
Ifil dx +
10f IDII dx ~ M
for i = 1,2, ... ,
then a subsequ~nce h; converges to a function f E Ll(O). Corollary 3.6.14 (Compactness) If O.C lllN is a bounded open set the boundar7J of which can locally be represented as the graph 0/ a Lipschitz function, then any set 0/ functions bounded in· the BV nonn has a subsequence that converges in Ll(O) to a function sn BV(O). . Proof: By the Approximation Theorem 3.6.12 we can choose functions hI. h 2 , ••• in COCCO) with
10f IIi -
hil dx
~~
and
z.
10f IDhil ~ M + 1.
Then Rellich's Theorem allows us to extract a subsequence hi; that converges in the Ll norm to fELl (0). A fortiori we know that h; converges to / in the Ll norm and we have f E BV(O) by the Semicontinuity Theorem 3.6.5.
•
3.7
Domains with Finite Perimeter
Definition 3.7.1 (i) If S is a Borel set and 0 C lllN is open, then the perimeter of S in o is denoted by peS, 0) and is defined by P(S, 0)
=
L
=
Sup{!s div(g) dCN : 9 E C:(O; lllN),
IDxsl Igl
~ I}, (3.84)
108
CHAPTER 3. MEASURES where Xs denotes the chamcteristic function
(ii) We say
S
is
01 locally finite
0/ S.
perimeter if
P(S,O) < 00 holds lor every bounded open set O. Sets also called Caccioppoli sets.
(3.85)
0/ locally finite perimeter are
(iii) If S is 01 locally finite perimeter, then there is a positive Radon measure p. and a p.-measumble]RN -valued function u, with lu(x)1 = 1 lor p.-almost every x, such that DXs = up. (see Corollary 3.6.10). It is customary to use the notation 118S11 for the Radon measure p. and to write· Vs lor the function u, so that (3.86)
Dxs = Vs 118811·
(iv) We write peS) = pes, JRN), and we say 8 is o/finite perimeter if P(8).< 00. Example 30702 II S is a C l dqmain, then 8 is of locally finite perimeter. To see this, suppose p is a Cl defining function for 8 so that
< O}
S = {x : p(x)
and Vp(x) =1= 0 for x E 8S. Let ifJ : JR -+ JR be any Coo function with ¢'(t) :5 0 for all t and ifJ(t) =
{Io
~ft:5
-1,
If 0:5 t.
Fixing a bounded open set 0 and a function 9 E C~(n;JRN) with we use integration by parts to estimate
is
div(g) deN
=
Igl
:5 1,
JRN ¢(p/t) div(g) d£N
lim ( t.j.o
JRN V [ifJ(p/t)] 0gd£N
-lim ( t.j.O
=
-lime l t.j.O
~
{ ¢'(p/t) (Vp) ogdCN JRN
- (inf ifJ') lim e t.j.O
l
{
IVpi dCN .
Jnn{IZ:-t$p(IZ)$O}
The last integral is finite because n is bounded and Vp is non-vanishing on 88.
3.7. DOMAINS WITH FINITE PERIMETER
109
In fact, the term (inf t/J') in the preceding argument can be made as close to -1 as we wish. Also, careful use of the change of variables formula for integration allows one to see that
lim t- 1 t.J,O
IVpi d.eN = 1lN-1 (0 n 8S).
{
Jnn{z:-tSp(z)SO}
We conclude that, for a Cl domain S, P(S,O) holds for each bounded open set
= 1lN - 1 (n n 8S)
(3.87)
n.
The following lemma is an immediate consequence of the fact that the integral of the divergence of a function with compact support is zero. Lemma 3.7.3 Suppose S has locally finite perimeter and open set. Then
118(JRN "and
\
S)II
n is a bounded
= 118SII,
pen \ s, n) = pes, n)
hold.
Proof: Note that, for 9 E C: (:JRN; ]RN), we have { div(g).e N = - (
15
div(g) .eN.
JRN\S
•
By Theorem 3.6.7, if S C ]RN "is a domain with finite perimeter, then the following formula holds for any 9 E (JRN ; JRN) :
C:
Is
div(g)d.e N = -
J
g·vsdIl8SII·
(3.88)
Equation (3.88) can be considered to be an abstract form of the classical Gauss-Green formula (or synonymously the divergence theorem) which we will discuss in more detail in Section 5.5. One aim of the study of sets of finite perimeter is to improve our understanding of the right-hand side of (3.88). As a first step, the next lemma extends the applicability of (3.88). Lemma 3.7.4 Suppose S C ]RN is of finite perimeter. If 9 E ~(JRN ; JRN) is differentiable .eN -almost everywhere and {
JRN then (9.88) holds for g.
Idiv(g)1 d.e N <
00,
(3.89)
CHAPTER 3. MEASURES
110
Proof: Let 1/JtT, 0' > 0, be a family of mollifiers as in Section 1.3. Then (3.88) applies to 1/JtT * g, so that
Is
tPtT*(divg)d£N
=
Is
div(tPtT*g)d.CN =-- !(t/J,.*9)'"sdIl 8SI1 (3.90)
Is
holds. As 0' .J.. 0 the left-hand side of (3.90) coriverges to div(g) deN and the right-hand side converges to I g'''sdIl8SII, both by Theorem 1.3.31. I
Remark 3.7.5 The preceding lemma admits the use of vector fields that are piecewise continuously differentiable and of those that have singular differentials on small sets in the Gauss-Green formula. The next lemma contains a particular instance of the Gauss-Green formula that will be useful later in this section. Lemma 3.7.6 Suppose S C JRN is of 10000ly finite perimeter and 9 E C:(JRN; JRN). For x E JRN, {
lsrot(z,r)
div(g)d.c N
={
g"lIsdIl8SII+ (
lsnsc2l,r)
llf(z,r)
g'I:=:I,crnN-1y, (3.91)
holds for £l-almost every r > O. Proof: Fix r > O. For
E
> 0, define g(y)
r+E-IY-2I1 ()
g.(.) - {
•
Iy - xl :5 r, if r :5 Iy - xl :5 r + E, if r + E :5IY - xl. if
9 Y
o We compute divg(y)
o
Iy - xl ::5 r, if r :5 Iy - xl ::5 r + E, if r + E::5 Iy - xl. if
2I 1dl"V ( ) _ ..Tl=.!.-. ( ) div .. (y) - { r+'-I,I• 9 Y 'llI-%1 9 Y By Lemma 3.7.4 we have
Is
div g. d.c N
=/
g. "115 d1l8SII·
Assume r is such that S n Sex, r) is tiN-I-measurable, which is true for .cl-almost every r > O. Then the result follows by letting E decrease to
O.
I
Area Minimization and the Isoperimetric Inequality Another reason for our interest in sets of finite perimeter is that they provide a context in which to study problems of area minimization.
3.7. DOMAINS WITH FINITE PERIMETER
111
Theorem 3.7.7 (Existence of Area Minimizers) Let U c JRN be a bounded open set, and let B be a set 0/ finite perimeter. Then there e:&ists a set S 0/ finite perimeter with S \ B \ U such that
u=
peS)
S P(T) lor any T with T \ U
= B \ U.
Proof: Let SI, S2, •.. be a sequence of sets of finite perimeter with S. \ U = B \ U, for i = 1,2, ... , and
.lim
• -too
P(S.)
= inf{P(T) : T \ U = B \ U} .
Let K be a compact set that contains U. Using the Compactness Theorem 3.6.14 to pass to a subsequence, but without changing notation, we may suppose that Xs,nK converges in £1 (K) to a function /. Passing to another subsequence if necessary, we can suppose that Xs,nK also converges almost everywhere to I, thus I is the characteristic function of a set C. We set S= CU(B\K), and the result follows from the Semicontinuity Theorem 3.6.5.
•
One of the central facts of classical Euclidean geometry is the Isoperimetric Inequality. A version of this result can be easily proved in the present context of geometric analysis. Later, in Sections 7.2 and 7:3, we will approach the Isoperimetric Inequality using Steiner symmetrization For this purpose we need a simple Sobolev type lemma. The general thrust of a "Soholev Inequality" is to use an integral norm of a higher order derivative to bound an integral norm of a lower order derivative. Actually, this was first done by Poincare on a bounded domain for the L2 norm of the first derivative of a function and the £~.Jlorm of the function, so that particular type inequality is often called Poincare's Inequality. Lemma 3.7.8 If / E BV(JRN
(klV III b
),
then the following inequality is valid:
lV-I deN) r r
S klV IV fl d£N.
(3.92)
Remark 3.7.9 This result can be generalized to other exponents and applies to functions in the appropriate function space, namely Soholev space (see Evans and Gariepy [1]). In the next chapter, we will give the formal definition of Soholev space and present the main theorems concerning such spaces.
=
Proof: We will give the proof in the case N 3. This case illustrates all the salient points without excessive notational complication. By Theorem 3.6.12, it will suffice to prove the result when / E C~(IRN). For any (x,y,z) E ]R3, we have
If(x,y,z)1
=
liZ
0.=-00
aal (u,y,Z)dU/ S x
1
uER
IV/(u,y,z)ldu,
CHAPTER 3. MEASURES
112
and similarly,
1/(x,lI,z)1 $
r
l/(x,lI,z)1
IV/(X,l1,Z)ldl1, J veR Accordingly we can estimate 1/(x,,I,z)li
$
~
1
weR
IV/(X,II,W)ldw.,
(leR'V/(U,,I,z)ldU? x (leR IV/(X,l1,Z)ldl1) t x (LeR IV/(X,,I,W)ldW) t.
Integrating with respect to x and using Holder's inequality, we have
1
:teR
I/(x,y;Z)lidX$l :teRN
[(1
ueR
IV/(U''I,Z)ldU)! x
(leR IVf(X,V,Z)ldV) ' x (LER IV/(x,y,w)ldW) '] dx
=
(leit IV/(U,y,Z)ldU)! x
1 (( ·1~f(X'V'Z)ldV)' (1 x
:teR J veR
$
weR
IV!(x,y,W)ldW) 1 dx
(leR1V/(U,y,Z)ldu)' x
(1 /, .
(1 1
IV!(X,V,Z)ldl1dX) , x IV/(x,y,W)ldWdX) , . :teR veR zER wER Then we integrate with respect to y, again use Holder's inequality, and estimate
11 1 (1
,1eR :telR
lIeR
~
I/(x,y,z)I' dxdy
ueR
~
(1 1 (1 1 (11
IV/(u,y,z)1dU)' x
zeR lIeR
zER wER
(rJzeR JveR { IV/(X,V,Z)ldl1dx)t x
1
IV/(x,v,z)1 dVdX) , x
IV/(x,y,W)ldWdx) , dy
lIeR uER
IV/(u,y,Z)ldUdY) 1
{ { IV/(x,y,W)ldWdxdfl)i. J ,IeR Jz eR weR Finally, we integrate with respect to z, use Holder's inequality, and estimate X (
J. 1 1
zeR lIeR zeR
1/(x,y,z)I' dxdydz ~
3.7. DOMAINS WITH FINITE PERIMETER
r }zeR r 1weR IV/(Z,Y,W)ldWdxdY)! (}"eR
1 [( r 1 (1 1 aeR
}zeR tieR
x
IV/(Z,V,Z)ldVdx)! x
"eR ueR
~
113
IV/(U,y,Z)ldUdY)!] dz
(r}"eR }zeR r }weR r IV/(z,Y,W)ldWdxdY) 1
x
r 1tieR IV/(Z,V,Z)ldVdxdZ)! x (1zER}zeR
(1 1 1
aeR "ER uER
.
=
(L IV/ldC s
IV/(u,y,z)1 dUdydZ) 1
i
3)
•
•
Corollary 3.7.10 (Isoperimetric Inequality) 1/ S c JRH is a set 0/ finite perimeter and finite Lebesgue measure, then the following inequality is valid: (3.93)
Proof: Apply Lemma 3.7.8 to Xs.
•
Remark 3.7.11 Recall that Remark 3.87 showed us that the perimeter of a C l domain in RN agrees with the (N - 1)-dimensional Hausdorff measure of the topological boundary of the domain. Thus, when N = 2, the Isoperimetric Inequality gives a lower bound for the length required to enclose a given area, and when N = 3, the Isoperimetric Inequality gives a lower bound for the area required to enclose a given volume. Similar interpretations apply in higher dimensions. The version of the Isoperimetric Inequality given in Corollary 3.7.10 is not sharp: Simply consider the unit ball in llt2 for which the left-hand side of (3.93) equals 7r l / 2 and is strictly less than the right-hand side of (3.93) which equals 27r. A fairly simple geometric argument, which we sketch, allows us to obtain the following inequality as a corollary of the above lemma. We will call this a Poincare Inequality, because it applies to the unit cube (even though it is not for the L2 norm).
CHAPTER 3. MEASURES
114
Corollary 3.7.12 Let Q denote the closed unit cube in aN and suppose n is open with Q c n. There exists a constant C, depending only on. N., such that N-l
(k II - IQI7P:r deN)
l¥
~ C kIV/ld£.N:
(3.94)
holds lor Ie BV(n) , where
(3.95) Proof: Again it follows from Theorem 3.6.12 that it suffices to prove the result under the stronger assumption that I e C!(O). We define a function g: [-1,3]N -+ ]RN by reflecting through the faces ofthe unit cube. That is, g(XloX2,'" ,XN) = l(x~,x2'" "xN)' where X~ =
{
-Xi
if -1
~ Xi ~
Xi
if 0 ~
Xi
~
1
if 1 ~
Xi ~
2
Xi
-1
0
Since I is continuous, there is a point Xo e Q such that IQ = /(xo). Let Uzo : RN -+ [-~, 3]N be any smooth transformation that is the identity on Q and at distance more than 1/2 from Q maps all points to Xo. The result follows by applying Lemma 3.7.8 to the function (g 0 U) - I Q and using the change of variables formula for multiple integrals. This yields a constant C(xo) for which (3.94) will hold whenever IQ = I(xo). FUrther supposing that the transformation Uzo is defined in a way that depends continuously on xo, we see the value of C needed to make (3.94) generally valid is the maximum of the values C(xo), as Xo ranges over Q. • Remark 3.7.13 We have used the unit cube because the notion of reflection through a face is easily understood. Poincare's Inequality 3.7.12 is typically stated with Q a ball of radius r. An elementary change of variables shows that the specific choice of Q as a cube, ball, etc. only affects the value of the constant, not the basic validity of the result. One caveat is that if non-symmetrical figures (for example, ellipsoids) are used, then the ratio of the smallest dimension to the largest dimension will need to be bounded above and below. Using our version of Poincare's Inequality, we can prove a local form of the Isoperimetric Inequality wl.ich will be useful for analyzing the fine structure of sets of locally finite perimeter. Corollary 3.7.14 (Local Isoperimetric Inequality) 11 S c RN is a set with locally finite perimeter, then lor each X e aN and each r > 0 the
3.7. DOMAINS WITH FINITE PERlMETER
115
following inequality is valid:
in! { .cN(B(x,r)nS), .cN(B(x,r) \S) } :5 C [ 1I0SIIB(x,r)]
N/(N-l)
. (3.96)
Here C is a COMtant depending only on N.
Proof: Apply Corollary 3.7.12 to Xs, but with Q replaced by the open ball of radius r about x (as we may b~ Remark 3.7.13). • Structure Theory for Sets with Finite Perimeter In an earlier example, we showed that a domain with a C 1 boundary is a set of locally finite perimeter. The main goal of our st~y of sets of locally finite perimeter is to show the extent to which the Converse of this statement is true. In other words, we wish to show that a set of locally finite perimeter has a Cl boundary in a measure-theoretic sense. The term "boundary" here cannot mean the topological boundary; point set topology is too crude a tool to isolate the relevant points. Instead we must use the "reduced boundary" as defined next. DeJlnition 3.7.15 Suppose S C ]RN is a set of locally finite ,perimeter. The reduced boundary of S, denoted by 0" S, is the set of:r: e]RN such that (i) 1I0SII(B(x,r») > 0 holds for r > 0, •••
(11) "S(X) (iii)
.
=l,$
dll OS II 1I0SII(B(x,r» exists,
fB(:r,r) liS
IVsl = 1.
Theorem 3.7.16 If S is of locally finite perimeter, then the reduced boundary of S contains lIaSIl-almost all of S. Proof: This is a consequence of the Lebesgue differentiation theorem for • general Radon measures on ]RN. At a boundary point of a smooth domain, the boundary will infinitesimally be an N -1 dinlensional ball that separates a half-ball in the domain from a half-ball in the complement of the domain. Such behavior can be interpreted in terms of densities at the boundary point. In particular, each element of the boundary is also an element of the reduced boundary. In the next lemma, we obtain some density estimates that apply at each point of the reduced boundary. These estimates are step toward showing the reduced boundary is much like the boundary of a smooth domain.
a
CHAPTER 3. MEASURES
116
Lemma 3.7.17 There exists a constant 0 < C = C(N) < ex) such that for any set S with locally finite perimeter the following inequalities hold Jor x e 8·S:
(ii)
C- 1 ~liminfr.j.O r- N .cN(B(x,r)
\S),
(iii) C-I~liminfr.j.O r-(N-I) 118SII(B(x,r») ~ limsuPr.j.O r-(N-I)
118SII(B(x,r»)
~ C.
=
Proof: For convenience of notation, let us suppose that 0 x E ao S and that lISCO) eN. We apply Lemma 3.7.6 with 9 chosen to be the constant vector eN in some neighborhood of O. Then, for all sufficiently small r > 0, the left-hand side of (3.91) vanishes. Thus we conclude that for .c1-almost all sufficiently small r > 0
=
eN' (
iJ1(o,r)
"5 dll8S11
=
l. eN'''sdIl8SII.
JB(o,r)
=
-
(
~
llN - 1 [Sn§(0,r)].
JSnS(O,r)
Dividing both ends of the preceding formula by Definition 3.7.15(ii) implies
eN' ..I/.. aHN-1y 1111 '
118S11 (B(O, r») , we see that
1lN-l[Sn§(O,r)] 1 < II· m -__...:=:....,.-....:..".~ -
r.j.O
II 8S II (B(O,
r»
(3.97)
Notice that limsup r-(N-l) 118SII(B(0,r») ~ C r.j.O
follows from (3.97), proving the second half of part (iii). Note that Lemma 3.7.6 also implies that
p(snB(O,r)} $ II as II (B(O,r») +ll N - 1 [Sn§(0,r)]
(3.98)
holds for .c1-almost every r > O. Combining the Isoperimetric Inequality of Corollary 3.7.10 and equations (3.97) and (3.98), we see that for each 1 < ~ there exists 0 < r(~) such that
3.7. DOMAINS WITH FINITE PERlMETER
117
holds for .e 1-almost every 0 < r < r(~). Equation (3.99) is actually a slightly disguised differential inequality: Notice that if we set /(r) = .eN(SnB(O,r», then J'(r) = ?iN-l[S n S(O, r)].
Thus (3.99) tells us that (after we make a choice of constants)
C:5 [J(r)]
~
and rename the
*-1 J'(r) = N ! [J(r)] *
(3.100)
holds for .e 1 -almost all sufficiently small r > O. Part (i) of the lemma now follows by integrating (3.100). As an additional bonus we also obtain the inequality (3.101)
°
We obtain part (ii) of the lemma by noting that if is in the reduced boundary of S, then it is also in the reduced boundary of ]RN \ S. Finally, the first half of part (iii)·follows from parts (i) and (ii) and the Relative Isoperimetric Inequality. • Example 3.7.18 The theorem tells us that 1{N-l(8*S) :5 118SII(]RN), so for a set of finite perimeter we have (3.102) Equation (3.102) is false if 8* S is replaced by the topological boundary: For example if S is a union of balls centered at the points ·of a countable dense subset of lR(O, 1) and with radii chosen so that the total .eN measure of S is strictly less than that of iB(O, 1) and the sum of the 1{N-l measure of the surfaces of the balls is finite, then S will be of finite perimeter, but 8S will contain 1(0, 1) \ S, and thus have positive .eN measure and infinite 1£ N -1 measure. Notation 3.7.19 For a set sCaN and for r > 0, we set
rS
= {rx : XES}.
An important tool in proving the Structure Theorem for sets of finite perimeter is blowing-up3. Blowing-up produces local information that relates the vector to the geometry of S. The easy part of blowing-up is in the following lemma.
Vs
3This is not the blowing-up of algebraic geometry (as discussed in Krantz and Parks [2], for example).
CHAPTER 3. MEASURES
118
Lemma 3.7.20 118 is of locally finite perimeter, 0 e 8-8, and ri is a ~ with r. -+·00 as i -+ 00, then there e:r:Ut a subsequence ril and a set H of locally finite perimeter such that (3.103) and such that ""'1 s 118(ril 8)11 converges weakly to IIHII8HII.
Proof: The existence of H as in (3.103) follows from the Compactness Theorem. 3.6.r4 and the density estimate in part (iii) of Lemma 3.7.17. The weak convergence is a consequence of the theorem of de la Vallee Poussin that states that any sequence of uniformly bounded Radon measures has a subsequence that wealdy converges to Radon measure (a consequence of the Banach-Alaoglu Theorem, Rudin [3],3.15, or see Appendix A of Giusti [1] for a direct proof). • The geometric import of blowing-up is shown in the following theorem. Theorem 3.7.21 (Blowing-Up) If 8 is of locally finite perimeter, 0 E 8-8, and IIS(O) = eN, then X,.S
-+
XII-
in Lloc
as r
r .cN[nn+ n (111(0, R) n 8)] ATo RN = lim ~o
.cN(nn- n (B(O,R)
(3.104)
-+ 00,
0
(3.105)
,
\ 8)] = 0
RN
(3.106)
'
lim 118811111(0, R) = 1.
(3.107)
~o WN,....lRN 1
Here JH[- is the lower half-space {x : x· eN half-space {x: x·eN > O}
< O} and
JH[+ is the upper
Proof: We concentrate on (3.104). The other conclusions will then follow easily. It suffices to show that any sequence Ti -+ 00 has a subsequence Tis such that (3.108) Xr;s S -+ XHin Lloc as j -+ 00. By the lemma, we know there is a subsequence Tis such that (3.109) for some set H of locally finite perimeter and that additionally
converges weakly to IIHII8HII. We must show that H
= JH[-.
"'Ii ',?
3.7. DOMAINS WITH FINITE PERIMETER
119
First, we show that VH = eN for 118HII almost every point. For q, E C:(lRN ;IIlN ), we use the change of variables r'iY = x and the notation t/J(y) q,(r'JY) to see that
=
= jRN f divq,(x)Xr'Js(x)dCNx = r~-l jRN f divt/J(y)Xs(y)dCNy = r~-l
f t/J'
d118811.
VS
We conclude that IIOr,; SIIB(O, R) and
1
Vr, S
B(O,R)
;
=rr., 1 1l8811B(0, r:', N R)
118r.,SII = rr.- 1
l
'B(o,r:J- N
Vs R)
(3.110)
118811·
(3.111)
Because 0 E 8* S, (3.110) and (3.111) allow us to conclude that •
hm
dll8r., SII . iB(o,r:- N R) ,; = hm; 118r., 81115(0, R) j--+oo 118SI115(0,
iB(o R) V r, S
j--+oo
Vs
r:
dll8S11
i N
R)
= eN.
(3.112)
Considering R > 0 for which 118HII8B(O, R) = 0, we see by the Semicontinuity Theorem 3.6.5, by (3.112), and by the weak convergence of vr "s!l8(r.,S)1I to vHII8HII that 118HIli8(0, R)
:5
=
lir.:r IIOr., 811!(0.. R) .lim
J-+OO
~
ji(O,R)
eN' Vr,.s dll8r.,SII ,
= ji(O,R) ~ eN'vHd1l8HII· Since IVHI
= 1 holds 118HII-almost everywhere, we must have (3.113)
118HII-almost everywhere. While it seems intuitively clear that (3.113) implies that H is a halfspace with its bounding hyperplane orthogonal to eN, the point of this theorem is to prove that the vector really does have that desired geometric significance. Thus we argue as follows: Let t/Jtr be a family of molIifiers as in Section 1.3. For rP E (lRN ; lRN), we have
Vs
C:
-f
grad(t/Jtr*XH)"rPdc.N
=
f
t/Jtr*XHdivrPdCN
120
CHAPTER 3. MEASURES =
/ X.Hdiv(1/1" *4»dC N
=
/ eN· (1/1" * 4»
=
/ 1/1,,* (4). eN) d1l8HII·
dll8HII
We conclude that the smooth function 1/1" * XH is independent of the first N - 1 coordinates and is a non-increasing function of the Nth coordinate. Since 1/1" * XH converges CN -almost everywhere to XH we conclude that H is some translate of 1HI- in the eN direction. Finally, since Xr/.s converges weakly to XH in Lloc we see that parts (i) 1 and (ii) ofLemma 3.7.17 imply that indeed .J! = 0- . • The following lemma shows that if the results of blowing-up are sufficiently uniform then significant global information can be obtained. The proof of the lemma· is highly technical, but it seems to be an unavoidable fact of geometric analysis that a highly technical construction lies at the heart of any argliment going from local information to global information. Lemma 3.1.22 Suppose that
(i) K C JRN is compact, (ii) ,,: K -+ S(O, 1) is continuous,
Then there exist countably many Lipschitz functions Ii : IRN orthonormal bases {en, ei2, •.. , eiN }, i = 1,2, ... , such that
1
-+ IR and
00
Kc
UG
i
(3.114)
i=1
where (3.115)
Proof: First, we observe that, for each j, since K is compact, K is contained in finitely many open balls of radius 2- i centered in K. Taking the union of all these finite families and replacing each open ball by its closure, we construct a countable set B of closed balls centered in K such that, for each to > 0, (3.116) { B E B : diam(B):::: to } is finite
3.7. DOMAINS WITH FINITE PERIMETER
121
and K C U{ B
e 8 : diam(B) < E }.
(3.117)
Next, we construct a countable family V of orthonormal bases of IIlN such that the set consisting of the Nth vector from each basis in V forms a countable dense subset of 8(0,1). Thirdly, we arrange B x V x N into a sequence (3.118) { (B(Pi,Ti), {eil,ei2, ... ,eiN}, M;) }~ .
.=1
n i"(Pi, Ti) satisfying y). em] eiNI for all x, y e Gi. (3.119)
For each i, let Gi be a maximal subset of K I(x - y). eiNI :5 M;I{x - y) - [(x The function
Ii is defined by setting
Ii (x • eil,X' ei2,···,X 'eiCN-l)} = The domain of Ii is
X·
eiN for x
Ai = {(x·eu,x·ei2, ... ,x·eiCN-l)} : x
e Gi •
e Gi}.
{3.120}
{3.121}
We see that Ii is Lipschitz from the defining condition (3.119) for G i , and that the Lipschitz constant of Ii does not exceed Mi' We can extend the domain of Ii to all of IIlN- 1 by using the following formula (due to H. Whitney): (3:122) fi(x} = SUP{Ji(Z) - Mi dist(x, z} : z e Ai}. It remains to show that (3.114) holds. To this end, we argue by contradiction and assume to the contrary that (3.114) does not hold. Suppose x· e K, but x· 'I. G i , for i = 1,2, .... By (3.116) and (3.117), there must be infinitely many balls in B that contain x·. Set 8' = {B e 8 : x· e B}. We now choose a sequence in 8' x V x N with some particular properties: Let i j be such that that x· e B(Pii' Ti;} and
= 0,
(3.123)
.lim ei;N = v(x·},
(3.124)
.lim
{3.125}
lim Ti'J
j .... oo
J .... OO J .... OO
MiJ
=
00.
'I. Gi;, there must be Yi e G iJ C K n i"(Pi;' Ti;}
Now, for each j, since x·
(3.126)
such that I(x· - Yi}' ei;NI > Mi; I{x· - Yi} - [(x· - Yi)' ei;N] eiJNI. Equations (3.123}-(3.127) contradict part (iii) of the hypotheses.
(3.127)
•
CHAPTER 3. MEASURES
122
Remark 3.7.23 The functions Ii defined in the proof of Lemma 3.7.22 can be extended to be Cl functions, instead of just Lipschitz functions, if we are willing to appeal to the Whitney Extension Theorem. While the Whitney Extension Theorem is not proved until Chapter 5, its proof is independent of the results of this section. Measure theory allows the local information provided by blowing-up to be made sufficiently uniform that the preceding lemma can be applied. The result is the following theorem.
Theorem 3.7.24 (Structure Theorem) If 5 c aN hal locally finite perimeter, then there cist countably many (N -1) -rectifiable seu, G1 , G2 , ••• and a set E fllith 118SII(E) = 0 such that 00
8*5 = Eu
UG•.
(3.128)
i=1
Proof: Egoroff's Theorem (see Evans and Gariepy [1], Section 1.2, or Federer [4], 2.3.7) tells us that, for each f > 0, pointwise convergence of measurable functions implies uniform convergence off a set of measure f. Applying Egoroff's Theorem countably many times, we see that there exist disjoint 118SII-measurable subsets of 8* S, covering 118SII-almost all of 8* S, on which the convergence in (3.105)", (3.106), and (3.107) is uniform. Similarly, Lusin's Theorem (see Evans and Gariepy [1], Section 1.2, or Federer [4], 2.3.5) tells us that, for each f > 0, a measurable function is equal to a continuous function off' a set of measure f. Applying Lusin's Theorem countably many times on each of the preceding subsets, we see that each is 118SII-almost the union of disjoint compact subsets on which the normal vector liS is a continuous function. We wish to apply Lemma 3.7.22 to each of these compact subsets. To do so, we need to verify that hypothesis (iii) of Lemma 3.7.22 is satisfied, but the density estimates of Lemma 3.7.17 insures that (3.105) and (3.106) imply that hypothesis. • Remark 3.7.25 As in Remark 3.7.23, the Whitney Extension Theorem allows one to conclude that the sets G. can be required to be subsets of the images of C 1 functions. The final result of this section gives a very general form of the GaussGreen Theorem. This result was discovered independently by E. De Giorgi and H. Federer.
Theorem 3.7.26·(Generalized Gauss-Green Theorem) If has finite perimeter, then for each l/J E C: (JRN ; aN) we have
1
divl/Jdr.N = [
S
18 • s
l/J. liS rIH. N -
1•
sCaN (3.129)
3.8. THE AREA FORMULA
123
Proof: We need to show that on 0° S the measure II oS II equals the N - 1 dimensional Hausdorff measure. This follows from (3.107) and the Structure Theorem 3.7.24.
•
Remark 3.7.27 lithe set S in Theorem 3.7.26 is not of finite perimeteror even locally finite perimeter-one can still use Equation 3.129 as the definition of the boundary. Of course, it is more interesting if both sides of the equation can be given by independent definitions .. We direct the reader's attention to just such a non-trivial extension of the Gauss-Green Theorem that has been given in Harrison and Norton [I].
3.8
The Area Formula
The main result of this section is the following theorem. Theorem 3.S.1 (The Area Formula) If f : function and M ~ N, then
]RM
-t ]RN is a Lipschitz
r JMf(x)dCMx= iRNr card(Anrl(y»dllMy
iA
(3.130)
holds for each Lebesgue measurable subset A of]RM.
Here 1£ M denotes the M -dimensional Hausdorff measure and JM f denotes the M-dimensional Jacobians of f which is defined in Definition 3.8.3. In case M = N, the M-dimensional Jacobian agrees with the usual Jacobian, Idet(Df)I· The proof of the area formula separates intt> three fundamental parts. The first is to understand the situation for linear maps. The second is to extend our understanding to the behavior of maps that are well approximated by linear maps. This second part of the proof is essentially multi-variable calculus, and the area formula for Cl maps follows readily. The third part of the proof brings in the measure theory that allows us to reduce the behavior of Lipschitz maps to that of maps that are well approximated by linear maps. In the next section we will treat the co-area formula that applies to a Lipschitz map f : ]RM -t ]RN, but with M 2': N instead of M ~ N. The proof of the co-area formula is similar to the proof of the area formula in that the same three steps of understanding linear maps, understanding maps well approximated by linear maps, and applying measure theory are fundamental. The discussion of linear maps in this section will be applicable to both the area formula and the co-area formula. Linear Maps
A key ingredient in the area formula is the K -dimensional Jacobian which
CHAPTER 3. MEASURES
124
is a measure of how K -dimensional area transforms under the differential of a mapping. Since a linear map sends one parallelepiped into another, the fundamental question is "What is the K -dimensional area of the parallelepiped determined by a set of K vectors in ~ 1" Of course the answer is known, but recently G. J. Porter [1] has discovered a particularly lucid derivation that we give below.
Proposition 3.8.2 If
Vii V2i
tt i = (
:
)
'
for i
= 1,2, ... ,K,
(3.131)
VNi
are vectors in ]RN, then the pamllelepiped determined by those vectors has K -dimensional area (3.132)
where V is the N x K matrix with
tt 10 112 , ••• ,11K
as its columns.
Proof: H the vectors ttl, tt 2, ... , tt A: are orthogonal, then the result is immediate. Thus we will reduce the general case to this special case. Notice t~at Cavalieri's Principle shows us that adding a multiple of tti to another vector tt j, j =I i, does not not change the K -dimensional area of the parallelepiped determined by the vectors. But also notice that such an operation on the vectors 11i is equivalent to multiplying V on the right by a K x K triangular matrix with l's on the diagonal (upper triangular if i < j and lower triangular if i > j). The Gram-Schmidt procedure is effected by a sequence of operations of precisely this type. Thus we see that there is an upper triangular matrix A with 1's on the diagonal such that V A has orthogonal columns and the columns of V A determine a parallelepiped with the same K -dimensional area as the parallelepiped determined by 111, 11 2 , ... , 11 N. Since the columns of V A are orthogonal, we know that Jdet (V A)t • (V A» equals the K-dimensional area of the parallelepiped determined by its columns, and thus equals the K-dimensional area as the parallelepiped determined by tt 10 11 2 , ••• ,11N. Finally, we compute
det(VA)t.(VA))
=
det(At.Vt.V.A)
= =
det
(At) det (V t •v) det(A) det (v t . V) .
•
3.B. THE AREA FORMULA
125
Definition 3.8.3
(i) SUppo$e U C RM and f : U ~ RN. We say J is differentiable at a e U iJ there exists a linear map L : RM ~ RN such that Ii IJ(x) - J(a) - L(x - a)1 - 0 :c~ Ix-al - .
(3.133)
In CtJ8e / is differentiable at a, we denote the linear map L such that (3.133) holds by DJ(a) and call it the differential of J at a.
(ii) Suppose that J : RM ~ RN is differentiable at a and K :5 M. We define the K-dimensional Jacobian of J at a, denoted JKJ(a), by setting
{1lK[D J(a)(P)] . J /( ) _ K a - sup 1l K [P] . P is a K-dimensional parallelepiped contained in RM}. (3.134)
The conventional situation considered in elementary multi-variable calculus is that in which K = M = N. In that case, it is easily seen from Proposition 3.8.2 that one may choose P to be the unit M -dimensional cube and that hJ(a) = Idet(D/(a»I. Two other special cases are of interest: They are when K = M < N and when M > N = K. When K = M < N, again one can choose P to be the unit M-dimensional cube in RM. The image of P under DJ(a) is the parallelepiped determined by the columns of the matrix representing DJ(a). It O
follows from Proposition 3.8.2 that h J(a) = Jdet [(DJ(a»t o· (D/(a»]. When M > N = K, then P should be chosen to lie in the orthogonal complement ofthe kernel of D J (a). This follows from Corollary 3.1.17 which tells us that orthogonal projection onto an affine subspace of IRN cannot increase Hausdorff measure. So, if we let L be the orthogonal complement of the kernel of DJ(a), then if pI is any parallelepiped, the image of the orthogonal projection of pI onto L is the same as the image of pI; but the orthogonal projection of pI has smaller K -dimensional area, giving a larger ratio of areas. It is plain to see that the orthogonal complement of the kernel of D J(a) is the span ofthe columns of (D J(a»t. If we begin with the parallelepiped determined by the columns of (DJ(a»t, then that parallelepiped maps onto the parallelepiped determined by the columns of (D /(a»· (D J(a»t. By Proposition 3.8.2, the N-dimensional area of the first parallelepiped is
J
det [(D J(a» • (D J(a»t]
CHAPTER 3. MEASURES
126
and the N-dimensional area of the second parallelepiped is det [{(DI(a». (D I(a»t) t • (DI(a». (DI(a»t)]
= det [(D I(a» • (D I(a»t] , so the ratio is JKI(a) = Jdet [(DI(a». (DI(a»t]. We summarize the above facts in the following lemma. Lemma 3.8.4 Suppose
(i)
I : lllM -+ lllN
II K = M = N, then
= Idet(DI(a))l.
(3.135)
= Jdet [(D!(a»t.(Df(a»].
(3.136)
JKf(a) = .jdet [(Df(a»·(Df(a»t].
(3.137)
JKI(a)
(ii) If K = M
~
N, then JKf(a)
(iii)
is differentiable at a.
II M
~
N = K, then
Remark 3.8.5 The Generalized Pythagorean Theorem from Porter [1] allows one to see that the right-hand side of either (3.136) or (3.137) is equal to the square root of the sum of the squares of the K x K minors of D f( a), where K = min{M,N}. This is the form one is naturally led to if one develops the K-dimensional Jacobian via the alternating algebra over lllM and RN as in Federer [4]. We will also need to make use of the polar decomposition of linear maps. Theorem 3.8.6 (Polar Decomposition)
(i) II M ~ Nand T : lllM -+ lllN is linear, then there exists a symmetric linear map S : lllM -+ JRM and an orthogonal linear map U : RM -+ lllN such that T = U 0 S. (ii) II M ~ Nand T : lllM -+ lllN is linear, then there exists a symmetric linear map S : JRN -+ RN and an orthogonal linear map U : JRN -+ RM such that T = S 0 U t . Proof: (i) For convenience, let us first suppose that T is of full rank. The M x M matrix Tt • T is easily seen to be symmetric and positive definite. So
3.8. THE AREA FORMULA
127
Tt •T has a complete set of M orthonormal eigenvectors ttl, tt 2, ••• ,11M associated with the positive eigenvalues ~1' ~2' ••• , ~M.
We define 8 : aM -+
aM
by settulg .
Using the orthononnal basis 111 , 112, ... ,11M , we see that 8 is represented by a diagonal matrix, thus S is symmetric. We define U : JRM -+ aN by setting 1
u(lI i ) = .;Ai T(lI i ). We calculate
Thus U is an orthogonal map. In case T is not of full rank, it follows that some of the ~i 's may be zero. For such i's we may choose U(tt.) arbitrarily, subject only to the requirement that U(lI 1), u(lI 2)"'" u(lI n) b~ an orthonormal set. (ii) We apply (i) to the mapping Tt to.2btain a symmetric 8 and orthogonal U so that Tt = U 0 8, but then T = (U 0 8)t = 80 Ut .
•
Remark 3.8.1 The equations in Lemma 3.8.4 can be obtained using the Polar Decomposition of linear maps. For example, if we are considering f : liM -4 liN with M :5 N, then we apply the Polar Decomposition to write D f(a) = U 0 S. Since an orthogonal map clearly preserves Hausdorff measure, the entire effect of D f(a) on the Hausdorff measure of a parallelepiped in liM must be due to the non-negative definite symmetric map 8, so we have JMf(a) = det(S). But then we can compute det(8)
=
Vdet(St)det(S)
=
Vdet(St. ut·U ·8) = Vdet [(Df(a»t • (Df(a))).
CHAPTER 3. MEASURES
128
The first application of the Jacobian is in the following basic lemma the behavior of Lebesgue measure under a linear map.
concernin~
Le~ 3.8.8 If A c aM is Lebesgue measurable and T : aM ~ aM is linear, then .eM (T(A» = Idet(T)I .eM (A).
Proof: Given ~ > 0, we can find an open U with A c U and .eM (U\A) < ~. We subdivide U into cubes and the image of each cube is a parallelepiped. So .eM (T(A)) :5 .eM (T(U» :5 Idet(T)I.eM (U) :5 Idet(T) I [~+ .eM (A)].
Letting
~
..l. 0, we see that .eM (T(A» :5 Idet(T)1 .eM(A).
Now we need to prove the reverse inequality. Note that if det(T) = 0, then we are done. Assuming det(T) :I 0, we apply the case already proved to T(A) and T- 1 to see that .eM (A) = .eM (T- 1(T(A))) :5 Idet(T- 1 )1 eM (T(A».
The result follows since det(T-1)
= (det(T»-l.
•
The Main Estinlates for the Area Formula Lemma 3.8.9 (Main Estinlates for the Area Formula) Suppose T : IRM ~ IRN is of full rank and 0 < ~ < ~. Let n be orthogonal projection onto the image of T. Set
A = inf{(T,v):
Ivl = I}.
(3.138)
If the Lebesgue measurable set A C IRM is such that
(i) Df(a) exists for a E A, (ii) IIDf(a) - Til <
~
holds for a E A,
(iii) If(y) - f(a) - (D f(a), y -
a)1 < ~ Iy - al
holds for y, a E A,
(iv) IIlf(A) is one-to-one, then (1 - 3~A _l)M . JMT· eM (A) :5 l£M (J(A»
:5 (1
+ 2~A-1)M ·JMT . .eM(A).
(3.139)
3.8. THE AREA FORMULA
129
Proof: First we bound l£M (t(A» from above. We use the polar decomposition to write T = U 0 S, where S : RM -+ RM is symmetric and U : RM -+ aN is orthogonal, and we note that S is non-singular with JMS = JMT and with ~-1 = liS-III. Set B = SeA) and 9 = loS-I. We know that £M(B) = JMSo£M(A)
We claim that
= JMLo£M(A).
Lip(gIB) $ 1 + 2£~ -1.
To see this, suppose z, be B. Then with a = S-l(b), y that Iy - al $ ~-llz - bl. Therefore we have
= S-l(Z), it follows
Ig(z) - g(b)1 $ Ig(z) - g(b) - (Dg(b), z - b)1 + I(Dg(b) - U, z - b)1 = If(y) - I(a) - (DI(a),y - a)1 +1«DI(a) - T) 0 5- 1 , Z - b)1 + Iz - bl $ t: Iy - al + liD I(a) - TlloliS-lUolz - bl + Iz - bl $ (1 + 2t:~ -1) Iz - bl.
+ I(U, z -
b)1
(3.140)
Finally, we haye
1lM (t(A»
=
'liM (g(B»
$
(1 + ~t:~-I)M o£M(B)
=
(1 +.2t:~-l)M oJMTo£M(A).
Next we bound l£M (t(A» from ·below. We continue to use the same notation for the polar decomposition. SeCV = IT(J(A» = IT(g(B» and h = (IT 0 9IB)-l. We claim that
Lip(hlc) $ (1- 3t:~-l)-l. To see this, suppose w, c e C. Let b e B be such that IT 0 g(b) = c and z E B be such that IT 0 g(z) = w. Arguing as we did to obtain the upper bound (3.140), but with some obvious changes, we see that Ig(z) - g(b)1 ;::: (1 - 2£~-l) Iz - bl.
Also, we have t:A-llz - bl
~
=
Ig(z) - g(b) - (Dg(b),z - b)l IIT(g(z) - g(b) - (Dg(b),z - b»
~
IIT.L(g(z) - g(b) - (Dg(b),z - b»)I
=
IIT.L(g(z) - g(b»I.
+IT.L (g(z) - g(b) - (Dg(b), z - b»
I
CHAPTER 3. MEASURES
130 Thus we have ID(g(z» - D(g(b»!
~ ~
Ig(z) - g(b)I-ID.L(g(z) - g(b» I (1- 2E~-1) Iz - bl- E~-llz - bl.
Finally, we have J MT • 1/. M (A)
=
CM(B)
::; (1- 3E~-1)M .£M(C) ::; (1- 3E~-I)M ''H.M(J(A»).
•
Rademacher's Theorem Theorem 3.S.10 (Rademacher's Theorem) A Lipschitz function 1 -+ JRN is differentiable C M -almost everywhere and the differential is a measumble function. "
aM
Proof: We may assume N = 1. We use inductiop on M. In case M = 1, the result follows from the classical theorem that an absolutely continuous function from IR to IR is differentiable C-almost everywhere. For the induction step M > 1, we see by Fubini's Theorem and the theorem about absolutely continuous functions that all M partial derivatives of 1 exist CM-almost everywhere and are "measurable functions. The goal is to show that these partial derivatives actually represent the differential at almost every" point. Let us write IRM = IRM - 1 X IR and denote points x E IRM by x = (y, z), 1/ E IRM-l, Z E R We consider a point xo = (1/0, %0) where the following two conditions are satisfied:
(i) As a function of the first M - 1 variables, f is differentiable.
(ii) All M partial derivatives of 1 exist and are approximately continuous (see Definition 3.5.7). For convenience of notation, we assume xo = (0,0) and that all the partial derivatives at xo vanish. It suffices to show that, for each E > 0, there exists r > 0 such that
I/(y,z)1 ::; ET holds whenever I(y, z)1 < T. Let E > 0 be arbitrary. We can choose TO > 0 so that, for Iyl < TO, I/(y,O)1 ::; Elyl. Choosing TO smaller if necessary, we can also assume that, for 0 < T < TO, the set of points in the cube [-T, r]M for which the absolute value of the Mth partial derivative exceeds E is less than EM TM •
3.B. THE AREA FORMULA
131
Inside any (M - I)-dimensional cube rr:!~l[Yi - fr,yd of side fr there must be a vertical line, say at y', in which the Mth partial derivative is bounded by f except for a set of length £r. "So I/(y',z)1
~ I/(Y',~)I + L% I:~
(y',()1
dC(
ely'l + elzl + Mer
~
and thus I/(y,z)1
~
I/(y',z)I+Mv'M-1fT
~
(2 + (1 +';M -l)M)fr.
•
Measure Theory of Lipschitz Maps
We need to know that the images of Lipschitz maps are measurable. Recall from the discussion of Borel sets versus Suslin sets in Section 1.3 that the properties of a set can be significantly altered by applying a function to it. Lemma 3.8.11 Suppose K < M and suppose I: JRM -+ JRN is a Lipschitz function. II A C IRM is llK -measurable and illl~ (A) < 00, then I(A) is llK -measurable.
Proof: Since A can be written as the union of countably many bounded measurable sets, it will suffice to prove the lemma under the additional assumption that A is bounded. Note that sets of llK measure zero are preserved under I. Since Hausdorff measure is Borel regular (by Lemma 3.1.2), it follows that A is contained in a Borel set B with llK (A) = llK (B). Note that (f(B) \ I(A» c I(B \ A) and llK (f(B \ A» = o. Thus, it will suffice to show that I(B) is ll K-measurable whenever B is a bounded Borel set with llK(B) < 00. Also because llK is Borel regular, a bounded Borel set B with llK (B) < 00 is llK-almost equal to a countable union of compact subsets of B. So the I-image of B is llK-almost equal to a countable union of I-images of compact sets. Of course, the I-image of a compact set is compact, hence ll K-measurable, and the result follows. • Lemma 3.8.12 Suppose K < M and suppose I : JRM -+ JRN is a Lipschitz function. II A c IRM is llK -measurable and illlK (A) < 00, then the function on IRN defined by
(3.141)
is llK measurable, and {
JRN holds.
card(A n 1-1 (y» d1l K
~ [Lip 11K llK (A)
(3.142)
CHAPTER 3. MEASURES
132 Proof: For each i > 0, write
where the sets Ai,j, j = 1,2, ... , are pairwise disjoint l£K-measurable sets all having diameter less than l/i. Then
is l£K-measurable by Lemma 3.8.11 and LX/(A;.;) (y):5 card(Anrl(y» for each y E JRN. j
Here LX/(A;,;)(Y)
E JRN, as i ~ 00,
t card(Anrl(y» for each Y
j
proving the 1l K-measurability of the function in (3.141). Finally, (3.142) follows from the Monotone Convergence Theorem and the estimate .
L RN
L
X I(A;,;) d1{.K
j
=
L
l£K[/(A i ,j)]
j
:5 [Lip I]K L l£K (Ai,j) j
= Corollary 3.8.13 For
[Lip I]K l£K (A).
•
I and A as in the preceding lemma,
1lK({y: rl(y)nA is infinite})
= O.
Proof: Equation (3.143) follows immediately from (3.142),
(3.143)
•
In the next few lemmas, we will show how to use the lexicographic ordering on JRM to divide a set into pieces on which the restriction of I is one-to-one. Definition 3.8.14 The lexicographic ordering, ~i, on JRM is given by defining (X},X2"",XM) -
if and only if, for some j E {I, 2, ... , M), Xl = Yl, X2 = Y2, ... , Xj-l = < yj. For X, y E aM, we will write X ~t Y if X -
Yj-l, and Xj
3.8. THE AREA FORMULA
133
Lemma 3.S.15 Suppose I : RM .-. R.N is a Lipschitz function. is eM -measurable, then, lor each i = t, 2, ... ,
Aj
II A c RM
i }
=
{x e A : card(rIV(x)] n A) =
=
I-I ({y e RN : card(f-I(y) n A)
= i })
is eM -measurable. Proof: It
will suffice to show that, for each positive integer i Ai = {x e A : card(rIV(x)] n A) ~ i
~
2,
}
is eM -measurable. It is also sufficient to prove this assertion under the additional assumption that A is bo\lDded. Fix i ~ 2. Define the Borel set S by setting
_ ==
(RM)2jn{(Ut.U2, ... ,Uj,X1.X2 ... ,Xj): Ui1 =1= Ui.. .XI
Define A C (RM)
for 1 ~ il < i2 ~
i,
= X2 = ... = Xj}.
2j by setting
A = {(Ul, U2,· .. , Uj'/(Ul), l(u2)"'" f(uj» : Ul> U2, ... , Uj e A}. Since A is the image of A under a Lipschitz mapping, it follows that A has . finite 1-lM measure and is 1-lM -measurable by Lemma 3.8.11. Now notice that Ai = II(S n A) where II : (RM) 2j -+ RM is projection onto the first factor. Since II is also a Lipschitz map, we see that Lemma 3.8.11 implies that Ai is eM-measurable. • Remark 3.S.16 Note that the preceding lemma gives us an alternative proof that if f : RM -+ RN is a Lipschitz function and A C RJI·[ is M _ measurable, then the function on RN defined by y 1-+ card(Anf-l(y» is 1-f.M -measurable.
e
Lemma 3.S.17 Suppose I : RM -+ RN is a Lipschitz function. If A C RM is eM -measurable, .then there exist £M -measurable sets Aj,k C A, i = 1,2, ... , k = 1,2, ... ,i such that j
Aj
=
UAj,k'
"=1 Here Aj is as in Lemma 3.8.15; also, for each k one-to-one, and I(A j ,")
e
{I, 2, ... ,i},
= {y e RN : card(rl(y) n A) = i }.
IIAj ...
is
CHAPTER 3. MEASURES
134
Proof: It is no loss of generality to assume that A = f- 1
({y e RN : card(r1(y) n A) = j
}).
We can then argue by induction on j. The case j = 1 is, of course, trivial. Suppose j ~ 2. Define the Borel sets 3., i == 1,2, ... ,M, by setting
-
(M)4n {(U,V,X,Y):U1=Vlt ... ,Ui_1 =Vi-1,U.
'::'i= R
x=y},
for i = 1,2, .. . ,M. Let n: (RMt _ RM be projection on the first factor. Define A C (RM) 4 by setting
A = {(u,v,f(u),J(v» : u,v e A}. Since A is the image of A under a Lipschitz map, A is ?tM-measurable by Lemma 3.8.11. Set !vi
B=
UII(:::. nA). i=l
Each set II(ZinA) is .eM-measurable because it is the orthogonal projection of a measurable set. Observe that; B consists of all points u e A for which there exists a point v e A that is larger than u in the lexicographic ordering on RM. Setting A;,l = A \B, we obtain an .eM -measurable set containing the lexicographically largest point of An f-1(y) for each y e f(A). Notice alSo that, on B, f is a (j - 1)-to-one function, so the induction hypothesis applies to B. • Lemma 3.8.18 Suppose f : RM - RN is a Lipschitz function and A C RM. If y e RN is such that f-1(y) n A is infinite, then every point of
f-1(y)
n {x:
Df(x) exists and is non-singular} nA
is an isolated point of that set.
•
Proof: Obvious. Lemma 3.8.19 Suppose f: RM is .eM -measurable and
B
=
-+
RN is a Lipschitz function. If A C RM
r1({y:r1(y)nA isinfinite}) \( {x: JMf(x) = O} U {x: Df(x) does not exist}),
has positive .eM measure, then there exists B' that fiB' is one-to-one.
c
B with .eM (B') > 0 such
3.8. THE AREA FORMULA
135
Proof: Choose a compact C C B with positive measure. Then, by Lemma 3.8.18, 1-1 [/(x)] is finite for each x e C. We can apply Lemma 3.8.17 to decompose C into countably many sets Ci ,,., at least one of which must have positive measure. • Proof of the Area Formula We first prove the result under the additional assumptions that DI(a) exists at every point a e A, that JMI(a) > 0 at every point of A, and that IIA is one-to-one. -By Lusin's Theorem'1.3.12 we may assume that DI(a) is the restriction to A of a continuous function. By Egoroff's Theorem 1.3.13 we may suppose that
I/(y) - I(a) - (DI(a),y - a)1
Iy-al converges uniformly to 0 as yeA approaches a e A. It is plain that, for any E > 0, every subset or" A with sufficiently small diameter satisfies conditions (i)-(iii) of Lemma 3.8.9 for some full rank linear T : JR.At _ JR.N -namely, - we can choose T to be D I at any point in such a sufficiently small set. Since we are assuming that D I on A is the restriction of a continuous function, we can find a positive lower bound for>. in (3.138). To see that condition (iv) of Lemma 3.8.9 is also satisfied on a subset of A of small enough diameter, we suppose II 0 I(y) = II 0 I(z); we show that, in this case, E > 0 can be chosen small enough compared to >. that conditions (i)-(iii) lead to a contradiction. Using (i)-(iii), we estimate I(T, y -
z)1 ~
~
III {T,y - z»1 III (T - Df(a),y - z)1 + III (Df(a) - DI(z),y - z)1 +III (DI(z),y - z)1 liT - DI(a)lIly - zl + IIDf(a) - DI(z) II Iy - zl +111 (Df(z),y - z)1
=
liT -
DI(a)lIly - zl + IIDf(a) - DI(z)lIly - zl +III(f(y) - fez) - (DI(z),y - z»1
~
liT -
DI(a)lIly - zl + IIDf(a) - DI(z)lIly - zl +I/(Y) - fez) - (DI(a),y - z)l·
By choosing a, y, z in a small enough set we can bound the right-hand side of the preceding inequality above by 3E Iy - zl, while the left-hand side is bounded below by >'Iy - zl. Choosing Esmaller than ~ gives a contradiction. Thus (iv) also must hold on subsets of small enough diameter. In case I is finite-to-one almost everywhere, we use Lemma 3.8.17 to decompose A into the sets Ai," on which the previous case applies.
1
136
CHAPTER 3. MEASURES
For the part of A on which I is infinite-to-one we know the right-hand side of (3.130) is zero by Corollary 3.8.13. H the left,:,~d side of (3.130) were not zero, we would contradict Lemma 3.8.19. Finally, to complete the proof, we need to shdW that the image of a set on which J M I = 0 has measure zero. That follows by defining I~ : RM -+ RM+N by "
Xt-+ (Ex,/(x»). This gives us the full rank hypothesis, but only increases the Jacobian by a bounded multiple of E. The image of I is the orthogonal projection of the image of IE and thus its Hausdorff measure is no larger than the Hausdorff measure of the image of I~. We conclude as E 1 0 that the Hausdorff measure of the image of I is O. •
3.9
The Co-Area Formula
The main result of this section is the following theorem. Theorem 3.9.1 (The Co-Area Formula) II I : RAt chitz lunction and M ~ N, then
-+
RN "is a Lips-
(3.144) holds lor each Lebesgue measurable subset A o/]RAt. Here JNI denotes the N-dimensional Jacobian of I which was defined in the previous section in Definition 3.8.3, and which was seen by "(3.137) to be given by
JNI(a) = "Jdet [(DI(a»· (DI(a»t]. In case M = N, the N-dimensional Jacobian agrees with the usual Jacobian, Idet(Df) I, and the area and co-area formulas coincide. In case M > N, and I : RAt = RN X RM-N -+ RN is orthogonal projection onto the first factor, then the co-area formula simplifies to Fubini's theorem, thus one can think of the co-area formula as a generalization of Fubini's theorem to functions more complicated than orthogonal projection. The co-area formula was first proved in Federer [3]. As in the proof of the area formula, the proof of the co-area formula separates into three fundamental parts. The first is to understand the situation for linear maps. This was done in the previous section. The second part is to extend our understanding to the behavior of maps that are well approximated by linear maps. The third part of the proof brings in the measure theory that allows us to reduce the behavior of Lipschitz maps to that of maps that are well approximated by linear maps.
3.9. THE CO-AREA FORMULA
137
Main Estimates for the Co-Area Formula Lemma 3.9.2 (Main Estimates for the Co-Area Formula) Suppose M > N, U : RN -+ RM is orthogonal, and 0 < E < Lebesgue measurable set A C RM is such that
l.
If the
(i) Df(a) exists for a E A,
(ii) IIDf(a) - Utll <
E
holds for a E A,
(iii) If(Y) - f(a) - (D f(a), y - a}1 <
ElY - al
holds for y, a E A,
then (1- 2E)M {
JaN
1tl"I-N(Anrl(y))dCN y
:5 (
JRN
~
1 A
JMf(a)dCMa
'HM-N(Anf-l(y» dCNy.
.
(3.145)
Proof: Let V : RM-N -+ RM be an orthogonal map such th~t ker(Ut ) and ker(Vt) are orthogonal complements. Defule F : RM -+ RN X RM-N by setting F(x) = (i(x), vt(x)) , and let II : ]RN x RM-N to see that
-+
]RN be projection on the first factor. It is easy
Subsequently, we will show that FIA is one-to-one so that, by the area formula, CM[F(A)] =
L
JMFdC M
=
L
JNfdC M .
Thus, using Fubini's Theorem, we h!l.ve CM[F(A)] (
JRN
'HM-N[F(A)
n II-1(z)] dCN z
( 'HM-N[F(Anf-l(z»]dCNz. JRN To complete the proof, we show FIA to be one-to-one and estimate the Lipschitz constant of F on Anf- 1 (z) and the Lipschitz constant of F-l on F(A n f- 1 (z». Suppose a, yEA n r 1 (z). Then F(a) = (/(a), Vt(a» =
CHAPTER 3. MEASURES
138
(z, vt(a» and F(y) = (f(y), vt(y» = (z, vt(y». We should like to comla - yl and IF(a)·- F(y)l. But the first components are the same,
pare so
IF(a) - F(y)1
= IVt(a) -
vt(y)l.
On the one hand, V t is distance decreasing, so
IF(a) - F(y) I ~
la - yl·
On the other hand,
IWt,y - a}1
~
I(DI(a),y- a}l + IIDI(a) -
<
2Ehi -ai,
I/(Y) - I(a) - (DI(a),y -
Utll Iy - al a}l + IIDI(a) - utll Iy - al
and
so Thus we have
VI -
4E2
Iy - al ~ IF(y) -
F(a)1 ~
Iy - al·
•
Corollary 3.9.3 $uppose M > N, T : ]RM -+ ]RN is 01 rank N, and 0< E < II the Lebesgue measurable set A C ]RM is such that
!.
(i) DI(a) exists lor a (ii) IIDI(a) - Til <
E
E
A,
holds lor a E A,
(iii) I/(Y) - I(a) - (DI(a), Y - a)l < ElY - al holds lor y, a E A, then (1- 2E)M
f
JRN
llM-N(Anrl(y» dCNy
~!.
aN
~ f htf(a)dCMa
JA
1i M - N (A n 1-1 (y») dCN y.
(3.146)
Proof: By the Polar Decomposition Theorem 3.8.6, there exists a symmetric linear map S : ]RN -+ lItN and an orthogonal map U : ]RN -+ ]RM such
3.9. THE CO-AREA FORMULA
that T = So ut. Set 9 = S-1 obtain
(1- 2f)M
r
JRN
0
139
f. Then we apply the lemma to 9 and U to
1t M - N (Ang- l (z»d,CNz
~
~ f
JA
JMg(a)d,CMa
r 1tM-N(Ang-l(z» dCN z.
JRN
(3.147)
Notice that if y = S(z), then
so by the change of variables formula in R,N applied to the mapping S, we have
Also we have JNS JMg = JMf; so
L
JNgJMg(a)d,CMa=
holds. Thus if we multiply all three (3.146).
L
JMf(a)dCMa
te~.in
(3.147) by JNS, we obtain •
Measure Theory of Lipschitz Maps We need to verify that the integrand on the right-hand side of (3.144) is measurable. (The measurability of the integrand on the left-hand side of (3.144) is given by Rademacher's Theorem 3.8.10.) First we obtain a useful preliminary estimate originally proved by Ellenberg and Harrold. Lemma 3.9.4 Suppose 0 ~ N :$ M < 00. There exists a constant C(M, N) such that the following statement is true: II I : liM -+ liN is a Lipschitz /unction and A C liM is'c M -measurable, then
holds. Proof: We may assume the right-hand side of (3.148) is finite. Fix u > O. By the definition of Hausdorff measure, there exists a cover of A by closed sets S1. S2, ... , all having diameter less than u, such that
~ TM •
(diar;(Si»)
M
:$
llM (A)
+ u.
140 For y
CHAPTER 3. MEASURES E .]RN
we observe that
Note also that if pES;, then !(S;)
c i(!(P), Lip(f) diam(S;») ,
so
Thus we have
r
iRN ~
1i~-N (A n rl(y» dCN y 2N - M "fM_N E(diam(S;»)M-N {
iRN
i
~
2N - M "fM_N"f N [Lip(f)]N
~
2N
. .
X/(s;)dC N
E (diam(S;»)N i
™:r:''fN
The result follows by letting LeIDIDa 3.9.5 Suppose mapping
(1iM(A)
(T
.J..
f : JRM
+0)-
•
O. -+]RN
is a Lipschitz function. Then the
is l£M -measumble. Proof: By the previous lemma, we can ignore sets of arbitrarily small measure, hence we may assume that A is compact. If A is compact, then
{y: 1i M -
N
(A n
r
1 (y)
~
t} =
nVj, j
here Vj is the open subset of JRN consisting of all points y for which An f-l(y) has a finite cover by open sets U; of diameter less than Iii such that
~T L.J i
M-N
(diam(U;») M 2
-N
+ !... J
•
3.9. THE GO-AREA FORMULA
141
Proof ot the Co-area Formula By Theorem 3.8.10 an~ (3.148), we may assume that D J(a) exists at every point a E A. First we prove the result under the additional assumption that JNJ(a) > 0 at every point of A. By LUBin's Theorem 1.3.12 we may assume that D J(a) is the restriction to A of a continuous function. By Egorofl"s Theorem 1.3.13 we may suppose that IJ(y) - J(a) - (DI(a),y - a)1 . I!I-al
converges uniformly to 0 as !I E A approaches a E A. It is plain that, for any £ > 0, conditions (i)-(Ui) of Lemma 3.9.3 are satisfied in any subset of A that has small enough diameter. Finally, to complete the proof, we need to consider the case in which IN J = 0 holds on all of A. In that case, the left-hand side of (3.144) is o. We need to show that the right-hand side of (3.144) also equals o. To this end, consider J. : lR,M+N -+ lR,N defined by . (x, !I) t-+ I(x)
+ EX.
We can apply what has already been proved to the .set A x [-I,I]N C ntM x ~.
We have .eM+N(A x [-1, I]N)
r
JAX[-l,ljN
= 2N .eM(A), JNJ.
JNJ.d.e M+ N = (
JRN
~
£ [£ + Lip(J)]N-l, and
ll M [(AX [-I,I]N) nJ;l(z)]d.eNz.
By (3.148) observe that G(M, N) llM [(A x
iN 1
?: =
[-1, I]N) n I.-l(z~
ll M - N [(A x [-1, I]N) n I.-l(z) n n-l(y)] d.cNy
[-l,ljN
llM-N[Anrl(Z-fy)]d.eNy. .
Thus 2N .eM (A) f [£ + Lip(J)]N-l
?:
(
JAX[-l,ljN
?:
1 G(M, N)
=
1 G(M, N)
=
G(:; N)
JNJ.d.c M+ N
r 1[-l,lIN llM-N[AnJ-l(z-£y)]d.cN!ld.cNz
JRN
1 iN
[-l,ljN
(llM-N[Anrl(z-£y)]d.cNzd.cNy
JRN
ll M- N [A n rl(z)] d.e N z
142
CHAPTER 3. MEASURES
holds, where the last equation holds by translation invariance. Letting E ../. 0, we see that
f 1lM - N [Anj-l(z)]dC N z=O. JRN
•
Chapter 4
Sobolev Spaces 4.1
Basic Definitions and Results
This book has given substantial attention to the C" spaces. Such spaces are suitable classes from which to select the defining function for a domain, and the C" spaces are' natural in a number of other geometnc contexts. However, in the study of partial differential equations and Fourier analysis, the Sobolev spaces are more convenient. The definition of the Sobolev spaces is less near the surface than that of the spaCes, but theorems about Sobolev spaces are more accessible. In the end, the Sobolev imbedding theorem allows one to pass back and forth between the C" spaces and the Sobolev spaces (however one must pay with a certain lack of precision that is present in the imbedding theorem). We will need the following basic definitions.
c"
Definition 4.1.1
(i) V is used to denote the space of infinitely differentiable functions on ]RN having compact support and topologized by uniform convefYence on compact sets. (ii) The Fourier transform of f E L1(JRN) is denoted by j and is defined by (4.1)
(iii) The Fourier Inversion Theorem (e.g. see Stein and Weiss /1j) states that if f, j E L1 (]RN), then f(x)
= [ j«(,)e27fiz·(dCN(, iRN
holds for almost every x E
]RN •
143
(4.2)
CHAPTER 4. SOBOLEV SPACES
144
For simplicity, we begin our discussion of Sobolev spaces on the domain RN, or all of space. We follow the discussion in Krantz [4). Definition 4.1.2 II tP
eV
and 8
eR
then toe define the nonn
(4.3) We let H-(lRN
)
be the cl08ure olV with respect to II • 11_.
In the case that
8
is a non-negative integer, then
Therefore
tP e H- if and only if if, . [EIQI~.I~IQl e L2. This last condition means that if,~Q e L2 for all multi-indices a with By the Plancherel theorem (see Stein and Weiss [1]), we have
(!)
Q
tP e j} Va such that
101 $
lal
$
8.
8.
Thus we have Proposition 4.1.3 118 is a non-negative integer, then
Here derivatives are interpreted in the sense
0/ distributions.
Notice that if s > r then H· C Hr because
The Sobolev spaces turn out to be easy to work with because they are modeled on L2-indeed each H· is canonically isomorphic as a Hilbert space to L2 (exercise). But they are important because they can be related to the more classical spaces of smooth functions. That is the content of the Sobolev Imbedding Theorem: Theorem 4.1.4 (Sobolev) Let 8> N/2. 1// e H·CIR.N), then I can be corrected, uniquely, on a 8et 01 measure zero to be continuous. More generally, il k e {a, 1,2, ...} and i/I e H·, 8 > N /2 + k, then I can be corrected on a set 01 measure zero to be C".
4.1. BASIC DEFINITIONS AND RESULTS
145
Proof: For the first part of the theorem, let I E H8. By definition, there exist 4>; E V such that 114>; - JIIH' -+ o. Then
114>; - JIIL2 = 114>; -
Iilo ~ 114>; -
III~ -+ O.
(4.4)
Our plan is to show that {4>;} is an equibounded, equicontinuous family of functions. Then the Ascoli-Arzela theorem (e.g. see Rudin [1]) will imply that there is a subsequence converging uniformly on compact sets to a (continuous) function g. But (4.4) guarantees that a subsequence of this subsequence converges pointwise to the function f. So f = 9 almost everywhere and the required assertion follows. To see that {4>j} is equibounded, we calculate that
14>; (x) I =
~
II
e- 21fiz • (~;{{) c.l{1
I 1~;({)I(1
+ 1{12)B/2(1 + 1{1 2)-B/2 c.l{
~ (/1~j({)12{1+1{12)Bc.l{r/2. (/{1+ 1{12)-Bc.l{r/2 Using polar coordinates, we may see easily that, for s > N/2,
1(1
+ 1{1 2)-B c.l{ <
00.
Therefore
14>; {x)1 ~ CII4>;IIH- ~ C' and {4>;} is equibounded. To see that {4>j} is equicontinuous, we write
14>;{x) - 4>; {y)1 =
II ~;({)
(e- 21fiZ :(
-
e- 2,,-i ll ·() d{1·
Observe that le- 21fiZ • ( - e- 2,,-ill • (I
~ 2 and, by the mean value theorem, 21fiz 2 le• ( _ e- ,,-ill· (I ~ 211"Ix - yll{l·
Then, for any 0
< f < 1, le- 21riZ ·( _ ~
211"< Ix
e-21rill·(ll-'le-21riz.~
_
e-21rill·~I'
- YI
Therefore
14>;{x) - 4>; (y)1
< C/
1~;{{)lIx -
~ Clx -
~
YI'
ylfl{I' c.l{
I 1~;({)I(1
+ 1{12)'/2
d{
Clx - YI'II4>; IIH- (/ (1 + 1{1 2)-B+<
d{) 1/2 ,
CHAPTER 4. SOBOLEV SPACES
146
where the last inequality is a consequence of the Schwarz inequality. H we select 0 < E < 1 such that -8 + E < -N/2 then We find that
!
(1 + leI 2)-s+f
d{
is finite. It follows that the sequence {tPi} is eciuicontinuous and we are
done. The second assertion of the theorem may be derived from the first by a • simple inductive argUment. We leave the details as an exercise. Remark 4.1.5 8 = N/2, then the first part of th~ ~heorem is false (exercise). In fact the sharp statement that one make is that in this, case,
(i) H
can
I lies in the space of functions of bounded mean oscillation or that I is exponentially integrable (see Stein [1], Folland and Stein [1], Adams [1]).
(ii) Theorem 4.1.4 may be interpreted as saying that HS C C" for 8 > k + N /2. In other words, the identity provides a continuous imbedding of HB into C". A converse is also true. Namely, if HB c C" for some non-negative integer k, then 8 > k +.N/2. . . To see this, notice that the hypotheses Uj -+ U in HB and Uj -+ v in C" imply that U = v. Therefore the inclusion of H" into C" is a closed map. It is therefore continuous by the closed graph theorem. Thus there is a constant C such that (4.5) Now, for x E JRN fixed and a a multi-index with lOll ::5 k, the tempered distribution e~ defined by
e~(tP) = (::0) t/J(x) is bounded in (C")· with bound independent of x and a (but depending on k). (See Stein and Weiss [1] or Krantz [4] for a discussion of tempered distributions.) Hence, by (4.5), {e~} forms a bounded set in (Ha). == H-II. As a result, for lOll :5 k we have that
lIe~ IIH-' = =
:5
(! 1(e:) 12 (1 + leI2)-e d{) 1/2 (! 1(-27rie)Oe2"';"'·(12 (1 + leI2)-s~) 1/2 C (J (1 + leI 2)-e+ d{) (e)
10 1
1/2
4.1. BASIC DEFINITIONS AND RESULTS
147
is finite, independent of x and a. But this can only happen if2(k-s) -N, that is if 8> k + N/2.
<
We cannot proceed any further without giving a precise definition of Sobolev spaces on a domain. In fact we shall give three such definitions, and in the next section we will compare and contrast them. Definition 4.1.6
(i) Let 0
c
lllN be any domain. Let s ~ 0 be an.integer. Define the norm
1If1IH'(O) =
E 11:~112
loiS-
L (0)
.
Let 'PB consist of those f e C 8 (11) for which IIfIlH'(o) is finite. This is clearly a linear space. Define HB(O) to be the closure of pB in the II • IIH'(o) norm. (ii) Let 0, II • IIH'(o) be as in the last definition. Let W8(O) consist of those f e L2(11) such that (aO /axQ)! e L2 for all lal ~ s. Here derivatives are, perforce,. interpreted in the weak sense. (iii) Let 0, II • IIH'(o) be as in the preceding two definitions. We define W3(O) to be the closure in the II • IIH' topology of the space 9~(11). Remark 4.1.7 (i) It is clear from the definitions that W3 c WB and W3(O) C HB(I1). That these inclusions are proper when s is large enough is clear from the Sobolev imbedding theorem. For if s > N /2 then an element of W3(O) would necessarily vanish on al1. And density in the II • IIH'(o) nonn would imply uniform density. Thus W3(O) could not be dense in either of the first two spaces. However more is true. In fact if s> 1/2, then W3(O) is not dense in WB(O), nor is it dense in HB(I1). The reader may try his hand at this as an exercise, or consult Taylor
[1]. (ii) It is a striking fact, discovered as recently as 1964 (see Meyers and Serrin [1]) that, for any domain 11 C lllN, H8(O) = WB(I1). The proof of this statement is a standard, but tricky, adaptation of the usual approximation by cutting off and convolving with a family of Friedrichs mollifiers. We omit the details, but refer the reader to
Adams [1]. Exercise: Imitate the proof of the Sobolev bnbedding Theorem 4.1.4 to prove Rellich's lemma: H s > r and if O2 CC 111 then the inclusion map
given by
I
t---+
1102
is a compact operator (see Krantz [4] for the details).
4.2
Restriction and Trace Theorellls for Sobolev Spaces
In the preceding section, we discussed the most basic facts about the Sobolev spaces. In this section, we will adapt some of our earlier results to bounded domains iIi space. First, we discuss the significance of the so-called "restriction theorem:" Let S = {(Xl, ... ,XN-I,O)} C,RN. Then S is a hypersurface, and is the boundary of {(XI, ... ,XN) E,RN : XN > OJ. It is the simplest example of the type of geometric object that arises as the boundary of a domain. It is natural to want to be able to restrict a function f defined on a neighborhood of S, or on one side of S, to S. If I is continuous on a neighborhood of S, then the restriction of I to S is trivially and unambiguously defined, simply because a continuous function is well-defined at every point. If, instead of being continuous, I is an element of the Sobolev space H N /2+< , then we may apply the Sobolev imbedding theorem to correct I on a set of measure zero (in a unique manner) to obtain a continuous function . . The corrected I may then be restricted to S. By contrast, restriction prior to the correction on the set of measure zero is prima lacie ambiguous. This is because a Sobolev space "function" is really an equivalence class of functions any pair of which need only agree up to a set of measure zero. The set S itself has measure zero, and thus two different elements of the equivalence class of I may have different restrictions to S. We wish to develop a notion of calculating the restriction or "trace" of a Sobolev space function on a hypersurface that applies to Hr for r < < N /2 and is such that the restriction operation works naturally in the context of Sobolev classes without relying on the Sobolev imbedding theorem. Our first result is a bound on norms. This bound, given in estimate (4.6), shows us when the trace of a Sobolev function is defined, thus the theorem is known as the Sobolev trace theorem. Theorem. 4.2.1 Identify S with ,RN-I in the natural way. Let s
>
1/2.
Then the mapping from C.;"'(,RN} to C.;"'(JRN-l} defined by 4> t---+ 4> 1s
extends to a bounded linear operator from H·caN ) to H S - I / 2 (,RN-l). That is, there exists a constant C = C(s} > 0 such that (4.6)
Remark 4.2.2
. (i) Since H·(R.N) and H·-l/2(R.N-l) are defined to be the closures of ~(JlN) and O~(R.N-l) respectively, we may use the theorem to conclude the following: H T is any {N -1}- dimensional affine subspace ofJlN, then a function f E H8{JiN} has a well-defined trace in H·-l/2 on T, provided that 8 > 1/2. Conversely, we shall see that if 9 E H8-1/2(T},8> 1/2, then there is a function 9 E HS(RN) such that 9 has trace 9 on T.
(li) It is a bit awkward to state the theorem as we have (that is, as an a priori estimate on O~ functions). As an exercise, the reader should attempt to reformulate the theorem directly in terms of the H8 spaces to see that in fact the statement of Theorem 4.2.1 is as simple as it can be made.
Proof of TheoreDl 4.2.1: We introduce the notation (X',XN) = (Xl, ... ,XN)
for an element of]RN. HuE Cg"(JRN), then we will ~se the notation ur(x') to denote u(x', 0). Now we have
111.Irll~'_1/2(RJV_l)
2 1fR N-l IUr(e') 12(1 + le'1 },-1/2 d{'
=
=.f [- foo {}{} IUI(e',XN}1 2(1 + le'1 2y-I/2 dxN ] .1RN-l 10 XN
df.'. (4.7)
Here UI denotes the partial Fourier transform in the variable x'. The product rule yields that, if D is a first derivative, then D(lhI 2 ) $ 21hl . IDhl. Therefore (4.7) does not exceed -
foo IUI(e', xN}1
2 {
1RN-l10
0/ 2
Since 20/{j $
$
·1 {}{}
XN
UI(e', XN}
I· (I + le'1 2},-1/2
dxN df.'.
+ {32, the last line is f
foo
1RN-l 10
I~UI(e',xN}12 dxN(1 + le'1 2},-1 df.' {}XN
foo IUI(e', XN }1 2 dxN(1 + le'1 2 }, d{'
+ f
1RN-l10
_
1+11.
Now apply Plancherel's theorem to term II in the XN variable. The result is II
~
0
~
Ollull~'(RN).
f f 1u(e',eN}1 2 df.N(I+le'1 2 )'d{' 1RN-l1R
150
CHAPTER 4. SOBOLEV SPACES
Planchere1's theorem, applied in the XN variable to the term I, yields
I
~
C {
~
C {
~
C
=
CliuIlH'(RN).
2 . {Iaa ii1(e"XN)1 dxN(1 + le'1 2 )B-l df.' JRN-IJ R ZN
JRN-l
2 2 2 J{R lu(e', eN )1 ·leNl deN (1 + le'1 )B-l df.'
kN lu(e)1
2 •. (1
+ lel 2 )B de •
As an immediate corollary we have: Corollary 4.2.3 Let u e HB(JRN) and let Q be a mul~i-inde:z: such that s> 101 + 1/2. Then DQu has trace in H B -l ol-l/2(RN-i). The reader will note that, in the trace theorem, it is only necessary to have u e HB+1/2(JR~) in order to obtain a well-defined restriction to JRN-l. In other words, the function .to be restricted need only be defined on one side of the boundary. Now we present a converse to the Sobolev trace theOrem. This converse is the extension theorem for Sobolev spaces. We continue to use the notation S = {(z', 0) e RN}. In this theorem it is convenient to let DN denote· i(8/8zN). TheoreDl 4.2.4 Let k be a non-negative integer, and suppose s > 1/2 + k. If if>o, •. . ,
IIfIlH'(RN) ::; C •. 1t
L lI¢ljll~'-H/2(RN-l).
j=O
Proof: Let h We define
e
C~(R),
h
and
fez) =
== 1 in a neighborhood of the origin, 0 ::; h::;
f
e- 27ris '
•
1.
fii1({',XN) de'.
This is the function f that we seek. For if m is any integer, 0 ::; m ::; k, then
D~f(z',O)
=
f
e- 27ris '
'('
D~ii1(X',XN)I"'N=o~'
4.2. SOBOLEV SPACE RESTRICTION AND TRACE THEOREMS
=
JeX
2 ,..iz'
0 ('
im
t ~, E(m) ~(-iXN)j I j=O ,. l=O t
J
e -2,..iz'
0 ('
~ .!.-, ,·m L.J
. J.
8:-~; [h(XN(l + le'12)1/2)]
J
aN=O
. ~(e') df.'
(m) .'(_ .);
j=O ,. ,
8xN
=
ZN=O
8:i:'N
8:-~l [h(XN(l + IfI2)1/2)]1
8x N
=
151
t
IaN=O .~(e') df.'
e- 2 ,..iz' 0(' ~(e') df.'
= c,t;m(X'), verifying our first assertion. Now we need to check that f has the right Sobolev norm. We have
IIfll~2
=
kN li(eW(l + le1 2
=
kNI(Ui(e',.»)~(eN)12(1+leI2Yd{
)6 dS
Using the fact that 12:;=0 (j12 .$ C· (2:;=01(;1 2 ) , where the constant"C depends only on k, we see that (4.8) does not exceed
Now write df.
= df.' df.N. We make the change of variables eN t-+ eN(l + le'1 2)1/2.
152
CHAPTER 4. SOBOLEV SPACES
Noting that
1 + lel 2 = 1 + le'I 2 + leNI 2
1 + le'I 2 + leNl2(1 + le'1 2) (1 + le'1 2)(1 + leNI 2),
1-+
=
we see that (4.9) does not exceed
C
"
~ 1 r (IA(j)( )12 1 1'12)1/2 f;;t, 0!)2 JRN-l JR h £.N (1 + le'1 2)Hl (1 + e xl~(e')12 [(1 + 1£.'1 2)(1 + leNI2)]8 de' deN
=
"
r
C ~ _.1_ (]!)2 JRN-l
f;;t,
x
"
::;
C];
::;
C•."
l~j(£.'W(1 + le'1 2)S-j-l/2 de'
Llh
W (eN)1 2(1
0~)211t/>j IIH'-i-
+ leN12)BdeN .
1/
2(lR N- 1)
•
IIhll~'+k(R)
L" lIt/>jll~'-i-l/2(RN-l)'
j=O .
•
completing the proof of our extension theorem. Remark 4.2.5
(i) The extension Ui was constructed by a scheme based on ideas that go back at least to A. P. Calderon-see Stein [1] and references therein. (ii) Consider a smoothly bounded domain n c JRN. Let {l!j}~=1 be a covering of an with open subsets of JRN. We may choose these open sets in such a way that there are diffeomorphisms ~j : Uj -t Wj C JRN such that ~i(aO) = Wi n {x E JRN : XN = OJ. This claim follows from the implicit function theorem. We may assume that there is a universal finite constant C such that IIV~jIl8uP ::; C and IIV[~j]-lllsup ::; C for each j. Let,pj be a partition of unity subordinate to the covering {Uj} of an such that Ei,pi := 1 in a neighborhood of an in JRN. Suppose that f E H8(O). Fix an index j, 1 ::; j ::; k. We write ~j(n n Uj) := V;. Then 9j == ,pi . f 0 (~j lies in H8(V;). According to our trace theorem for Sobolev functions on the upper half space ~, the function 9i has a trace on ~j(an). Unraveling all the notation, we find that ,pj . f has a trace on an. Summing over j, we conclude that f itself has a trace on an. We shall not examine
llvi)
4.3. EXTENSION THEOREMS FOR SOBOLEV SPACES
153
minimal hypotheses on the smoothness of 80 that will guarantee the validity of the trace theorem for elements of HB, 8> 1/2. Clearly it suffices to aSsume that the boundary is Cle with k > 8. Remarks similar to those in the preceding two paragraphs apply to show that the extension theorem has a valid fonnulation for Sobolev functions on the boundary of a domain with sufficiently smooth boundary. We shall say no more about the matter here.
4.3
Domain Extension Theorems for Sobolev Spaces
Next we turn our attention to domain extension theorems for Sobolev functions. Such theorems are related to the converse of the trace theorem, but they have a slightly different formulation. We will be concerned here with extending a Sobolev function from a given domain in JRN to all of space. It is easy to see that a successful extension theorem will necessitate some regularity condition on the boundary of the domain in question. For let us consider the situation in JR2 • Fix an integer k > O. Let
Define f(xI, X2) = I/Xl. If k 2: 2 then f E L2(Ok). If k 2: 4 then f E Wl(OIc). More generally, if k 2: 28 + 2 then f E W8(Ok). But for s > 1 we cannot expect f to extend to a function in W 8 (JR2). If it did, then the extended function would be (after correctioE. on a set of measure zero) continuous-by the Sobolev theorem. Since f is unbounded at (0,0), that conclusion is clearly impossible. Very sharp conditions on the boundary of a domain in Euclidean space, in order to guarantee a valid extension theorem, are known (see Jones [1]). We shall not attempt such a precise discussion, but shall instead treat some of the standard results. Definition 4.3.1
(i) Let 0 C JRN be a domain. A linear operator
is said to be a simple m-extension operator for 0 if 1. Eu(x) = u(x) for almost every point x E OJ 2. IIEullw'"(RN) $ K ·lIullw'"(o).
CHAPTER 4. SOBOLEV SPACES
154 (ii) With 0 as above, we say that
E : {functions on O} --+ {functions on RN}
is a total extension operator il the restriction 01 E to wm(o) is a simple m-eztension operator lor each non-negative integer m. We shall present three extension theorems for Sobolev spaces (and this three part unity is quite typical of extension questions for other function spaces too). In our discussion, we shall restrict attention to hounded domains in Euclidean space. This choice makes the statements of the theorems a bit cleaner. A treatment of unbounded domains may be found in Adams [1] and in Stein [1]. It is also possible to develop Soholev space theory for domains in a manifold-see Hormander [1]. .' . Theorem 4.3.2 II 0 C]RN is a bounded domain with cm boundary, then there exists a simple m-extension operator lor non-negative integer m.
each
Proof: First let us assume that 0 is the upper half space ~
= {(X17X2, ••• ,XN) E aN : XN > OJ.
(Of course this domain is not bounded, but that will be seen not to matter in what follows.) It is convenient to denote a point of 1R~ by (X',XN), where x' is a shorthand for (Xl, ••• ,XN-t). Set u(x) if XN > 0 Eu(x) = { m+1 ,. • Lj=l ~ju(x, -JXN) 1f XN :5 0,
where Ah ... , Am+1 are the unique solutions to the m+ 1 equations in m+ 1 unknowns given by m+1
L: (_j)A: Aj = 1
k
= 0,1, .. . ,m.
;=1 Also define the auxiliary operators
E u(x) _ { u () x a -
if
r::;::-+/ (- j)a
N
Aju(x', - ixN)
XN
>0
if xN:5 0,
for a = (ah ... ,aN) any multi-index with lal :5 m. IT u E C m (~) then it is straightforward to calculate that
Eu
E
Cm(]RN),
lal:5m,
4.3. EXTENSION THEOREMS FOR SOBOLEV SPACES {
ID'"Eu(xWdx=
JRN·
(
JR~
155
ID'"u(xWdx
density result Remaark 4.I.5(ii», i.e., Hm = This completes the proof of the theorem when the ·domain is the half space. IT now n is a bounded domain with em boundary, then we follow a by now familiar procedure of covering an with finitely many open sets Uj, each diffeomorphic to a ball, and each equipped with a mapping i)j : Uj ~ Wj C ]RN such that i)(Uj nan) = Wj n {x E ]RN : XN = O} and i)j(Uj n n) = Wj n ~. This, in effect, reduces the problem on n to that already treated on the upper half space. To wit, if I E wm(n) then one extends each of the functions 10 i)j -1 from Wj n ~ to wj, then pulls the extension back via i)j and patches all these together. Details are left to the reader. • By the previously
DOted
wm, these inequalities extend to all of W m.
A refinement of the proof of the last theorem yieids the following stronger result. We omit the proof, and refer the reader to Adams [I] for details. Theorem 4.3.3 If (l c aN is a bounded domain with there is a total extension operator lor n.
em
boundary, then
Using deeper techniques, related to the Cafder6n-Zygmund singular integral theory (for which see Stein [1]), it is ac~ually possible to prove the following still stronger result. Again we omit the proof. Proofs may be found in Adams [I] and Stein [I]. As previously noted, an even sharper result appears in Jones [I]. The result of Jones is precisely sharp in dimension two. Theorem 4.3.4 If n c aN is a bounded domain with Lipschitz boundary, then there is a total extension operator for n.
Chapter 5
Smooth Mappings 5.1
Sard's Theorem
Sard's Theorem is an important tool in differential topology and in Morse theory. We shall begin our discussion of Sard's theorem by treating the most elementary form of the result, and then shall later develop generalizations and variants of the theorem. It is worth noting that the basic idea here was discovered by A. B. Brown in 1935 (see Brown [1]). Later, in 1939 and 1942, the result was rediscovered by A. P. Morse (see Morse [1]) and A. Sard (see Sard [1]). The method of proof of Sard's theorem is quite robust and can be modified to yield a number· of interesting results. We shall close the section with a discussion of one of these variants that is known as "hard Sard." ]iN, V C:: JRM be open sets and let f : U -+ V be a C 1 function. We say that y E V is a critical value of f if there is "an x E U with f(x) = y and V' f(x) has rank less than M. The point x is of course called a critical point.
Definition 5.1.1 Let U C
We begin with the following special case of the theorem of Brown, Morse, and Sard: TheorelU 5.1.2 Let U c ]iN, V C]RM be open sets and let f : U -+ V be a C 2 function. Assume that 0 < N < 2M. Let C denote those points in U where V' f = O. Then the image of C under f has Lebesgue tJolume measure
equal to zero. Proof: For each N-tuple of integers k
= (k1 , ..• , kN) let
= {X=(Xb ... ,XN) E]iN :"kj $Xj $kj+1,j=1, ... ,N}. Then ]iN = UIcQIc and the QIc have disjoint interiors. Set QIc
Sic = {y E V: 11 E fCC) and 3x E QIc such that f(x) = y}.
157
CHAPTER 5. SMOOTH MAPPINGS
158
Then U"S" equals I(c), and it is enough for us to show that each S" has rp.e~ zero. Given this reduction, we may as well suppose that I has as its domain the cube Q = {z e ItN : O:oS zi:oS 1,; = 1, ... ,N}. Now the function V I is uniformly continuous on Q. Better yet, V 2 I is continuous on Q, and hence is bounded. It follows from the mean value theorem that V I itself is Lipschitz. Let K be the Lipschitz constant:
IV fez) - V I(z + h)1 :oS Klhl·
(5.1)
Let E > O. Choose a 6 > 0 such that if z, t E Q and Ix - tl < N5, then IV/(z)-V/(t)1 < E. In fact 6 = E/(KN) will do. We may certainly assume that 5 < E. Select an integerp, (1/6)+1 ~ p > 1/5. Note that 1/5,..., (KN)/E. Break Q up into rP closed subcubes, with side length equal to lip, sides parallel to the axes, and disjoint interiors. Consider anyone of the ~ sub cubes, call it P, that contains a critical point. The diameter of this cube is less than .JN5. And there is a point in the cube where V I vanishes. It follows that IV fl < E on the entire cube. We conclude, from the mean value theorem applied to I restricted to one dimensional segments in P (for instance), that the image of the cube P under I has diameter at most E·.JN5 < E· .IN/p. Thus it has volume at most (5.2)
Summing over all subcubes P that contain critical points (there are at most fCC) has volume at most
rP of these) we find that the set pN.
[2v'NE/P]M =
(2.JNE)M
.pN~M.
(5.3)
But now recall that p"" KN/E. Thus (5.3) is essentially
(2vNf)M (KN)N-MEM-N :oS (2.JN)M (KN)N-M 2M-N. f
Since E > 0 was arbitrary, and since 2M - N set I(C) has measure zero.
> 0, we conclude that the •
In fact Sard's theorem is true with no restrictions on the dimension of the domain and of the range, and it is true for the set of all critical values-not just the image of the points where the gradient vanishes. However a certain smoothness of the function f is necessary; examples that demonstrate the necessity of the smoothness requirements may be found in Federer [4], p. 317 JJ. The complete statement of the classical Sard theorem is this: Theorem 5.1.3 Let U C ]RN, V C IRM be open sets and let I : U -+ V be a C" function, where k ~ N/M. Then the set 01 critical values of I has Lebesgue M -dimensional volume measure equal to zero.
5.1. SARD'S THEOREM
159
Proof: In this proof we shall assume that the map is Coo, and we shall not give the more delicate proof that is valid for Cle functions, k ~ N / M. We refer the reader to Federer [4], p. 316 for a full account of these matters. The proof that we present comes from Milnor [1] and Sternberg [1]. We proceed by induction on N. Notice that the statement of the theorem makes sense, and is obviously true, for N = o. That is the starting point of our induction. Assume now that the result has been proved for dimension N - 1. Let C be the set of points where the rank of V f is less than M. Let C1 C C be the set of points where V f = 0 (this set was denoted by C in Theorem 5.1.2). More geneJ;ally, let C i denote the set of points where V f = 0, V 2 f = 0, ... Vi f = o. Of course each C i is closed and we have
We shall prove the following three assertions: 1. The image of C \ C1 under f has measure zero. 2. The image of Ci \
CHI
3. The Jmage of Cj under
under f has measure zero, i = 1,2, ....
f has measure zero if j is sufficiently large.
Step 1: If M = 1 then C = C 1 and there is nothing to prove. So assume that M ~ 2. Choose t E C \ C 1 • By definition, th~re is some partial derivative 8ftl8xj(t) that does not vanish. After juggling indices, we may assume that 8fd8xl(t) I: o. Define h(x) =·(ft(X),X2' ... '_~N).
Then the inverse function theorem implies that there i~ a neighborhood V of t on which h is a diffeomorphism onto an open set V' C RN. Now the composition 9 = f 0 h- 1 will map V' into RM, and the set C' of critical points of 9 is nothing other than h(V n C). In other words, the set g(C') of critical values of 9 is equal to f(V n C). So it is enough for us to determine the measure of g(C'). For a point of the form (S,X2, ... ,XN) E V', observe that the image g(S,X2, ... ,XM) belongs to the hyperplane {s} x RM-l. Therefore 9 maps hyperplanes into hyperplanes. Now define g. : [{s} x lIlN -
1]
n V' -+ {s}
to be the restriction of 9 to the hyperplane {s} Jacobian matrix of 9 has the form
(!
0 ) (8gf/8 . X t)
X
X
lIlM -
1
lIlN -1. Notice that the
160
CHAPTER 5. SMOOTH MAPPINGS
Thus a: point of {8} X jRN-1 is critical for g8 if and only if it is critical for the map 9 itself. By the inductive hypothesis, the set of critical values of g8 has M - 1 dimensional measure zero. By Fubini's theorem (taking the union over s), it follows that the set of critical values of 9 itself has M dimensional measure zero.
Step 2: Fix j, and consider a point t E C j \ Cj +1. By definition, some U + 1}1\ derivative of some component of 1 is not zero at t. Thus there is some ktb order partial derivative D, and index t, and another index m, such that w(t) == Dft(t} = 0 but 8/8x m Dft(t) ::F O. We may take m = 1. Then the map h : U -+ jRN defined by
is a diffeomorphism of some neighborhood V of t onto an open set V' C JR.N. The map h maps C j n V into the hyperplane {OJ x lIlN - 1 • Proceeding as in Step 1, we consider
We let" g* :
[{O}
X
jRN-l] n V'
-+
JR.M
denote "the restriction of 9 to the hyperplane {OJ x lIlN - 1 • By induction, the set of critical values of g* has (M - 1) dimensional measure zero. But each point of h( C j n V) is a critical point of g* (because aU deriVatives of order not exceeding j vanish by definition). Thus [g*
0
h](Cj
n V)
= I(C n V) j
has measure zero. Since we may cover Cj \ C j +1 by countably many such sets V, we may conclude that I(Cj \ Cj+1) has measure zero. Step 3: Refer to the proof of Theorem 5.1.2 above. For a point x E Cj we may replace line (5.1) in that proof by I/(x
+ h) -
l(x)1 ~
K'lhl j -
1•
Thus the estimate for the volume of the image of a cube in line (5.2) becomes
Since 6 < t, in the final calculation the term t 2M - N is replaced by t jM H j > N/M then the rest of the proof goes through. •
N •
5.1. SARD'S THEOREM
161
Implicit in the proof that we have just presented of Sard's theorem is the following remarkable lemma of A. P. Morse. We record it here for interest's sake, and invite the reader to g;.ve· a proof (details may also be found in Sternberg [1], p. 49). LeDlDlB 5.1.4 (Morse) Let A c RN and let q ~ 0 be an integer. Then we may decompose A fI6 A U~Ai such that: Let U be a neighborhood 0/ A and / : U -+ RM a C' mapping. Suppose that every point 0/ A is a critical point all. Then there are functions ei(f) (these junctions depend on f) such that each e, satisfies ei(f)·~ 0 monotonically fI6 f decreases to o and
=
lor any x, 11 E Ai. Remark 5.1.5 We note explicitly that the sets Ai in the decomposition of A in Lemma 5.1.4 are independent of the choice of the function I. Here is a typical application.of Sard's theorem. Example 5.1.6 Let I: ]RN ~ ]R be a C lc function, k ~ N, that is proper, where "proper" means that the inverse image of any compact set is compact. Then for almost every c E ]R the set Se
== {x
E]RN :
f(x) =c}
is either empty or is a C Ic compact manifold. To see this, notice that Sard's theorem guarantees that the set C of critical values for I has measure zero. If c is in the complement of C, and is in the image of I, then every point of f-l(cjis not a critical point. Thus V I "I. 0 in a neighborhood of each such point, and therefore 1-1 (c) is a CIc hypersurface in a neighborhood of each such point. We have established that if c ¢ C and if f-l(C) is nonempty, then it is a CIc manifold. The properness of f guarantees that it is compact. Now we give a formulation of a version of "hard Sard." We shall only sketch the proof, and refer the reader to Hirsch [1] or Federer [4] for a more detailed treatment.
Theorem 5.1.7 Let U C]RN, V C ]RM be open sets and let I: U -+ V be a C lc junction. Let us say that a critical value w is 01 rank l i/ w = lex) lor at least one point x at which the Jacobian matrix has rank not exceeding l. Denote by Ct the set 0/ points in V that are critical values of rank l. Then 'H.l+(N-l)/Ic(Cl)
In other words, the set 0/ critical l1alues not exceeding l + (N -l) / k.
= O.
0/ rank l has Hausdorff dimension
J.U":'
UHAP'l'ER 5. SMOOTH MAPPINGS
The spirit of the proof is that, near a point at which the Jacobian of / has rank l, the mapping will collapse a small N-dimensional ball in U to an essentially l- dimensional ball in the image. Since the latter balls cover the set Cl, it is clear that this puts a bound on the Hausdorff dimension of
.
~.
5.2
Extension Theorems
We have already (in Section 1.1) defined what it means for a function / : U -+ ]RM to be C" when U is an open subset of ]RN. A difficult question arises when / is not a priori defined on an open set. What should it mean in that case for / to be C"? An easy definition, which is very often used, is to require the existence of an extension of / to a larger open domain that contains the original domain and to demand that the extension be C" on that larger domain. While this definition is expedient, it really begs the question. One should be able to determine from information about / on 'its given domain whether or not / is in a particular smoothness class. It should not be necessary to appeal to an extension "oracle" to learn the answer. The replacement for an extension "oracle" is an extension theorem. The following classical extension theorem is well-known and can be found in many references (e.g. see Dugundji [1]). Theorem 5.2.1 (Tietze's Extension Theorem) Let S be a closed subset 0/ RN. A function / : S -+ JRM is continuous i/ and only if / has a continuous extension to all 0/ JRN . Tietze's Extension Theorem is a model for what we would like to accomplish: A characterization of the functions that are restrictions of those in a certain category. The preceding extension theorem is more topological than analytical, requiring little from the geometry of JRN • To move into the arena of analysis we deal first with Lipschitz functions. Recall from Definition 3.1.15 that / : X -+ Y, with X and Y subsets of Euclidean spaces, is a Lipschitz function if there exists C < 00 such that
and the smallest such C is called the Lipschitz constant of /. The following theorem is known as Kirzbraun's Extension Theorem. Theorem 5.2.2 (Kirzbraun [1]) Suppose S C ]RN. A function / : S -+ RM is Lipschitz if and only if / has a Lipschitz extension to all ofntN. The extended function may be taken to have the same Lipschitz constant. Proof: It is clear that the restriction of a Lipschitz function is Lipschitz. The heart of the matter is proving the existence of an extension. Let us
5.2. EXTENSION THEOREMS
use C to denote the Lipschitz constant for f: S-+ aM. In case M can simply use the followiilg fonnula, due to H. Whitney, to set F(x)
= sup{J(z) -
Cdist(z,z) : z E S},
= 1, we (5.5)
and easily check that this defines an extension of / with the same Lipschitz constant. In case M > 1, the fonnula (5.5) can be applied to each component to obtain a Lipschitz function, but the Lipschitz constant is generally larger than C. To obtain a Lipschitz extension having the same Lipschitz constant C requires a more elaborate construction. Consider the family :F of Lipschitz extensio~ of / (to some set T with SeT) which also have the Lipschitz constant C. This family is non-empty because·it at least contains the original function /. We define a partial ordering on :F as follows: Suppose 91 : T1 -+ aM and 92 : T2 -+ aM are both elements of :F, then we write 91 ~ 92 if and only if 92 is an extension of 917 i.e. S C T1 C T2 and 91(X) = 92(X), for all x E T 1 • (The same partial ordering is defined if we recall that a function from a subset of aN to lllM is a set of elements of lllN x lllM, and we partially order :F by inclusion.) By Hausdorff's Maximal Principle, :F has a maximal totally ordered sub-family !t. Let S be the union of the domains of the functions in !t. We define a function F : S -+ lllM by setting F(x) = 9(X) where 9 E !t and z E dom(g). One readily sees that Fis Lipschitz and that Lip(F) = C. We claim that S = lllN. IT not, then we can· fix Xo E lllN \ S. A contradiction will be reached if we can show that there is Yo E lllM such that
Iy -
whenever 11 = l"(x)
Yol :::; Clx - xol
and
xES,
that is, if we can show
o=F
n B(F(z),Clz - xol).
(5.6)
zes Since (5.6) involves an intersection of compact sets, it suffices to show that any such finite intersection is non-empty. Accordingly, let Xl, X2, ... ,X n E S be fixed. Set Yi = F(Xi)' Tj = IXi - %01, for i = 1,2, ... , n, and r* = sup{r1,r2, ... ,rn}. We know that for any sufficiently large value of 'Y n
K.., ==
nB(Yi, "Y7"i) ¥ 0. i=l
Set 'Yo
= inf{-y : K.., =F 0}.
Since (5.7)
164
CHAPTER 5. SMOOTH MAPPINGS
and the intersection of any finitely many of the sets on the right-hand side of (5.7) is non-empty, we see that
K'YO =/:0. It will suffice to show 'Yo S C. We may, of course, assume 'Yo > O. Note that K..,o must contain exactly one point, for if y', y" e K..,o, then we can use the following convexity argument:
I(y' + y")/2 - y,1 2
= Itl + 1/"1 2 /4 + ly,1 2 - (y' + y") • y, = I1l12/2 + ly"1 2 /2 - Iy' - y"1 2 /4 + ly,1 2 -y' . y, - y" • Yi
=
(ly' - y.1 2 + lIy" 2 2
S 'Yor, -
Iy' - y"1 2
2 _
'Yo
/2 - Iy' - y"1 2 /4
2
4(r*)2 ri'
holds for i = 1,2, ... , n, and (1/' + y")/2 'Y=
Yil 2 )
e
K.., with
Iy' - y"1 2 4(r*)2
< 'Yo,
contradicting the definition of 'Yo. By translating the coordinate system if necessary, we may assume to} = K..,o' Consequently, we have ly.1 S 'YOTiJ for i = 1,2, ... , n. We now claim that 0 is in the convex hull of {Yi : ly.1 = 'YoT'}' Were that not the case, there would exist an M - 1 dimensional plane separating the origin from {y. : IYil = 'YOTi}, but then for all sufficiently small E > 0 we would have B(O, E) on the opposite side ofthe plane from {Yi :"IYil = 'YOTi}, again contradicting the definition of 'Yo. Thus we can choose non-negative A1, A2, ... , An with and n
1 =
LAi, i=1 n
0
= LAW•. .=1
It follows that
5.2. EXTENSION THEOREMS
165
n
=
2
E
Ai Aj1li '1Ij
i,j=1 n
= E
AiAj [l1Ii1 2 + 11Ijl2 -11Ii -1IiP]
i,j=1 n
E
~
AiAj h~r1
+ 'Y~r~ -
~Ixi - xjl2]
i,j=1 n
= E
AiAj [2(Xi - xo)· 'Yo(Xj - xo)
i,j=1
=
I
n Ai (Xi - XO) 2 'Yo ~
+ h~ -
C 2)lxi - xjl2]
r
n AiAi IXi + (~- C 2 ) .J;1
xjl2.
(5.8)
If there were but one non-zero Ai, then the second term in (5.8) would vanish and the first would be positive, a contradiction. Thus there are at least two non-zero Ai'S and the second term in (5.8) must be non-positive, • forcing 'Yo $ C, as desired.
To discuss the extension of smoother functions, we first need to assess what the characterizing condition might be. To this end we consider Taylor's Theorem. It is elementary to see that, for a real-value~ function
~!
11
[
= _ ..!..¢(k) (0) k!
+
1 (k - 1)!
t [¢(k-1~~0) _
10
¢(k-1) (t)] d(l _ t)k-1
must hold. Induction leads to
~!
11
[¢(k) (0) - ¢(k) (t)] d(l _ t)k
(1 E ~¢(j)(0) + 10 [¢'(O) j=2 }. k
= -
¢'(t)] d(l - t)
0
k
= - E ~¢(j)(0) -
.p'(O) - ¢(O) + 1(1),
j=2}·
which (after rearranging the terms) is the standard Taylor formula with remainder. Now suppose U is a domain in ]RN that contains the points a and x and the line segment connecting them, and suppose f : U -+ ]R is C k on an open set containing that line segment. Define the real-valued
CHAPTER 5. SMOOTH MAPPINGS
166
function of one variable, 4>, by setting
It is easy to see that
for j = 0,1, ... ,k. Thus we obtain Theorelll 5.2.3 (Taylor's Theorelll) If /: U -t 1R is C le on an open set containing the closed line segment joining a and x, then Ie
/(x)
=
lea)
+~
L
(
)<> 8 101 / x :!a 8xo (a)
1=1101=i
+
[1
10
L 101=1e
(x - ,a)<>
°
[~o~ (a) 0_ ~o~
Q.
X
«1 _t)a + tX)] d(l _ t)le.
X
0
Definition 5.2.4 When
is considered as a polynomial in X1o ••• , XN it is called the Taylor polynOlllial for / at a or, synonymously, the k-jet of / at a. Relllark 5.2.5 It is easy to see that if j IPI = j, then 8(3 Po ( ) 8x(3 x is the (k - j)-jet of
at a. From Taylor's theorem follows:
~
k and
P is a multi-index with
5.3. PROOF OF THE WHITNEY EXTENSION THEOREM
167
e"
Corollary 5.2.6 II I: U ~ R. is on the open set U, A is a compact subset 01 U, P,.(x) denotes the k-jet 01 I at 0, and P is a multi-index with IPI :5 k, then lim 1{fJI (x) - {fJp,. (x) 1Ix 8xfJ 8x/J
z-+,.
al lPI - k =
0
holds unilormly on A. In the early 1930's H. Whitney proved the converse of this corollary to Taylor's theorem:
TheoreJD 5.2.7 (Whitney Extension Theorem) Suppose k is a nonnegative integer, A is a closed subset oj aN, and to each 0 E A there corresponds a polyriomiol in X1o ••• ,XN 01 degree not exceeding k such that Jor each multi-index P with IPI :5 k lim 1{fJ P" (b) - atJ Po (b)\lb 8x P 8xP
"-+0
al lPI - k =
0
holds unilormly. on compact subsets oj A. Then there exists a -+ 1R with 801 g 8OtPo 8xOl (a) = 8xOl (a)
(5.9)
e le
map g :
]RN
Jor each multi-index a with
lal :5 k
(5.10)
and each a E A.
We will give the proof of the Whitney Extension Thecirem in the next section.
5.3
Proof of the Whitney Extension Theorem
The Whitney Extension Theorem 5.2.7 allows one to extend a given function on a closed set A to a e le function on all of ]RN , provided there exist prospective Taylor polynomials for the function that satisfy the a priori bounds that Taylor's Theorem requires of a C le function. The fundamental geometric tool needed for the proof of the Whitney Extension Theorem is the Whitney decomposition of aN \ A. A Whitney decomposition of an open set U is a family of closed cubes, with disjoint interiors, the union of which equals U. In addition, each cube in a Whitney decomposition of U must have its diameter bounded above and below by constant multiples of the distance from the cube to the complement of U. Whitney's original construction of what is now called a Whitney decomposition was a paragon of simplicity (see item 8 on page 67 of Whitney [1]), but not all the cubes had their diameters bounded from below by a multiple of the distance from the cube to the complement of the open set. Our version of the proof retains the simplicity of Whitney's construction while obtaining the lower bound for all cubes.
CHAPTER 5. SMOOTH MAPPINGS
168
Theorem 5.3.1 (Whitney Decomposition) Let 1:5 r 1 and 2r1 + 1:5 r2 be fixed. For any non-empty closed subset A of R,N, there exists a family W of closed cubes with sides parallel to the coordinate axes such that
= UcewC, en 0' = 0, if C, c' E W with C i: C',
(i) R,N \ A (ii) (iii)
r 1· diam(C) :5
dist(C,A) :5
r2 . diam(C),
for C E W.
Proof: Assume without loss of generality that 0 E A. Let 'Ro denote the family of 3 N - 1 cubes, aU of side length obtained by taking all possible cartesian products of N intervals chosen from the set
i,
{[-I, -1], [-1,1], [1,1]}, but omitting the N-fold cartesian product of [-1, 1] with itself. For k any integer, let 'R" denote the family
{3"C: C
E
'Ro}
of cubes. For k > 0, the cubes.in 'R" are dilations of the cubes ·in 'Ro; for k < 0, the cubes in 'R" are contractions of the cubes in 'Ro. Let 'R denote the union of all the families R".(see Figure 5.1), so 'R=
U'R".
"eZ Note that if C is a cube in Ric; then diam(C)
=
2VN3"-1,
dist(C,A)
:5
dist(C,O):5 VN3"-t,.
thus any cube C E 'R satisfies dist(C, A) :5
! diam(C) :5 r 2 • diam(C).
(5.11)
Of course, the cubes in 'R have pairwise disjoint interiors and sides parallel to the coordinate axes. Let Wo denote the family consisting of all cubes C in 'R that satisfy
r 1 • diam(C) in addition to (5.11). Set
:5 dist(C,A),
(5.12)
Bo = 'R \ Woo
Let m be a non-negative integer. Suppose the family of cubes Bm has already been formed, and suppose that for each cube C E Bm the inequalities (5.11) and (5.13) dist(C,A) < r 1 ·diam(C)
5.3. PROOF OF THE WHITNEY EXTENSION THEOREM
169
~
Figure 5.1: Initial Decomposition both hold. This is the case for 80. Let 8!,. be the set of cubes formed ·by subdividing each cube of 8 m into 2N congruent subcubes with pairwise disjoint interiors and sides parallel to the axes. If C' is one of the cubes of 8!,. formed by subdividing the cube C of 8 m , we see that diam(C')
=
~diam(C),
dist(C', A)
:5
dist(C,A)
+ diam(C')
< r 1 • diam(C) + diam(C') =
2r1 • diam(C') + diam(C')
:5
r 2 • diam(C').
So every·cube in 8!,. satisfies (5,11). Let Wm+l be the family consisting of all the cubes in 8!,. that satisfy (5.12), and let 8 m +l be the family consisting of all the cubes in 8!,. that satisfy (5.13). Thus 8 m + 1 satisfies the same condition assumed for 8 m , and the construction can continue inductively. Set 00 W=
U Wm · m=O
It is clear from the construction that every cube in W satisfies the required inequalities, but also UW exhausts RN \ A because
and liN \ A is open.
•
Definition 5.3.2 For a non-empty closed set A c aN, a collection, W, of closed cubes with sides parallel to the coordinate axes and satisfying (i)-(iji)
CHAPTER 5. SMOOTH MAPPINGS
170
0/ Theorem 5.3.1 will be called a Whitney decomposition of aN \ A or a Whitney family for A. Whitney's definition of the.extension of a function from A to aN \ A is given by a summation over all the cubes of the Whitney decomposition of JIlN \ A : For each cube the summand is a cut-off function times one of the hypothesized polynomials Pdx), where e A is a point as near as possible to the cube. The cut-off function associated with a given cube is a translation of one fixed function with support scaled to match the diameter of the cube. To see that the extension formula comes out correctly, we will need to establish some facts about that cut-off function and its derivatives. Such is our next task.
e
Definition 5.3.3 For C c JlN" a cube, let center as C, with sides parallel to the sides that o/C.
a denote the cube with the same 0/ C, but with side-length double
Lemma 5.3.4 Let 1 ~ r 1 and 2r 1 + 1 :$ r 2 be fi:xed. There e:z:ists a constant K = Krl.r2 such that whenever A is a non-empty closed subset of liN, W is a Whitney decomposition of JIlN \ A. and p e JIlN \ A, then
(i) C E W ==>
an A = 0,
(ii) card{ C E W : p E C} ~ K. Proof: (i) Arguing by contradiction, if we suppose a E anA, then we can consider the line through a and the center, c, of C and let x be the point of BC nearest to a and x' the point of BC farthest from a. Then we have
la-xl + Ix- cl = la':""cl ~ 2lx-cl = Ix-x'i <
diam(C),
hence dist(A, C)
~
la - xl < diam(C)
:$ rldiam(C),
contradicting condition (iii) of Theorem 5.3.1.
(ii) Fix p e JIlN \ A and set 6 = dist(p, A) > O. We will eatablish the existence of constants Cl and C2 such that if pea, then C
c B(p, Cl 6)
and
diam( C) 2:: C2 6.
Since the cubes in W are pairwise disjoint, we see by considering the Ndimensional volumes that it suffices to take K
= U(N) (~~)
N
NN/2,
5.3. PROOF OF THE WHITNEY EXTENSION THEOREM
171
where r N denotes the N-dimensional volume of the unit ball in JRN. Assuming pEe, we easily estimate dist(P, C} ~ ~ diam(C).
(5.14)
Also, it is clear that dist(p, A) - dist(p, C)
~
dist(C, A) ~ dist(p, A)
+
dist(p, C)
holds. By condition (iii) of Theorem 5.3.1, we also have diam(C)
~
r 1 diam(C) ~ dist(C,A) ~ r 2 diam(C).
First we obtain diam(C)
~ ~
~
dist(C, A) dist(p, A) + dist(p, C) 6 + ~ diam(C) ,
so that .diam(C)
~
26
and, by (5.14), dist(p, C)
~
6.
Clearly if r = diam(C) + dist(p, C), then C c i(p, r), so we can take Cl = 3 and obtain C C i(p, c 1 6). N ext we estimate
6
so we can take
C2
= ~
= ~
dist(p, A) dist(p,C)
~
~diam(C)
=
(~+r2)diam(C),
+ dist(C, A) + r 2 • diam(C)
•
+ r 2•
Definition 5.3.5
(i) For the remainder of this section, fix some Coo function ¢> : JR with the following properties
o ~ ¢>(t) ~ 1 for all t E IR, [-~ - E, ~
+ E] C {t: ¢>(t) =
supp¢> C (-1,1).
~
JR
(5.15) I}
for some
E
> 0, (5.16) (5.17)
:>M.UU'l'l1 M.APPlNG::;
l..I.DJU· H!df. l>.
,1/'"
(ii) For a cube C with sides parallel to the coordinate axes centered at p ='(Pl,P2, ... ,PN) and with side length s > 0, that is N
C=
II[-!+p;,!+pd, i=1
we set
-
,pc{x)
'
= lIN ,p (Xi --Pi) - . ;=1
s
Based on Definition 5.3.5, the following result 'is plainly true. Lemma' 5.3.6
(i) For any closed cube C with sides parallel to the coordinate axes and having non-empty interior, the Junction ~C is supported in the interior of C and is identically 1 on C. (ii) .If W is a Whitney family for the
non-e~pty
closed subset A, then
L:0 ~c(x) ~ 1 > 0 for all E JRN \ A, L: ~c(x) =Oforall~EA, X
(5.18) ,
CeW
(5.19)
GeW
and there exists a family of constants {I(o} (determined by the par'ticular choice of,p) such that (5.20)
holds for each multi-index Q. Lemma 5.3.7 If W is a Whitney family for the non-empty closed subset A and
then there exists a family of constants
{Ko} such that (5.21)
holds on JRN \ A.
5.3. PROOF OF THE WHITNEY EXTENSiUN 'l'Hl!JUHEM
173
For the proof of Lemma 5.3.7, we will use the formula for the derivatives . of the composition of two functions. In the one variable case, this is known as the Formula of Faa. de Bruno (see Krantz and Parks [2], Lemma 1.3.1).· For the situation at hand, we will not need the actual values of the coefficients. We leave it to the reader to use induction to verify the formula. Proposition 5.3.8 (Generalized ForlDula of Faa de Bruno) There are positive integer coefficients qA,o) such that, for each multi-index A E AN, iff and 9 are functions of class CI).1 and h = 9 0 f, then D).h(x)
=L A
II
qA,o)
g{k) [f(x)] (DO f(x»A{o) ,
oEAN
holds, where the sum extends over functions A : AN
L
-t
Z+ such that
A(a) a =A,
OlEAN
and where k = k(A) is defined by
Proof of LelDlDa 5.3.7:· Set get) = lIt and <1>(x) =
L
¢c(x),
CEW
then h = 9 0 <1>. Now, since <1> is bounded below on IR \ A by 1, we see quite simply that Ig{k) [<1>(x)] k!
I: :;
holds on IR\A. By the upper bound (5.20) and the bound on the number of non-vanishing ¢c at any point of IR \ A, we have a bound on D°<1>(x), and the lemma follows from the Generalized Formula of Faa de Bruno. • In the next definition, we construct a partition of unity for ]RN \ A. It is crucial that we can estimate the size of the partial derivatives of the functions making up this partition of unity and that all those functions vanish on A.
Definition 5.3.9 (Whitney Partition of Unity) Suppose W is a Whitney decomposition for the non-empty closed set A.
(i) For C E W, we set ¢c(x)
= ¢c(x) /
(
L C'EW
¢c' (X») for x E IRN \ A,
CHAPTER 5. SMOOTH MAPPINGS
174
so
E
q,c(x) = 1 for all x E RN
\ A,
(5.22)
CeW
E
q,c(x)
= 0 for all x EA.
(5.23)
CeW
There e:Eists a family of constants {Ko} such that (5.24)
holds for each multi-index Q.
(ii) Let
e: W
~
A be such that dist(C, A) = dist ({(C), A)
(5.25)
holds for each C E w. Of course, the choice of e(C) need not be unique; any point meeting the condition may be chosen. Remark 5.3.10 The constants Ko in (i) depend not only on the choice of 4>, but also on the value of Krt.r2 from Lemma 5.3.4. Now that we have dealt with the preHrninaries regarding the Whitney decomposition and the associated partition of unity, we can pr~ceed to the actual proof. Proof of the Whitney Extension Theorem 5.2.7: Let k be a nonnegative integer and let A be a non-empty closed set. Suppose that to each a e A there corresponds a polynomial
of degree not exceeding k, such that the limits in the hypotheses of the Whitney Extension Theorem are attained uniformly on compact subsets of A. Recall those limits are Eq. (5.9) lim 1 8PP" (b) - aPPo (b)llb- al lPH
"-'0
8xP
8xP
= o.
Choosing r 1 > 2 in Theorem 5.3.1, we let W be a Whitney decomposition for liN \A. The desired function g can simply be written down by setting
g(x) = {
P.,(x)
forxEA,
Lcew 4>c(x) p(c) (x)
for xE RN
\A.
(5.26)
5.3. PROOF OF THE WHITNEY EXTENSION THEOREM
175
The difficulty lies in showing that 9 is C le and has the polynomial P" as its Taylor polynomial of degree k at a E A. Obviously, 9 is infinitely differentiable on RN. \ A and D"'g(x)
=
L L (~)D,8cf>c(X)D"'-,8P«C)(X)
(5.27)
CEW,8S'"
for x E RN \ A and any multi-index a. For any multi-iildex a, with lal $ k, we define a function T", : RN ~ R by setting for x E A, for x E RN \A. Also, for any any multi-index a, with lal <:: k, and any x E RN, we define a linear function L",(x) : RN ~ R by setting N
(L",(x), v}
= ~)v. ei) T",.+ (x), i=1
where
for j = i, forj # i.
We will show that, if a is a multi-index with lal $ k, then T", is continuous at every point of A, and if lal < k and a E A, then lim IT",(x) - T",(a) - (L",(a), x - a}1 = O.
Ix - al
"-:>,,
This will verify that 9 is C le on all of RN and that the Taylor polynomial of 9 at any point a E A coincides with Pa. For notational convenience, in that which follows let us assume that A is compact. Then we may set
€(8) = sup {
ID"'Pa(a)-D",p,,(a)1
la _ ap"'l-k
• • 0 < la -
: a, a E A,
al $
8, 0 $
lal
$ k
}
.
By hypothesis, €(o) .j. 0 as 0 .j. O. We have broken the details of the proof up into six steps. The first step will show that T", is continuous on A. Step 1: Suppose We have
lal
$ k, and consider
a ~ a with a,a E A.
IT",(a) - T",(a)1 = ID'" Pa(a) - D'" P,,(a)1 $
ID'" Pa(a) - D'" p,,(a)1
+ IDa P,,(ii) -
DO P,,(a)l.
(5.28)
CHAPTER 5. SMOOTH MAPPINGS
176
As a ~ a, the first term in line (5.28) approaches 0 because of (5.9) while the second term approaches 0 because of the continuity of DOt Pa. The second step will show that the difference quotient goes to zero when attention is restricted to A. The estimate is almost as simple as in Step 1. Step 2: Suppose lal < k, and consider Similarly to Step 1, we have
a ~ a with a, a E A.
(5.29) (5.30) As a ~ a, the term in line (5.29) approaches 0 again because of (5.9) while the term in line (5.30) approaches 0 by Taylor's Theorem applied to DOt Pa. When we consider approaching the point a E A through points lying outside of A, the situation becomes more complicated. In Steps 3 and 4 we establish a pair of preliminary estimates that will facilitate. the rest of the argument. Step 3: Suppose lal ~ k, and consider a,a E A and x E lRN. We apply Taylor's Theorem to the polynomials DO: Pa and DOl P a at the point a to estimate
IDO: Pa(x) -
<
no: Pa(x)1
L
ID1'+O:Pa(a)_D'Y+O:Pa(a)llx-~lbl
O:5I1'I:5k-lo:l
~
€(lii - aD
'Y.
L 0:511'19-10:1
Iii _ alk-I1'I-lo:llx - ~Ibl . 'Y.
(5.31)
Comparing the summation in line (5.31) with the binomial formula, we conclude that there exists a constant r 3 such that (5.32) holds. Step 4: Suppose lal ~ k, and consider x E
]RN \
Ix - iii = dist(x, A).
A and ii E A such that
5.3. PROOF OF THE WHITNEY EXTENSION THEOREM
177
Noting that
DOt Pa(x)
=
DOt
[L
rJ>c(x) Pa(X)]
CEW
L L (~)DPrJ>C(X)DOt-PPa(X)'
=
CEW P$Ot
we see that
ITOt(x) - DOt Pa(x)1
(~)DPrJ>C(X)DOt-pP(C)(X)
L L
=
- DOtPa(x)
CEW 13$0.
~
L L (~) IDprJ>c(x) I IDo.-pp(C) (x) -DOt - i3Pa(x)l· (5.33)
CEWp$Ot
.
We will estimate the term in line (5.33). Consider
IDprJ>C(x) I IDo.-p p(C) (x) - Do.-p Pa(x)1 ' . where we may assume x E C, since otherwise DprJ>c(x) have
= O. By (5.24), we (5.34)
We now need to show that Ix - iii is comparable to diam(C) and that and I~(C) -iii can be bounded by a constant multiple of diam(C). To see this, let p E C be such that I~(C) - pi = Qjst(C, A). We have I~(C) -xl
dist(C, A) ~ Ip-dl ~ Ip-xl+lx-iil, so Ix - iii
-Ip - xl
~
dist(C, A)
~
dist(C, A) - diam(C) dist(C, A) - 2diam(C) (r 1 - 2) diam(C).
~ ~
(Recall we chose r 1
Also, we estimate
r2
•
diam(C)
~
dist(C, A)
=
Ip-~(C)I
~
Ix - ~(C)I-Ip-xl Ix - ~(C)I- 2diam(C) Ix - iii - 2 diam(C).
~
~
> 2.)
CHAPTER 5. SMOOTH MAPPINGS
178 Thus we have
(f2 + 2)diam(C) ~ 1% - iii ~ (rt - 2)diam(O)." "
(5.35)
Next, we can estimate le(O) - iii
$
le(O) - pi + l1>-zl + Iz - iii
$
dist(O,A) +diam(C) + dist(%,A)
$
f
$
(3 + 2r2 ) diam(C).
2 •
diam(C) + 2diam(C) + dist(%, C) + dist{C,A)
Thus we also have I{{C) - %1 $ I{(C) - iii + Iii - zl $ (5 + 3r2)diam(O).
(5.36)
Returning to the estimation of the term in line (5.33), we can apply (5.32) and (5.36) to conclude that
IDa-.B p~(C)(z) -
Da-.B PA(Z)
I
$ r3 E({(C),ii) [diam(C)] k-1al+I.BI.
Combined with (5.34) this shows us that there exist constants r 4 and rs such that "
In Step 5, we will complete the "demonstration of continuity of Ta.
Step 5: Suppose 101 $ k, and consider z -+ a with z E Let a E A be such that
]RN \
A.
1% - al = dist(z, A). Clearly, we have 1% - al $ 1% - ai, so Iz - iil-+ 0 and a -+ a as % -+ a. We estimate ITa(z) - Ta(a)1
=
ITa(%)-DapA(%)+DapA(z)
+ D a Pa{z) D a PA(%)I + IDa PA(z) _Da Pa(z)
$
ITa{%) -
+ IDa Pa(%) -
D a Pa{a)l.
Ta(a)1 D a Pa(%)1
(5.38)
(5.39)
5.4. APPLICATION OF THE WHITNEY THEOREM
179
As z -. a the term in line (5.38) approaches 0 by the estimate (5.37) in Step 4, the second term in line (5.38) approaches 0 by the estimate (5.32) in Step 3, and the term in line (5.39) approaches 0 by the continuity of Dapa• Step 6 will complete the demonstration of differentiablity of Ta. Step 8: Suppose lal < k. Again let ii e A be such that Iz -
iii = dist(z, A).
We estimate
ITa(z) - Ta(a) - (La (a), z - a)1 Iz-al
< ITa(z) - DOl Pc\(z)1 -
Iz-al
·IDa Pc\(z) - DOl p .. (z) I Iz-al
+~--~~--~~~
+
IDa Pa(z) - DOl Pa(a) - (La(a), Z - a)1 Iz-al··
(5.40) (5.41) (5:42)
The term in line (5.40) approaches 0 by the estimate (5.37) in Step 4, the term in line (5.41) approaches 0 by the estimate (5.32) in Step 3, and the term in line (5.42) approaches 0 by Taylor's Theorem applied to the polynomial DOl P a . •
5.4
Application of the Whitney Extension Theorem
The so-called Cantor function provides a well-known example of a function on the unit interval that is non-decreasing and non-constant, but for which the derivative vanishes almost everywhere. By a similar construction H. Whitney was able to construct a function on an arc in the plane that can be extended to a C 1 function in the whole plane with the amazing property that the entire arc consists of critical points of the function, but the function is not constant on the arc. The extension of the function from the arc to the whole plane is accomplished by the Whitney Extension Theorem 5.2.7. We base our construction on Whitney [2]. The Cantor Set We form a Cantor set C in the closed unit square [0,1] x [0,1] as follows: Fix 0 < ~ < 1/2. Let C 1 consist of four closed squares of side length ~
CHAPTER 5. SMOOTH MAPPINGS
180
-[I]jj]
-~
(a) First Stage
(b) Second Stage
Figure 5.2: Forming the Cantor Set centered in the four quart«;!x:s of the unit square, as in Figure 5.2(a). Let us denote these squares by So, Sl, S2, S3 in positive cyclical order and, to be precise, let So be the square in the bottom left quarter of the unit square. Notice that the distance from any square Si to the exterior of its quarter of the unit square is ~ (~ - ~) so, if i 'I i, then dist(Si, Sj) ~ (! - A). Let C 2 consist of sixteen closed squares of side length A2 formed by applying the preceding co~struction with each square Si taking the role of the unit square. These siXteen squares are illustrated in Figure 5.2(b). The subsquares of Si will be denoted by SiO,Sil,Si2,Si3 again labeled in cyclic order, but we will specify later whether this is to cycle in the positive or negative direction and we will specify later which subsquare is designated Sm. As before, we can estimate the distance between subsquares:. if i2 'I i~, then dist (Sit i2 , Sil i;,) ~ A (~ - ~). Continue this construction to form Cs , C4 , ••• where Cle consists of 4i squares of side length Aj, with those sqllares denoted by Sil i2 ... i.' The distance estimate extends to dist(Si,i2 ... i._, i., Sili2 ... i._l;~) ~ ~Ie-l The Cantor set is then
(! -~)
if
ile =1= i~.
(5.43)
00
C= nCIe. 10=1
Any point p E C is associated with a unique sequence (iI, i2, ...) such that 00 {p} =
n
S;I;2 ...;.·
(5.44)
1e=1
Thus we can easily define a function
i : C ~ III by setting 00
i(P) = Likpk.
(5.45)
10=1
Lemma 5.4.1 Suppose p, q E C with corresponding sequences (il, i2,"') and (il,§2, ... ). Iiil =il' i2 =§2, ... , ik=i" andile+! =l=ile+!, then
(5.46)
5.4. APPLICATION OF THE WHITNEY THEOREM and I/(P) - f(q)1
~
181
31",,.+1 •
(5.47)
-'"
•
ProoC: This follows easily from (5.43) and (5.45). Corollary 5.4.2 There exist constants 0 < a and p,q E C with P:F q, then
I/(P) - l(q)1 ~
r <
riP - ql""·
00
such that
il
(5.48)
ProoC: The corollary follows by solving for k in both (5.46) and (5.47) and comparing the results. We find log", logA'
a=--
r
•
= 1 ~'" (~ _ A)"" •
Remark 5.4.3 The estimate in Corollary 5.4.2 shows that, if we let K be the greatest integer strictly less than a, then I can be extended from C to ]R2 as "a C K function with every point of C a critical point. This is a consequence of the Whitney Extension Theorem 5.2.7 applied with the polynomial Pa at a E C chosen to be the constant polynomial 1(0). While K can be made as large as we wish by choosing IJ small, the resulting set of critical points is totally disconnected. Also, if we consider the set of numbers I(C), all of which are critical values, this is also generally a disconnected set. A crucial observation is that jf IJ then I(C) [0,1]. This choice of '" is governed by fact that we have put four subsquares in each square at every stage of the construction. ~"
=h
=
The Arc From now on we set 11.-
1
r-4
and choose
(5.49)
We describe how to run an arc A given by I{) : [0,1] -t ]R2 through the Cantor set C. The construction proceeds in stages in concert with the formation of the Cantor set. Simultaneously, we will extend I from the Cantor set C to the arc. First Stage: The following construction is illustrated in Figure 5.3(a).
(i) Starting at the point (0,0), run a line segment to the bottom left comer of the square So. Define I{) on the interval [0, II to map linearly
onto this line segment, with 0 mapping to (0,0). The value of I on this part of the arc will equal the minimum value of I on the set So n c, i.e. O.
CHAPTER 5. SMOOTH MAPPINGS
182
(a) First Stage
(b) Second Stage
Figure 5.3: Running an Arc through the Cantor Set
(ii) From the bottom right corner of 8 0 run a line segment to the bottom left comer of SI. Define I{J on the interval [i, i] to map linearly onto this line segment, with i mapping to the bottom right corner of So. The value of 1 on this segment will be the maximum of 1 on the set 8 0 n C, whicl& also equals the minimum "alue 01 1 on the set SI n C. That value is ~. (This coincidence of the maximum of 1 on So n C and minimum of Ion 8 1 n C is a result of choosing JJ = ~.) (iii) From the top left left corner of 8 1 run a line segment to the bottom left corner of S2. Define I{J on the interval [~,~] to map linearly onto this line segment, with .~. mapping to the top left corner of SI. The value of 1 on this segment is t which is the maximum of 1 on SI n C and the minimum of f on 82 n C. (iv) From the top left corner of 8 2 run a line segment to the right corner of S3. Define cp on the interval [f,~] to map linearly onto this line segment, with \ mapping to the top left corner of S2. On this segment, set 1 equal to i' the maximum of 1 on 8 2 n C and the minimum of f on S3nC. (v) From the top left corner of 8 3 run a line segment to the point (0,1). Define cp on the interval [J, 1] to map linearly onto this line segment, with mapping to the top left corner of S3. On this segment, set 1 equal to 1, the maximum of f on 8 a n C.
I
The arc cp has been defined on ~,I]u[i,i]u[t,~]u[f,~u[J,~.
Define CPl : [0,1] -+
]R2
by extending
[I,
n [I,
I{J
linearly to the missing intervals
~], [~,
f),
[~,
lJ.
Second Stage: Scale the previous construction by a factor of ~ and rotate and translate to fit it in the subsquares Si and to connect with the part of
5.5. A MULTIDIMENSIONAL FUNDAMENTAL THEOREM
183
the arc already constructed. At this time, we can determine the ordering of the labeling of the squares Sij, so the second index increases as we proceed along the arc. In the process, the definition of
81'
1+91] 81
'
[2+91 ~] 81'
81
,
[4+91 5+91] 81'
81
'
[6+91 7+91] 81'
81
'
[8+91 9+91] 81'
81
which are subsets of I', 1¥], for l = 1,3,5,7. This is illustrated in Figure 5.3(b). The values of I are again set by the maximum and minimum process used in the first stage. . Define
Theorem 5.4.4 Let A be the arc constructed above and let I be the function defined on A as above. Then the function I can be extended to a C1 function on all of]R2 with every point 01 A a critical point and with . I(A) = [0,1]. Proof: Since I-' < >. we have 1 < u in Corollary 5.4.2. The distance between points of A \ C that have differing values is at least as large as for points of C, so the estimate (5.48) extends to all of A.__ •
5.5
Multidimensional Versions of the Fundamental Theorem of Calculus
Introductory Remarks A cornerstone of multi variable real analysis is the generalization, in its various guises, of the Fundamental Theorem of Calculus from a bounded interval in ]R1 to a (smoothly) bounded domain in ]RN. Already in this informal statement we can begin to sense a profound difference between the one dimensional setting and the multi-dimensional setting. For in one dimension any open set is a disjoint union of intervals; therefore it is reasonable to take a bounded open interval as a "typical" open set. In several real variables there is no typical open set. In many applications, the Whitney decomposition (Theorem 5.3.1) serves in several variables as a substitute
184
CHAPTER
o.
SMOOTH MAPPINGS
for the easy structure theorem for open sets in R1. In the present section, however, we will deal with a smoothly bounded domain in {} c a.N just as it is given. . Let J be a C1 function defined on O. The goal of any multidimensional FUndamental Theorem of Calculus is to relate the integration of I on the boundary of {} to the integral of some derivative of J on the interior of O. Recall that, in dimension one, the first half of this syllogism is almost trivial: It consists ~f evaluating J (with appropriate signs, or orientation) at the endpoints of the interval. In higher dimensions, the boundary will be a smooth manifold of dimension at least one, and the boundary integration becomes a substantive consideration. At the end of the section we shall briefly describe de Rham's theorem for a smooth manifold. In a sense, de Rharp's theorem puts into the natural context of cohomology theory what the multidimensional Fundamental Theorem only begins to suggest: Namely that the operation 8 of calculating the boundary of a geometric object and the operation 8/8x; of taking the partial derivative of a smooth function are in fact obverse facets of the very same phenomenon. One is the topological notion of differentiation and the other is the analytic notion. The multidimensional Fundamental Theorem of Calculus-in its many different manifestations-makes the connection between these two concepts both explicit and ~ncrete. In some sense, Stokes's theorem is the paradigm that sits above all the versions of the Fundamental Theorem in any dimension. Thus, speaking strictly logically, one should first derive the theorem of Stokes and then derive the Gauss-Green theorem, the divergence theorem, and other versions of the Fundamental Theorem from it. 1 However the strictly logical approach is not always the most didactically useful. In fact, it will be more convenient for us to first describe, state, and prove the divergence theorem in any dimension. This is so because this particular version of the Fundamental Theorem may be formulated and proved with a minimum of notation and fuss. And all the essential ideas are already present in this result. One of the key issues in formulating the Fundamental Theorem is that of orienting the domain and the boundary. Recall that, in one real dimension, the domain is an interval that is oriented according to the natural ordering of the real numbers. The boundary is oriented in (what we will see to be) a compatible fashion by assigning to the right-hand endpoint of la, b] a plus sign and to the left-hand endpoint a minus sign. Thus the one dimensional Fundamental Theorem takes the form
1"
f(x) dx = J(b) - I(a).
lThe GaUSII-Green Theorem can also be generalized in another direction by considering sets that are not smoothly bounded; that topic was treated in the earlier discussion of domains with finite perimeter in Section 3.7.
5.5. A MULTIDIMENSIONAL FUNDAMENTAL THEOREM
185
Of course orienting a boundary of dimension one or greater, in a fashion that is compatible with the orientation of the domain that it bounds, is a more complex matter. Fortunately, we will be assisted by the fact that we are dealing with a domain sitting in space. The Concept of Orientation The standard way to impose an orientation on an N-dimensional manifold M is to do so pointwise: H P E M and U is a small topologically trivial neighborhood of Pin M, then choosing an-orientation at P amounts to selecting a generator for the homology group HN(U\ {P}) (the rigorous definition involves relative homology, but what we have said captures the spirit of the idea). In a two dimensional manifold M, for example H 2 (U \ {P}) is just Z and there are two choices for the generator (corresponding to ±1)geometrically, choosing an orientation for M at P E M then amounts to choosing a direction of rotation. Since we will be orienting a domain, and its boundary, in Euclidean space, we can dispense with technical machinery such as homology groups. However it will be useful to keep the ideas in the last paragraph in mind as motivation. Now, orienting the domain n itself amounts to choosing a "standard" order for the differentials dx 1 , dx 2 , ••• , dx N. Of course we will follow custom and select dXl dx 2 •• ·dxN, or more precisely dx 1 1\ dx 2 1\ .• ·l\.dxN, as our standard order. 2 The compatible orientation at any boundary point Pis determined by selecting the outward unit normal vector v to the boundary at P. IT p is a Cl defining function for ao near P then this normal is given by gradp v(P) = Igrad pl· For example, if the dimension N = 3, then specifying the outward unit normal v determines, by way of the "right hand rule," a direction of rotation in the two dimensional boundary of O. According to our earlier discussion, this is the way that we specify an orientation. IT nee liN is a domain with C 1 boundary, then we let dO' denote the N - 1 dimensional Hausdorff measure on the boundary. We let dV be the standard Lebesgue N-dimensional volume measure in liN. Now if
2The differentials d:i:hd:i:2, ..• ,dxN are the dual basis to the standard basis el ,e2, ... ,eN for liN. The symbol "A" used here denotes exterior multiplication which is a part of the subject of Grassmann algebra. A thorough treatment of Grassmann algebra can be found in Federer [4], Chapter 1. Differentials will be essential for our discussion of Stoke's Theorem later in this section.
186
CHAPTER 5. SMOOTH MAPPINGS
is a vector field, defined on n and having Cl coefficients Xj, then the divergence of X is defined to be N
~8Xj • X = L..J--. d IV j=1
8zj
It is a straightforward exercise with the chain rule to see that the divergence is invariant under coordinate changes. Now we can state our theorem. Theorem 5.5.1 (The Divergence Theorem) Let n c lIlN be a bounded domain with a Cl defining function. Define the unit outward normal vector field to the boundary v as above. Let (5.50) be a ·vector field, defined on
n
and having Cl coefficients. Then
r X.vdu =.r10 divXdV.
(5.51)
iBn
Remark 5.5.2 We have already noted that the divergence divX of a vector field X is a coordinate-free quantity. Now let us look at the left hand side of (5.51). It is not true that v is coordinate-free, nor is it true that du is coordinate-free. However the expression vdu (which is essentially a volume frame) is coordinate-free. These observations will be useful in the proof of the divergence theorem, for we will perform the ~culations on local coordinate patches. These patches will be chosen in such a fashion as to simplify the proof. We follow the ideas in Loomis and Sternberg [1]. Proof of the Divergence Theorem: To simplify things a bit we shall assume that the boundary is C 2 (however only Cl is really necessary). Let U be a tubular neighborhood of the boundary of n. Choose 1] > 0 such that V'I = {z € aN : dist(z,80) < 1]} lies in U. Cover with open Euclidean balls of radius 1]. Pass to a finite sub cover, and let
n
n.
=
5.5. A MULTIDIMENSIONAL FUNDAMENTAL THEOREM
187
The Case Ui C n: This case is relatively trivial, and is not at all the point of the theorem. Clearly the left side of (5.51) is zero in this case. As for the right hand side, recall that if t/J(z) any any C~(JR) function then
L!
t/J(z) dz
=0
(5.52)
(just use the Fundamental Theorem of Calculus). Now we analyze the term
{ ~(yi)jdV
10 ax;
by integrating in the Xj variable first. But then (5.52) tells us that the integral is zero. This completes the treatment of the first case.
The Case Ui n an "1= 0 : We use the ideas in Section 1.2 to choose a local coordinate system on Ui so that Ut n n = {x E Ui :
XN
> OJ.
In particular,
Uinan = {x E Ui: XN = OJ. Let ·us begin our study with the right side of (5.51). The integrals. of 8~i (yt); for 1 $ j < N vanish just as in the first case. Using the Fundamental Theorem of Calculus in the variable XN, we see that the integrai of 8~N (yi)N equals - IRN_l(yl)N dXl dx2·· ·dxN-l. But we have normalIzed coordinates so that (yi)N = V· yi. Thus we see that, starting from the right side of (5.51), we have obtained the left side. •
Differential Forms and Stokes's TheoreIIl Differential forms were invented for the purpose of performing oriented integrations on manifolds. Masterful treatments appear in Federer [4] and Whitney [3]. A more summary treatment is in Krantz [2]. For the present discussion, suffice it to say that a differential form of degree k (or a k-form) in ]RN is a linear combination, with coefficients that are smooth functions, of expressions of the form dxOl = dX 01 /\dx 02 / \ · •• dX Olk ; hence 01. 02, ••. , Ok are elements of {1,2, ... ,N}. Differential forms are alternating in their indices, so that repetition of an index causes the form to be zero. We integrate a form f3 on a coordinate patch U in a k-dimensional manifold M (or, more particularly, on an open subset U of ]RN) by using a smooth parametrization 4>:W~U
from an open set We RN to U. Write f3 = L bo(x)dxOl • We assume that each multi-index 0 has length k, corresponding to the dimension of U C M. Then
CHAPTER 5. SMOOTH MAPPINGS
188
where q,*P is the canonical pullback of {3 under q, (whose nature is forced by the functoriality of exterior algebra). To put it more prosaically, if we let WhW" ••• ,WN be·the coordinates on W then z = q,(w} and dz; =
az' E aw' dwp; N
p=1
p
the form (being made up Qf objects such as dz;) transforms accordingly. Of course ,p* P ends up being a linear combination of expressions dw a of length Ie with Coo coefficients. One performs the resulting integration by explicitly calculating the orientation:
f
b(w)dwQ
==
(_1)£
f
b(w)duli .,
where e is the sign of the p~rmutation of a = (aI, a2, ... , Qi) to a degree Ie multi-index Q with entries in increasing order. The final notion that we need is that of exterior differentiation. H {3 = La badza is a differential' form, then we define its exterior derivative, dP, by setting3 . N
d{3
= E E :!~dz;" dza. Q
;=1
(5.53)
,
A simple calculation using the alternating property of exterior algebra establishes the fact that, for any form {3, d(d{3) = O. With these preliminaries out of the way, we may now formulate a version of Stokes's theorem. Theorem 5.5.3 (Stokes's TheoreIn) Let n c JRN be a bounded domain with C1 defining function. Let {3 be a differential form of degree N - 1 with C1 coefficients defined on n. Then
in In {3 =
d{3.
(5.54)
We shall not provide a proof of Stokes's theorem. The steps are just the same as in the proof of Green's theorem, and it is arguable that these two results are the same theorem stated with two different notations. The ambitious reader may wish, for instance, to derive Green's theorem from Stokes's. A reasonable facility with exterior algebra will prove to be requisite for this exercise. Details may be found in Krantz [2]. ' De Rham's TheoreIn We conclude this section with a quick description of de Rhaw's theorem. 3See Federer [4] or Whitney [3]) for a coordinate-free definition of the exterior derivative.
5.5. A MULTIDIMENSIONAL FUNDAMENTAL THEOREM
189
This·theorem has proved to be a major influence in modem mathematical analysis and geometry. Although the theorem is correctly attributed to G. de Rham, it is the culmination of ideas instigated by E. Cartan and others. And it has also proved to be a step along the way to other great insights, such as the Atiyah-Singer Index Theorem (see Gilkey [1]). Let M be a compact manifold of dimension k. Let us denote the differential forms of degree p on M with the notation N'(M). Then, with the aid of the exterior differentiation operator d, we have a complex (in the sense of algebraic topology): l\o(M)
~ 1\1(M) ~ ... ~ I\P(M) ~ ...
(5.55)
By this we mean that the image of any given map is contained in the kernel of the map following. Thus it makes sense to form quotient groups as follows. Let d" denote the exterior differentiation operator acting on Jrforms. Then for p = 1,2, ... , we define the de RhaJll cohomology group, ll P (M) by setting ll P (M) = kerdp
/
imdp _ 1 '
(5.56)
Here the quotient is taken in the sen!)e of modules. It seems plain that differential forms taken as a whole, and these quotient modules in particular, contain a lot of information about the differential geometric structure of M: As an example, suppose that M were the annulus in ]R2 (this is not a .compact manifold, to be sure, but suits our temporary purpose). Consider the differential form -y x· {3 = 2 2 dx + 2 2 dye x +y x +y Then {3 is well-defined on M, dfJ = 0, yet fJ is not in the image of do (exercise). As we know frolQ advanced calculus, the only reason that such a fJ can be constructed is because M has a hole. And this hole is one that is detected by the first homotopy group, and also by the first singular homology group. De Rham's theorem turns this example into a general phenomenon. We use HP(X) to denote the standard singular cohomology group of a topological space X (see Greenberg and Harper [I)). Theorem 5.5.4 (de Rbam) Let M be a compact manifold. each p = 1,2, ... we have the group identity HP(M) = ll P(M).
Then for (5.57)
In short, the de Rham theorem says that the structure of M, calculated from topological considerations, equals the structure of M when calculated from analytical considerations. We shall provide no further details about de Rham's theorem, but instead refer the reader to de Rham [1] and Whitney [3].
Chapter 6
Convexity 6.1
The Classical Notion of Convex.ity
Let scaN be any set. In classical geometry, S is said to be convex if, whenever P and Q are points of S then the entire segment PQ lies in S. We will refer to· this concept as geOlnetric convexity. Convex sets originally arose in Archimedes's axiomatic treatment of arc length. (Refer to Fenchel [1] for an authoritative discussion of this and other historical· matters.) While geometric convexity was treated sporadically in the mathematics of the eighteenth and nineteenth centuries, it did not receive systematic treatment until the twentieth. Some of the modem treatises are Valentine [1], Bonneson and Fenchel [1], and Lay [1]. The hallmark of all the existing treatises on convexity is a strictly geometric approach. For tone, we record here the.statements of two results that are part of the metier of classical treatments: TheoreDl 6.1.1 (Helly) Let S = {B 1 , ••• , BAl} be a family of geometrically convex sets in lllN with k ~ N + 1. If every subfamily of N + 1 geometrically convex sets in S has non-empty intersection, then the intersection of all the sets in S is non-empty. An interesting application of Helly's theorem is the following: TheoreDl 6.1.2 (Kirchberger) Let K and L be non-empty compact subsets of aN. Then K and L can be strictly separated by a hyperplane if and only if for each subset T c K U L containing at most N + 2 points there exists a hyperplane separating T n K from T n L. Our viewpoint in this book is that the geometric approach to convex sets is not expedient for the purposes of analysis. The reason is that the geometric approach is not quantitative and not formulated in the language 191
CHAPTER 6. CONVEXITY
192
of functions. Our purpose here is to provide such a quantitative, functiontheoretic development. We begin by considering a domain n, i.e. a connected open set. We assume that n has C 2 boundary, and thus is given by a C 2 defining function p: n = {ic E ~ : p(z) < O}. As usual, we require that V p =I 0 on an. What does it mean for such a set to be convex? Let us approach this question in the language of freshman calculus. Let p be a two dimensional plane that intersects the interior of n. Since the present discussion is for motivational purposes only, let us suppose that at each point Q of p n an the intersection is transversal. This means that the tangent space to an at Q does not contain the tangent space to p. In other words, the two tangent spaces span all of aN. IT n is geometrically convex, then it follows a lortiori that n n p is also geometrically convex. Now let us think of this intersection as a planar domain. Imagine that the planar coordinate system in n n p has been rotated and translated so that the tangent line to an n p at Q is in fact the z-axis and that the point Q corresponds to the origin. With these normalizations, we may think of the boundary, near the origin, as the graph of a function of the single variable z, that is, the boundary is given as the graph . y
= I(z),
with 0 = 1(0) = I'{O). IT n n p lies in the upper half-plane, then the restriction of p to this planar coordinate system has the form r(z, y) = fez) - y, because p is negative on an n p. Also the graph of I is "concave up," so calculus teaches us that I" ;:: 0, hence that
B2r
ax2 (x,y) ;::
o.
(6.1)
Similarly, if n n p lies in the lower half-plane, then the restriction of p is r(x,y) = y - fez), and since the graph of f is "concave down," we have f" ::; o. Again (6.1) holds. What we have just discovered, using elementary geometry, local coordinates, and calculus, is the following: if n has C 2 boundary and is geometrically convex, if Q E an, and if ~ is a tangent direction at Q, then
;:
~2P(Q + t~)Lo o. Using the chain rule, this condition may be rewritten as N
B2
L a:
j,k=1
Xj Xk
(Q)·~j~k;::
O.
This calculation will be the basis for our analytic definition of convexity.
6.1. THE CLASSICAL NOTION OF CONVEXITY Definition 6.1.3 Let 0 = {x E function for the domain O.
193
JR.N : p(x) < O}, where p is a C 2 defining
(i) Let Q E 80. We say that 80 is analytically convex at Q if for every choice of e = (el,"" eN) in the tangent space to 80 at Q it holds that
(6.2) (ii) If every point of 80 is analytically convex, then we say that the domain 0 is analytically convex.
(iii) If it further holds that whenever 0 :F e= (el>"" eN) and the tangent space to 80 at Q it follow that N
2:
j,le=l
e lies in
a2 p
aa(Q) ejele > 0, Xj Xle
(6.3)
then we say that Q is a point oj strong analytic convexity. (iv) If every point" oj ao is strongly analytically convex, then we say that the domain O. is strongly analytically convex. H a smooth boundary point is a point of analytic convexity, but not of strong analytic con.vexity, then we sometimes (for emphasis) refer to it as a point of weak analytic convexity.
LelDlIla 6.1.4 Letn C RN be strongly analytically convex. Then there is a constant C > a and a defining function p Jor 0_ such that
N 2: j,le=l
a
a2p
a (P)WjWIe Xj Xle
~
2
Clwl ,
.
N
Proof: Let p be some fixed C2 defining function for O. For A > ()
P>. x =
(6.4)
VP E 80,w E R .
a define
exp('\p(x» - 1 ,\ .
We shall select ,\ in a moment. Let P E an and set X = Xp =
{
W E RN : Iwi = 1 and
L: a ~p} .a (P) WjWIe ~ 0 . j,le x Xle J
Then no element of X could be a tangent vector at P, hence
Xc {w:
Iwi = 1 and
L:ap/aXj(p)Wj j
=I- a}.
CHAPTER 6. CONVEXITY
194
Since X is defined by a non-strict inequality, it is closed; it is of course also bounded. Hence X is compact and
is attained and is non-zero. Define - minwex L J• I: 8~8e_ (P)WjWI: ' -, -10 1'2
). =
Set p = p><. Then for any WEaN with 1) that
L a aap
Iwi =
+ 1.
1 we have (since exp(p(P» =
2
j,le
Xj
(P)WjWI:
Xle
" { -a.a a2p (P) = 'L...J j,le
x,
X4:
ap. (P)-a ap (P) } + ).-a
x,
WjWIe
Xle
2
=
L {ax,a.~ j,le
If
W
(P)}WjWIe
+).
L j
Xle
:p. (P)Wj x,
.'
(6.5)
¢ X, then the expression in (6.5) is positive by definition, while if Since {w e JRN :
e X then the expression is positive by the choice of ).. Iwi = I} is compact, there is thus a C > 0 such that
W
2 '" L...J a aap j,lc
Xj
(P) WjWIc ~ C, Vw
e
JRN such that
Iwi =. 1.
Xle
This establishes our inequality (6.4) for P e an fixed and W in the unit sphere of JRN. For arbitrary w, we set W = IwltV, with tV in the unit sphere. Then (6.4) holds for tV. Multiplying both sides of the inequality for tV by IwI 2 and performing some algebraic manipulations, we obtain the result for fixed P and all W e JRN. Finally, notice that our estimates-in particular the existence of C, hold uniformly over points in an near P. Since an is compact, we see that the constant C may be chosen uniformly over all boundary points of n. I Notice that the statement of Lemma 6.1.4 has two important features: (i) that the constant C may be selected uniformly over the boundary and (ii) that the inequality (6.4) holds for all wE JRN (not just tangent vectors). In fact it is impossible to arrange for anything like (6.4) to be true at a weakly convex point. Also note that our proof shows that (6.4) is true not
6.1. THE CLASSICAL NOTION OF CONVEXITY
195
ao
just for P e but for P in a neighborhood of 80. It is this sort of stability of the notion of strong convexity that makes it a more useful device than weak convexity. . .- Proposition 6.1.5 Let 0 ~ {x e ntH : p(z) < O} hafJe C2 bounda",. II o is analytically confJez, then we may write 0 as an increasing union 01 strongly analytically confJez domains. Proof: To simplify the proof we shall assume that 0 has at least C3 . boundary. Assume without loss of generality that N ~ 2 and 0 e O. For ~ > 0 and MeN to be selected, let P.(x)
Ixl = p{z) + ~ ~ 2M
and
O. = {x : P.{x) <
OJ.
Then O. CO.' if~' < ~ and U.>oO. = O. If MEN is large and ~ is small, then O. is strongly analytically convex. This explicitly exhibits the required exhaustion. • Proposition 6.1.6 I/O is connected and strongly analytically confJez, then o is geometrically contlez. Proof: We use a connectedness- argument. Clearly OxO is connected. Set S = ({P},P2) E OxO: {1-~)PI +~P2 e 0, all 0 ~ ~ ~ I}. Then S is plainly open and non-empty. To show S is closed, we fix a defining function p for 0 as in Lemma 6.1.4 and argue by contradiction. If-S is not closed in 0 x 0, then there exist. PI, P 2 e 0 such that the function .-
assumes an interior maximum value of 0 on [0,1]. But the positive definiteness of the real Hessian of p contradicts that assertion.
•
Proposition 6.1.7 110 is connected and weakly analytically confJez, then o is geometrically confJez. Proof: By Proposition 6.1.5, 0 is exhausted by strongly analytically convex domains OJ. By Proposition 6.1.6, each OJ is geometrically convex, so 0 is geometrically convex. • Proposition 6.1.8 Let 0 CC lllN hatle (J2 bounda", and be geometrically confJez. Then 0 is weakly analytically contlez.
CHAPTER 6. CONVEXITY
196
Proof: Seeking a contradiction, we suppose that for some P E .flome W E Tp(aO) we have
'"' IJ2p L..J ~(P) WjWIe j,1e
Xj
Xle
an
= -2K < O.
and
(6.6)
Suppose without loss of generality that coordinates have been selected in aN so that P = 0 and (0,0, ... ,0, 1) is the unit outward normal vector to an at P. We may further normalize the defining function p so that ap/aXN(O) = 1. Let Q = Qt = tw + E • (0,0, ... ,0,1), where E > 0 and t E III Then, by Taylor's expansion, we have p(Q)
Thus if t = 0 and E > 0 is small enough then p(Q) > O. However, for that sam~ value of E, if It I > J2E/K then p(Q) < O. This contradicts the definition. of geometric convexity. • Remark 6.1.9 The reader can already see in the proof of Proposition 6.1.8 how useful the quantitative version of convexity can be. The assumption that an be C2 is not very restrictive, for convex func~ tions of one variable are twice differentiable almost everywhere (see Zygmund [1] as well as the discussion later in this section). On the other hand, C 2 smoothness of the boundary is essential for our approach to the subject. Exercise: Prove that if no is any bounded domain with C2 boundary, then there is a relatively open subset U of ano such that every point of U is strongly analytically convex. (Hint: Fix Xo E no and choose P E ano that is as far as possible from xo). Convex Functions We next turn our attention to convex functions. We begin our discussion by restricting our attention to functions of one real variable. Definition 6.1.10 Let I = (s, t) be an open intenJal in III A function ~ R is called convex if, for all a, bEl and all 0 $ A $ 1, we have that (6.7) f(Aa + (1- A)b) $ Af(a) + (1 - A)f(b).
f :I
6.1. THE CLASSICAL NOTION OF CONVEXITY
197
This definition is plainly equivalent to requiring the set
((x,y) E R x R: s < x < t, I(x) ::; y} to be geometrically convex. Convex functions are remarkably regular, as we shall now demonstrate. Let us begin by making a small change in notation. Ifa < c < b are in the domain of the convex function I, then by rewriting (6.7) with c = .\a + (1 - .\)b so that .\ = (b - c)/(b - a), we have
b-c c-a I(c) ::; b _ aJ(a) + b _ al(b).
(6.8)
In turn, (6.8) may be rewritten as
b-c c-a b _ a [/(c) -/(a)] :::; b _ a [J(b) -/(c)] or
< f(b) - f(c) . (6.9) c-a b-c This says, quite simply, that the slopes of secants to the graph of 1 increase as we move from left to right along the graph. In particular, if [a,.8) is a compact subinterval of the domain I = (8, t) of 1 tlien every quotient " f(x + h) - f(x) I(c) -/(a)
h for x E [a,.8) and 0 < h < .8 - x, is bounded below by [/(q) -I(P»)/(q - p) for any 8 < p < q :::; a and bounded above by [I (v) -/(u»)/(v - u) for any .8 :::; u < v < t. In particular, it follows that f is absolutely continuous. Thus 1 is differentiable" off a set of Lebesgue "~easure zero, and it follows from (6.9) that the derivative function is monotone increasing. We may take this first derivative function to be corrected so that it is monotone increasing everywhere. Being a monotone function, the derivative is in turn differentiable except on a set of Lebesgue measure zero (see e.g. Folland [1)). We conclude that a convex function of a single variable is twice differentiable except on a set of measure zero and that second derivative is non-negative wherever it exists. N ow a convex function of several real variables may be defined in a manner similar to that in the first paragraph:
Definition 6.1.11 Let n c RN be geometrically convex; that is, assume that whenever a, ben and 0 :::; .\ :::; 1 then .\a + (1 - .\)b E n. A function f:n~1R
is termed convex if, for all a,b E satisfied, that is,
n
and all 0 :::; .\ :::; 1, Eq. (6.7) is
f (.\a + (1 - .\)b) :::; .\f(a) + (1 - .\)/(b).
198
CHAPTER 6. CONVEXITY
By the discussion of one variable convex functions, we see immediately that if f is, in addition, (fJ then I will have non-negative second derivative in every direction ~ E aN at each point of its domain. An elementary calculation, similar to the one we performed when deriving the condition for analytic convexity of a domain in space, then shows that this observation entails the explicit inequality:
for all points PEn and for all direction vectors ~ E ]RN. In other words, the Hessian matrix of I must be positive semi-definite. Definition 6.1.12 II the Hessian matrix 01 the convex function I is positive definite at a point PEn, then we say that f is strongly convex at P.
In the converse direction, a C2 function f of several real variables that has positive semi-definite Hessian matrix at each point of its (convex) domain will satisfy condition (6.7). This is true because, once points a,b in the domain of f and a number 0 :5 ..\ :5 1 have been fixed then we may restrict attention to points in the.domain lying on the line passing through a and b. Then the positive semi-definiteness of the Hessian translates to non-negativity of the second derivative of this restricted function along this line segment. Thus, by elementary calculus, (6.7) will hold. The consideration of smoothness of an arbitrary convex function of several real variables is rather more subtle than that for functions of a single variable. A proof that a convex function of several variables is twice differentiable at almost all points of its domain (the Aleksandrov-Buseman-Feller theorem) may be found in Evans and Gariepy [1]; see also Bianchi et al. [1]. Its full explication would require the development of considerable auxiliary machinery. We shall forgo the details at this time. A detailed discussion of convex functions of a single variable, from a traditional point of view, appears in Zygmund [1], pp. 21-26. Of particular interest in that context are the applications to integration theory. We shall sketch some of those applications now. Suppose that q,,1/J are two non-negative functions defined on an interval [0, D], each continuous, strictly increasing, and vanishing at the origin. Further assume that q, = 1/J-l. Define
4>(x) =
1"
q,(t)dt
and
\}I(x) =
1'"
1/J(t)dt.
(It is not difficult to see, in light of the preceding discussions, that 4> and \}I are convex functions.) Such a pair of functions is said to be complementary in the sense of Young. An examination of Figure 6.1 shows that
6.1. THE CLASSICAL NOTION OF CONVEXITY
199
b
a
Figure 6.1: Complementary Functions if 0 $ a,b $ D, then
ab $ CI(a) + iP(b).
(6.10)
To derive an interesting and familiar example, we fix .\ > o. Set ~(t) = t>' and 1/J(t) = tIl>'. Now set p = 1 + A and p' = 1 + 1/.\. Then calculation of CI, iP and application of (6.10) yields the inequality ab $
aP
1/"
p + Ii.
(6.11)
It is important to note that p and 'P are connected by the equation 1 1 .
];+p;=1.
IT I, 9 are measurable functions on a measure space (X, Ji), then we may apply (6:10) to a = I/(x)1 and b = Ig(x)l. Thus
II . gl $
CI(lfD + iP(lgl)·
In particular, by (6.11), 11, I/(x) . g(x)1 $ ];If(x)I P + pi I/(x)I P .
Integrating both sides gives
Ix
I/(x) . g(x)1 dJi(x) $
~
Ix
If(x)I PdJi(x) + ~
Ix
Ig(x)lP'dJi(x). (6.12)
This inequality tells us that if f e V and 9 E V' then I . 9 E LI. However the inequality is not in the most standard and useful form that comes up in measure theory. In particular, the inequality (6.12) would not stand up to a dimensional analysis. We remedy this matter by normalizing I,g so that
Ix
I/(x)IP dJi(x)
=1
and
Ix
Ig(x)IP' dJi(x) = 1.
CHAPTER~
200
CONVEXITY
Then inequality (6.12) becomes
Ix
I/(x) . g(x)1 dp(x)
~ 1.
This may be rewritten as ( I/(x)· g(x)1 dp(x)
Jx
~
{ I/(x)IP dp(X)l/P. { Ig(x)IP' dp(X)l/P'. (6.13)
Jx
Jx·
This is the familiar inequality of HOlder. The theory of Orlicz spaces is inspired by the treatment of Holder's inequality that we have just given. We now outline its principal features. We let f/J, t/J, ii, W be as above and, following Zygmund [1], p. 170 ff., we declare a measurable function I on the measure space (X,p) to be in L. if iI 0 III is integrable. Now suppose that the function I on (X,p) has the property that I· 9 is integrable for every function 9 E L>1I. We then say that I E L•. We set
II/IIL. ==
II/II~ = sup {11x I(t) . get) dp(t) I:
Ix
W(lg!)(t) dp(t)
~ I} .
Certainly the inequality (6.12) or (6.13) implies that L. C L•. Of course i is a normed linear space, and it is called an Orlicz space. One can prove that if f E L4> then 11/11. is finite, but this assertion is not entirely obvious. It turns out that Orlicz spaces are very natural generalizations of V spaces, and they often arise in questions of harmonic analysis and partial differential equations. As an example, the Hardy-Littlewood maximal function is bounded on V for p > 1 but is definitely unbounded on Ll. It is reasonable to seek function spaces X such that V C X C Ll for all p > 1 and for which the Hardy-Littlewood maximal function is bounded (in some strong sense) on X. The sharp result is that if f satisfies
Ix
I/(t)l·log+ I/(t)! dp(t)
<
00,
(6.14)
where 1
+ I/(t)1 = {1/(t)!
og
-
0
if if
I/(t)1 ~ 1 I/(t)1 < 1,
then M I is Ll (here M I is the Hardy-Littlewood maximal function-see Section 3.5). The class of functions that satisfy (6.14) is commonly denoted by L(log+ L). In the study of convergence of Fourier series, one studies function spaces of the form L(log+ L)°(log+ log+ L)" in an effort to find the sharp function space on which pointwise convergence obtains.
6.2.
OTHER CHARACTERIZATIONS OF CONVEXITY
201
Obversely, in the study of partial differential equations it is common to . encounter spaces of functions that satisfy an integrability condition of the form
Ix
exp(c '1/(t)la) dJ.&(t).
Because 00
e' =
t;
L1' ;=0 1 .
it is clear that a function satisfying the exponential integrability condition will be in every £P class, and that the £P norms blow up at a certain rate. It turns out that functions of bounded mean oscillation are exponentially integrable in a certain sense. The reference Krantz [6] discusses these ideas in some detail. In a certain sense, which we shall not specify in any detail (but which fits naturally into the Orlicz duality of Lit and Lq,), functions in the exponential integrability classes are dual to functions in the logarithmic integrability classes. The source Zygmund [1) contains further information on this topic.
Other Characterizations of Convexity
·6.2
Supporting Hyperplanes Definition 6.2.1 Let S C JRN be' any set and let TI C JRN be a hyperplane. We say that TI is a supporting hyperplane for S at PES if P E TI and S lies in one or the other 01 the two closed-half-spaces determined by TI.
One can use induction on N to show that every boundary point of a geometrically convex set S has a supporting hyperplane (see Theorem 6.2.5). The geometrically convex sets in JRI are simply the intervals, so the first non-trivial case is JR2, and interestingly this case is the key to the entire argument. LelllIlla 6.2.2 Su.ppose n c ]R2 is open and geometrically convex. For each (J, set R«(J) = n n {(tcos(J,tsin(J) : 0 < t < oo}.
(i) If then
CHAPTER 6. CONVEXITY
202
(ii)
II R(8)
#: 0,
R(8 + 11") #: 0,
then the origin is a point 01 o.
•
Proof: Obvious.
Lemma 8.2.3 110 is a geometrically conl1ez, open subset 0/)R2 and the origin is a boundary point 0/0, then there ezists 8* such that
R(8*) = R(9*
+ 11") = 0,
where R(9) is as in Lemma 6.f!.f!. That is, L(8*) == {(tcos9*,tsin8*):
-00
< t < oo}
is a supporting line lor 0 at the origin.
Proof: Since 0 is non-empty, there exists 90 so that R(90)
91 = in£{ 9 : R(9)
and 92
#: 0,
= sup {9 : R(9) #: 0,
90
-
11"
#: 0.
Set
< 9 < 80 }
80 < 8 < 80
+ 11"}.
By Lemma 6.2.2(i), -we have R(9) #: 0 for 8 e -(8}, 80 1U [90 ,92 ) = (91 ,92 ). Arguing by contradiction, we can see that we must have 92 - 81 ~ 11". H not, then there would exist 9 such that 91 < 8 < 8 + 7r < 92 • But then, by Lemma 6.2.2(ii), -the origin would be a point of 0, contradicting the assumption that the origin is a boundary point of n. Because 0 is open, we have 91 < 90 < 82 , _and thus
(6.15) Also, because 0 is open and by the inequalities in (6.15), we have R(91 ) R(92 ) = 0. Thus, we may take 8* = 9 1 . •
=
Remark 6.2.4 In the proof of Lemma 6.2.3, we could equally well have chosen 9* 82 -11". Of course, it may happen, and does when the boundary is smooth, that 91 = 82 - 7r.
=
Theorem 6.2.5 110 C JR.N is open and geometrically convex, then every
boundary point 01 0 has a supporting hyperplane.
=
Proof: We argue by induction on N. The result is trivial in case N 1 2. Suppose now that N ~ 3, P and is shown by Lemma 6.2.3 in case N is a boundary point of 0 c JR.N, and the result holds in JR.N-l.
=
6.2. OTHER CHARACTERIZATIONS OF CONVEXITY
203
We may suppose an orthonormal coordinate system (X1,X2, ••• ,XN) has been chosen with P at the origin. We will now show that there is a line L passing through P that does not intersect O. sei .
0- = 0 n {
(X17X2'0,0, •• :,0)
: X1,X2 E R}.
IT 0- is empty, then the xl-axis can be chosen as L. H 0* is non-empty, then, by Lemma 6.2.3, there is (J- so that
= {(tcos(J.,tsin(J.,O,O, ... ,O): -00 < t < oo} is a supporting line for 0-. In this case, we take L = L(e-). L(e-)
Rotating coordinates if necessary, we may assume that L is the Xl-axis. Let ot be the orthogonal projection of 0 on the hyperplane II1
= { (0, X2, X3, ••• ,XN) : X2,X3, •• • ,XN E R}.
Identifying the hyperplane II1 with ]RH-1, we may apply the induction hypothesis to obtain an (N - 2)-dimensional plane
lIt C
{(0,X2,X3"",XN) :X2,X3, ••• ,XN
that contains the origin and is such that ot
II = {
(Xl> X2, X3,'"
"is a supporting hyperplane to
n lIt
ER}
= 0. Then
,XN) : (0,X2,X3,'" ,XN)
E lIt}
n at P.
•
Corollary 6.2.6 If S is a geometrically convex set, then every boundary . point of S has a supporting hyperplane. Proof: It is easy to see that if the interior of S is
non-~pty,
then
Thus if the interior of S is non-empty, then the corollary follows from Theorem 6.2.5. The only way that the convex set S can have no interior point is if the entire set S is contained in a hyperplane. In the case that S is contained in a hyperplane, then that hyperplane is a supporting hyperplane • for every point of S, and the conclusion holds. Corollary 6.2.7 liS is a geometrically convex set, then the Lebesgue measure of the boundary of S is zero. Proof: By Corollary 3.5.6, almost every point of 8S has density 1. But by Corollary 6.2.6, every point of 8S has density less than or equal to 1/2. Thus 8S must have Lebesgue measure zero. •
It turns out that, for non-empty open sets, the existence of a supporting hyperplane at each boundary point is a characterization of geometric convexity.
CHAPTER 6. CONVEXITY
204
Theorem 6.2.8 If 0 C ]RN is open and every boundary point of 0 has a supporting hyperplane, then 0 is geometrically convex. Proof: Set
n
s=
H(P),
PeOO
where H(P) is the closed half-space that is determined by the supporting hyperplane to 0 at P and that contains O. Clearly, 0 C S. To see that S C 0, we argue by contradiction. IT Q E S \ 0, then let Po E ao be such that
IPo -
QI
= dist(Q,O) ..
Since dist(Q,O) > 0 and, by definition, n C H(Po), we have Q ¢ H(Po ). Since S C H(Po ), we have Q ¢ S, contradicting the assumption Q E S \ o.
•
Theorems 6.2.5 and 6.2.8 give us another method for recognizing convexity analytically, a method that applies when the defining function for a . domain is only C 1 . Corollary 6.2.9 Suppose 0 = {x E ]RN : p(x) < O}, where p is a Cl defining function for the domain O. Then n is geometrically convex if and only if, for each P E a~,
'lp(P).· (Q - P)
<0
(6.16)
holds for all Q E O. Families of Functions Definition 6.2.10 Let.c be the collection of affine functions on f E .c means that
]RN.
Thus
f(Xl, ... ,XN) = b+alxl +···+aNxN·
(i) If K
C]RN
then we define
K == {x E The set
K
]RN :
f(x) ::; maxf(y) for every f IIeK
is termed the hull of K with respect to
E
.c}.
.c.
(ii) We say that a set S is convex with respect to the family whenever K C C S then K C C. S.
.c
if,
Proposition 6.2.11 Let 0 C ]RN be a connected, open set. Then 0 is geometrically convex if and only if it is convex with respect to the family .c.
6.2. OTHER CHARACTERIZATIONS OF CONVEXITY
205
Proof: It follows easily from the existence of supporting hyperplanes, Corollary 6.2.6, that if n is geometrically convex, then it is convex with respect to £'. On the other hand, if n is convex with respect to £, and P, Q e n, then we can take K = {P, Q}. We note that K equals the segment PQ, so PQ C {}. Thus {} is geometrically convex. • The notion of convexity with respect to a family of functions is a useful one, especially when that family is either a linear space or forms a cone. The function theory of several complex variables is another subject in which this circle of ideas arises (see Krantz [3] and Hormander [2]). Nearest Point Property Theorem 6.2.12 II scaN is closed and geometrically convex, then lor each P E JRN there exists a unique Q e S with IQ -
PI
= dist{P, S),
(6.17)
and the mapping P t-+ Q is a Lipschitz function with Lipschitz constant 1. Proof: It is easy to see that if there were two distinct points Q and Q' that both satisfied (6.17), then the midpoint {Q + Q')/2 would be strictly closer to P and would be a point of S, which is impossible. In showing that the Lipschitz constant of the mapping is 1, we may assume that S is not contained in an {N - I)-dimensional hyperplane. Suppose PI and P2 are distinct points of aN \ S, and Qi is the point of S satisfying (6.17) with P replaced by Pi. Let n be the 2-dimensional plane determined by PI, Q1, and Q2. Let Li be the line of-intersection of n with the supporting hyperplane for S at Qi. (We leave -it to the reader to consider the special case in which n is contained in the supporting hyperplane at Q2). We also let P2 be the orthogonal projection of P2 onto n. Note that the line from P2 to Q2 must be orthogonal to L 2. Because of the convexity of S, the situation must be as illustrated in Figure 6.2. It is now easy to verify that IP2 - PII ~ IQ2 - QII· Since 1P2 - PII ~ IP2 - PII, the result follows. • Interestingly, the following converse of Theorem 6.2.12 is also true. Theorem 6.2.13 II S is a non-empty closed set such that for each P E there is a unique Q e S satisfying IQ - PI = dist{P, S),
aN
(6.17)
then S is geometrically convex. Proof: Arguing by contradiction, suppose there are points PI, P2 E S such that there is a point Q on the line segment between them with Q f/. S. We
CHAPTER 6. CONVEXITY
206
Figure 6.2: The Nearest Point Configuration can then choose r
> 0 so that . B(Q,r)nS=0,
and can then let B(Q',r') be the ball with largest possible radius such that
B(Q, r) C B(Q',r')
and B(Q' ,r') n S = 0.
(6.18)
Because there is a unique point P' of S nearest to Q', we must have
B(Q',r') n S = {P'}. Heuristically, the argument proceeds as follows: While keeping the ball I( Q, r) on the inside, expand a ball so that its radius is as large as possible with only the surface of the ball making contact with S. Presumably the center of the ball might shift as the radius increases, but finally a center Q' and radius r' will have been chosen so that r' is as large as possible subject to the condition (6.18). We note that the nearest point property tells us that the ball B( Q', r') can only be touching S at one point. Physical experience indicates that contact with a ball at only one point is not enough to hold the ball fixed; there is a lot of freedom to move the ball B(Q',r') away from contact with S by moving its center. The difficulty is in keeping i( Q, r) inside while moving the ball. Nonetheless, if we succeed in moving the ball away from contact with S while keeping B(Q, r) on the inside, then we can increase the radius of the ball. This contradicts the maximality of the radius r'. The rest of the formal argument makes this precise. Notice that if v is any unit vector with
v· (Q' - P') > 0,
6.2. OTHER CHARACTERIZATIONS OF CONVEXITY
207
Figure 6.3: Moving the Largest Ball then, for all sufficiently small
f
> 0,
B(Q' + fv,r')
n s = 0.
H we were to also have
B(Q,r) CB(Q'+fv,r'), then a larger value of r' could be chosen and still satisfy (6.18); which is the desired contradiction. In fact, one can choose v = Q' - P' if B( Q, r)e: B( Q' , r'), and ·otherwise chooSe v = Q" - P' where Q" is the unique point of B( Q, r) \ B( Q' , r') (see Figure 6 . 3 ) . · • The concept of a set of positive reach was introduced in Federer [3]: Definition 6.2.14 Let E C ]RN be a closed set. We call E a set of positive reach if there is an open neighborhood U of E so that each element x E U has a unique nearest point in E. The reach of a set E is the supremum of the numbers r so that every point in
{P: dist(P,E) < r} has a unique nearest point in E. Theorems 6.2.12 and 6.2.13 show us that the only sets of infinite reach are the convex sets. In Section 1.2 we showed that a domain with C2 boundary is a set of positive reach. The following example shows that a domain that is less smooth than C2 need not be a set of positive reach.
208
CHAPTER 6. CONVEXITY
EXBJDple 6.2.15 For 6
> 0 consider the domain
n = n6 = {(x,tI) EJR2 : y > IxI2-6, z2 + y2 <
I}.
The second inequality in the definition of (} is provided only to make the domain bounded; it is really superfluous. ConsiderpointsP P E (0, e) e n, e > o. We claim that if e is small enough then P has two nearest points in 80. In particular, (0,0) is not the nearest boundary point to P. One can see this by direct computation; this is a tedious approach, but it is educational and we sketch it. We assume here that 0 < 6, e < < 1. Now let 4>(t) = (t,ltI 2 - 6 ) be a point of the boundary curve. Then the distance squared of P to 4>(t) is
=
Then
I'(t) = 2t + 2(ltI 2- 6
Notice the following about
-
e) . (2 - 6)(sgn tWI 1 -
=
6•
I :
I' < 0 when t > 0 is so small that (t 2 - 6 - e) < -e/2 and t- 6 > 8/e; (ii) I' is positive when t is large enough that (t 2 - 6 - e) is positive. This discussion already shows that I does not have a local minimum at 0 (recall that I is even). It does have a miirlmum at a unique positive value to (the point where I' vanishes) and a cor.responding local minimum at -to (i)
by evenness. These are the abscissas of the two nearest points on
an to P.
ExtreIDe Points and the Krein-Milman TheoreID Definition 6.2.16 A point x in a set E C liN is called extreIDe il whenever a, beE and >.a + (1 - >.}b = x lor some 0 < >. < 1, then a = b = x. . IT E is the closed unit ball in jRN, then each boundary point is extreme (exercise). Interior points are never extreme. The example of the closed unit ball is potentially misleading, for one might infer that an extreme point is one where all curvatures are positive. While the converse of this statement is true, the domains Em = {(XhX2) e R2 : IXll2 + IX212m < I} for m = 2,3, ... have all boundary points extreme while the curvature fails to be positive at the points (±1,0). A domain for which every boundary point is extreme is called strictly convex. Compare this concept with the notion of strong convexity of a domain. A strongly convex domain is surely strictly convex, but the converse is not true as the example at the end of the last paragraph shows. All of our various characterizations of convexity may be applied to detect extreme points. Consider the point of view of affine, or linear functions. A
6.2. OTHER CHARACTERIZATIONS OF CONVEXITY
209
point x in the boundary of a convex domain n C ]RN is extreme if and only if there exists a linear function a on aN such that a assumes its maximum value on at x and only at x. Now consider the point of view of defining functions. A boundary point x of a (fl, bounded, convex domain (} = {x E ]RN : p(x) < O} is extreme provided that for each tangent vector ( to on at x there is a positive integer m m«() such that the mth directional derivative of p in the direction ~ is non-zero. Unfortunately, this condition is not quite necessary, as the example n = {(Xl,X2) : IX112 + 2e-l/lz212 < I}
n
=
and the boundary point (1,0) shows. We shall not provide detailed verification of these various descriptions of extreme points, as" we shall not be using them in what follows. A characterization of extreme points that is germane to our discussions is as follows: Let n C liN be a bounded, convex domain. A point x e on is extreme if there exists a hyperplane P such that P n n = {x}. In the case that n has C2 boundary, then this hyperplane is unique and will coincide with the tangent plane at x. H the boundary is not smooth, then this plane need not be unique (consider; for instance, the case that n is a truncated half cone and the point x is the vertex). Note that the converse of this assertion is not true. In functional analysis it is useful to characterize convex sets in terms of linear functionals. For example, a closed set E in a Banach space is convex if and only if, for every point y that is external to E, there is a linear functional I{J such that
210
CHAPTER 6. CONVEXITY
We have already encountered the notion of convex combination in our ·discussion of extreme points: a point % e E is extreme if it cannot be written as a convex combination of points a, beE with at least one of a or b different from %. Notice that if a, beE and if c is a convex. combination of a and b, then any convex combination of a and c, for instance, is also a convex combination of a and b. It is plausible that if E is a bounded set, then its convex hull can be generated by adjoining to E all possible convex combinations of points of E. In fact this is the case: Proposition 6.2.18 Let E c JR.N be a bounded set. The closed convex hull E of E coincides with the smallest ·closed let containing all points of the form E;=l ).jej with ej e E, ).j 2: 0, and Lj).j = 1. Proof: Clearly any geometrically convex set containing E will contain any convex combination of points of E. Therefore the closed convex hull will contain all these convex combinations. That proves one direction. Conversely, suppose that % is a point that is not in the closure of the set S of all convex combinations of elements of E. Then there is a hyperplane P that separates % from S. But then % cannot be an element of the convex h~ of E, for the half space determined by P and containing E is a convex • set that excludes %. That completes ·the other half of the proof.. Informally stated, the Krein-Milman Theorem asserts that a compact, convex set is the closed convex hull of its extreme points. In point of fact the heart of the matter is the following more technical result, also due to Krein and Milman [1]: . Proposition 6.2.19 Let E C JR.N be a non-empty, compact, convex set. Then E possesses an extreme point. Proof: Let P be a point not in the set. Let B(P,r) be the smallest closed ball that is centered at P and contains E (just take the intersection of all such balls). Then any point in BB(P, r) n E must be extreme for E since it is extreme for B(P, r). • In fact this proof may be used to show that E must have more than one extreme point. As an exercise, give a lower bound for the number of extreme points this proof guarantees for a compact, convex set in]RN (your answer should be a function of N). The example of the unit simplex will be useful to consider. Theorem 6.2.20 (Krein-Milman) Let E C RN be a non-empty, compact, convex set. Let S be the set of extreme points of E, which we know by the proposition is non-void. Then E iIJ the closed convex hull of s. Proof: Clearly the closed convex hull of S will be contained in E.
6.3. EXHAUSTION FUNCTIONS
211
For the converse direction, let x be a point of E that is not the closed convex hull of S. Then, just as in the proof of Proposition 6.2.18, there is a hyperplane separating x from the convex hull of S. Therefore surely there is an extreme point of E that lies on the same side of the.hyperplane as does x. That is a contradiction. •
6.3
Exhaustion Functions
Definition 6.3.1 Let (} c RN be a domain. A function I : (} -+ III is called an exhaustion function lor 0 if, whenever c is a real constant, {x e (}: I(x) < c} is relatively compact in O. The purpose of this section is to examine the following result: Proposition 6.3.2 Let 0 C RN be a domain. Then (} is geometrically conve.x if and only il it possesses an exhaustion function that is convex. In fact there is a stronger result which is extremely useful:
Theorem 6.3.3 Let 0 c JRN' be a domain. Then 0 is geometrically conve:t if and only il it possesses a Coo exhaustion function that is strongly convex. Before proceeding to proofs, we record some additional results that are in the spirit of exhaustion functions. The functions that we describe in these next two results are sometimes called bounded exhaustion functions. Theorem 6.3.4 Let 0 C RN be any bounded geometrically convex set with c/o boundary, k ~ 2. Then there is a c/o function_uon n such that (i)
'II.
is con~ex on
(ii)
'II.
< 0 on OJ
(iii)
'II.
== 0 on
n;
ao.
Finally, we have: Theorem 6.3.5 Let 0 C RN be a bounded geometrically convex domain with c/o boundary, k ~ 2. Then there is a c/o convex function'll. defined on all of ItN such that V'll. =F 0 on ao and
0= {x
e R.N
: u(x) <
OJ.
Now we shall establish the first result of this section. Theorem 6.3.3 will be a consequence of that one. We begin by constructing a function that behaves like an exhaustion near the boundary, and then we modify it elsewhere to suit.
CHAPTER 6. CONVEXITY
212
LelJlDUl 6.3.6 Let n C]RN be a geometrically convex domain (bounded or
unboundetl). For zen, let 6(x) denote the Euclidean distance from x to the complement of n. Then the function I'(x)
==
is a convex exhaustion function for
-log6(x)
n.
Proof: That the function I' is an exhaustion function is obvious. The claim of convexity amounts to checking that if a, ben and if 0 ::; ~ ::; 1, then -log(~o(a)
+ (1- A)6(b»
::; -A log 6(a) - (1- ~)logo(b).
Some elementary calculations show that this is equivalent to [6(a)] A . [6(b)](1-A) ::; o(Aa + (1 - A)b).
The concavity of the logarithm function shows that [o(a)] A
•
[6(b)]
(I-A) ::;
M(a)
+ (1- ~)6(b).
So it suffices for us to check that ~c5(a)
+ (1 -
A)6(b) ::; 6(Aa
+ (1 -
A)b) ..
But this inequality is obvious from the definition of geometric convexity.
•
Lenuna 6.3.7 If ~ is a convex exhaustion function for the domain n (bounded or unbounded), then there exists a Coo exhaustion function that is strongly convex.
We defer the proof of this lemma for a moment and turn to a result that may be of separate interest: Proposition 6.3.8 Let f be convex on an open set U C and let r > 0 be sufficiently small. Then
f(P) ::; _1_
r
f(P
+ rt;) d1l N - 1 t;.
]RN.
Let P e U
(6.19)
WN-l JS(O.l)
Proof: We certainly know that if t; is any direction vector of unit length, then f(P) ::; f(P + rt;) ; f(P - rt;) . (6.20) Now integrate both sides of (6.20) over { in the unit sphere with respect to the (N - 1)-dimensional Hausdorff measure. The result follows. •
6.3. EXHAUSTION FUNCTIONS
213
It should be noted that an inequality such as (6.19) can be used to characterize subharmonic functions. There is no such characterization for convex functions. All we are observing here is that a convex function is subharmonic. While this assertion may at first be surprising, we may dispel the surprise by noting that if u is convex on U C ]RN, then {]2u/8xj ~ 0, j = 1, ... N, hence fl.u = Lj {)2u/8xl ~ o. The reader should see Hormander [2] for more on these matters. A corollary of the Proposition 6.3.8 is the following:
Corollary 6.3.9 Let f be convex on an open set U C ]RN. If l/J is a mdial function on JRN (this means that ¢>(x) = l/J(x / ) if Ixl = Ix'l) and if, furthermore, l/J E C c and I l/J = 1, then
Proof: The corollary is proved by introducing polar coordinates and invoking the already-proved sub-averaging property over spheres.
•
Proof of LeInIna 6.3.7: Let!lc == {x e- 0: .4>{x) + Ixl2 <:: c}, c E III Then each Oc cc 0, by the definition of exhaustion function, and c' > c implies that Oc eeOc'. Let 0 :5 l/J E cgo (JRN ), J ¢> = 1, l/J radial. We may assume that l/J is supported in B(O,I). Pick €j > 0 such that €j < dist{!lj+1>8!l). For x E !lj+l, set
Then 4>.t..is Coo and strongly convex on OJ+l. We know that 4>j{t) > ~(t) + 1fi2 on OJ. Let X E COO(JR) be a convex function with X(t) = 0 when t :5 0 and X'(t),X"(t) > 0 when t > o. Observe that Wj(x) == X(4)j{x) - (j -1» is positive and convex on OJ \ OJ-l and is, of course, Coo. Now we inductively construct the desired function ~/. First, 4>0 > 4> on !lo. If al is large and positive, then 4>~ = ~o + alwl > 4> on 0 1 • Inductively, if aI, ... , at-l have been chosen, select at > 0 such that 4>[ 4>0 + L~=l ajwj > ~ on Ot. Since Wl+k = 0 on !It, when k > 0, we see that 4>l+k = 4>l+k' on !It for any k, k' > o. So the sequence ~l stabilizes on compacta and 4>' == limt-+oo ~l is a Coo strongly convex function that majorizes 4>. Hence 4>' is the smooth, strongly convex exhaustion function that we seek. •
=
A nice consequence of Lemma 6.3.7 is the following corollary: Corollary 6.3.10 Any geometrically convex domain is the increasing union of smooth, strongly analytically convex domains.
CHAPTER 6. CONVEXITY
214
Proof: Fix a geometrically convex domain n and let \II be a smooth, stroDgly CIOIlftlI'i exhaustion function. By Sard's theorem (see section 5.1), almost every sublevel set Oc == {x en: \II(x) < c} has smooth boundary. or c:ourae it will also be strongly analytically convex. • A useful tool in treating bounded exhaustion functions is the Minkowski functioDal. We describe that functional now; the reader is referred to Rudin [3] and Yoshida [1] for further reading on this topic.
Definition 6.3.11 Let 0 C]RN be a domain. Assume that 0 is a point of If x e JtN' then we define the Minkowski functional, Mg, by setting
n.
M(~)
=Mn(x) =
infP
e lR : .\ > 0
and (1/'\)x E a}.
In case n is the unit ball then M(x) is just IIxli. For more general n, the Minkowski functional is a device for putting a quasi-norm on RN. We are particularly interested in the case when a is geometrically convex.
Proposition 6.3.12 Let 0 be a geometrically convex, bounded subset of lRN containing the origin. Then the Minkowski functional for a is a convex function. Proof: Because the convex set can be intersected with the plane determined by any two points and the origin, it will suffice to prove the result in case N = 2. First suppose that an is smooth-at least C 2 • L~t p be a defining function for a .. We define a function f by setting
f(x,y,z)
= p (~,~).
We will denote the partial derivatives of P with respect to its first and second arguments by DIP and D 2 P, respectively. Similar notation also will be used for the second partial derivatives of p. We have
6.3. EXHAUSTION FUNCTIONS
215
Since the Minkowski functional satisfies l(x,y,M(z,y» ferentiate implicitly to find
81 8x
+ 818M 8z 8x
=0
8 21 8x8y
81 + 818M =0 8y 8z 8y ,
'
8 21 +21P1 8M 8x8z 8z 8x2 1P18M 8y
+ 8z8z
= 0, we can dif-
+ (J21
8z 2
(8M)2 8x
(6.26).
+ 81(J2M = 0 8z 8x2
(6.27)
'
8 21 8M
+ 8y8z 8x + 8 21 8M 8M + 81 8 2M 8z 2 8x 8y
=0
8z 8z8y
8 21 21P18M (J21 (8M)2 8182M =0 8y2 + 8y8z 8y + 8z 2 8y + 8z 8y2 ' Multiplying equations (6.27), (6.28), and (6.29) by (¥Zf ~2,
,
(6.28)
(6.29)
(¥Zf 2~'1,
and (¥Zf'12 , respectively, using (6.26), and collecting all terms involving second derivatives of M on one side of the equation, we have
_ (8 1 8z
)3 (IPM e+2 82M ~71+ 8 2M 71 8z 8x8y 8y2 2
2)
=v-H,-v
(6.30)
where H, is the Hessian of I and v is the vector
!~
v= (
.
~~ +71U
)_.
(6.31)
We note that, because the gradient of p is an outward pointing normal vector, we have (6.32) and hence
81 8z < 0,
when we evaluate at z
= M(z,y). Thus the sign of
IPM 2 82M 8 2M 2 8z2 ~ + 2 8x8y ~71 + 8y2 71 will agree with the sign of v - H, - v. Using (6.22)-(6.25), we have
v-Hr v
= ~z (~81 +71 (1 ) 8x 8y
Vp-w+w-Hp-w
(6.33)
CHAPTER 6. CONVEXITY
216 where
(.~ (eM+'1U) -~M) 81 . . ~ (!!.1.!!.1.) e8:e+'18J/ -;8%.
= w
We have w·Hp·w that 2 ~
~
.
0 by the convexity of O. Finally, using (6.21), we see .
(8e8x1 + '1lJy( 1) V p. w = Z52 (xD1P + yD2P)(eD1P + '1D2p)2.
(6.34)
By (6.32), the right-hand-side of (6.34) is non-negative. The proof in full generality follows by approximating 0 by an increasing family of smoothly bounded analytically convex domains and passing to the limit. • Corollary 6.3.13 Let 0 be a convex, bounded subset ol[l.N containing the origin. II80 is C k , k ~ 1, then the Minkowski functional for 0 is C k , except at the origin. Proof: Let p be a Ck defining function for
feu, v) =
o. For u E [l.N
and 0 < v, set
p(~).
As in the proof of Proposition 6.3.12, we note that the Minkowski rw;.ctional satisfies l(u,M(u» = 0, so the result follows from the Implicit Function Theorem.
. •
Notice that if 0 is a convex domain that contains the origin and if M is its Minkowski functional, then the function u(x) == M(x) -1 is a bounded convex exhaustion function. By Corollary 6.3.13, it will be as smooth as the boundary of 0 (except at the origin). Notice further that this u is defined and convex on all of ]RN. Thus we have established Theorems 6.3.4 and 6.3.5. Now the Minkowski functional M(x) can be used to establish a result on approximation from the outside when 0 is a bounded, convex domain. IT c > 1, then O~ == {x E ]RN : M(x) < c} is a convex domain that contains o and nco" = 0 exhibits 0 as a decreasing intersection of convex domains. Further note that 0 is relatively compact in Oc for each c > 1. Since Oc can be exhausted by smooth, strongly convex domains, we may find a strongly convex domain U such that 0 cc U cc Oc. It follows that 0 is the decreasing intersection of smooth, strongly analytically convex domains.
217
6.4. CONVEXITY OF ORDER k
6.4 Convexity of Order k Let n be a smoothly bounded domain (Ck+l
will be sufficiently smooth) and P E an a point of analytic convexity. We say that P is a point of convexity of order k if the tangent hyperplane to an at P has order of contact precisely k with the boundary. Let. us flesh out this definition.
Definition 6.4.1 The following are equivalent definitions of the order of contact of a hyperplane II with a hypersurjace M C ]RN at a point P EM:
(i) We say that II has order of contact k with M C ]RN at P if there are orthonormal coordinates (t}, ... , tN-b u) = (t, u) on]RN and a smooth function f(t) = f(t}, t2, ... , tN-d on ]RN-l such that, in these coordinates, P = (0,0), II = {(t,u) : u = O}, and there are neighborhoods U of P in]RN and U' of-O-in ]RN-l and a constant o < C < 00 for which
UnM = {(t,J(t») : t
E
U'},
and
(6.35) If(t) - f(O)1 ~ Cltl k Jor t E U', and further, for no choice oj C does (6.35) hold with k replaced by k+ 1.
(ii) We say that II has order of contact k with M C]RN at P if, whenever M is expressed as M = {x E ]RN : p(x) = O} with p a smooth function satisfying V' p # 0 on M, then there is a constant 0 < C < 00 such that _ Ip(x)l:::; Clx - Plk for x E II, (6.36) but Jor no choice of C does the inequality (6.36) hold with k replaced by k + 1. This definition tells us that all tangential derivatives of p of order not exceeding (k - 1) vanish; but some tangential derivative of p of order precisely k does not vanish. (iii) We say that II has order of contact k with M C ]RN at P if, whenever M is expressed as M = {x E JRN- : p(x) = O} with p a smooth function satisfying V' p # 0 on M, there are local curvilinear coordinates tl,t2, ... ,tN on]RN such that, in these coordinates, P = 0, II = {(tl, t2, ... ,tN) : tN = O} and P(tb t2,"" tN) = tN where, for some constant 0 < C <
+ R(tl, t2,···, tN-d, 00,
R satisfies
IR(tl, ... ,tN-dl ~ C(ltllk
+ .. , + ItN_llk),
(6.37)
and for no choice of C does the inequality (6.37) hold with k replaced by k + 1.
CHAPTER 6. CONVEXITY
218
The equivalence of these definitions is easily proved: That (i)~(ili) follows by setting p(t,1.&} = U - f(t}. That (ili}~(ii) is immediate by changing coordinates. Finally, the implication (ii}~(i) is a consequence of the Implicit Function Theorem. Notice that a strongly analytically convex point is a point that is convex of order 2-the canonical example is a boundary Point of the unit ball. The point (1,0) in the boundary of
is a convex point of order k = 2m. The last example raises the question of whether an analyiically convex point can have odd order. The answer is "no", for the simple reason that, if the lead term (afte~ ·the linear term) in the Taylor expansion of p has odd order, then the boundary near P locally looks like the graph of an odd degree polynomial, therefore cannot be geometrically convex, and hence cannot be analytically convex (by Proposition 6.1.8). A special feature of strongly convex points is that they are stable. This assertion can be formulated more precisely as follows.
Proposition 6.4.2 Let Q C ]RN be a domain with C2 boundary. Let P E be a point of strong· convexity. The~ any x E an that,s sufficiently near P is also a point of strong convexity.
an
Proof: By Lemma 6.1.4, we can suppose the defining function for n has a Hessian that is positive definite on all vectors, not just the tangent vectors at P. But the second derivatives of the defining function ar~ continuous. Thus the Hessian at nearby points (whether in the boundary or not) will • also be positive definite. The same proof establishes the following: Proposition 6.4.3 Let 0 = {x E ]RN : p(x) < O} be a domain with C 2 boundary. Let P E ao be a point of strong analytic convexity. If ifJ E C~ (]RN) has support in a small neighborhood of P and has sufficiently small C2 nonn and if we set p(x) = p(x) + q,(x) then
fi == {x
E ~ : p(x)
< O}
will also be strongly analytically convex at points that are near to P. The analogue of Proposition 6.4.2 is the following result. Theorem 6.4.4 Let 0 C ]RN be a smoothly bounded analytically convex domain. Assume that P E ao is a convex point of order k. Then boundary points that are sufficiently near to P are convex of order at most k.
6.4. CONVEXITY OF ORDER k
219
Proof: In fact the assertion has nothing to do with convexity. Our assertion amounts to proving that if the defining function for a domain n vanishes at P e an to order precisely k, then it vanishes at nearby points to order at most k. This is really just an assertion about Taylor expansions. To make the preceding paragraph precise, note that Definition 6.4.1(ii) tells us that if n bas the smooth defining function p then, at P, which is a point of convexity of order k, there are vectors Vi = (Vi,l, Vi,2, ••• ,Vi,N), i = 1,2, ... ,k, which are tangent to an at P and are such that N
O"p
N
N
L E ... E
it=l h=l
i.=l
VI ,it t12,h
•.. v",i.
a a Xl
alc p
X2···
a
(P).
(6.38)
X,.
But the inequality (6.38) is an open condition, so it will hold for all points pI sufficiently near to P and all vectors v~, v~, ... ,v'Ic sufficiently near to the vectors VI, V2, ••• , Vic, respectively. Thus for pI sufficiently near to P and with the vi defined to be the orthogonal projections of the Vi onto the tangent plane to an at pI, we have
proving the result.
•
Definition 6.4.5 Let n = {x e aN : p(x) < O} be a smooth, analytically convex domain. If U C aN is open, U n an will be said to be convex of order k if every point of U n an is a point of convexity of order l ::; k. Theorem 6.4.4 tells us that any point bf an that is a point of convexity of order k has a neighborhood in an that is convex of order k. Finding an analogue of Proposition 6.4.3 for convex points of order k with k > 2 is somewhat subtle. Such points are only weakly analytically convex, and an arbitrarily small perturbation in any smooth topology can turn the point into a concave point. Thus we need to restrict attention to perturbations of a certain kind. Figure 6.4 illustrates the sort of perturbation we wish to consider. Theorem 6.4.6 Let n = {x e aN : p(x) < O} be a domain. Let P e an be a point of convexity of order k. Then there is a neighborhood U of P and a function ¢ e C:+1 such that
(i) ¢ is supported in a small neighborhood of P (ii) ¢::; 0
(iii) ¢ < 0 near P
CHAPTER 6. CONVEXITY
220
Figure 6.4: Perturbing at a Point Convex of Order k and iJ i'i is given by
n=
{x E aN
: p < OJ,
an
P = p + q" then U n is non-empty and is convex oj order k. In short, points oj order k can be "bumped" outward. with
Remark 6.4.7 Notice that this theorem has real content. Consider for exrup.ple, the unit box B = {(x,y) E]R2 :
Ixl < 1,lyl < I}.
Then P = (1,0) is a point in aBo But there is no outward perturbation of aB near P that will still be analytically convex. The following lemma will be crucial. Lemma 6.4.8 Suppose unan is convex oj order k. The set W oj points in Un an that are not strongly analytically convex Jorm a (relatively) closed, nowhere dense set in
an.
Proof: Since the condition for strong analytic convexity is open, it is clear that the set of strongly analytically convex points is open. Therefore its complement is closed. H W has interior, then there is a relatively open set U in an on which the Hessian vanishes. But that means that, on U, the defining function is linear. But then, because it is infinitely flat, that piece of the boundary is • not convex of type k for any k. That is a contradiction. Proof of Theorem 6.4.6: We shall present the proof only in dimension 2, where it is quick and easy. The higher dimensional case involves interesting but nasty geometric details. We may as well suppose that the defining function for n has the form p(x,y) =-y+p(x)nearP. LetP= (p,p(P». ThenI'" ~Onearx=p. By Lemma 6.4.8, choose points a < p < b such that an is strongly analytically convex at (a,p(a» and at (b,p(b». In other words, p"(a) > c > 0 and I'''(b) > c > o. Select 0 > 0 so that these inequalities persist on (-0 + a,a + 6) and on (-0 + b, b + 0).
221
6.4. CONVEXITY OF ORDER k
Let t/> be a
ego function supported in [a - 6, b + c5]. Assume further that {%: 1/>'(%) ~ O} C [a - 6/2, a] U [b+ 6/2]
and that 1/>(%) == -1 for a + 6/2 Now consider the function
~ % ~ b-
6/2.
P=P+Et/>. IT E > 0 is chosen small enough then sup IEQ>"I < c/2.. And of course 1/>" will be supported in the set where Ip"l > c. It follows that the domain fi == ((%,y) : p(%,y) < O} is still convex of order k. And it is an outward perturbation of the original domain n as required. •
Chapter 7
Steiner Symmetrization 7.1
Basic Properties
The Swiss geometer Jakob Steiner (1796·- 1863) devoted much of his energy to seeking simple principles from which theorems in geometry could be derived in a natUI:al way. The Isoperimetric Theorem provided one venue for this process and symmetrization was the simple principle he discovered. Mathematical mythology attributes the classical Isoperimetric Theorem (stating ·that the circle is the plane figure enclosing the greatest area for a given. perimeter or equivalently the figure of least perimeter enclosing a given area) to Queen Dido, founder of Carthage. More reliable authority (see Knorr [1] and [2]) credits Zenodorus with a proof in the second or third century B.C. Steiner himself refers to another Swiss mathematician SimonAntoine-Jean L'Huilier (1750 - 1840) as the one who provided the first definitive exposition of the Isoperimetric Theorem. L'Huilier published a treatise on the subject (in Latin) in 1782. Though·an admirer of L'Huilier, Steiner was not satisfied with the status of the Isoperimetric Theorem, for Steiner felt that while L'Huilier's work was often cited, L'Huilier's methods were not rightfully appreciated or understood. Steiner thus sought a deeper and clearer understanding of the classical Isoperimetric Theorem and its variants. Steiner's first results using symmetrization to prove the Isoperimetric Theorem were presented to the Berlin Academy of Science in 1836 (see Steiner [1]). Ultimately, Steiner gave five proofs showing that i/there is a figure of least perimeter enclosing a given area, then it must be a circle. That Steiner did not successfully address the problem of existence of such a figure is not surprising, especially in view of the well-known history of Dirichlet's Principle (see the introduction of Courant [1]).1 Definition 7.1.1 Let V be an (N -I)-dimensional vector subspace o/RN. 1 Dirichlet
and Steiner were contemporaries at Berlin University.
223
224
CHAPTER 7. STEINER SYMMETRIZATION
Figure 7.1: Steiner SYlDlDetrization Steiner sYlDlDetrization with respect to V is the operation that associates with each bounded subset T of IRJV .the subset S of]RN having the property that, for each straight line l perpendicular to V, inS is a closed line segment with center in V or is empty and the conditions
and
inS = 0 i/ and only if l n T = 0 hold. Figure 7.1 illustrates the application of Steiner symmetrization to an ellipse T in the plane. Symmetrization is performed about the vertical line V. The result of symmetrization is another ellipse S that is, of course, symmetric about V and is such that any horizontal line, that is, any line perpendicular to V, intersects T and S in line segments of the same length. For Steiner's work the significant facts were that the area of a plane domain is unchanged by symmetrization about a line (by Cavalieri's Principle or, in modem terms, Fubini's Theorem) and that the perimeter is decreased (unless the figure was already symmetric about some line parallel to that with respect to which the Steiner symmetrization is performed). We will need to know more about Steiner symmetrization than just those two facts. In particular, we will need to know the many ways in which Steiner symmetrization is well-behaved. The sequence of propositions below will document this behavior.
225
7.1. BASIC PROPERTIES
Proposition 7.1.2 If T iB a bounded £N -measurable subset of]iN and if S is obtained from T by Steiner symmetrization, then S is £N -measurable and £N (T) = £N (S). Proof: This is a consequence of Fubini's Theorem. LelDma 7.1.3 Fi:I: 0 < M aN n 8(0, M) such that
then
•
< 00. If A and Al,A2' ... are closed subsets of .
lim sup 1(.N (Ai) ~ 1£N (A). i
Proof: Let and
E
> 0 be arbitrary. Then there exists an open set U with A c U .
1(.N (U) ~ 1£N (A)
+ E.
A routine argument shows that, for all sufficiently large i, Ai C U. It follows that lim sup 1(.N (Ai) ~ 1£N (U), i
and the fact that
E
was arbitrary implies the lemma.
•
Proposition 7.1.4 If T is a compact subset fllaN and if S is .obtained from T by Steiner symmetrization, then S is compact. Proof: Let V be an (N - I)-dimensional vector subspace of aN, and suppose that S is the result of Steiner symmetrization of T with respect to V. It is clear that the boundedness of T implies the boundedness of S. To see that S is closed, consider any sequence of points PltP2, ... in S that converges to some point p. Each Pi lies in a line li perpendicular to V, and we know that dist(Pi, V)
~
41£I(ti n S) = 41(.I(l; n T).
We also know that the line perpendicular to V and containing P must be the limit of the sequence of lines t 1, t 2 , •••• Further, we know that dist(p, V)
= .lim
dist(p;, V).
1-+00
The inequality limsup1(.l(t; n T) ~,1(.I(l n T) i
(7.1)
CHAPTER 7. STEINER SYMMETRIZATION
226
would allow us to conclude that dist(p, V)
== i-+oo lim diSt(pi, V):S
11£I(lnT), -21Iimsup1l1(li nT):S -2 i-+oo
and thus that peS. To obtain the inequality (7.1), we let qi be the vector parallel to V that translates li to l, and apply Lemma 7.1.3, with N replaced by 1 and with l identified with Ii, to the sets Ai = Tq; (li n T), which are the translates of the sets li n T. We can take A = l n T, because T is closed. • Because the Steiner symmetrization is constructed using closed line segments, Steiner symmetrization does not send open sets to open sets. For example, the Steiner symmetrization of the open unit square in JR2 (0,1) x (0,1)
= {(x,1/) : 0 < x
< 1, 0 < 1/ < I}
about the x-axis is the set (0,1) x
[-t, tJ = {(x,y) : 0 < x
< 1,
-t :S 1/ :S H·
The following proposition tells us that the Steiner symmetrization of an open set is a G 6 set. . Proposition 7.1.5 If T is a bounded open subset of JRN and if S is obtained from T by Steiner symmetrization, then S is a G 6 set. Proof: Let V be an (N - I)-dimensional vector subspace of JRN, and suppose that S is the result of Steiner symmetrization of T with respect to V. For each positive integer i, let Si be' the subset of JRN having the property that, for each straight line l perpendicular to V, the set l n Si is an open line segment with center in V or is empty and the conditions
and
l n Si
=0
if l n T = 0
hold. First, we claim that each Si is open. Consider i fixed. IT a point p is an element of Si and if l is the line perpendicular to V passing through p, then dist(p, V) <
~ (~ + 1£1 (l n T»)
must hold. Define rl > 0 by 2rl =
~ (~ + 1£1(l n T»)
- dist(p, V).
7.1. BASIC PROPERTIES
227
. By general measure theory, in T must contain a compact subset C with rl
<
~ (~+1ll(C») - dist(p, V).
Since T is open, there is r2
> 0 such that
{x E ]RN : dist(x,C) < r2} C T. Now, if i' is any line perpendicular to V and at a distance less than r2 from i,then holds, which implies dist(p, V)
+ rl < ~
G+ 1l
l
(i'
n T»)
.
We conclude that, for any point q, if q - p resolves into components in the i direction and in the hyperplane V of magnitude-less than rl and r2, respectively, then q is in Si. This condition is satisfied by all points in the open ball centered at p and haying radius min {rIo r2 }. It is clear that S c niSi. To complete the proof, we must show that niSi c S. If p E niSi and if l is the line perpendicular to V passing through p, then for each i we must have
.
dist(p, V)
1(1 1l (l n T) ) ,
< '2 i +
so passing to the limit as i approaches
00,
1
we conclude that
1 .dist(p, V):5 '21l1(lnT),
and thus that pES.
•
Remark 7.1.6 It is natural to ask whether or not the Steiner symmetrization of a Borel subset of]RN is also a Borel set. The answer is "no", but we will only sketch the argument, since a complete, self-contained discussion would take us rather far afield. Let N" be the cartesian product of infinitely many copies of the discretely topologized positive integers, as in Federer [4], §2.2.6, and let 11"1 : ]R(N-l) x N -. R(N-l) be projection on the first factor. By definition, a Suslin set2 is any image 11'1 (F) where F is a closed subset of ]R(N-l) x N. The classical fact (see Federer [4], §2.2.11) discovered by Suslin is that there exists a Suslin set that is not a Borel set. Applying Federer [4], §2.2.9, we can find a Borel subset r of the standard, middle thirds removed, Cantor set C C [0,1] such that N is homeomorphic to r. Let 'Y be the homeomorphism. If F is a closed subset of 2 Suslin
sets are also discussed in Section 1.3.
CHAPTER 7. STEINER SYMMETRIZATION
228
IIl(N-l) xN, then (idx1')(F) is a closed subset oflll(N-l) xr, where id is the identity map on IIl(N-l). Thus (id x 1')(F) is a Borel subset of R(N-l) x R. As stated above, we may and shall moose the closed set F so that 11'1 (F) is not a Borel set. Let 11'0 : lR(N-l) x r -+ R(N-l) x to} be orthogo~al projection, and let inj : lR(N-l) -+ IIl(N-l) x to} be the homeomorphism sending x to (x,O). Then the diagram lR(N-l)
x
r
1"0x to}
IIl(N-l)
is commutative. Because C has I-dimensional measure zero, the Steiner symmetrization with respect to IIl(N-l) x to} of any subset T of IIl(N-l) x C is the same as the orthogonal projection of T on IIl(N-l) x to}. In particular, we set T = (id x 1')(F) and conclude. that if S is the Steiner symmetrization ofT with respect to IIl(N-l) x to}, then S =
11'0 0
(id x 1')(F) = inj
0 11'1 (F).
Since 11'1 (F) is not a Borel set and inj is a homeomorphism, we see that S is ·not a Borel set. • ·Proposition 7.1.7 If T is a bounded geometrically convex subset of RN and S is obtained from T by Steiner symmetrization, then S is also a geo-
metrically convex set. Proof: Let V be an (N - I)-dimensional vector subspace of ]iN, and suppose that S is the result of Steiner· symmetrization of T with respect . to V. Let x and y be two points of S. We let x' and y' denote the points obtained from x and y by reflection through the hyperplane V. Also, let is and ell denote the lines perpendicular to V and passing through the points x and y, respectively. By the definition of the Steiner symmetrization and the convexity of T, we see that is n T must contain a line segment, say from pz to qs, of length at least dist(x, x'). Likewise, '-II n T contains a line segment from PII to qll of length at least dist(y, y'). The convex hull of the four points pz, qz,PII' qll is a trapezoid, Q, which is a subset of T. We claim that the trapezoid,Q', which is the convex hull of x,x',y,y' must be contained in S. Let x" be the point of intersection of '-z and V. Similarly, define y" to be the intersection of '-II and V. For any 0 ~ T ~ 1, the line i" perpendicular to V and passing through (1 - T}X"
+ TY"
intersects the trapezoid Q c T in a line segment of length (1 - T) dist(ps, qz}
+ T dist(PlI' qll)
(7.2)
7.1. BASIC PROPERTIES
229
and it intersects the trapezoid Q' in a line segment, centered about V, of leJlgth (1 - T) dist(z,z') + Tdist(U,y'). (7.3) But S must contain a closed line segment of i", centered about V, of length at least (7.2). Since (7.2) is at least as large as (7.3), we see that
i" n Q' c f' n S. Since the choice of 0 ::; T ::; 1 was arbitrary we conclude that Q' c S. In particular, the line segment from z to y is contained"in Q' and thus in S .
•
Proposition 7.1.8 If T is a bounded geometrically convex subset of aN and S is obtained from T by Steiner symmetrization, then diam(S) ::; diam(T). Proof: The argument proceeds similarly to the proof of Proposition 7.1.7. Let V, z, y, Z/, y', I .. , and III be as in that proof. Since I .. n T has 1l1_ measure at least dist(z, x'), it follows that I .. n T contains two points P.. and q.. with dist(p.. , q.. ) ~ dist(x, x'). Similarly, ly n T contains two points PlI and qy with dist(Pll,qy) ~ dist(y,y/). It is elementary to check that the" longest diagonal of the trapezoid that has vertices P .. , q.. , Py, qll is at least as long as either of the diagonals of the trapezoid that has vertices z, x', y, y'. The basic fact needed is that the diagonals of the symmetric trapezoid with vertices z, x', y, y' are shorter than the longest diagonal of an unsymmetric trapezoid" with the same height and base lengths. This is readily checked if one sets notation as in Figure 7.2, where X and Y are signed quantities satisfying X + Y = A + B, and considers the problem of minimizing P + Q '" the sum of the lengths of the diagonals, as a function of X (here A and B are fixed and Y is constrained by the equation). • We will be using the Hausdorff distance topology on the set of nonempty compact sets. The definition and relevant facts can be found in the Appendix, Section A.1. One key to the utility of Steiner symmetrization is the fact that any collection of compact subsets of lllN that is closed under Steiner symmetrization with respect to all (N - I)-dimensional vector subspaces of lllN and is closed in the Hausdorff distance topology must contain a ball. Toward proving that fact we first present the following lemma: LeIIllDa 7.1.9 Consider a sequence (finite or infinite) T 1 , T2, . .. of compact subsets ofRN and a sequence Vi, 2, ... of (N -I)-dimensional vector subspaces of lllN such that Ti+1 is the result of Steiner symmetrization with respect to Vi applied to T i • IfTi CB(O,r) andp E 8B(O,r), f> are such that B(p,f) n8i(O,r)nTi = 0,
v
°
CHAPTER 7. STEINER SYMMETRIZATION
230
Figure 7.2: Diagonals of a Trapez.oid
then lR(p,e:) n8'i(0,r) nTj
=0
and lR(p',e:) n 8'i(0,r) n T j = 0
hold for j
~
i, where p' is the reflection of p through Vi.
Proof: It suffices to prove the conclusion for j = i + 1. Let i be the line perpendicular to V passing through p and p'. For any line i' parallel to t. and at distance less than e: from i, the Hausdorff measure of the intersection of i' with Ti must be strictly less than the length of the intersection of [' with 'i(O,r), so the intersection of [' with 8'i(O,r) is not in Ti+l. • Theorem 7.1.10 lfe is a non-empty family of non-empty compact subsets of JR.N' that is closed in the Hausdorff distance topology and that is closed under the operation of Steiner symmetrization with respect to any (N -1)dimensional vector subspace of JRN, then e contains a closed ball (possibly of radius 0) centered at the origin. Proof: Let
e be such a family of compact subsets of JRN and set r = inf{s: there exists Tee with T c '1(0, sn.
If r = 0, we are done, so we may assume r > O. By Theorem A.1.5, any uniformly bounded family of non-empty compact sets is compact in the
7.1l. ISODIAMETRIC INEQUALITY ETC.
231
Hausdorff distance topology, so we can suppose there exists aTe C with
T c iiI(O, r). We claim that T = iiI(O, r). Hnot, there exists p e iB(O, r) and f > 0, such that T c iiI(O, r) \B(p, f). Suppose TI is the result of Steiner symmetrization of T with respect to any arbitrarily chosen (N - I)-dimensional vector subspace V. Let i be the line perpendicular to V and passing through p. For any line i' parallel to i and at distance less than f from i, the Hausdorff measure of the intersection of i' with T must be strictly less than the-length of the intersection of i' with iiI(O,r), so the intersection of i' with 8i1J(O,r) is not in T 1 • We conclude that if PI is either one of the points of intersection of th~ sphere of radius r about the origin with the line i, then
lB(plt f) n 8B(O, r) n TI = 0. Choose a finite set of distinct additional points P2,P3, ... ,P,. such that
88(0, r) C U~IB(pi' f). For i = 1,2, ... , k - 1, let Ti+1 be the result of Steiner symmetrization of
Ti with respect to the (N -I)-dimensional vector subspace perpendicular to the line through Pi and PHI. By
t~e
lemma it follows that
lB(pi,f)n81(0,r) nT; = 0 holds for i :5 j :5 k. Thus we have
T,. n81(0,r)
= 0,
so .
Tic holds for some
7.2
8
c 1(0,8)
•
< r, a contradiction.
The Isodiametric, Isoperimetric, and Brunn-Minkowski Inequalities
To obtain the main results of this section, we need to investigate the behavior of Lebesgue measure as a function on the collection of compact sets topologized by the Hausdorff distance. It would be convenient if the Lebesgue measure were a continuous function on the non-empty compact sets equipped with the Hausdorff distance topology, but this is just not true. Example 7.2.1 H we set
An = {i/n: i = 0, 1, ... , n}, then we see that HD(An, [0, 1])
£([0,1)) = 1.
= 2~
-+
0 as n
-+ 00,
while £(An) = 0 and
232
CHAPTER 7. STEINER SYMMETRIZATION
However, the Lebesgue measure is upper-semi-continuous, and if we restrict the domain to the non-empty compact, geometrically convex sets, then it is continuous. This assertion will follow from the next several propositions. Proposition 7.2.2 Let An C JR.N be a non-empty -compact set for each n = 1,2, .... Suppose C C JR.N is such that HD(An' C) -+ 0 as n -+ 00. Then CN(C) ~ limsupCN(An) n~oo
holds. Proof: This is a corollary of Lemma 7.1.3.
•
Lemma 7.2.3 Let An C JR.N be a non-empty geometrically convex set for each n = 1,2, .... Suppose that Xo, Xl. ... ,XN are affinely independent points (i.e. not all contained in a hyperplane) such that sup{ dist(xi, An), i = 0,1, ... N} -t 0 as n -t 00. If X is a compact set such that X is contained in the interior of the convex hull of {xo, Xl, ••. ,XN}, then, for all sufficiently large n,
XcA n . Proof: Let M be the matrix that has Xl - Xo, X2 - Xo, ... , XN - Xo as its rows. For X E X, let Al{X),A2{X), ... ,AN{X) be the components of
M-l{x - xo).
(7.4)
In (7.4) we think of X - Xo as a column vector. Setting N
AO{X) = 1 - I>i(X), ;=1
we have ;=0
so, by the hypothesis that X is contained in the interior of the convex hull of {XO,Xl, ... , XN}, we have
0< A;(X)
< 1,
for i = 1,2, ... ,N,
and N
LAi(X) < 1 i=l
7.2. ISODIAMETRIC INEQUALITY ETC.
233
hold. Since X is compact and the ~i(Z) are continuous functions of z, we conclude that there exists E > 0 so that E$~i(z)$I-E,
and
fori=I,2, ... ,N,
(7.5)
N
L~i(z) $1-E
(7.6)
i=l
hold for all z E X. For each n = 1,2, ... , let af be a point of An nearest to Z;, for i = O,I, ... ,N, and let Mn be the matrix such that Mn has af - a8,a~ a8, ... , a~ - a8 as its rows. For large enough n, Mn is non-singular and we can let '\f(z), ~~(z), ... , ~~(z) be the components of M;;l(Z -
a8).
Since matrix inversion is continuous on the set of non-singular matrices, and because of (7.2) and (7.6), we conclude that, for all sufficiently large n,
o< and
~f(z)
< 1, for i =
1,2, ... ,N,
N
L~f(z) < 1 i=l
hold. Then the conclusion of the lemma follows from the fact that An is geometrically convex. • Proposition 1.2.4 Let An C JR.N be a non-empty, compact, geometrically convex set for each n = 1,2, .... Suppose C is such that HD(An, C) -+ 0 as n -+ 00. Then C is also geometrically convex and
holds. Proof: It is easy to see that C must be geometrically convex. The more difficult question concerns the Lebesgue measure of C. Suppose without loss of generality that eN(C) > o. Since C is geometrically convex, 8e has eN-measure zero (by Corollary 6.2.7). Thus, in view of Proposition 7.2.2, it will suffice to show that for any compact ACe,
holds.
234
CHAPTER 7. STEINER SYMMETRIZATION
Let A be any compact subset of C. Let x e A be arbitrary. We claim that there exist %0, %1, ••• ,%N in C which are affine1y independent and sucli that % is an interior point of the convex hull of {xo, %1, ••• , %N}. This is actually obvious since we can suppose coordinates are chosen with % at the origin and then set %0 = (-r; -r, •.. ,-r) and, for i = 1,2, ... , N, set Xi equal to r times the i tb standard basis vector, where r·is chosen sufficiently small but positive. Let Uz be an open neighborhood of % chosen so that U z is contained in the interior of the convex hull of {%O, %1, ••• , % N}. By the preceding lemma, there exists an integer n~ such that n ~ n z implies U z CAn.
Since A is compact, finitely many of the sets Uz • x e A, cover A, so we conclude that, for all sufficiently large n, A C An. from which the required inequality follows. • We will also need the following general topoiogicallemma. LeDlD1a 7.2.5 Let T be a topological space, let f : T -+ III be continuous, and let {g")'} ")'Er be a family of functions g., : 7 ~ 7 such that
f[g")'(x)]
~
f(x)
holds for all % E T and all 'Y e r. If SeT and S is the smallest closed subset of T that contains S and is closed under g.,. for all 'Y e r, then sup{J(x) : xeS} ~ sup{J(x) : xeS}
holds. Proof: Set a
= sup{J(x) : XES}.
Then
S' = {x
E
7: f(x) $ a}
is closed in the topology of T and is closed under g." for all 'Y E
ScS'.
r, so
•
Now we can apply some of the tools we have collected. Consider any nOD-empty bounded T C )RN. Since T C B[p,diam(T)] holds, for any choice of pET we obtain the inequality eN(T) ~ TN [diam(T)t
(in case T is not eN -measurable we use the outer measure of T). Presumably a sharper inequality could be obtained by making a more careful
7.2. ISODIAMETRIC INEQUALITY ETC.
235
choice of p, with the choice of the center of a closed ball indicating the best for which one could hope. In fact, that best hope is fulfilled in the following general result: Theorem 7.2.6 (Isodi8D1etric Inequality) For any non-empty bounded
TeaR, hold3. Proof: It is no loss of generality to assume that T is compact and geometrically convex. Now let be the space of non-empty compact, geometrically comex subsets of JRN with the Hausdorff distance topology. Let the function I : T -+ JR be defined by
.r
I(A) = TN
[! diam(A)]N -
.cN(A).
By Proposition 7.2.4, I is continuous. For each (N -I)-dimensional vector subspace of ]RN, we let gv be Steiner symmetrization with respect to V. It is a consequence of Proposition 7.1.8 that Lemma 7.2.5 can be applied to the set S containing the single element T. By Theorem 7.1.10 we know that the corresponding S contains a ball. But I vanishes on any ball, so jeT) ~ 0 holds as required. • Next, we will use Steiner symmetrization to give a proof of the BrunnMinkowski Inequality. This inequality is concerned with the Lebesgue measure of the vector sum of two subsets of Euclidean space. The original result goes back to t~e Inaugural-Dissertation and Habilitationsschrift of Hermann Brunn (Brunn [1] and [2]) in the late..l880's. Minkowski's work on the topic, Minkowski [1], appeared in the late 1890's. Many proofs of the Brunn-Minkowski Inequality have been given, so we will not assert that our proof is new (but it is new to us).
Definition 7.2.7 For subsets A and !3 of]RN the vector B is the set A + B = {a + b: a E A, bE B}.
SUDl
of A and
The line of argument ahead of us would have been far shorter but for the unfortunate fact that the Steiner symmetrization of the vector sum of two sets is not necessarily equal to the vector sum of their Steiner symmetrizations. Example 7.2.8 Working in
]R2,
A = [0,1] x {OJ so
A
+B
we set and
B
= {OJ x [0,1],
= [0,1] x [0,1].
CHAPTER 7. STEINER SYMMETRIZATION
236
Consider Steiner symmetrization with respect to any I-dimensional vector subspace V other than either of the coordinate axes. The Steiner symmetrizations, A' and B ' , of A and B, respectively, are both closed line segments in V, so A' + B' is also a closed line segment in V. But the Steiner symmetrization, (A+B)" of A+B must have area 1,80 (A+B)' ,:; A' +B'. The very simple case dealt with in the following theorem is the basic fact on which our proof of the Brunn-Minkowski Inequality rests. LemDla 7.2.9 Suppose the non-empty sets A, B C IR are both disjoint unions 0/ finitely many open internals. Then
£1(A + B) ~ £1(A) + £1(B)
holds. Proof: We write
m
A = U(Oi,.oi) i=1
and
n
B = Uh;,dj), j=1
where . and 1'1 < dl ::; 1'2 < ~ ::; ... ::; 1'n < dn hold, and argue by induction on m + n. The conclusion of the lemma is obviously true when m + n = 2. Assume now m + n ~ 3. Without loss of generality, we may assume that m ~ 2. Set m-l
A' =
U (Oi' .oi).
i=1
By the induction hypothesis, we have
£1 (A'
+ B)
~ £1 (A')
+ £1 (B).
Noting that (Om
+ 1'n, 13m + dn ) C
(A + B) \ (A'
+ B),
we conclude that £1 (A
+ B)
~
~
= as needed.
(13m - Om) + £1 (A' + B) (13m - Om) + £1 (A') + .c,l (B) £1 (A) + £1(B) •
7.2. ISODIAMETRlC INEQUALITY ETC.
237
Lemma 7.2.10 Let A and B be non-empty subsets oj III Then
£"1 {A + B) ~ £1 (A)
+ £1 (B)
holds. Proor: We may and shall assume that both A "and B are sets of finite measure. Let 10 > 0 be given. We can find compact sets A' c A and B' C B such that " £1 (A')
+ £1 (B')
~
£1 (A)
+ £1 (B) _ 10.
Let and
l!i:::>V2 :::> ••• :::>B' be decreasing sequences of open sets such that
nUi =A' 00
"
00
and
"
n~=B'. i=l
i=l
Since A' and B' are compact, we may assume each Ui and each Vi is a finite union of bounded open intervals. We note that
n(Ui + Vi) = A' + B' i=l 00
so, using the result of Lemma 7.2.9, we have
£l(A+B)
~
= ~
= ~
£l(A' +B') .lim [£l(Ui + ~)]
'-'00
.lim [£1 (Ui)
'-.00
+ C} (~)]
£l(A') + £1 (B') £1(A) + £1 (B) _ 10.
Since 10 > 0 was arbitrary, the result follows.
•
Lemma 7.2.11 Let A and B be non-empty subsets oJ]RN. Let V be an (N - I)-dimensional vector subspace oj JRN". IJ X, Y, and Z are the result oj Steiner symmetrization with respect to V applied to A, B, and A + B, respectively, then
X+YCZ
CHAPTER 7. STEINER SYMMETRIZATION
238
and
[.eN(X
+ y)]l/N _
[.eN(X)]l/N _ [,CN(y)]l/N
(7.7)
$ [.eN(A + B)]l/N _ [.eN(A)]l/N _ [.eN(B)]l/N hold.
Proof: First we show that X + Y c Z. Let z + 11 E X + Y be arbitrary. Set z = x + y. Let lz be the line perpendicular to V, passing through x. Similarly define i. and la. By the definition of Steiner symmetrization, lz n A is non-empty and
holds. Likewise, i.
n B is non-empty and lI!(l.
n B)
;::: 2dist(1I, V)
holds. H a E iz nA and b E l.nB, then a+b E l .. n(A+B). It follows that l .. n (A + B) is non-empty, and by Lemma 7.2.10, isometrically identifying lz, i., and i .. with lR, we have
1£1[l.. n(A+B)]
;::: ;:::
1£1(lznA)+1l 1 (l.nB) 2 dist(x, V) + 2 dist(y, V) = 2 dist(z, V).
Thus we have z E Z, as required. The inequality (7.7) follows, since .eN (X) and .eN (X + Y) $ .eN (Z) .eN (A + B).
=
= .eN (A), .eN (Y) = .
.eN (B), •
We could now prove the Brunn-Minkowski Inequality in the limited context of compact, geometrically convex sets, but the result is true more generally, so we will introduce another family of compact sets on which the Lebesgue measure is a continuous function. Definition 7.2.12 For f > 0, let e. and c. be the maps sending the compact set C to the compact sets e. (C) and c. (C) defined by
e.(C) = {z: dist(x,C) $ f}, c.(C) = {x: dist(x, RN \ C) ;::: f}. We will say that a compact set, C, is f-thick if C
= e.
0
c.(C).
The collection of non-empty f-thick sets will be denoted by IC. and will be topologized by the Hausdorff distance.
239
7.2. ISODIAMETRIC INEQUALITY ETC.
Remark 7.2.13 The E-thick sets were used in Parks [1] to study the problem of area minimjzation in arbitrary dimensions and codimensions. We thiDk of e. (C) as being the "expansion" of C and ~(C) as being the "core" of C. Acute corners are among the things that the hypothesis that a set is E-thick rules out. Thus a triangle with interior in the plane cannot be t:-thick. More generally, an N-simplex in JllH, N ~ 2, is not E-thick. The following lemma is an almost immediate consequence of the definition of E-thick sets; its two corollaries are clear. Lemma 7.2.14 For E > 0, A E IC. holds there ezists yEA with x E B(y, t:) C A.
i/ and only if/or each
Corollary 7.2.15 If A is a non-empty compact Bet and t: IC•. Corollary 7.2.16 IfE
x E A
> 0, then e.(A)
E
> 0 and A,B E IC" thenA+B E IC•.
Verifying the continuity of the Lebesgue measure with respect to the Hausdorff distance topology on IC. requires more work. Proposition 7.2.17 Fix t: > O. The collection 0/ non-empty t:-thick sets is closed .in the Hausdorff distance topology, and the Lebesgue measure is a continuous function from IC~ to R.
l.:l
Proof: Suppose that {Ai C IC. and C is a non-empty compact set such that HD(Ai, C) ~ 0 as i ~ 00. Let x E C be arbitrary. Let Xi E Ai be such that Ix.- Xii = dist(x,Ai). By Lemma 7.2.14, for each i there exists Yi sucn that Xi E B(Yi, t:) C Ai.
Passing to a subsequence if necessary, but without changing notation, we may assume that y. converges to some Y, and it follows that
x E B(y, t:) C C. Since x E C was arbitrary, we see by Lemma 7.2.14 that C E IC•. Set JJ = lim inf eN (Ai). Pass to a subsequence, but without changing notation, so that Let DeC be countable and dense in C. Applying the argument of the previous paragraph to each point in D, and using a diagonalization argument, we can pass to a subsequence, again without changing notation, such that for each d ED there exist Xi(d) E B(Yi(d),t:) C Ai and y(d) E C with
y.(d)
~
y(d),
de i(y(d),t:) C C.
CHAPTER 7. STEINER SYMMETRIZATION
240
In the remainder of the proof we will use this subsequence.
For I
= 1,2, ... , set
B,
= n Ai,
00
and E
= U B,. '=1
i~'
We will show that c. N (C \ E) = O. Suppose x e C \ E is arbitrary. Choose any 0 < r < We can find a point d e D with Id - xl < r 13. There exists 10 so that i ~ 10 implies IYi(d) - y(d)1 < r/3. Then, for i ~ 10, we have
if.
B(y(d),f-r/3) CB(JIi(d),f) C Ai and Iy(d) -
xl
:$ Iy(d) - dl
+ Id - xl < f + r/3. IT I
B(y(d), f
-
~ 10
we have
r/3) C B,.
Assuming y(d) - x =1= 0, let u be the unit vector in that direction; then we have
iR(x,r)nB,
~
~ ~
iR(x,r)ni(y(d),f-r/3) B(x,r) n i[x + (f + r/3)u, f iR[x+(SrI6)u,rI6].
-
r/3]
H on t4e other hand y(d) = x, then
B(x,r) nB, ~ B(x,r) ni(y(d),f- r/3)
= iR(x,r).
In either case, we see that
holds for I ~ 10 , We conclude that C \ E cannot have density 1 at x, but this implies that C.N(C \ E) = O. Finally, C \ E is the decreasing intersection of the sets C \ B" so C.N(C \ B,) -+ 0 as 1-+ 00. But B, is a subset of A" so C.N(C \ A,) -+ 0 as I -+ 00. It follows that
and the proposition follows by Proposition 7.2.2. Proposition 7.2.18 Let A and B be f-thick subsets oj RN. Then
•
7.2. ISODIAMETRIC INEQUALITY ETC.
241
= /C. x /C•. The function I
Proof: Consider the topological space T II defined by
:T
~
I(A,B) = [,CN(A + B)]l/N _ [CN(A)]l/N _ [CN(B)]l/N is continuous. For each (N -I)-dimensional vector subspace V of aN, let the map !IV : T ~ T be simultaneous Steiner symmetrization with respect to V, by which we mean Uv(A,B) (A',B') where A' is the Steiner symmetrization of A and B' is the Steiner symmetrization of B. As a consequence of Lemma 7.2.11, the conditions of Lemma 7.2.5 are satisfied. Let S consist of the single pair (A, B). By Theorem 7.1.10, we know that if S is as in Lemma 7.2.5, then among the elements of S is a pair of balls. But I vanishes on any pair of balls, so the result follows by Lemma 7.2.5 . •
=
We can now prove the general result: Theorem 7.2.19 (The Brunn-Minkowski Inequality) Let A and B be non-empty subsets of aN. Then
Remark 7.2.20 Following our usual convention, if T is a non-measurable set, then C(T) denotes the outer measure of the set. Proof: Let A and B be given. Let Aa and Ab be any constants satisfying with and . CN (B) > Ab ~ O. Find non-empty compact A' C A, B' C
E! with
and For f > 0, by Proposition 7.2.18, we have [.cN(e.(A')
+ e.(B,»]l/N
Taking the limit as
f
~
[CN(e.(A,»]l/N
~
[Aa]l/N + [Ab]l/N.
+ [.cN(e.(B'»]l/N
..L. 0 we obtain
Since Aa and Ab were arbitrary, the theorem follows.
•
The classical application of the Brunn-Minkowski Inequality is a proof of the Isoperimetric Inequality in aN comparing the N-dimensional measure of a set to the (N -I)-dimensional area of its boundary. This will bring us almost to the goal Steiner originally sought: We will have shown that no
CHAPTER 7. STEINER SYMMETRIZATION
242
:6.gureenclosesgreater N-dimensional volume, given the (N-l)-dimensional area of its boundary, than does a sphere. The Bnmn-MiDkO~ski Inequality is the perfect tool for the· proof of the Isoperimetric Inequality, if the (N -I)-dimensional boundary surface area is measured using the Minkowski content. Recall Definition 3.3.1 from Section 3.3. Definition Suppose A C ]RN and 0 S K S N. The K-dimensional upper Minkowski content of A, denoted by M·K(A), is defined by. .eN {x: dist(x, A) < r} M .K(A) -1· - unsup T N-K . r.l.O
N-Kr
Similarly, the K -dimensional lower Minkowski content of A is denoted by M~ (A) and defined by
M!, (A)
= lim inf .eN {x : dist(x, A) r.l.O TN_Kr N - K
< r} .
In case the K -dimensional upper Minkowski content and the K -dimensional lower Minkowski content of A are equal, then their common value is called the K -dimensional Minkowski content of A and is denoted by MK (A).
Remark 7.2.21 The Minko~ski content is traditionally used for integer values of K. By Federer [4], §3.2.29, for any classical K-dimensional smooth surface, the Minkowski content is equal to the standard value for the Kdimensional measure of the surface. Theorem 7.2.22 (The Isoperitnetric Inequality) Suppose that SeliN and.e N [S] < 00. Then N-l
M~-l[aS] ~ NTN (.e~~) r r
(7.8)
Proof: We may and shall assume that S is closed. If .eN (8S) > 0 held, then the Minkowski content of as would be infinite, so (7.8) would trivially hold. So we may also assume .eN (as) = o. Considerr > o. Set T = {x: dist(x,IRN \ S) ~ r}. Observing that {x: dist(x, S) < r} J T
+ JIi(O, 2r),
and applying Theorem 7.2.19 (the Brunn-Minkowski Inequality), we conclude that .eN({x: dist(x,S)
< r})
~
[(.eN(T»l/N +2TNl/Nr]N
~
.eN(T) + 2N (.eN (T») (N-l)/N TNl/Nr.
7.3. EQUALITY IN THE ISOPERIMETRIC INEQUALITY Since
. {x: dist(x,8S)
243
< r} :::> ({x: dist(x,S) < r} \ T),
we have ,CN{X:
~~,8S) < r} ~ NTNI/N(,CN(T»(N-I)/N.
The result follows since our assumption that ,CN (8S)
= 0 implies
• Rem8l"~ 7.2.23 The results in this section have shown that the sphere mininiizes area among surfaces enclosing a given volume, but we have not shown that the sphere is the unique such surface. We will give a proof of uniqueness in the next section. Steiner symmetrization also can be used to show that the sphere is the unique shape making the Isoperimetric Inequality an equality. An exposition of this important fact by G. Talenti, in the context of Caccioppoli sets, can be found in Gruber and Wills [1].
The Isoperimetric Inequality arises in an extraordinary number of mathematical ·contexts. In this' book, we have alre.ady proved and used a form of it in Section 3.7. A survey about isoperimetric inequalities appears in Osserman [1], and we call the reader's attention to a proof using elementary complex variable theory that appears in Gamelin and Khavinson [1].
7.3
Equality in the Isoperimetric Inequality
In considering the uniqueness of the solution to the Isoperimetric Problem, we will change context to the category of C2 surfaces. This will save us from considering some rather messy technical issues, and will allow us to present the beautiful method of Aleksandrov. Theorem 7.3.1 If S C]RN is a C2 surface enclosing a region R stich that any isotropic defonnation of]RN that preserves the N -dimensional volume of R increases the (N -I)-dimensional area of S, then the mean curvature of S is constant. Proof: Consider any two distinct points p and q in S. Since S is a C 2 surface, we can find coordinate systems (Xl, X2,··., XN) and (YI, Y2,"" YN) with origin at p and q respectively, and such that the plane XN = 0 is the tangent plane to S at p, while the plane YN 0 is the tangent plane to S at q. Let f : IRN - 1 --+ IR be the C 2 function such that
=
CHAPTER 7. STEINER SYMMETRIZATION
244
gives S in a neighborhood of p. Let 9 : that
aN - 1
-+ III be the (J2 function such
gives S in a neighborhood of q. For simplicity also suppose that the xN-axis and the YN-axis point toward the exterior of R. Let 5 > 0 be small enough that the open 5-ball about the origin of the x-coordinate system and the open 5-ball about the origin of the ycoordinate system do not intersect. Let 77(Xl,X2, ... ,XN-l) be smooth and compactly supported in the open 5/2-ball, and bounded by 5/2. Let (Y1J1I2, ... ,YN-d have the same properties. For (s, t) E B2 [(0, 0), 1], define a comparison surface S(B,t) which agrees with S outside the 5-balls about the origin of the x and y coordinate systems, but in those balls is the graph of J+8TJ and g+t(, respectively. Let 6. V(s, t) denote the difference between the volume enclosed by S(B,t) and the volume enclosed by S. Likewise, let 6.A(s,t) denote the difference between the area of S(s,t) and the area of S. It is elementary to see that 6.V(s,t)
=
6.A(s, t)
=
J
S77(x) deN-Ix
J{\/1 +
+
!
IDJ(x) +sDT](x)1 2
-VI + IDJ(x)j2 } +
J{'1'1
t(y)deN-Iy,
deN-Ix
+ IDg(y) + tD(y) 12
-VI + IDg(y)1 2 }
deN-Iy,
where the integrals are over the open 5-balls. To find comparison surfaces that fit the requirement of enclosing the same volume, we need to find choices of (s, t) for which 6. V(s, t) = O. In general one applies the Implicit Function Theorem, but in our case we can simplify matters by choosing 77 and ( to be the same function and choosing t = -s. With these choices, J(s)
= 6.A(s, -s) =
J{'1'1 +
IDf(x)
+ sDT](x)12
-VI + IDf(x)12 } deN-Ix +
J{'1'1 +
IDg(x) - sDT](x)12
-VI + ID9(x)1 2 } deN-Ix.
7.3. EQUALITY IN THE ISOPERJMETRIC INEQUALITY
245
Since AA(O,O) must be minimal, we can differentiate J(s) with respect to s and conclude that J'(O) = O. The result of this computation is 0= J'(O) =
f {V1 +
D1J(z) . D I(z) . _ D1J(z)· Dg(z) } dCN-Iz. IDI(z)1 2 IDg(z)1 2 .
V1
*
The Gauss-Green Theorem reveals that
o=
f( 1/ x
){eli ( v
) Vi +D I(z) IDI(z)12
-
d· ( IV
) Vi +Dg(z) IDg(x)12
} dCN -
l
x.
As is usual, since the function 1J is essentially arbitrary, the other factor in the integrand must vanish identically, so we conclude that eli ( v
DI(z)
v'1 + IDI(z)12
)
d· (
-
IV
Dg(z)
)
0
Vi + IDg(z)12 =.
Evaluating at the origin we find that b./(0) = b.g(O).
By Lemma 2.3.5, we see that N:'I b. 1(0) and N:'l b. g(O) are the mean curvatures of Sat p and q, respectively. • In the context of C 2 surfaces, the question of the uiliqueness of the .solution to the Isoperimetric Problem has been reduced to the characterization of the surfaces of constant mean curvature in Euclidean space. The physical analogue of a 2-dimensional surface of constant mean curvature in IR3 is a soap bubble, because the difference in air -pressure between the inside and outside of the bubble is proportional to the mean curvature of the bubble. The question in those terms is whether or not a soap bubble enclosing just one region must be spherical.· This was answered in the affirmative in 1958 by A. D. Aleksandrov. We present his elegant argument below.
Theorem 7.3.2 If S c ]RN is a connected C2 surface of constant mean curvature enclosing a bounded region R, then S is a sphere. Proof: Choose a coordinate system (Xl, X2, ... ,ZN) such that the origin is a point of S, the plane z N = 0 is tangent to S at the origin, and S is contained in the half-space XN ~ O. Consider a plane V = {x: ZN = d> OJ. We reflect the surface Sn{(Zl,Z2, ... ,XN)
:XN
::;d}
in the plane V to form the surface Sd. We claim there is a largest positive number do such that Sd is contained in SuR for all 0 < d < do. To verify the claim it will suffice to show there is some t5 > 0 such that Sd is contained
CHAPTER 7. STEINER SYMMETRIZATION
246
in SuR for all 0 < d < 6. Because S is a C2 surface, there is 6 that orthogonal projection of
. Sn {(XI,X2, ••• ,XN)
: XN ::;
> 0 such
26}
into the plane XN = 0 is injective. Let D be the image of this injection and for (Xt.S2, •.. ,SN-I) E Diet /(XI,X2, ••• ,XN-t> be the unique number less than or equal to 6 such that
ForO
D' = {(Xl, X2, ••• , XN-t>
: /(X1.X2,'" ,SN-l) ::;
d}
and let 9 : D' -+ III be defined by
Then SIl is the graph of 9 and is clearly contained in SUR. We know that Silo lies on one side of S and it either makes contact with S above the plane X N = do or it makes first order contact at some point of the intersection of S with the plane SN = do. Either way the maximum principle of E. Hopf (see Gilbarg and Trudinger [1], Chapter 3) guarantees that Silo = S n {(X1.X2,'" ,XN) : XN ~ do}. That is, S is symmetric in the plane. Now the plane can be chosen orthogonal to any direction, so the family (with exactly one element) C = {SUR} is closed under Steiner symmetrization and thus is a closed ball by Theorem 7.1.10.
•
Aleksandrov's Theorem answers a very natural question, but almost immediately another question arises: H the surface S is not the boundary of a region, that is, if it is not imbedded, 3 then can it have constant mean curvature and fail to be a sphere? A partial answer to this question was given even before Aleksandrov's work. H. Hopf (in Hopf [1]) showed that if a surface of constant mean curvature is topologically a sphere, then that surface is, in fact, a Euclidean sphere. Quite remarkably, H. Hopf's result cannot be extended to higher topological types. In 1984, H. Wente constructed a constant mean curvature torus in Jll3 (see Wente [1] and Walter [1]). Later, using perturbation techniques, N. Kapouleas produced constant mean curvature surfaces of any genus greater than or equal to 2. 3Imbedded surfaces with constant mean curvature in 1113 were discussed earlier in Section 2.3.
Chapter 8
Topics Related to Complex Analysis 8.1
Quasiconformal Mappings
The subject· of quasiconformal mappings exists for at .least four reasons. • First, there·is the remarkable theorem of Liouyille (which we shall prove later): except for isometries, dilations, and inversions (to be defined below), there are no smooth conformal mappings in dimension three or greater. Thus one seeks a natural generalization of conformal mappings to higher dimensions. • Second, many properties that are commonly p'roved for conformal mappings (in dimension two) are really properties of quasiconformal mappings. Mathematical facts, particularly geometric ones, are often most clearly understood when the fewest hypotheses are used. • Third, taking the first remark into account, any basic geometric property that distinguishes among dimensions-particularly among Euclidean spaces-is of a priori interest. • Fourth, the theory is used in the study of differentials on, and moduli for, Riemann surfaces-work pioneered in Teichmilller [1]. Let U, V C JIlN be open sets. Recall that a Cl mapping F : U -+ JIlN is said to be conformal if, at each point P E U, the Jacobian matrix is an orthogonal transformation or an orthogonal transformation composed with a dilation or contraction. (Using the language of Riemannian geometry, a coordinate-free rendition of conformality may be formulated. Also, the notion may be formulated for arbitrary Riemannian metrics. But the formulation we have given will suffice for our purposes.) Both the orthogonal translation and the dilation will, in general, depend on the point P. 247
248
CHAPTER 8. TOPICS RELATED TO COMPLEX ANALYSIS
Note that, in lilt, any Cl mapping is conformal. For the role of the ortbogonal transformation is moot, and the dilation is simply given by the scalar value IF' (P) I. In lIl2 , it is a standard result of complex function theory that the sense preserving conformal mappings are precisely the complex analytic mappings and conversely. Of course it is clear from the definition we have given that any isometry of -aN, or any dilation, or any composition of these, is a conformal mapping. There is a third type of conformal mapping-in any dimension-which we shall now describe. By an inversion we shall mean a mapping of the form X 1-----+
x-P IIx _ PII 2 •
Here P is a fixed point in space and
II II
denotes the standard norm on
JllN. Such a mapping is undefined at the point P, but we may compactify
JllN by adjoining a point at infinity that will serve as the image of P under this mapping. In practice, this topological detail is of no interest. Let us check that inversion is a conformal mapping. We may assume that P = 0 (in other contexts, the resulting mapping is called the Kelvin inversion-see Krantz [3]). H we use y to denote the variable in the target space then it is straightforward to calculate that
if i = k ifi-:Jk.
One checks by hand that the.matrix
:J =
(a
Yk )
aXj
is a dilation of an orthogonal matrix. In fact, the dilation factor is 1/lIxII2. So we see that in one dimension the theory of conformal mappings is meaningless, and in two dimensions (by the lliemann Mapping Theorem) it is rich. We have given three examples of smooth conformal mappings in JllN, and later we shall see that these are all that there are. Now we turn our attention to quasi conformal mappings in IR2. The theory of quasiconformal mappings is currently an active field of research. Results in the subject tend to become quite technical rather quickly. In the present treatment we shall content ourselves with a survey of the elementary ideas. It is both convenient and appropriate to first study quasiconformal mappings in dimension two. For here the notation and the statements are particularly elegant, and we may clearly see the connections with the more classical notion of conformality. At the end of the section, we shall make some remarks about the higher dimensional case.
8.1. QUASICONFORMAL MAPPINGS
249
The books Ahlfors [1] and Lehto and Virtanen [1] are excellent sources for an introduction to quasiconformal mappings in dimension two. Equivalent Definitions of ComorUlal Mappings Grotzsch, who invented quasiconformal mappings (in Grotzsch [1]) but did not call them by that name l , considered the fact that a square S (together with its interior) in the plane cannot be mapped conformaIly to a rectangle R (together with its interior) that is not a square-in such a fashion that vertices go to vertices. He wished to study the most nearly conformal, vertex-preserving map of the square to the rectangle, and thus desired a measure of "almost conformality." Alternatively, one can consider a mapping F of a complex variable to be simply a mapping of (a region U in) JR.2 into JR.2. Considered in this way, the Jacobian of F is just a 2 x 2 matrix-or a linear mapping-and it will take circles to ellipses. If we mandate a bound on the eccentricity of the ellipses that arise, then the result is a quasi conformal mapping. Thus we may think of quasiconformal maps in terms of either distortion of squares or distortion of circles. We will also have a differential characterization of quasiconformal mappings. . We will develop the habit of identifying JR.2 with C. When we are thinking of real analysis, we write our variables as (x, y); in complex analysis we write them as z and z. Of course, these two notations are related by z Z
tz
= =
x +iy x - iy.
(8.1)
tIl
The vectors and are a basis for the tangent space of ]R2. In complex notation, it is customary to take .. ()
-
{)z ()
-
Oz
!2
1
~~() -i~~ .() {)x
{)y
(8.2)
-2 -+z{)x {)y
as basis vectors. Notice that ()
.
-z=1
~z=O ()z
-z=O
~z=l. Oz
{)z ()
Oz
tz
:z
These four conditions uniquely determine and as first order linear differential operators with constant coefficients. It is convenient to write, for a C l diffeomorphism F : U -+ V of planar domains, F(x,y) = u(x,y)+iv(x,y) ~ (u(x,y),v(x,y». Then the Jacobian 1 Ahlfors
is credited with the terminology
250
CHAPTER 8. TOPICS RELATED TO COMPLEX ANALYSIS
determinant of F is
:J =
U z l11/ -l1z
Uv·
(Here we use subscripts to denote derivatives.) An easy calculation shows that
:J = IF,z12 -1F'i"12 =F o.
Of course the Jacobian never vanishes (because F is a diffeomorphism). H we assume in advance that F is sense-preserving at each point, then :J > 0 at each point and therefore
1F,z1 > IF'i"I·
(8.3)
We take (8.3) as a standing hypotheses in what follows. In fact the columns of the Jacobian matrix are Fo: and F,I • In complex notation, we may say that the columns are (F,z + F z) and (Fz - F z ). Thus any directional derivative of f in the direction T = (I',~) is
VTF = p(F,z where 1', ~ E C and
Ipl2 + 1~12
+ Fz ) + ~(F,z - Fz),
= 1.
Clearly
1F,z1-1F'i"1 :5 IVTFI :5lFzl + IFzl· Indeed the upper extremum is realized when
The lower extremum is realized when
We think of the Jacobian of F, frozen at a point, as a linear map; and we think of this matrix operating on a circle, centered at the origin, in the tangent plane. Then the image is an ellipse and the ratio of its major axis to its minor axis is, according to our calculations, (8.4) We c:all the number D F the dilation factor of the mapping F (also called the dilitation factor). Already we can see that F is conformal if and only if D F = 1 at each point, and this in turn is true precisely when Fz = 0 at each point. This last equation, Fz = 0, is equivalent to the Cauchy-ruemann equations
In conclusion, the following are equivalent:
251
8.1. QUASICONFORMAL MAPPINGS • F is conformal
• circles go to circles
• DF=1 • the Cauchy-Riemann equations are satisfied • F is holomorphic
It is sometimes convenient to formulate our condition in terms of the quantity
(8.5)
Clearly
DF = 1 + IFTI/lFzl = 1 + dF 1 - IFTI/IFzl 1 - dF· In particular, F is conformal at P E U if and only if dF(P)
= O.
Definition 8.1.1 Let. K ~ 1 be a constant. The mapping F is said to be K-quasiconformal i/ DF :::; K at each point 0/ its domain u.. 1/ K is not specified then we simply s.ay that F is qu~iconformal. Setting k = (K -I)/(K + 1) we see that F is K-quasiconformal if and only if dF :::; k, that is, if and only if
Ihl :::; klfzl· Thus, F is conformal precisely when K == 1 or
(8.6) .~_ ==
O.
Quasiconformal Mappings of Plane Domains It is now instructive to return to Grotzsch's formulation of quasiconformality. Again, we follow the exposition in Ahlfors [I]. Let Q be a closed rectangle with sides 0,/3, and let Q' be a closed rectangle with sides 0',/3'. We assume that the sides 0, a' are horizontal and that /3, /3' are vertical. We consider a continuously differentiable mapping F from Q to Q'. The mapping is assumed to take the side a to a' and the side /3 to /3'; thus we assume that the origin and neighboring vertices go to vertices. If a 1/3 =I: 0'1/3' then the mapping F cannot be conformal; we wish to measure how far from conformality it may be, and we wish to determine that mapping F that is most nearly conformal (by whatever yardstick we choose to measure proximity). We may assume that al/3 < a' 1/3' (otherwise rename the rectangles and vertices). Assume that each of Q, Q' has lower left corner at the origin. Now we perform an elementary calculation, using nothing more than calculus: 0'
:::;101 Id(F(x + iy)1 :::;101 (lFzl + 1FT!) dx.
(8.7)
252
CHAPTER 8. TOPICS RELATED TO COMPLEX ANALYSIS
Multiplying both sides of (8.7) by
a'P
/3,
we obtain
·L (lFzl + IFzl) Q
$
P
=
L{J LQ (lFz I + IFzl) dx dy
=
If [~:~:: ~ :~::] .[v'lFz 12 - IFzI2]
dx
dx dy .
As a result, we have
(a')2p 2
(Sch~arz)
II :~:: ~ :~;: 11 IFzl2 -IFzI2 II 11 dxdy·
=
DFdxdy·
J:Fdxdy
Q
$
dxdy
Q
Q
Q
[(area of Q). SUpDF]·(area of Q'). Q
Dividing through by [(area Q) . (area Q')] = ap . a' /3', we obtain 01'
1/3'
-/~ $SUpDF. Q
OIp
. .
In fact this minimum is actually attained by the linear mapping of Q to Q':
L(z)
/3') z +"21 (a'~ - 7i/3')_z.
="21 (01' (; + 7i
We have proved the following proposition: Proposition 8.1.2 Let Q and Q' be rectangles as above, with side lengths a, P and /3' respectively. There exists a K -qv.asiconformal mapping of Q to Q' if and only if 1 01' 1/3'
01',
-<--
011/3 -
An important problem is to characterize when two domains in the plane are quasiconformally equivalent. A complete answer can be given to the more restrictive problem of characterizing which domains are quasiconformally equivalent to an open metric ball (or disc). Theorem 8.1.3 A domain n in the plane is qv.asiconformally equivalent to an open metric ball if and only if ao is a nondegenerate continuum.
B.1. QUASICONFORMAL MAPPINGS
253
For details, we refer the reader to Gehring [2] and to the references therein. Liouville's TheoreIn Having set the stage, by placing conformal mappings in the more general context of quasiconformal mappings, we shall prove Liouville's theorem which says, in effect, that there are no interesting conformal mappings of Euclidean space in dimensions 3 and higher. We state the result for C 4 mappings, but weaker hypotheses suffice (see Gehring [1] and Reshetnyak [I». TheoreIn 8.1.4 (Liouville [I)) Let U c]iN be an open set, and let F : U -+ aN be a C4 mapping that is con/tmnal. Then F is the composition 0/ an isometry, a dilation, and an inversion. Proof: For the proof, we follow closely the argument in Dubrovin, Fomenko, and Novikov [1]. We restrict our attention to N = 3. Let x be a point in the domain of F and 1/ the corresponding image point. It follows from conformality that if two vectors ~ and TJ are orthogonal with respect to the standard Euclidean inner product in the x-coordinates, then their images will certainly be orthogonal in the standard Euclidean inner product in the 1/-coordinates. Let us denote the dilation factor at each point by '\(x). We shall assume that A never vanishes. Thus, if 3 denotes the Jacobian matrix of F, .. 1131]11 2 = A(x)II1]1I 2 at each point x. Now fix three mutually orthogonal vectors TJI,TJ2,TJ3. Let {TJH1=1 be the coordinates of TJ" in the x-coordinate system, k = 1,2,3. Of course the three vectors 3 TJl, 3 TJ2, and are then mutually orthogonal in the y-coordinate system. We differentiate the equation (3TJl , 31]2) = 0 with respect to the coordinate Xi to obtain
.11]3
~(3TJ1, 3TJ2 ) OXj
0
0
= \/ Lt x,O~;Xt 1]L31]2) + \ /31]I'L . t x,O~;Xt TJ1) = O. We multiply both sides of (8.8) by
rd and sum over j
to obtain
02y 13 2) + (.11]1, ~ 02y 23) (~ L...J 0 .ox TJtTJi' '"' TJ L...J ax .ox TJtTJj • t,i x, t t,j , t '7
(8.8)
(8.9)
Now plainly we may permute the indices in (8.9) to obtain three such equations. Adding two of these, and subtracting the third, yields the equation (8.10)
254
CHAPTER 8. TOPICS RELATED TO COMPLEX ANALYSIS
There are two other equations, similar to (8.10), that may be obtained by permuting indices. Because we are in dimension 3, and since the .:ITJl are mutually perpendicular, equation (8.10) implies that .
Lj,l 8x~: '1}'11 = J.&(x)[.:ITJl] + v(x)[.1TJ , Xt
2]
(8.11)
for some coefficients J.&(x) and ,,(x). Taking the inner product of this equation with .:ITJ 1 or with .1rt enables us to solve for I' and v. Thus we have I'
=
1I.1~1112 (t.1 8X~~Xt TJ}TJ1, .:ITJ
v
=
1I.1~1I2 (t.1 8X~~Xt TJ}TJ1, .:ITJ
1
2
)
,
(8.12)
)
•
(8.13)
But we know that 11.1'1112 = ~IITJII2. It is convenient in this argument to replace ~ with A = 1/~. With this notation, the equations (8.12) and (8.13) may be rewritten as I'
= = =
[1"" 8 ( A [1 8 (IITJ 112)] 11711112 2 ~
11'1A 1112 2 ~ TJm2 8xm
1
I
2
TJm 8xm
_..!..."
2
1 ]
.:ITJ ,.1TJ )
A(x)
8A
2x~"'m8xm
and, by a similar calcul.ation,
" =
_..!... "" 1 8A 2A ~TJm8xm·
Substituting these last two expressions for I' and" into (8.11), and performing some manipulations, we obtain that
1 2 = -1
82 y
-::--~-TJ''''l 8xj8xt J
If we set p = 8 2 (py)
2X
~,
(L
L
oX 8y 1 2+ --"'tTJ' 8X 8y 1 2) --'1t71' j,t OXj 8xt J j,t 8xt 8xj J
(8.14) •
then
8p oy
op 8y
8 2y
82p
- = -8xj -+ -8xt -+ p 8xj8xt +8xj8xt' y-8xj8xt 8xt OXj
= _1_ (8A !!JL + oA 2~
8xj 8 xt
8y
OXt 8xj
+ 2A
82 y ) 8xj8x t
+y
82p. 8xj 8x t
(8.15)
8.1. QUASICONFORMAL MAPPINGS
255
Equation (B.15), together with (8.14), gives us that
~
82
(
1 2)
L..J 8x.8x P!Jf1j''11 j,l
I
J
, ( ~ P 1 2) = ,L.J 8x.8x "Ij"ll Y· j,l J I
(B.16)
Differentiating (8.16) with respect to xI' and then multiplying through by "I: and summing over p, we obtain ~ 8 3 (py) 1 2 3 L..J 8x ·8x 8x "Ij"llTJp
j,l,p
t
J
I'
2 12) ",'13+ ~ ( 83P 123) "8.17). L..J 8x.8x 8x TJjTJITJ
_ ~ ( 8 P - L..J 8x.8x'1jTJt ·t J,
J
t
'7
. _
.,
J
J, ,P
t
I'
p
IA
Now notice that the first and last expressions in (8.17) are invariant UDder permutations of the indices 1,2,3. It follows that the middle expression is likewise invariant. Therefore we have
82 P I 2) '7 3 ( 82P 2 3) '7 I ( 8Xj8xlTJjTJt "'TJ.= 8xj 8x,TJj "ll ",TJ· Recall that none of .:JTJI , :JTJ2 , :J'13 is zero (since our transformation is sup- . posed to have non-vanishing Jacobian· at each point).· Thus the mutual . orthogonality of these expressions implies that " L.J t ,I'
~P 8 8 '1t1TJp2=_ O. Xt XI'
(8.18) .
In short, the bilinear form represented by the left side of (8.18) vanishes· on any pair of orthogonal vectors. It follows that the coefficient matrix of such a form must be scalar, that is, a multiple of the identity. As a result,
8 2p _ {u(X) 8x j 8xp 0
if if
j = p j =1= p.
(8.19)
We shall next demonstrate that u is identically constant. Let el , be fixed vectors at the point x. We differentiate (8.19) three times, and then multiply by e]{1e:; finally we sum over the repeated indices. The result is
e, e
L ·t
J,
8 3
p 8xj 8:&t 8xp ,p
e}de; = EI (8u e:) <e,e)· 8xt
We obtain a similar equation by interchanging ing that new equation from (8.20). The result is
(8.20)
e1 and e, and subtract(8.21)
256
CHAPTER 8. TOPICS RELATED TO COMPLEX ANALYSIS
Since (8.21) holds for all vectors stant. As a result·,
tPp
e
i,
= U6lj =
(JXllJzj
it follows that
{iT0
if l if l
u is identically con-
=j
=/: j.
It follows that with aI, b1 constant. Of course the inverse mapping to F is also conformal, with conformality factor 1/),. It follows by similar reasoning that
Ii = 1
a211y -
2
QII + b2,
with a2, ~ constants and Q the image of P under F. In summary, we have ( all1x -
PII 2 + b1 )
(
a211y - PII 2 + ~) = 1.
(8.22)
Now we reason as follows. Our transformation maps any sufficiently small sphere centered at P to a sphere centered at Q. Take any line segment L that passes through P. Of course this line segment is perpendicular to all the spheres centered at P. Thus the curve that is the image of Lunder F must likewise (by conformality) be perpendicular to all the spheres centered at Q. As a result, this image curve is also a line segment-passing through
Q. Consider now a fixed line passing through P: let p' be another fixed point on that' line and x a variable point. Let Q' and y be the corresponding image points under F. By (8.22), we see that
a211y - QII2
+~
=
a2
[UY - Q'1I 2 + 211Y - Q'IIIIQ' - QII + IIQ' - QII2 + b:z 1
(8.23)
So we see that lIy - Q'II is an algebraic function of IIx - PII. Now if we parametrize the segment yQ' by a parameter T, tl :5 then the formula Lm(dYm)2 = (l/X) Lm(dxm)2 teaches us that
lIy- Q'II
=
it h
E (ddTYm )2 dr = it :\1 L tl
m
T
(dxdT )2 dT. m
m
IT we select T to be arc length along the segment Px, with tl = liP' and t = IIx - PII, then this last line becomes
, l"z-PI
lIy-QII=
IIP-PII
dT= -
vy..
l"z-P" IIP'-PII
:5 t,
dT 2· (alT +bt}
PII
8.1. QUASICONFORMAL MAPPINGS
257
H neither al nor b1 is zero, then it follows from calculus that this last line is a transcendental function of liz - PII, contradicting what we learned from (8.23). Thus either al = 0 or bt = O. Hal = 0 then Xis a constant function and our transformation is plainly· an isometry followed by a dilation. H instead b1 = 0 then al i: 0 and we may follow the transformation y = F(z) with the inversion
z· to see that
z-p = -::------=~ IIz-PII 2
a2i1y _P1I 2 = all1z·11 2.
This brings us back to the other case of an isometry followed by a dilation. Thus, in this second instance, our mapping is the composition of an isometry, a dilation, and an inversion. • Concluding RelDarks We conclude with a brief discussion of a metric characterization of quasiconformality. Let C be a Jordan curve in the plane. We call C quasiconforlDal if there is a quasiconformal mapping of a domain U that contains C which carries C into a circle. .. Just as, in complex function theory, we wish to know which curves can be carried by a conformal mapping to a circle, so it is of interest to identify the quasiconformal curves. With that goal in mind, we define an auxiliary concept. Definition 8.1.5 Let C be a.Jordan curve in the plane. Let z,w be two points on C and consider the smaller (in diameter) of the two arcs determined by z and w. If the quotient 0/ this diameter by the distance Iz - wi is bounded over all pairs z, w then we say that C is of bounded turning. The following theorem (which we shall not prove, but see Lehto and Virtanen [I], p. 100) gives a complete metric characterization of quasiconformal curves: TheorelD 8.1.6 A Jordan CUn1e in the plane is quasiconformal if and only if it is of bounded turning. It is enlightening to look at a simple example that illustrates the theorem. Define -y(t) = (t, v'iii), -1:::; t :::; 1.
Of course this is not a closed curve, but it is easily closed up; and we are only interested in behavior near the origin. Let zi = (-1/i, 1/..!J) and Wi = (1/i, 1/..!J) for i = 1,2, .... Then IZi - Wj I = 2/i while the diameter
258
CHAPTER 8. TOPICS RELATED TO COMPLEX ANALYSIS
of the smaller part of the curve between zi and Wi is on the order of 1/../J. Of course the ratio in the definition of "bounded turning" .~. unbounded, so the curve is not quasiconformal. It is somewhat less intuitive to see directly from the definition that the curve is not quasiconformalj we leave the details for the interested reader. In dimensions higher than two and for functions that are not necessarily differentiable, we have the following definition: Definition 8.1.7 Suppose 0 and 0' are bounded domaim in JRN and F : 0' ill a homeomorphiMn. We say that F is quasiconCormal if the following function has a finite usential supremum with respect to Lebesgue measure:
o -+
r})
T( ) -Ii (max{lF(x) - F(y)1 : Ix -yl = x - mr.j.O sup mm . {IF(x) - F(y) I : Ix - y I = r } .
.r
(8.24)
As we have already seen in the 2-dimensional case, there are several equivalent definitions of quasiconformal mapping. The reader may see Caraman [1] for the details of such definitions and proofs of their equivalence. Theorem 8.1.6 is only valid for Jordan curves in the planej it does not generalize to surfaces in JR3 that are homeomorphic to the sphere.. We illustrate this with the following interesting example in dimension three, for which we are grateful to A. Baernstein. Exalllple 8.1.8 Let ¢l: IR -+ lR be given by ¢let)
={
0 (t - 1/2)2
if t ~ 1/2 if t > 1/2
Now set
fh
= {x E Jlti : Ixl
< 1, Ix - (lxl,O,O)1 > ¢l(lx!)}
and
fh
= {x E Jlt3 : Ixl < I} U {x E
as : Ix -
(lxi, 0, 0)1 < ¢l( -Ixl + 7/4)}.
Then 0 1 is not quasiconformally equivalent to the unit ball, but O2 is quasiconformally equivalent to the ball. Intuitively these assertions are true because O 2 is easier to flatten than a ball with a right circular cone protruding while 0 1 is harder to flatten than a domain with a right circular conical indentation (the point being that a ball with a right circular cone added or deleted has Lipschitz boundary and thus is obviously quasiconformal equivalent to a ball). Of course, both the analogous curves 0 1 and O 2 that would result if JR3 were replaced by JR2 are conformally (not just quasiconformally) equivalent to the disc-by the Riemann mapping theorem. The theory of quasiconformal mappings is a lovely mixture of geometry and analysis. Quasiconformal mappings have proved to be natural tools in both pure and applied applications. It is a rich theory that is still under development.
8.2. WEYL'S THEOREM
8.2
259
Weyl's Theorem on Eigenvalue Asy~ptotics of a Domain-in -Space
If one considers the modes of vibration of a plane domain, one observes that the resonant frequencies for the wave equation2 are precisely the eigenvalues of the Laplace operator (see Courant and Hilbert [1], Chapter V); thus the spectrum (meaning the set of eigenvalues) of the Laplace operator for a domain is the matheIllaticai equivalent of the sound of a drum head in the shape of the domain. (Of course, a real drum is much more complicated, but we ignore those complications.) The article Kac [1] formulated the question "Can you hear the shape of a drum?" This query captures the spirit of the present section: Which geometric features of a domain (in the plane, for instance) determine the spectr&:l_ properties for the Laplace operator on that domain and, conversely, if one has complete knowledge of the spectral properties, then can one give a geometric description of the domain? It is now known that the answer to Kac's original question is "no:" In fact, there are two (isometrically) distinct 2-dimensional drums with the same eigenfrequencies. (It is not know~ -whether there are two such drums that are also convex.) It would take us far afield to develop the machinery necessary for presenting the famous eJC;ample of Gordon, Webb, and Wolpert [I]. But what we can, and shall, do is to acquaint the reader with some of the basic ideas of the spectral theory of a domain in space. Our discussion will culminate with a presentation of Weyl's theorem for the Dirichlet eigenvalues of the domain. In fact Weyl's theorem gives an estimate on the behavior of eigenvalues of the Dirichlet problem in terms of the volume of the domain. Volume is perhaps the most elementary, and the coarsest, geometric property of a domain. More sophisticated results derive spectral properties of a domain in a manifold from (for instance) curvature considerations. We refer the reader to Chavel [1] for more information on such advanced results.
Basic Terminology
Fix a domain n in real Euclidean N -space. For convenience, we assume that the domain has smooth boundary; sometimes we will consider domains with piecewise smooth boundary, where the pieces have transversal crossings. The main fact that we shall need is that, on either of these two types of domains, Green's Theorem is valid. We also assume the domain to be connected. Let b. denote the usual Laplace operator in Euclidean space. We consider the partial differential equation b.u+ ~u = o. (8.25) 2The wave equation is the standard equation for a vibrating membrane.
260
CHAPTER 8. TOPICS RELATED TO COMPLEX ANALYSIS
Any value of ~ such that (8.25) has a non-trivial solution is called an eigenvalue and any such non-trivial solution is called an eigenfunction. Because of the connection with the wave equation, eigenvalues are sometimes called eigenfrequencies. Definition 8.2.1 (i) The Dirichlet eigenvalue problem is the problem of finding all values of ~ for which there exist non-trivial solutions of (8.~5) with
ulao == o.
(8.26)
(ii) The Neumann .eigenvalue problem is the problem of finding all values of ~ for which there exist non-trivial solutions of (8.!!5) with
au/avlao == o.
(8.27)
Here v is the unit outward nonnal vector field to
an.
(iii) A mixed eigenvalue problem is the problem of finding all values of ~ for which there .exist non-trivial.solutions of (8.!!5) ~th
au/avlunao Here U is an open set such that
=
o.
(8.28)
an \ U i- 0 and U n an i- 0.
Lemma 8.2.2 Let f/J and'f/; be smooth junctions. If either ¢ satisfies the Dirichlet condition (8.!!6), 'f/; satisfies the Neumann condition (8.!!7), or both satisfy the mixed condition (8.~8), then
10 ¢ l:l.'f/;dV = - 10 grad¢·grad,pdV
(8.29)
holds. Proof: By Green's Theorem, we have
o
=
= =
/00 ¢grad,p·v du 10 div(¢grad,p) dV 10 grad ¢ • grad ,p dV + 10 ¢ l:l.,p dV.
•
The following corollary is an immediate consequence of Lemma 8.2.2.
B.2. WEYL'S THEOREM
261
Corollary 8.2.3 Let 4> and", be smooth junctions. If both 4> and", satisfy either the Dirichlet condition (8.26), the Neumann condition (8.27), or the mixed condition (8.28), then
(8.30)
holds. The content of Corollary 8.2.3 is summarized by saying that equation (8.25) is self-adjoint. Considering (8.29) with 4> an eigenfunction corresponding to the eigenvalue A and with f/> = 'I/J, we see that
>. =
10 Igrad 4>12 dV
>0
Iof/>2 dV
.
(8.31)
Since the problem is elliptic, it is known (see Widom [1]) that the eigenvalues do not accumulate at the origin, nor in any finite part of the half line. In conclusion, we typically write
(8.32) where each eigenvalue is listed according to multiplicity. We define the associated sUInlIlatory functions: For the Dirichlet eigenvalue problem, we set N(>')
=
the number of eigenvalues of (8.25), subject to the Dirichlet boundary condition (8.26), which do not exceed >.
(8.33)
and for the Neumann eigenvalue problem, we set
M(>.)
the number of eigenvalues of (8.25), subject to the Neumann boundary condition (8.27), which do not exceed A.
(8.34)
The standard notation N(>') for the Dirichlet summatory function must be distinguished by context from the dimension of JRN. Hermann Weyl's celebrated theorem says, with TN denoting the volume of the unit ball in JRN , that N(>')
>.N/2:::::
T Nvol(n)
(21r)N
as A -+ +00.
(8.35)
We shall derive separately the estimate on N(A)j AN / 2 from above and from below. Each of these will involve elegant geometric ideas related to measure, content, and capacity. The estimate from above will use, as an incidental tool, certain estimates on M(>.). In fact we shall proceed as follows:
262
CHAPTER 8. TOPICS RELATED TO COMPLEX ANALYSIS
A. We give some preliminary background about Hilbert space.
B. We begin by enunciating certain maximum-minimum and related monotonicity properties of the eigenvalues in question.
c.
We perform certain explicit calculations for the case of n a rectangular parallelepiped in Euclidean space.
D. We apply the monotonicity properties in B, together with the calculations in C, to derive the estimates from above and below that lead to (8.35). E. We provide the proofs of the results in B. For some particulars of the arguments, we shall take the liberty of referring the reader to Chavel [1], and to the detailed literature cited therein. A. Hilbert Spaces In our discussion of B, we shall require knowledge of certain elementary properties of Hilbert space. Here we enunciate those properties, but we refer the reader to Rudin [2] for details of the proofs. Consider a complex vector space 1£ that is equipped with an inner product ( ., .) that is conjugate symmetric in its entries. The inner product induces a norIll on 11. by IIxll == (x, x)l/2. We say that two elements x, y E 11. are orthogonal if (x, y) = 0; in this circumstance we write x ..L y. The space 1/. is called a (complex) Hilbert space if it is complete in the topology induced by the norm II . II. Note that it is also useful to consider inner product spaces over the ground field IR.. In this. case we of course do not require the inner product to be conjugate symmetric. When the space is -complete, then it is called a real Hilbert. space. A finite or infinite collection {XO}OEA of elements in 1/., where A is some index set, is said to be orthonormal if (xo, XIt) = 60 ,1t, where 6a ,It is the Kronecker delta. It is an exercise with Zorn's Lemma (or the Hilbert maximum principle) to see that any Hilbert space contains a complete orthonormal system {Xa}aEA. This means that if z E 1/. and z..L Xa for every 0 E A then z 0, i.e., no nontrivial element may be added to the complete orthonormal system. Note that the index set A could, in principle, be uncountable. If the Hilbert space 1£ happens to be separable then 11. possesses a complete orthonormal system that is countable; in fact, under these circumstances, any orthonormal system for 1/. will be countable. If Xl, X2, • •• are orthonormal in 11. and if x E 11. then we set
=
OJ
= OJ(X) = (X, OJ).
We sometimes call OJ the jth Fourier coefficient of x with respect to the
8.2. WEYL'S THEOREM
263
orthonormal system. Bessel's inequality then says that
.L 10;1
2
$lIxll 2
;
and Parseval's identity says that 00
L~; = IIx1l 2 • ;=1 For 0 C JRN a domain, we can think of L2(0) with the inner product
as a Hilbert space. We may consider eigenfunctions {
and the result is immediate. If K is the closed subspace of L2 spanned by the {
L
grad
= Aj 6i ,k •
(8.36)
B. MaxhnUID-MinilDUID Methods With the requisite Hilbert space background in place, we now begin to examine maximum-minimum methods. Suppose that X and Y are continuous vector fields on an open set 0 C ]RN. For us the most important type of vector field will be the gradient of some C1 function on O. We define the inner product
(X, Y) =
L
(X(s), Y(s»dVs,
264
CHAPTER 8. TOPICS RELATED TO COMPLEX ANALYSIS
where (X(s), Y(s» is the usual inner product of vectors. The associated norm is
IIXII =
L
IX(s) 12 dVs
=
L
(X(s),X(s» elVs.
Thus the space of (square integrable) vector fields on n is a linear space equipped with an inner product. Passing to the completion, we obtain an L2 space. We denote the completed space, which is of course a Hilbert space, by £2(0). As a special case of the discussion in the preceding paragraph, consider a Cl function 9 on 0 and a C I vector field X on O. Then, as is easily checked, (8.37) (gradg,X) = -.~,divX). This equation is just the integrated form of the assertion that the divergence operator is the (negative of the) adjoint of the gradient operator. In analogy with standard distribution theory, we now define a weak form of (8.37). If 9 e L2(0) and Z e L2(0) then we say that Z is a weak derivative or weak gradient of 9 if
= '--:(g,divX) Notice that, in case 9 is CI, the' new definition.
(Z,X)
for any CI vector field X. is justified by the formula (8.37). If the weak derivative exists then it is unique. The standard, and most important, justification for using the weak derivative is that it extends the operator grad to a larger domain so that it is now a closed operator. For use in the present context, we recall from Section 4.1 some of the notation and terminology used in the study of Sobolev spaces. We let HI(O) denote the subspace of L2(0) consisting of those vector fields that have a weak derivative in L2(0). We call HI(O) the Sobolev space (of order 1). A useful inner product on HI(n) is (g,hh
= (g,h) + (gradg, grad h).
It is easy to check that, equipped with this inner product, HI (0) is a Hilbert space. The associated norm is
It is straightforward to check that, when 0 has smooth boundary, Coo (n) is dense in HI (0). The space HI (n) is appropriate for the study of the Neumann eigenvalue problem, but for the Dirichlet eigenvalue problem we need to restrict ourselves to the completion of the Coo functions with compact support in n in the HI topology. This space is sometimes called HJ (n), and it is a proper subspace of HI (see Krantz [4] or Taylor [1] for details). For the
8.2. WEYL'S THEOREM
265
mixed eigenvalue problem, the appropriate function space is the completion of the Coo functions with compact support in {} U U, where we recall from Definition 8.2.I(iii) that the mixed eigenvalue problem also involves an open U such that a!l \ U I- 0 and Un 00 ':F To cover all three cases, we let 11. denote the appropriate function space, so
e.
for the Dirichlet problem,
HJ(!l) { 11.= Hl(!l)
for the Neumann problem,
HI completion of C8"(!l U U)
for the mixed problem.
(8.38) One of the basic ideas in this subject is to extremize a certain quadratic form known as the Dirichlet integral or energy integral. This form is given by D[J, g] = (grad J, grad g) for J, 9 e HI. It is a classical observation that the Dirichlet integral is connected with the Laplace operator. For instance, the calculus of variations (e.g. see Courant and Hilbert [I], Chapter IV) tells us that a C2 function J that minimizes the Dirichlet integral, subject only to a boundary condition
D[J,J], must satisfy the equation 6.J =
o.
Our first main concern will be to determine the validity of the formula
(b.rp, J)
= -D[rp,/],
(8.39)
where rp is an eigenfunction for some eigenvalue problem, and J lies in a subspace of HI. The construction that we study is a variant of the standard Hilbert space construction of the dom.ain or the adjoint of an unbounded Hilbert space operator. To wit, by Lemma 8.2.2, equation (8.39) is certainly valid for J smooth and compactly supported in!l. Now the identity (8.39) shows that, if we set T.p(J) = (6.rp, J), then
IT.p (J) 1:5 ID[rp,J]1
(Schwarz)
:5
IIrpII1I1JII1·
Thus T.p is a bounded linear functional, which may be extended to 11.. We now state our results. The proofs will be deferred until Subsection E. Theorem. 8.2.4 (Rayleigh) Let !l C JRN be a domain with either Coo boundary or piecewise Coo boundary. Consider anyone oj the eigenvalue problems from Definition 8.2.1, and suppose that the eigenvalues are
266
CHAPTER 8. TOPICS RELATED TO COMPLEX ANALYSIS
where each eigenvalue is repeated accorrling to its multiplicity. Then lor any I e 1£, I =I 0, toe have
D[/,J)
.\1
S 11/112
•
(8.40)
Equality holds in (8.40) il and only ill iI actually an eigenfunction corresponding to the eigenvalue .\1. II {4>1, 4>2, ... J is a complete orthonormal basis 01 L2({}) such that 4>j is an eigenfunction corresponding to the eigenvalue .\j lor each i, then any I e 1£ that satisfies the additional condition (1,4>1)
= (1,4>2) = ... = (I, rPlc-1) = 0
will also satisfy the inequality (8.41)
Equality holds in (8.41)' if and only if f iI an eigenfunction corresponding to the eigenvalue .\Ic. TheoreDl 8.2.5 (The MaxilDulD-Minimum TheoreDl) Let there be given functions U1, U2, ••• , UIc-1 in L2(S1), and set ." f" D[j,fl JJ=ln ~'
where
I varies over the subspace of functions
Ulo U2,
• •• , UIc-1
in the
L2
(8.42)
in 1£ that is orthogonal to inner product. Then, with.\1c as in Rayleigh's
Theorem 8.2.4, it holds that Moreover, if we consider JJ = JJ(Ulo U2, ••• ,UIc-l)
in (8.42) to be a function 01 Ul, U2, ••• , Uk-I, then the maximum of JJ is .\Ic and it is attained if and only if each Ut is an eigenfunction corresponding to the eigenvalue .\l, l 1,2, ... , k-l.
=
RelDark 8.2.6 Note that JJ is defined to be the infimum as I varies over the appropriate subspace and that the conclusion of the second part of the theorem refers to the maximum that this infimum can obtain; hence the name "maximum-minimum theorem" .
The next result will be fundamental for our analysis that leads to Weyl's theorem. Our strategy will be to compare the eigenvalue enumeration for a domain {} with the enumeration for a system of rectangular parallelepipeds either sitting inside S1 or containing S1 (this technique continues to be one of the most important tools in the subject). "
8.2. WEYL'S THEOREM
267
Figure 8.1: ExalDple SubdolDains for the Monotonicity TheorelD TheorelD 8.2.7 (DolDain Monotonicity) Let 0 c aN be a domain with either Coo boundary or piecewise Coo boundary. Consider anyone of the eigenvalue problems from Definition 8.!!.1, and suppose that the eigenvalues are 0< Al $ A2 $ .... ,
where each eigenvalue is repeated according to its multiplicity. Suppose that fh, n2 , ••• ,On are pairwise disjoint domains contained in Omega. Also suppose that the boundary of each OJ, j = 1,2, ... , n, meets the boundary of o transversally (e.g. see Figure 8.1). For each r = 1,2, ... , m, consider the eigenvalue problem on Or which imposes vanishing Dirichlet data on aOrnO a'{ld imposes the original data on aO r n an, be it Dirichlet, Neumann or mixed; further, arrange all the eigenvalues of all the Or as a single set in an increasing sequence: O$TI$T2$ ..··.,.
Then for all j = 1,2, ... we have Aj $
Tj.
The theorem has the following immediate corollary. Corollary 8.2.8 Let 0 be an open domain as usual and V C 0 a subdomain. Consider the Dirichlet eigenvalue problem on V and anyone of the eigenvalue problems from Definition 8.!!.1 on O. Then we have Ak(V) ~ Ak(O).
If in fact 0 \ V is open in 0, then the inequality is strict. TheorelD 8.2.9 (Monotonicity for Vanishing Neumann Data) Let 0 C RN be a domain with either Coo boundary or piecewise Coo boundary. Consider anyone of the eigenvalue problems from Definition 8.!!.1, and suppose that the eigenvalues are
0< Al $ A2 $ ... ,
where each eigenvalue is repeated act:OJ'ding to its multiplicity. Suppose that
n1 , n2 , ••• , nn are pairwise disjoint domains that partition n, i.e. n = 0 1 U 02U ••. UOr , and are such that the boundaries o/the nj meet the boundary 0/
n transversally. For each r = 1, 2, ••• , m, consider the eigenvalue problem on nr which imposes vanishing Neumann data on an r n n and imposes the original data on an r n an, be it Dirichlet, Neumann or mixed; further, arrange all the eigenvalues 0/ all the Or as a single set in an increasing sequence: 0~'T1 ~'T2~···.
Then lor all j
= 1,2, ... we have
C. Eigenvalue Problems on Rectangular Parallelepipeds
We now sketch some basic, and well-known, ideas about eigenvalue problems on (i) an interval in IR and (ii) a rectangular parallelepiped in JR.N. These special cases will be important building blocks in the arguments that follow, but they also illustrate the key geometric insight in eigenvalue analysis. Let us begin with an interval [0, T). The differential equation analogous to (8.25) in dimension one is
1/>" + ~I/> = o.
(8.43)
The Dirichlet boundary condition is 1/1(0) = I/>(T) = O. Of course the general solution-which one can find by sepa,ration of variables-to (8.43) (for A> 0) is 1/>9{S) = Acos~s+Bsinv:\s. Here A and B are arbitrary real constants. However, only A = 0 and ~ = (1rk/T)2, k = 1,2, ... give solutions that satisfy both boundary conditions. We thus devolve upon the sequence of eigenvalues
and corresponding eigenfunctions
Note that the constant preceding the sine function is chosen so that the eigenfunction will have norm 1. We see that the solution of the eigenvalue problem depends on the length of the interval (the volume of the domain) in a manner that is predictable from physical considerations. For example,
solving the wave equation by separation of variables leads precisely to this eigenvalue problem. And we know that, in the vibrating string problem, making the string longer forces the natural modes of vibration to be slower (Le. produces lower notes, as on a guitar). It is now a straightforward exercise to see that, on the domain
n=
(O,Td x (0,T2) x ... x (0, Tm),
the Dirichlet eigenvalue problem may be solved using separations of variables (try it for yourself using just two variables). We shall not replicate the analysis here. The resulting eigenvalues are
{rr2 (~t + ... ~) }. Now let us do a calculation to verify, by hand, that Weyl's theorem is true on the rectangular parallelepiped. A useful way to organize our information about the eigenvalue asymptoties on a rectangular parallelepiped is by way of dualization. To this end, we think for a moment about tori. Let r be a lattice, that is a discrete additive subgroup of ]RN. We shall treat only r of rank N: this means that there is a linearly independent set of vectors VI, V2, .•• ,Vn (called a basis for r) such that
r
=
{t{3jVj :{3j E Z Vi}. }=I
Then ]RN /r is an N-dimensional torus, which we denote by T. To obtain a set of eigenfunctions for the Laplacian on T, we consider the dual lattice: r* A natural dual basis
j,k
= {z
E
]RN :
(x, z) E Z for all x E r}.
{WI, W2, ..• , W N }
for r* is determined by the equations
(wi> Vk) = 6j ,k,
where 6 is the Kronecker delta. Now for each z E r* we may define the function
c/>z (x) = e21ri (z,%). (Note that, in this discussion of tori, we shall be considering complex valued solutions of the equation 6u + AU = 0 and complex valued eigenfunctions. This is for convenience only, and should lead to no confusion.) Note that the function c/>% is invariant under the action ofr. This means that if g e r and Tg denotes translation by g, i.e. Tg(X) = x + g, then c/>% 07g = c/>%. So 41% is a well-defined function on T. Moreover,
l::!.c/>z =
-4rr2IzI241%.
270
CHAPTER 8. TOPICS RELATED TO COMPLEX ANALYSIS
The corresponding eigenvalue is
(8.44) In other words, 4>11 is an eigenfunction of the Laplacian on the torus T. It is not difficult, using some elementary Fourier analysis, to see that the functions 4>11 span L2(T). Moreover, if %1, %2, ... , %" are distinct, then the corresponding functions tPlll' 4>112' ••• ,4>11,. are linearly independent (exercise, or see Chavel [1]). Thus the functions 4>: account for all the eigenfunctions of the Laplacian on T. To summarize what we have just learned, for a given .\ > 0, the eigenfunctions corresponding to .\ are just the functions 4>: for values of %that satisfy (8.44). In other words, the value of the summatory function N('\) is just the number of elements of r* that lie inside'the ball 'i(D, v>'/(21r». Thus, with N denoting the summatory function on T,
'N(>') = N(V)./(27r».
By the preceding calculations, N(>')
= N(V>./(21r» = ~
N(V)./7r)/2 m T m( V)./7r)n(vol 0) ~ T m>.m/2(vol oy 2m (21r)m
This is the Weyl formula in the special case when the domain 0 is a rectangular parallelepiped. D. Eigenvalue Problems on Arbitrary Smooth Domains
Let us use these last two results, together with our monotonicity results, to derive the Weyl formula for an arbitrary smoothly bounded domain o in JRN. To this end, let R 1 , R2 ... , Rm be pairwise disjoint, open Ndimensional rectangles, each relatively compact in O. Fix a positive number >.. For j = 1,2, ... , m, let N;(>') be the number of Dirichlet eigenvalues for R; that do not exceed >.. Recall that in the domain monotonicity theorem we numbered all the eigenvalues of all the subdomains of 0, and then we compared that enumeration to the enumeration of eigenvalues for 0 itself. As a result, we can be sure that m
N(>') ~
L N;(>.).
(8.45)
;=1
The inequality (8.45) implies TN ~ I(R) · . fN(>.) > ~l' ·nfN;(.\) I tmlD >.N/2 _ ~ lIDl .\N/2 = (27r)N ~ vo ; . Fl
3=1
(846) .
8.2. WEYL'S THEOREM
271
Notice that (8.46) holds for all possible choices of parallelepipeds lying in O. Taking the supremum on the right over all such choices, we find that l"N 1(1"\) liminf N(~) ~N/2 ~ (211')NvO H •
This is half of the Weyl asymptotic formula. For the other half, we must estimate the limit supremum from above. Let R 1 , R2, ... ,Rm be pairwise disjoint open rectangles with
Let M(~) denote the summatory function for the Neumann eigenvalues for the domain int{Rl U R2 U ... URm ). Also, for each j, let Mj(.\) be the summatory function for the Neumann eigenvalues on the individual domain Rj.
Now the Domain Monotonicity Theorem 8.2.9 for th~ Neumann problem tells us that m
N('\) ~ M(~) ~
L Mj(.\). j=1
. Therefore . N(>') l"N ~ hmsup .\N/2 ~ (211')N vol(Rj).
£;;r.
Taking the infimum on the right over all possible choices of rectangular . parallelepipeds, we find that
Combining this inequality for the limit supremum with the earlier inequality for the limit infimum yields Weyl's asymptotic formula. E. Proofs of the Results in B Proof of Rayleigh's Theorem 8.2.4: By our earlier discussion, if tP is an eigenfunction and / E 11. (recall 11. denotes the function space appropriate to the eigenvalue problem under consideration as defined in (8.38», then (~tP,f) = -D[tP,/). For any such /, we set (as before) Qj = (f,tPj). We are assuming, in the second part of the theorem, that Ql = Q2 = ... = Ok-l = O. Imitating the proof of Bessel's inequality and using (8.36), we have for 1 ~ k :$ r that
272
CHAPTER 8. TOPICS RELATED TO COMPLEX ANALYSIS r
r
=
D[/, J] - 2 L ajD[/, 4>j] + L aj alD[4>j, 4>l] j=1I j,l=1I
=
D[/,J]+2Laj(f,~4>j)+ L aja,D[4>j,4>l] j=1I j;l=lI.
=
D[/, I] - 2
r
r
r
L
j=1I
r
aj(f, ,\j4>j) + L ,\jaj j=1I
r
=
D[/, I] -
L
,\jaj.
j=1I
We thus conclude that '\1111/11 2
00
00
j=l
j=1I
= '\11 L aj = '\11 L
00
aj :5
L ,\jaj :5 D[/, I] <
00.
j=1I
This establishes the desired inequality. In the case of equality, all the coefficients aj must equal 0 for >'j > '\11 and then it follows that I is an eigenfunction corresponding to the • eigenvalue '\11. Proof of the Maximum-MiniInUDl TheoreDl 8.2.5: Consider a function 1 that is orthogonal to the span of {Ul' U2, ••• , Uk} and that, in addition, has the special form II
1= Laj4>;. j=l
with each 4>j an eigenfunction corresponding to the eigenvalue >'j, j 1,2, ... ,k. Thus
=
II
E aj(4);. Ut) = 0
(8.47)
j=l
for l = 1,2, ... , k - 1. We think of the numbers (4)j, Ul) as given, for j = 1,2, ... , k and l = 1,2, ... , k - 1. And we think of the aj as unknowns. Then the system in (8.47) has more unknowns (namely, k of them) than it has equations (namely, k - 1 of them). So it certainly must have a non-trivial solution. But then, for these values of aj, we see that k
1S1I1I12 :5 D[/, I] =
L
,\jaj :5 '\1111/11 2 •
j=l
This inequality gives the desired conclusion.
•
273
B.2. WEYL'S THEOREM
Proof of the Domain Monotonicity Theorem 8.2.7: Of course we shall apply the maximum-minimum method from the last theorem. Let ,pI, tP2, ••. ,,pll be the eigenfunctions that correspond to the eigenvalues ~1' ~2' ••• ,~II on the full domain o. Corresponding to j = 1,2, ... , II: let .,pj : 0 -+ R be an eigenfunction corresponding to the eigenvalue Tj on the appropriate subdomain (recall that each Tj comes from one of the subdomains OJ) and identically zero elsewhere. Note in particular that each vanishes on the boundary of its particular subdomain. It follows that e 1£. Moreover, it is straightforward to check that ihe are orthogonal in L2(0); after normalization, they may be taken to be orthonormal. Thus we may select numbers aI, a2, ... , ak, not all zero, such that
""j ""j
""j
II
L aj (""j , ,pi) = 0 j=1
for i = 1,2, ... , II: - 1 (as in the proof of the maximum-minimum theorem). As a result, the function k
1= Laj""j j=1
is orthogonal to
,pI, tP2, ... , tPk-1
in L2 (0). Therefore k
~kIIf1l2:5 D[/,/]
=L
vja~ :s;; Tkllfll2.
j=1
•
That is what we wished to prove.
Proof of the Monotonicity Theorem for Neumann Data 8.2.9: Let
""t :n -+ R be the eigenfunction corresponding to the eigenvalue when ""i is restricted to the corresponding subdomain; let ""i be set identically Tt
equal to zero otherwise. If f is any function in 1l, then I e 1l(Or), a lortiori for each T, where 1l(Or) is the function space for the eigenvalue problem on the sub domain Or. Therefore if f is orthogonal to ""2, ... , ""k-I in L2(0) then
""x,
But there is a non-trivial function
""1, ""2, ... ,"""-1 in L2(0); as a result, This implies the theorem.
f
= :E;=1 (zj,pj that is orthogonal to
•
Appendix A.1
Metrics on the Collection of Subsets of Euclidean Space
Let S be the collection of all subsets of JRN' • It is often useful, especially in geometric applications, to have a metric on S. In this section we address methods for achieving this end. In practice, S (the entire power set of aN) is probably too large a collection of objects to have a reasonable and useful metric topology (see Dugundji [1], Section IX.9 for several characteriza-' tions of metrizability). With these considerations in mind, we shall restrict attention to B, the collection of bounded subsets of aN. We have: Definition A.I.l Let Sand T be elements of B. We set
HD(S, T) =
max{ sup dist(s, T) , sup dist(t, S) }. _eS
(A.1)
teT
This function is called the Hausdorff distance. To obtain the most satisfactory results, we further restrict attention to' compact) subsets of aN.
B, the collection of both closed and bounded (i.e.
Lemma A.I.2 Let S, T E B. Then there are points s E Sand t E T such that HD(S, T) = Is - tl· Proof: By the triangle inequality, the function s t-+ dist(s, T) is a continuous function on S. So it aChieves its maximum at some point s' E S. Likewise, the function t t-+ dist(t, S) achieves its maximum at some point t' E T. Let d = maxi dist(s',T), dist(t', S)}, and suppose without loss of generality that this maximum is achieved by s'. Since T is compact, there is a point t* E T such that d = It* - s'l. Thus s' ,t* are the points that we seek. •
Proposition A.I.3 The function HD is a metric on
B.
Proof: Clearly HD ~ 0 and, if S = T, then HD(S, T) = O. Conversely, if HD(S, T) = 0 then let s E S. By definition, there are points tj E T such that Is - tj I ~ O. Since T is compact, we may select a subsequence {tj.} such that tj. ~ s. But, since T is compact, we then conclude that sET. Hence SeT. Similar reasoning shows that T C S. Hence S =T.
276
APPEND;
Finally we come to the triangle inequality. Let S,T,U S, t e T, u e Then we have
u.
Is-ul
$
Is - tl
+ It -
ul
Is - tl
+ It -
ul
e 8.
Let s
.\J. dist(S,u)
s .\J.
dist(S,u)
s
dist(S,u)
s
dist(S,u) dist(S,u)
dist(S, t)
+ It -
ul
HO(S, T)
+ It -
ul
s
HO(S, T)
+
.\J. $
HO(S, T)
+ sup
.\J. .\J.
dist(T, u) dist(T, u)
uEU
.\J.
sup dist(S,u) uEU·
S
HO(S,T)
+ sup
dist(T,u).
uEU
By symmetry, we have sup dist(U, s) S HO(U, T) BES
+ sup dist(T, s) BES
and thus maxi sup dist(S,u) , sup dist(U,s) } uEU
BES
$ max{ HD(S, T)
+ sup
dist(T, u), HO(U, T)
uEU
.
+ sup dist(T, s)}. BES
We conclude that HD(U,S) $ HO(U,T)
+
HO(T,S).
•
There are fundamental questions concerning completen~ss, compactness, etc. that we need to ask about any metric space. Theorem A.1.4 The metric space
(8,
HO) is complete.
Proof: Let {Sj} be a Cauchy sequence in the metric space (8, HD). We seek an element S E 8 such that Sj -+ S. Elementary estimates, as in any metric space, show that the elements Sj are all contained in a common ball B(D, R). We set S equal to the closure of
A.1. METRICS ON SUBSETS OF EUCLIDEAN SPACE
277
Then S is closed and bounded so it is an element of B. To ~. t;hat Sj -+ S, select E > o. Choose J large enough so that if j, k ~ J then HD (Sj , SIe) < E. For m > J set Tm = UT:.JSl. Then it follows from the definition, and from Proposition A.1.3, that· HD(SJ, T m) < E for every m > J. Therefore, with Up = U~pSl for every p > J, it follows that HD(SJ, Up) :5 E. We conclude that that HD(SJ,n:f=J+1 Up) :5 E. Hence, by the continuity of the distance, HD(SJ,S) ~ E. That is what we wished to proved. • The next proposition informs us of the seminal fact regarding the Hausdorff distance topology. Theorem A.I.5 The set 0/ non-empty compact subsets O/]RN with the Hausdorff distance topology is boundedly compact, i.e. any bounded sequence has a subsequence that converges to a compact set. Proof: Let AI, A 2 , ••• be a bounded sequence in the HausdQrff distance. We may assume without loss of generality that each Ai is a subset of the closed unit N -cube, Co. We will use an. inductive construction and a diagonalization argument. Let Ao,i = Ai for i = 1,2, .... For each k ~ 1, the sequence Ale,;, i = 1,2, ... , will be a subsequence ofthe preceding sequence AIe-1,i, i = 1,2, .... Also, we will construct sets Co :J C 1 :J ... inductively. Each Cle will be the union of a set; of subcubes of the unit cube. The first set in this sequence is the unit cube itself Co. For each k = 0,1, ... , the sequence Ale,;, i = 1,2, ... , and the set Cle are to have the properties that
Cle n Ale,; '" 0 holds for i
= 1,2, ...
(A.2)
and
Ale,; C C,.
holds for all sufficiently large i.
(A.3)
=
It is clear that (A.2) and (A.3) are satisfied when k o. Assume Ale-I,;, i = 1,2, ... and CIe-l have been defined so that
CIe-l n AIe-l,i
'"
0 holds for i = 1,2, ...
and
Ale-I,; C C,.-l
holds for all sufficiently large i. For each integer k ~ 1, subdivide the unit N-cube into 2 leN congruent subcubes of side-length 2- 1e • We let Cle be the collection of subcubes of side-length 2- 1e which are subsets of Cle-i. A subcollection, C C CIe, will be called admissible if there are infinitely many i for which D
n AIe-I,i '" 0 holds for all DeC.
(AA)
Let Cle be the union of a maximal admissible collection of subcubes, which is immediately seen to exist, since Cle is finite. Let Ale,!, AIe,2, ... be the
278
APPENDIX
subsequence of AII-I,lt AII- I ,2, • •• consisting of those AII-I,i for which (A.4) is true. Observe that AII,i C CII holds for sufficiently large i, else there is another subcube which could be a,dded to the maximal collection while maintaining admissibility. We set 00 C=
n
Cle
Ie=O
and claim that C is the limit in the Haus!iorff distance of AII,II as k -+ 00. Of course C is non-empty by the finite intersection property. Let ~ > 0 be given. Clearly we can find an index ko such that Clio C {x: dist(x,C)
There is a number io such that for i
~
< ~}.
io we have
Allo,i C Clio C {x : dist(x, C)
So, for k
~
<
~}.
leo + i o, we know that
<
AII,II C {x : dist(x, C)
holds. We let kl
~
~}
1.0 + io be such that
../Fi 2- le1 < ~. Let c E C be arbitrary. Then c E Clel so there is some cube, D, of sidelength 2- 111 containing c and for which DnAlel,i
holds for all i. But then if k
~
kl' we have D
dist(c,Ak,k) $ It follows that HD(C,Ak,k)
#·0
n Ak,k # 0, so
../Fi 8- k
< ~ holds for all k
<~. ~
k1 .
•
Next we give two more useful facts about the Hausdorff distance topology. Proposition A.1.6 Let C be the collection of all closed, bounded, convex sets in JR.N. Then C is a closed subset of (8, HD). Proof: There are several amusing ways to prove this assertion. One is by contradiction. IT {Sj} is a convergent sequence in C, then let S E iJ be its limit. IT S does not lie in C then S is not convex. Thus there is a segment t with endpoints lying in S but with some interior point p not in S. Let ~ > 0 be selected so that the open ball U(p,~) does not lie in S. Let a, b be the endpoints of t. Choose j so large that HD(Sj, S) < ~/2. For such
A.I. METRICS ON SUBSETS OF EUCLIDEAN SPACE
279
i, there exist points aj,bj E Sj such that laj - al < e/3 and Ibj - bl < e/3. But then each point Cj(t) == (1- t)aj + tbj has distance less than e/3 from c(t) == (1 - t)a + tb, 0 $ t $ 1. In particular, there is a point Pj on the line segment lj connecting aj to bj such that Ipj - ple/3. Noting that Pj must lie in Sj, we see that we have contradicted our statement about U(p,e). Therefore 8 must be convex. • Proposition A.I. 7 Let {Sj} be a sequence of elements 0/ B, each 0/ which is connected. Suppose that Sj -+ S in the Hausdorff metric. Then 8 must be connected. Proof: Suppose not. Then S is disconnected. So we may write 8 = Au B with each of A and B closed and non-empty and An B = 0. Then there is a number 11 > 0 such that if a E A and b E B then la - bl > 11. Choose i so large that HD(Sj,S) < 7]/3. Define
Aj = {s E Sj : dist(s, A) :5 1'//3} and
B j = {s E 8 j
:
dist(s, B) $ 11/3}.
Clearly Aj n B j = 0 and Aj , B j are closed and non-empty. Moreover, Aj U Bj = Sj. That contradicts the connectedness of Sj and completes the proof. . • Remark A.I.8 It is certainly possible to have totally disconnected sets E j , i = 1,2, ... , such that E j -+ E as i -+ 00 and E is connected (exercise). The proof of the last result may be modified to show that if {Sj} c B, if 8 j -+ S in the Hausdorff metric, and if each Sj has at most M connected components then S has at most M connected components. It is not true that the collection of simply-eonnected sets in B forms a closed subset. For instance, let N = 2 and let
Sj
=
{(rcos8,rsin8) E
]R2 :
1 $; r :5 2,
Iii $
8:5 21r}.
Then each Sj is simply connected but the 8 j converge to an annulus. Using elementary PL topology, one can construct counterexamples for every homotopy group or every Betti number. Although the Hausdorff metric is the most far-reaching notion of distance among sets, it is by no means the only metric on sets. It is frequently useful to put a topology on the collection of smoothly bounded domains. We should like to briefly discuss some of the methods for doing that. Definition A.I.9 Fix an integer k ~ 1. Consider the collection [k 0/ all bounded domains in aN with Ck boundary. Fix one such domain no. Associate to it the imbedding tPo into]RN given by ¢a(x) = x. In other words, ¢a is the identity mapping. Now let e > O. 8et U. = {n E [k : there is a surjective imbedding
ifJ 0/0.0 to 0. such that lIifJ - ¢aile .. < fl·
APPENDIX
280
Here we use the standard e" topolOf/fl on /unctions and mappings. Then the sets Ue lorm a sub-basis lor /I topology on ell. This is called the e" topology on domains.
e"
The topology is equivalent to considering the boundary M of each element 0 of ell as a regularly imbedded compact· manifold and declaring OJ -+ 0 0 if the boundariesMj of OJ converge to the boundary Mo of 0 0 in the sense that the imbedding maps converge. It is convenient, but not necessary, to formulate a topology on smoothly bounded domains by considering an associated topology on imbeddings of the .domains. Instead, one could look at a local boundary neighborhood and consider the boundary of 0 0 as locally the graph of some function 10 of N - 1 variables. Then a sequence of domains OJ converges to 0 0 in the e" topology if each boundary is locally given as the graph of a function I; and the I; converge to 10 in e". Properly formulated, we would have to cover ao with finitely many open neighborhoods U l and demand that If converge to IJ in a suitable sense. When one is studying perturbation theory for partial differential equations, or other perturbation problems in geometric analysis, it is important to have a suitable topology on domains.
A.2
The Constants Associated to Euclidean Space
There are many special constants of mathematical analysis that arise from the consideration of the geometry of Euclidean space. A simple example is the constant 1r, which is the ratio of the circumference of any circle to its diameter. Similar considerations, in higher dimensions, give rise to the gamma function and the beta function. The elementary representation theory of the orthogonal group also gives rise to many of the special functions of mathematical physics; these, in turn, are the source of the constant e, the Bessel functions, and so forth. In this section, we shall touch on just a few of these topics. The book Vilenkin [1] is an excellent source for some of the topics mentioned in the first paragraph. Another fine source is Erdelyi [1]. We begin by introducing two of the special functions of mathematical physics. Our point of view in the present section is that one "discovers" these functions by considering geometric questions in space; nonetheless, it is best to be informed in advance.
> 0, we'define the gaDlIDa function
Definition A.2.1 For Rez ting
r(z) =
1
00
tz-1e- t dt.
by set-
(A.5)
It is easy to see that the integral in (A.5), called an Eulerian integral of the second kind, converges absolutely for the range of z specified in the definition. By Morera's theorem, r is a holomorphic function on the right half plane. One checks that
r(1) = 1,
(A.6)
and, by integration by parts, that the functional equation
xr(x) = r(x + 1)
(A.7)
holds. As a result of (A.6) and (A.7),
r(k + 1) = k!
(A.8)
holds for positive integers k. The following teChnical fact concerning the so-called Gaussian integral will occur repeatedly: Lemma A.2.2 It holds that
{
iRN
e- 1z12 dx =
1rN/2.
APPENDIX
282
Proof: By Fubini's theorem, it suffices to treat the case N = 1. Let
=
8
1
00
e-~ 2
dz.
-00
Then
82
=
1
=
1 e-~2-112
00 -00
e-~2 dx
1
00 -00
e- II 2 dy
dxdy
(A.9)
e- r2 rdrdD
(A.tO)
R2
11 2
= =
o
#
0
00
1r,
where we have changed to polar coordinates to go from (A.9) to (A. 10). Therefore 8 = .;:;, as desired. • Now we may evaluate r(1/2). Using the change of variable see that
. =1
00
r(I/2) .
C
l/2
0
e- t dt = 2
1
00
0
e-· 2 ds =
S2
"fii.
= t, we
(A.Il)
Using integration by parts, we may analytically continue the gamma function to C \ {O, -I, -2, ... }. The function is meromorphic, and has a simple pole at each of 0, -I, -2, etc. Good sources of information on the gamma function are Carrier, Crook, and Pearson [I] and Greene and Krantz [I]. Next we have Definition A.2.3 For Rez > tion by setting
°and Rew > 0, we define the beta func(A.12)
The integral in (A.12) is called an Eulerian integral of the first kind. The beta function is holomorphic in each variable separately. We have the following connection between the gamma and beta functions: Proposition A.2.4 If Rez > a( /J
°and Rew > 0, then it holds that
) _ r(z)r(w) z,w - r(z+w)·
(A.13)
A.2. THE CONSTANTS OF EUCLIDEAN SPACE
283
Proof: By analytic continuation, it suffices to verify the identity for z, w real and positive. Now we have
r(z)r(w)
=
L
oo
sz-l e-· ds L
=
L
oo
e-·s z -
=
L
oo
sZ+"'-21°O e-t (; -1) ",-I dtds
=
L
OO
sz+w-l
=
1
L
=
1
(t - 1)"'-lt-Z - ' " dt· r(z + w)
=
Ll t"'-I(1 - t)"'-l dt· r(z + w)
=
(3(z, w) • r(z
00
00
oo
1
1
00
1
00
oo
t'-l e -t dt
e-t+·(t - S)"'-1 dtds
e-tB(t - 1)"'-1 dtds
SZ+",-l e- Bt-"-"'(t - 1)"'-1 dsdt
+ w).
Next we will calculate the surface area of the unit ball in IRN :
•
Proposition A.2.5 Let B = {x E ]RN : Ixl < I} and E ;, 8B its boundary. Then the N - 1 dimensio.nal Hausdorff measure of E (i.e. the rotationally invariant area) is WN-l
== u(E)
21rNf2 = r(N/2)·
(A.14)
Proof: By Lemma A.2.2 and using spherical coordinates, we have 1= [
1RN
e-1rizi2 dx = [ do
1E
roo e-1rr2r
10
1
and therefore
__ 1 = 00 e- 1rr2 r N - 1 dr. WN-l 0 Using the change of variable 8 = r2, we find that 1 WN-l
=
1
=
-1r
00
o
e
-1rB
s
N/2 d8 28
1 -N/21°O e -B 8 N/2 -ds 2 0 s
= ~1r-N/2r(N/2).
N- 1
dr
APPENDIX
284
•
This is the desired formula.
Using our elementary facts about the gamma function (A.6), (A.7), and (A.H), the reader may check that the formula in Proposition A.2.5 gives the expected value for the area of the sphere in dimensions 1, 2, 3. Now we will calculate the volume of the unit ball in aN. Proposition A.2.6 Let B = {x E volume of B is TN
== vol(B}
aN : Ixi <
I}. Then the Euclidean
1rN/2
= r«N + 2} /2} .
Proof: Using spherical coordinates, we calculate that TN
= WN-l
11
r N- l dr
= W~-l.
But then Proposition A.2.5 tells us that 21rN/2 TN
= N. r(N/2}
1rN/2 = r«N + 2}/2}"
•
Using our elementary facts about the gamma function (A.6), (A.7), and (A.ll), the reader may check that the formula in Proposition A.2.6 gives the expected value for the volume of the ball in dimensions 1, 2, 3. An immediate but interesting corollary of our calculations is the following: PropositioOn A.2.7 Let BN be the unit ball in aN and EN-l its boundary. Let TN denote the volume of BN andwN_l the surface area olEN-I. Then lim TN = lim WN-l =
N-+oo
N-+oo
o.
We next note an alternative recursion formula for the volume of the ball. It gives rise to the beta function in a natural way. In ]RN, let us write x' = (Xl, ... ,XN-t} and x = (X',XN). Then
=
1 1 1 z'EBN_l
=
1 dVl(XN)dVN-l(X'}
IZNI
(I-lx'1 2 )dVN_l(x')
z'EBN_l
(A.I5)
A.2. THE CONSTANTS OF EUCLIDEAN SPACE
285
The change of variable 8 = r2 converts (A.15) to a beta integral, and Proposition A.2.4 turns that into an expression in terms of gamma functions. We leave the details for the interested reader. A similar technique may be used to calculate the value of
( Iz ol2 dVN(Z)
JBN
for any multi-index Q. These numbers are useful in various spherical harmonic expansions (see, for instance, Krantz [4]).
Guide to Notation DESCRIPTION Area of the (N - I)-dimensional unit sphere in RN
NOTATION
Boundary of the set A
8A
Cardinality of the set S
card(S)
Characteristic function of the set A Closed ball of radius r .. centered at x
B(x,r)
Closure of A is a compact subset of the interior of B
ACCB
Closure of the set A
A
Compactly supported functions of class CDt Co~plement
of the. set A
Complex numbers
C
Determinant of the matrix M
det[M]
Diameter of the set S
diam(S)
Differentiability classes k times cOntinuously differentiable infinitely differentiable analytic k times continuously differentiable with kth derivative Holder continuous of order a
Distance between sets AandB
dist(A,B)
Divergence of the vectorfield v
div(v)
Euclidean space
Rm
Extended reals
i: i(E.) =
Fourier transform
J f(x)e- 27ri:r:.( dx
GUIDE TO NOTATION
288
K
Gaussian curvature Gradient of the function
I
grad I
Hausdorff distance between sets A and B Hessian of the function
HD(A,B)
I
Hess(f)
Interior of the set A
A
Laplacian of the function u
~(u)
Length of the arc
= VI
I:'
len(r)
Lipschitz constant of the map
I
Lip (f)
Mean curvature
H
Multi-indices
A(m)
functions of
= (z+)m
JL! = JLl! •••.• JLm!
IJLI
= JLl + ... + JLm
Ixl = IX11"1 ..... IXm I"... (x)" = (Xd"l ..... (x m )"..
<:) = (:~e operator based on relations on
JL
< v iff JLl < vI, ... , JLm < 11m
JL :$ v iff JLl :$ vI, ... , JLm ~
Non-negative integers
z+
Normal space to S at p
NSp
Normal bundle over S
NS
Open ball of radius r centered at x
lB(x, r)
Positive integers (same as the
natur~
numbers)
N
Real numbers
lR
Scalar curvature
S
Second fundamental form
TI(v,w)
Set difference
A\B
Vm
289
GUIDE TO NOTATION Sphere of radius r centered at x Support of the function
S(x,r)
I
Tangent space to S at p
suppl
Tangent bundle over S
TS" TS
Trace of the matrix M
tr[M]
Transpose of the matrix M
Mt
Volume of the N-dimensional unit ball in )RN
TN
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Index a-dimensional density 72 lower density 71 upper density 72 adjoint 265 admissible cover 57 analytic sets 18 analytically convex 193 approximate continuity 89 approximately continuous 89 approximating measure 57 approximation of BV functions 104 arc 66 area minimizers 111 . of the unit ball 281 barycentric coordinates 68 Besicovitch covering Theorem 91 Bessel's inequality 263 beta function 281 blowing-up 118 Borel regular 17 set 17 bounded exhaustion function 211 turning 257 variation 101 variation on n 101 Brunn-Minkowski Inequality 241 Cle boundary 8 topology on domains 279
Caratheodory construction 58 measure 64 closed operator 264 co-area formula 136 compactness of BV(n) 107 complementary in the sense of Young 198 completeness of BV(n) 102 of V'(m) 23 cone 31 conformal mapping 247 conjugate exponents 23 continuously differentiable 1 convex analytically 193 combination 209 function 196, 197 hull 209 .-- integral geometric measure 64 of order k 219 strongly analytically 193 convexity of order k 217 geometric 191 with respect to a family of functions 204 convolution 24 countably m-rectifiable set 70 (c/J, m )-rectifiable set 70 critical point 157 value 157 curvature 44
Coo varieties theorem 4, 5 Caccioppoli sets 108 305
INDEX
306 cycloid 50 DeRham cohomology 189 theorem 189 defining function 8 Delaunay surface 46 differentiable 125 dilation factor 250 direction of curvature 39 directionally limited 93 Dirichlet eigenvalue problem 260 integral 265 distance function 12 divergence 186 theorem 109, 186 €-thick set 238 eigenfrequencies 260 eigenfunction for the Laplacian 260 eigenvalue of the Laplacian 260 ends 46 energy integral 265 epicycloid 50 Eulerian integral of the first kind 281 of the second kind 280 exhaustion function 211 extension theorem for Sobolev spaces 150 exterior derivative 188 extreme point 208 face of a simplex 68 Fatou's Lemma 21 finite perimeter 108 Fourier coefficient 262 inversion iheorem 143 transform 143 Frenet formulas 45 Friedrichs mollifier 25 functions whose partial derivatives are measures 103 gamma function 279 gauge 57 Gauss map 27
Gauss-Green formula 109 generalized 122 Gaussian curvature 39 integral 280 general position 67 generalized Gauss-Green Theorem 122 genus 46 geometric convexity 191 Gillespie measure 64 Gross measure 63 hard Sard 161 Hardy-Littlewood maximal function 86 Hausdorff dimension 62 distance 274 measure 61, 63 Helly's Theorem 191 Hilbert space 262 hull 204 hypocycloid 50 Holder's Inequality 23 infinitely differentiable 1 inner regularity 18 integrable function 20 integral geometric measure 63 Isodiametric inequality 235 Isoperimetric Inequality 113, 242 local 114 Jacobian, K-dimensional 125 k-jet 166
Kirchberger's Theorem 191 Kirzbraun's Extension Theorem 162
LP norm 22 space 22 Laplacian 40 Lebesgue area 68 covering theorem 95 differentiation theorem 87 measure 17, 22 space 22
INDEX Lebesgue's covering Theorem 6 dominated convergence theorem 21 monotone convergence theorem 21 Leibniz's rule 2 length 66 Lipschitz constant 65 function 65 local graph 14 locally finite perimeter 108 integrable function 20 lower integral 20 lsc domination principle 5 m-rectifiable set 70 Marstrand's Theorem 73 maximal function 86 with respect to 94 mean curvature 39 measurable function 19 set 16 measure 15 minimal surface 50 Minkowski content, lower 74 content, upper 74 dimension, lower 78 dimension, upper 78 functional 214 inequality 23, 25 measure, lower 78 measure, upper 78 mixed eigenvalue problem 260 multi-index 1 nearest point property 205 Neumann eigenvalue problem 260 normal bundle 27 space 28 normality 96 order of contact 217
,)VI
Orlicz space 200 outer· measure 16 (,p, m)-rectifiable set 70 packing dimension . lower 81 upper 81 parametric hypersurface 15 Parseval's indentity 263 partition of unity 7 perimeter 107 Poincare's Inequality 111, 113 positive reach 207 principal curvature 39 direction 39 .. . product measure 21 proper map 161 purely (,p, m )-unrectifiable 70 quasi-linear function 68 quasiconformal homeomorphism 258 Jordan curve 257 mapping 251 Rademacher's Theorem 130 Radon measure 17 rank of a critical value 161 reach 207 real analytic 2 rectifiable 66 . reduced boundary 115 refinement 95 regular measure 17 Rellich's Theorem 107 roulette 50 u-algebra 16 Sard's Theorem 158, 161 scalar curvature 39 second fundamental form 38 semicontinuity in BV(S1) 102 separation of sets 95 signed distance function 12 simple m-extension operator 153 simplex 68 simplicial complex 68
INDEX
308 Sobolev imbedding theorem 144 space on a domain 147 spaces 144 spherical measure 58, 63 Steiner symmetrization 224 step function 20 Stokes's Theorem 188 strictly convex 208 strongly analytically convex 193 convex function 198 structure theorem 71 subitem for sets of finite perimeter 122 summatory functions 261 support 1 supporting hyperplane 201 surface 46 . Suslin set 18, 227 symmetrized Hausdorff measure 64 tangent bundle 28 cone 31 vector 8 Taylor formula 165 polynomial-166 Taylor's TheOrem 166 Tietze's Extension Theorem 162 topology compact-open C" 2 fine 2 strong 2 weak 2
Whitney 2 total curvature 39 extension operator 154 trace of order K 39 theorem for Soholev spaces 148 triangulable 68 triangulation 68 upper integral 20 Urysohn Lemma 4, 6
valence 6 vector sum of sets 235 vertices 68 Vitali Covering Theorem 99 volume of the unit ball 282 weak
analytic convexity 193 derivative 264 gradient 264 type (1,1) 86 Weingarten map 35 Whitney decomposition 168, 170 Extension Theorem 167 family 170 formula 163 Wiener Covering Lemma 84
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