Manifolds, Tensor Analysis, and Applications, Second Edition (Applied Mathematical Sciences 75)

~:t

defined by
The product topology on S x T consists of all subsets that are unions ofsets which have the form U x Y, where U is open in S and Y is open in T. Thus, these open rectanglesform a basis for the topology.

Products of more than two factors can be considered in a similar way; it is straightforward to verify that the map «u. v). w) >-> (u. (v. w» is a homeomorphism of (S x T) x Z onto S x (T x Z). Similarly, one sees that S x T is homeomorphic to T x S. Thus one can take products of any number of topological spaces and the factors can be grouped in any order; we simply write SI x ... X Sn for such a finite product. For example, iRn has the product topology of ::t x ... x iR (n times). Indeed, using the maximum metric d(x, y) = max lSiSn ( I Xl _ yi I ), which is equivalent to the standard one, we see that the E-disk at x coincides with the set lx I - E, X I + E[ x ... x jx n - E, xn + E[. For generalizations to infinite products see Exercise

Section 1.4 Subspaces, Products, and Quotients

19

I.4K, and to metrie spaces see Exercise I.4N. 1.4.3 Proposition Let S and T be topological spaces and denote by PI: S x T Sx T

-7

-7

Sand P2:

T the canonical projections: PI(S, t) = s, and P2(s, t) = t. Then

(i) PI and P2 are open mappings; and (ii) a mapping
maps PI 0

-7

Sand P20

-7

T are continuous.

Proof (i) follows directly from the definitions. (ii)

=
T)

n
V)

= (PI

0
n (P20
the assertion

•

In general, the maps Pi' i = 1, 2, are not closed. For example, if S = T = lR the set A = {(x, y) I xy = 1, x> o} is closed in S x T = ]R.2, but pI(A) = ]0, oaf which is not closed in S.

1.4.4 Proposition A topological space S is Hausdorff iff the diagonal which is defined by = {(s, s)

I sE

~s

S} c S x S is a closed subspace of S x S, with the product topology.

Proof It is enough to remark that S is Hausdorff iff for every two distinct points P, q exist neighborhoods Up' Uq of p, q, respectively, such that (Up x U q)

n ~s =0.

E

S there

•

In a number of places later in the book we are going to form new topological spaces by collapsing old ones. We define this process now and give some examples.

1.4.5 Definition Let S be a set. An equivalence relation - on S is a binary relation such that for all u, v, W E S, (i) u - u (reflexivity);

(ii) u - v iff v - u (symmetry); and (iii) u - v and v - w implies u - w (transitivity). The equivalence class containing u, denoted [uJ. is defined by [u] equivalence classes is denoted S/-, and the mapping the canonical projection.

7t :

S

-7

= {v E S I u -

v}. The set of

S/- defined by u ...... [ul is called

Note that S is the disjoint union of its equi valence classes. The collection of subsets U of S/- such that rr-I(U) is open in S is a topology because (I)

7t- 1(0) = 0, 7t- I (S/-) = S;

(2)

7t- I (U I

(3)

7t- I

n U 2) =

7t- I (U I )

(U ,.v,,) = U

n

7t- I(U 2);

a 7t- I (U a )·

and

20

Chapter 1 Topology

1.4.6 Definition Let S be a topological space and - an equivalence relation on S. Then the collection of sets {U

C

S/-I7t""I(U) is open in S} is called the quotient topology on S/-.

1.4.7 Examples A The Torus Consider]R2 and the relation - dermed by

(Z denotes the integers). Then ']['2 = ]R2/_ is called the 2-torus.

In addition to the quotient

topology, it inherits a group structure by setting [(a!' az)] + [(bl' b2)] = [(aI' az) + (bl' b2)]. The n-dimensional torus 1I'" is defined in a similar manner. The torus 1I'" may be obtained in two other ways. First, let D be the unit square in jR2 with the subspace topology. Define - by x - Y iff any of the following hold: (i) x =y;

(ii) xI=YI' x2 =0, Y2=I; (iii) XI = Yl' x 2 = 1, Y2 = 0; (iv) x2 = Y2' XI = 0, YI = 1; or (v) x 2 = Y2' XI = 1, YI = 0, as indicated in Figure 1.4.1. Then ']['2 = D/-. Second, define ']['2 = SI

X

SI, also shown in

Figure 1.4.1.

Figure 1.4.1 B The Klein bottle is obtained by reversing one of the orientations on D, as indicated in Figure 1.4.2. Then K = D/- (the equivalence relation indicated) is the Klein bottle. Although it is realizable as a subset of ]R4, it is convenient to picture it in ]R3 as shown. In a sense we will make precise in Chapter 6, one can show that K is not "orientable." Also note that K does not inherit a group structure from ]R2, as did ']['2.

Section 1.4 Subspaces, Products, and Quotients

21

1+------1

L---J Figure 1.4.2

C Projective Space On JR"\{ O} define x - y if there is a nonzero real constant A such that x = Ay. Then (JR"\{O})/- is called real projective (n - I)-space and is denoted by JR1P" - I. Alternatively, lRlP'n - 1 can be defined as sn - 1 (the unit sphere in JRD) with antipodal points x and - x identified. (It is easy to see that this gives a homeomorphic space.) One defines complex projective space

elF"' - 1

in an analogous way where now A is complex. •

1.4.8 Proposition Let - be an equivalence relation on the topological space S and 1t: S ~ s/the canonical projection. A map cp: s/- ~ T, where T is another topological space, is continuous iff cp 0 1t : S

~

T is continuous.

Proof

r

= {(s, s') I s - s'} c: S x S is called the graph of the equivalence

relation -. The equivalence relation is called open (closed) if the canonical projection 1t: S ~ S/is open (closed).

We note that - is open (closed) iff for any open (closed) subset A of S the set 1t-I(1t(A» is open (closed). As in 1.4.8, for an open (closed) equivalence relation - on S, a map cp : S/- ~ T is open (closed) iff cp 0 1t : S ~ T is open (closed). In particular, if - is an open (closed) equivalence relation on Sand cp: S/- ~ T is a bijective continuous map, then

22

Chapter 1 Topology

Assume that points [x], [y]

E

r is closed and - is open. If S/- is not Hausdorff then there are distinct S/- such that for any pair of neighborhoods Ux and Uy of [x] and [y],

respectively, we have Ux n Uy '" 0. Let Vx and Vy be any open neighborhoods of x and y, respectively. Since - is an open equivalence relation, 7t(V x) = Ux and 7t(V y) = Uy are open neighborhoods of [x] and [y] in S/-. Since Ux n Uy '" 0, there exist x' E Vx and y' E Vy such that [x'] = [y']; i.e., (x', y') E r. Thus (x, y) E cl(r) by 1.1.9(i). As r is closed, (x, y) E r, i.e., [x] = [y], a contradiction. _

Exercises 1.4A Show that the sequence xn = lin in the topological space ]0, 1] (with the relative topology from JR) does not converge. l.4B If

r f = {(s,

l.4C Let X and Y be topological spaces with Y Hausdorff. Show that for any continuous maps f, g : X -+ Y, the set {x E X I f(x) = g(x)} is closed. (Hint: Consider the mapping x 1-+ (f(x), g(x» and use 1.4.4.) Thus, if f(x) = g(x) at all points of a dense subset of X, then f= g. l.4D Define a topological manifold to be a space locally homeomorphic to JRn. Find a topological manifold that is not Hausdorff. (Hint: Consider JR with "extra origins.") 1.4E

Show that a mapping
l.4F

r f (give r f the

Show that every subspace of a Hausdorff (resp., regular) space is Hausdorff (resp., regular). Conversely, if each point of a topological space has a closed neighborhood that is Hausdorff (resp., regular) in the subspace topology, then the topological space is Hausdoff (resp., regular). (Hint: use Exercise 1.lF and 1.lG.)

l.4G Show that a product of topological spaces is Hausdorff iff each factor is Hausdorff. 1.4H Let S, T be topological spaces and -, = be equivalence relations on Sand T, respectively. Let
Section 1.4 Subspaces, Products, and Quotients

1.41

Let S be a Hausdorff space and assume there is a continuous map 0 : S/7t

1.4J

23

0

0

~

S such that

= is/ _' the identity. Show that S/- is Hausdorff and o(S/-) is closed in S.

Let M and N be metric spaces, N complete, and cp: A

~

N be unifonnly continuous

(A with the induced metric topology). Show that cP has a unique extension cP: cl(A)

~

N that is unifonnly continuous.

I.4K Let S be a set, T a a family of topological spaces, and CPa: S

~

T a a family of

mappings. Let 'E be the collection of finite intersections of sets of the fonn CPa-1(Ua ) for U a open in T a. The initial topology on S given by the family CPa: S ~ Ta has as basis the collection 'E. Show that this topology is characterized by the fact that any mapping cP : R

~

S from a topological space R is continuous iff all CPa 0 cP : R

~

T a are continuous.

Show that the subspace and product topologies are initial topologies. Define the product of an arbitrary infinite family of topological spaces and describe the topology.

1.4L Let T be a set and CPa: Sa

~

T a family of mappings, Sa topological spaces with

0a. Let 0 = {U C T I CPa-1(U) E 0 for each IX}. Show that 0 is a topology on T, called thejinal topology on T given by the family CPa: Sa ~ T. Show

topologies

that this topology is characterized by the fact that any mapping cP: T

cP 0 CPa : Sa

~

~R

is continuous iff

R are all continuous. Show that the quotient topology is a final topology.

1.4M Show that in a complete metric space a subspace is closed iff it is complete.

1.4N Show that a product of two metric spaces is also a metric space by finding at least three equivalent metrics. Show that the product is complete if each factor is complete.

~1.5

Compactness

Some basic theorems of calculus. such as "every real valued continuous function on [a.b] attains its maximum and minimum" implicitly use the fact that [a.b] is compact. The general theory of compactness is described in this section. I.S.1 Definition Let S be a topological space. Then S is called compact iffor every covering of S by open sets Ua (that is UaUa =S) there is afinite subcollection of the Ua also covering S. A subset A C S is called compact if A is compact in the relative topology. A subset A is called relatively compact if cl(A) is compact. A space is called locally compact ifit is Hausdorff and each point has a relatively compact neighborhood.

I.S.2 Proposition (i) If S is compact and A C S is closed. then A is compact. (ii) If Cjl : S ~ T is continuous and S is compact. then Cjl(S) is compact. Proof (i) Let {Ua } be an open covering of A. Then {Ua • S \ A} is an open covering of S and hence contains a finite subcollection of this covering also covering S. The elements of this collection. except S \ A. cover A. (ii) Let {Ua } be an open covering of Cjl(S). Then {Cjl-I(Ua )} is an open covering of S and thus by compactness of S a finite subcollection (Cjl-I(Ua(i» I i = 1•...• n } • covers S. But then {Ua(i)}. i = 1•...• n covers Cjl(S) and thus Cjl(S) is compact. • In a Hausdorff space. compact subsets are closed (exercise). Thus if S is compact. T is Hausdorff and Cjl is continuous. then Cjl is closed; if Cjl is also bijective. then it is a homeomorphism. 1.S.3 Proposition A product space S x T is compact iff both S and T are compact. Proof. In view of I.S.2 all we have to show is that if Sand T are compact. so is S x T. Let {Aa} be a covering of S x T by open sets. Each Aa is the union of sets of the form U x V with U and V open in S and T. respectively. Let {UJ3 x VJ3} be a covering of S x T by open rectangles. If we show that there exists a finite subcollection of UJ3 x VJ3 covering S x T. then clearly also a finite subcollection of {Aa} will cover S x T. A finite subcollection of {UJ3 x VJ3} is found in the following way. Fix s E S. Since the set {s} x T is compact. there is a finite collection

Section 1.5 Compactness

covering it. If Us =

nj = I..... i(s)U~;,

25

then Us is open. contains s. and

usxV"PI ..... Usx V". PI(S) covers (s)xT. Let Ws=Usx T; thenthecollection (Ws) is an open covering of SxT and if we show that only a finite number of these W s cover S x T. then since Ws =Uj=l ..... i(s)(UsXV~). it follows that a finite number of U~ x V~ will cover S x T. Now look at S x (t), for t E T fixed. Since this is compact. a finite subcollection W SI ..... W... covers it. But then

• As we shall see shortly in 1.5.8. [-1. 1] is compact. Thus ']['1 is compact. It follows from 1.5.3 that the torus ']['2. and inductively 'lI"'. are compact. Thus if 1t :]R2 ~ ']['2 is :he canonical projection we see that ']['2 is compact without ]R2 being compact; i.e .• the converse of 1.5.2(ii) is false. Nevertheless it sometimes occurs that one does have a converse; this leads to the notion of a proper map discussed in Exercise 1.5J. 1.5.4 Bolzano-Weierstrass Theorem If S is a compact Hausdorff space, then every sequence has a convergent subsequence. The converse is also true in metric and second-countable Hausdorff spaces. Proof Suppose S is compact and (un) contains no convergent subsequences. We may assume that the poins of the sequence are distinct. Then cl( (un)) = (un) is compact and each un has a neighborhood Un that contains no other urn' for otherwise un would be a limit of a subsequence. Thus (Un) is an open covering of the compact subset (un) which contains no finite subcovering. a contradiction. Let S be second countable. Hausdorff. and such that every sequence has a convergent subsequence. If (Ua) is an open covering of S. by the Lindelof lemma there is a countable COllection {Unl n=1.2 ... } also covering S. Thus we have to show that {Un I n=I.2 .... } contains a finite collection covering S. If this is not the case. the family {S \ Ui=I ..... nUJ consists of closed nonempty sets and S \ Ui=I .....nUi ::::> S \ Ui=I ..... rnUi for m ~ n. Choose Pn e S\Ui=I ..... nUi. If {Pn I n = 1. 2 .... } is infinite. by hypothesis is contains a convergent SUbsequence; let its limit point be denoted p. Then PES \ Ui=I .... nUi for all n. contradicting the fact that {Un I n = 1. 2 .... } covers S. Thus {Pn I n = 1.2.... } must be a finite set; i.e .• for all n ~ N. Pn = PN' But then again PN E S \ Ui=I ..... n Ui for all n. contradicting the fact that {Un I n = 1. 2 .... } covers S. Hence S is compact. Let S be a metric space such that every sequence has a convergent subsequence. If we

26

Chapter 1 Topology

show that S is separable. then since S is a metric space it is second countable (Exercise 1.2C). and by the preceding paragraph. it will be compact. Separability of S is proved in two steps. First we show that for any E > 0 there is a finite set of points {p\' ...•Pn} such that S = U i = 1•... ,nDt(p). If this were false. there would exist an E> 0 such that no finite number of E-disks cover S. Let PI E S be arbitrary. Since Dt(PI) *" S. there is a point P2 E S \ Dt(PI)· Since DipI) U Dt (P2) *" S. there is also a point P3 E S \ (Dt(PI) U Dt (P2»' etc. The sequence {Pn I n=1.2 •... } is infinite and d(Pi' Pj) ~ E. But this sequence has a convergent subsequence by hypothesis. so this subsequence must be Cauchy. contradicting d(Pi' Pj) ~ E for all i. j. Second. we show that the existence for every E> 0 of a finite set (Pl •...• Pn(t)} such that S = U i = 1,..n(tPt(Pi) implies S is separable. Let A.. denote this finite set for E = lin and let A = Un~A... Thus A is countable and it is easily verified that c1(A) = S. • A property that came up in the preceding proof turns out to be important. 1.5.5 Definition Let S be a metric space. A subset A c: S is called totally bounded iffor any E > 0 there exists afinite set {p\' ...• Pn} in S such that A c: U i = 1,..nDt(P)' 1.5.6 Corollary A metric space is compact iff it is complete and totally bounded. A subset of a complete metric space is relatively compact iffit is totally bounded. Proof The previous proof shows that compactness implies total boundedness. As for compactness implying completeness. it is enough to remark that in this context. a Cauchy sequence having a convergent subsequence is itself convergent. Conversely. if S is complete and totally bounded, let {Pn I n = 1.2 •... } be a sequence in S. By total boundedness. this sequence contains a Cauchy subsequence. which by completeness. converges. Thus S is compact by the Bolzano-Weierstrass theorem. The second statement now readily follows. • 1.5.7 Proposition In a metric space compact sets are closed and bounded. Proof This is a particular case of the previous corollary but can be easily proved directly. If A is compact. it can be finitely covered by E-disks: A = U i = l,..nDt(P). Thus n

diam(A) :5

L diam(Dt(Pi»

= 2nE

i=1

•

From 1.5.2 and 1.5.6. if S is compact and
27

Section 1.5 Compactness

Proof By 1.5.2(i) it is enough to show that closed bounded rectangles are compact in JRn. which in tum is implied via 1.5.3 by the fact that closed bounded intervals are compact in R To show that [-a. a]. a> 0 is compact. it suffices to prove (by 1.5.6) that for any given can be finitely covered by intervals of the form

]p - E. P + I'.

[.

completeness of R Let n be a positive integer such that a < n E. Let t E the largest (positive or negative) integer satisfying k (k + 1) E. Thus any point t k = - n •...• 0 •...• n

where

E [-

I'. ~

t. Then - n

~

k

~

> O. [-a. a]

[-

a. a] and k be

n and

k I'.

~

t<

I'. -

E. kE + 1'.[.

{]kE - E. kE + 1'.[ I k = O. iI •.... in}

is a finite

a. a] belongs to an interval of the form

and hence

I'.

since we are accepting

]k

covering of [- a. a] . • This theorem is also proved in virtually every textbook on advanced calculus. As is known from calculus. continuity of a function on an interval [a. b] implies uniform continuity. The generalization to metric spaces is the following. 1.5.9 Proposition A continuous frUlpping
~

M 2• where M, and M2 are metric spaces

and M, is compact, is uniformly continuous. Proof The metrics on M, and M2 are denoted by d, and d2. Fix I'. > O. Then for each p E M" by continuity of

0 such that if d,(p. q) < Sp' then di
f/2. Let

cover the compact space M, and let S =min {Sl2 ..... S/2}. Then if p. q

E

M, are such that

d,(p. q) < S. there exists an index i. 1 ~ i ~ n. such that d,(p. Pi) < SJ 2 and thus

A useful application of Corollary 1.5.6 concerns relatively compact sets in C(M. N). for metric spaces (M. dM) and (N. d N) with M compact and N complete. 1.5.10 Definition

A subset ;F e C(M.N) is called equicontinuous at rna E M. if given

I'.

> O.

there exists S > 0 such that whenever dM(m. rna) < S. we have dN(
and N complete. A set ;Fe C(M. N) is relatively compact iff it is equicontinuous and all the sets :F(m) = {
28

Chapter 1 Topology

Proof If 1" is relatively compact. it is totally bounded and hence so arc all the sets 1"(m}. Since N is complete. by Corollary 1.5.6 the sets 1"(m} are relatively compact. Let {
:r.

Then there exists 0 > a such that if dM(m. m'} < O. we have

dN(
This shows that 1" is equicontinuous. Conversely. since C(M. N} is complete. by Corollary 1.5.6 we need only show that 1" is totally bounded. For E > O. find a neighborhood Urn of mE M such that for all m' E Urn' dN(
a: {1 •...• n}

1"=

UaE .st1"a.

~

{1 •...• k}.then Jl is finite and

where 1"a = {
But if
+ dN('!'(m(i}}. '!'(m» < E; i.e .• the diameter of 1"a is

:0:;;

E. so 1" is totally bounded. •

Combining this with the Heine-Borel theorem. we get the following. 1.5.12 Corollary

If M is a compact metric space, a set 1" C C(M,lRn} is relatively compact iff

it is equicontinuous and uniformly bounded (i.e., lI
:0:;;

constant/or all

The following example shows how to use the Arzela-Ascoli theorem. 1.5.13 Example Let fn : [a, 1)

~

lR be continuous and be such that I fn(x} I

:0:;;

100 and the

derivatives fn' exist and are uniformly bounded on )0. 1[. Prove fn has a uniformly convergent subsequence. We verify that the set {fn} is equicontinuous and bounded. The hypothesis is that I fn'(x}1 5 M for a constant M. Thus by the mean-value theorem,

Section 1.5 Compactness so given

E

29

we can choose 0 = F/M, independent of x, y, and n. Thus {fn } is equicontinuous.

It is bounded because II fn II

=sUPo s x S 1 I fn(x) I ~ 100.•

Exercises 1.5A Show that a topological space S is compact iff every family of closed subsets of S whose intersection is empty contains a finite subfamily whose intersection is empty. 1.5B Show that every compact metric space is separable. (Hint: Use total boundedness.) 1.5C Show that the space of exercise 1.11 is not locally compact. (Hint: Look at the sequence

(lIn, 0).) 1.5D (i) Show that every closed subset of a locally compact space is locally compact. (ii) Show that S x T is locally compact if both S and T are locally compact. 1.5E Let M be a compact metric space and T : M d(ml' m2) for m 1

"#

~

M a map satisfying d(T(m 1), T(m 2» <

m2. Show that T has a unique fixed point.

1.5F Let S be a compact topological space and - an equivalence relation on S, so that S/- is compact. Prove that the following conditions arc equivalent (cf. 1.4.10): (i) The graph C of - is closed in S x S; (ii) - is a closed equivalence relation; (iii) SI- is Hausdorff. 1.5G Let S be a Hausdorff space that is locally homeomorphic to a locally compact Hausdorff space (that is, for each u E S. there is a neighborhood of u homeomorphic, in the subspace topology. to an open subset of a locally compact Hausdorff space). Show that S is locally compact. In particular, Hausdorff spaces locally homeomorphic to JRn are locally compact. Is the conclusion true without the Hausdorff assumption?

1.58 Let M3 be the set of all 3 x 3 matrices with the topology obtained by regarding M3 as JR9. Let 80(3) = {A E M3 I A is orthogonal and det A = I}. (i) Show that 80(3) is compact. (ii) Let P = (Q E 80(3) I Q is symmetric} and let
30

Chapter 1 Topology

Prove that Fn has a uniformly convergent subsequence. 1.5J Let X and Y be topological spaces, Y be first countable, and f: X

~

Y be a continuous

map. The map f is called proper, if f(x n) ~ y implies the existence of a convergent subsequence (xn(i»), xn(i) ~ x such that f(x) =y. (i) Show that f is a closed map. (ii) Show that if Y is locally compact, f is proper if and only if the inverse image by f of every compact set in Y is a compact set in X. (iii) Show that if f is proper and Y is locally compact, then X is also locally compact. (We have defined properness only when Y is first countable. The same defmition and properties of proper maps hold for general Y if in the definition "sequence" is replaced by"net.") (iv) Show that the composition of two proper maps is again proper. (v) Show that any continuous map defined on a compact space is proper.

§1.6 Connectedness Three types of connectedness treated in this section are arcwise connectedness, connectedness, and simple connectedness. 1.6.1 Definition Let S be a topological space and 1= [0, 1] c: lIt. An arc

arc wise connected if every two points in S can be joined by an arc in S. A space S is called locally arcwise connected iffor each point

XES

and each neighborhood U of x, there is a

neighborhood V of x such that any pair ofpoints in V can be joined by an arc in U. For example, lItn is arcwise and locally arcwise connected: any two points of lItn can be joined by the straight line segment connecting them. A set A c: lItn is called convex if this property holds for any two of its points. Thus convex sets in lItn are arcwise and locally arcwise connected. A set with the trivial topology is arcwise and locally arcwise connected, but in the discrete topology it is neither (unless it has only one point). 1.6.2 Definition A topological space S is connected if 0 and S are the only subsets of S that are both open and closed. A subset of S is connected ifit is connected in the relative topology. A

component A of S is a nonempty connected subset of S such that the only connected subset of S containing A is A itself; S is called locally connected if each point has a connected neighborhood. The components of a subset T c: S are the components of T in the relative topology of T in S. For example, lItn and any convex subset of

]Rn

are connected and locally connected. The

union of two disjoint open convex sets is disconnected but is locally connected; its components are the two convex sets. The trivial topology is connected and locally connected, whereas the discrete topology is neither: its components are all the one-point sets. Connected spaces are characterized by the following. 1.6.3 Proposition Thefollowing are equivalent: (i) S is not connected;

(ii) there is a nonempty proper subset of S that is both open and closed; (iii) S is the disjoint union of two nonempty open sets;

(iv) S is the disjoint union of two nonempty closed sets. The sets in (iii) or (iv) are said to disconnect S.

32

Chapter 1 Topology

Proof To prove that (i) implies (ii), assume there is a nonempty proper set A that is both open and closed. Then S = A U (S\A) with A, S\A open and nonempty. Conversely, if S = A U B with A, B open and nonempty, then A is also closed, and thus A is a proper nonempty set of S that is both open and closed. The equivalences of the remaining assertions are similarly checked. _ 1.6.4 Proposition If f: S ~ T is a continuous map of topological spaces and S is connected (resp., arcwise connected) then so is f(S). Proof Let S be arcwise connected and consider f(sl)' f(S2) E f(S)

C

T. If c : I ~

s,

c(O) = s\'

c(1) = ~ is an arc connecting Sl to s2' then clearly f 0 c : I ~ T is an arc connecting f(sl) to f(S2); i.e., f(S) is arcwise connected. Let S be connected and assume f(S) C U U V, where U and V are open and un V = 0. Then r-l(U) and r-l(V) are open by continuity of f, r-l(U) U r-I(V) = r-I(U U V) :::> f- 1(f(S» = S, and r-I(U) n f-I(V) = r-1(0) = 0, thus contradicting connectedness of S by 1.6.3. Hence f(S) is connected. _ We shall prove that arcwise connected spaces are connected. We shall use the following. 1.6.5 Lemma The only connected sets of]R are the intervals (finite, infinite, open, closed, or hIllj-open). Proof Let us prove that [a, b[ is connected; all other possibilities have identical proofs. If not, [a, b[ = U U V with U, V nonempty disjoint closed sets in [a, b[. Assume that a E U. If x = sup(U), then x E U since U is closed in [a, b[, and x < b since V'" 0. But then lx, b[ c V and, since V is closed, x E V. Hence x E un V, a contradiction. Conversely, let A be a connected set of R We claim that [x, y]

C

A whenever x, y E

A, which implies that A is an interval. If not, there exists Z E [x, y] with z e A. But in this case ]-00, z[ U A and ]z, 00 [ U A are open nonempty sets disconnecting A. _ 1.6.6 Proposition If S is arcwise connected then it is connected. Proof If not, there are nonempty, disjoint open sets Uo and U I whose union is S. Let Xo E U o and XI E Uland let cp be an arc joining Xo to XI. Then Vo = cp-I(U O) and V I = cp-I(U I) disconnect [0, I]. _ A standard example of a space that is connected but is not arcwise connected nor locally connected, is {(X,y)E ]R21 x>O and y=sin(1/x)} U {(O,y)I-I
Section 1.6 Connectedness

33

Proof Fix XES. The set A = {y E Sly can be connected to x by an arc} is nonempty and open since S is locally arcwise connected. For the same reason, S\A is open. Since S is connected we must have S\A = 0; so A = S, i.e., S is arcwise connected. _ 1.6.8 Intermediate Value Theorem

Let S be a connected space and f: S

~

lR be

continuous. Then

Proof Suppose feu) < a < f(v) and f does not assume the value a. Then U = {uo I f(u o) < a} is both open and closed. _ An alternative proof uses the fact that f(S) is connected in lR and therefore is an interval.

1.6.9 Proposition Let S be a topological space and B (i) If B cAe cl(B), then A is connected.

C

S be connected.

(ii) If Ba is afamily of connected subsets of S and Ba n B "# 0, then B U (U a Ba) is

connected.

Proof If A is not connected, A is the disjoint union of U 1 n A and U2 n A where U 1 and U2 are open in S. Then from 1.1.9(i), U 1 n B "# 0, U2 n B "# 0, so B is not connected. We leav~ (ii) as an exercise. _ 1.6.10 Corollary The components of a topological space are closed. Also, S is the disjoint union of its components. If S is locally connected. the components are open as well as closed.

1.6.11 Proposition Let S be afirst countable compact HausdorjJspace and closed. connected subsets of S with An

C

("l

a sequence of

An _ I' Then A = nn '" I " is connected.

Proof As S is normal, if A is not connected, A lies in two disjoint open subsets U 1 and U2 of S. If An n (S\U 1) n (S\U 2 )"# 0 for all n, then there is a sequence un E An n (s\U 1) n (s\U2) with a subsequence converging to u. As An' S\UI' and S\U2 are closed sets, u E An (S\U 1) n (s\U2 ), a contradiction. Hence some An is not connected. _ Finally we consider the notion of simple connectivity. 1.6.12 Definition Let S be a topological space and c : [0, 1] ~ S a continuous map such that c(O) =c(1) = pES. We call c a loop in S based at p. The loop c is called contractible if there is a continuous map H: [0, 1] x [0, 1] ~ S such that H(t, 0) = c(t) and H(O, s) = H(l, s) = H(t, 1) = P for all t E [0, 1]. (See Figure l.6.l.)

34

Chapter 1 Topolngy

Figure 1.6.1 We think of cs(t) = H(t, s) as a family of arcs connecting Co = c to c\' a constant arc; see Figure 1.6.1. Roughly speaking, a loop is contractible when it can be shrunk continuously to p by loops beginning and ending at p. The study of loops leads naturally to homotopy theory. In fact, the loops at p can, by successively traversing them, be made into a group called the fundamental group; see Exercise 1.6F. 1.6.13 Definition A space S is simply connected if S is connected and every loop in S is contractible.

In the plane ]R2 there is an alternative approach to simple connectedness, by way of the Jordan Curve Theorem; namely, that every simple (nonintersecting) loop in ]R2 divides ]R2 (that is, its complement has two components). The bounded component of the complement is called the interior, and a subset A of lR,2 is simply connected iff the interior of every loop in A lies in A. We close this section with an optional theorem sometimes used in Riemannian geometry (to show that a Riemannian manifold is second countable) that illustrates the interplay between various notions introduced in this chapter. 1.6.14 Alexandroff's Theorem An arcwise connected locally compact metric space is separable and hence is second countable.

Proof (Pfluger [1957]) Since the metric space M is locally compact, each mE M has compact neighborhoods that are disks. Denote by r(m) the least upper bound of the radii of such disks. If r(m) = 00, since every metric space is first countable, M can be written as a countable union of compact disks. But since each compact metric space is separable (Exercise 1.58), these disks and also their union will be separable, and so the proposition is proved in this case. If r(m o) < 00, since r(m) ~ r(m o) + d(m, mo)' we see that r(m) < 00 for all m E M. By the preceding argument, if we show that M is a countable union of compact sets, the proposition is proved. Then second countability will follow from Exercise I.2e.

Section 1.6 Connectedness

35

To show that M is a countable union of compact sets, define the set Gm by Gm = {m' E M I d(m', m) $; r(m)!2}. These Gm are compact neighborhoods of m. Fix m(O) E M and put Ao = Gm(O)' and, inductively, define

Since M is arcwise connected, every point m E M can be connected by an arc to m(O), which in tum is covered by finitely many Gm . This shows that M = Un 1! o~. Since Ao is compact, all that remains to be shown is that the other An are compact. Assume inductively that An is compact and let {m(i)} be an infinite sequence of points in ~ + I. There exists m(i)' E ~ such that m(i) E Gm(iY. Since An is assumed to be compact there is a subsequence m(ik)' that converges to a point m' E ~. But d(m(i), m')

$;

d(m(i), m(i)') + d(m(i)', m')

$;

r(m(i)') d( (.)' ') 2 + m 1 ,m

$;

r(m') 3d(m(i)', m') -2- + 2

Hence for ik big enough, all m(ik) are in the compact set {n E M I d(n, m') $; 3r(m')/2}. so m(ik) has a subsequence converging to a point m. The preceding inequality shows that mE ~+I. By the Bolzano-Weierstrass Theorem, An+1 iscompact. _

Exercises 1.6A Let M be a topological space and H : M --t lR continuous. Suppose e E H(M). Then show H-I(e) divides M; that is, ~-I(e) has at least two components. 1.6B Let 0(3) be the set of orthogonal 3 x 3 matrices. Show that 0(3) is not connected and that it has two components. 1.6C Show that S x T is connected (locally connected, arcwise connected, locally arcwise connected) iff both S and T are. (Hint: For connectedness write S x T = U 1 E {tj) U ({so) x T)] for So E S fixed and use 1.6.9(ii).)

T

[(S x

1.6D Show that S is locally connected iff every component of an open set is open. 1.6E

Show that the quotient space of a connected (locally connected, arcwise connected) space is also connected (etc.). (Hint: For local connectedness use Exercise 1.6D and show that the inverse image by 1t of a component of an open set is a union of components.)

Chapter 1 Topology

36

1.6F (i)

Let S and T be topological spaces. Two continuous maps f. g : T

homotopic if there exists a continuous map F: [0. 1] x T f(t) and F(I. t) (ii)

=g(t)

~

~

S are called

S such that F(O. t)

=

for all t E T. Show that homotopy is an equivalence relation.

Show that S is simply connected if and only if any two continuous paths Ct. c 2 : [0. 11

~

S satisfying c](O)

=ciO).

c](I) = ci1) arc homotopic. via a homotopy

which preserves the end points. i.e .• F(s. 0)

= c](O) = c2 (0)

and F(s. 1) = c](l)

=

c 2(1 ). (iii)

Define the composition c]

*

c 2 of two paths Ct. c 2 : [0. 11

~

S satisfying c] (1) =

c 2(0) by if tE [1.1/21 if t E [1/2. 11 . Show that this composition. when defined. induces an associative operation on endpoints preserving homotopy classes of paths. (iv) Fix So

E

S and consider the set

1t 1(S.

so) of endpoint fixing homotopy classes of

paths starting and ending at so. Show that

1t 1(S.

so) is a group: the identity element

is given by the class of the constant path equal to So and the inverse of c is given by the class of c(l- t). (v) Show that if S is arcwise connected. then s

E

S.

1t 1(S)

1t 1(S.

so) is isomorphic to

1t 1(S.

s) for any

will denote any of these isomorphic groups.

(vi) Show that if S is arcwise connected. then S is simply connected iff

1t 1(S) =

O.

§1.7 Baire Spaces The Baire condition on a topological space is fundamental to the idea of "genericity" in differential topology and dynamical systems; see Kelley [1975) and Choquet [1969) for additional information. 1.7.1 Definition Let X be a topological space and A C X a subset. Then A is called residual if A is the intersection of a countable family of open dense subsets of X. A space X is called a Baire space if every residual set is dense. A set B C X is called ajirst category set if B C Un
°

1.7.3 Baire Category Theorem Complete pseudometric and locally compact spaces are Baire spaces. Proof Let X be a complete pseudometric space. Let U c::::: X be open and A = nn~IOn be residual. We must show UnA i; 0. Since cl(0n) = X, un On i; 0 and SO we can choose a disk of diameter less than one, say Yl' such that c1(Y I) C U n 0 1, Proceed inductively to obtain cl(Yn) C un On n Yn- I, where Yn has diameter
38

Chapter 1 Topology

*

nn2Icl(Vn} and so un (nn210n) 0; i.e .• A is dense in X. If X is a locally compact space the same proof works with the following modifications: Vn are chosen to be relatively

E

compact open sets. and {xn } has a convergent subsequence since it lies in the compact set cl(V 1}· • To get a feeling for this theorem. let us prove that the set of rationals Q cannot be written as a countable intersection of open sets. For suppose Q = nn210n. Then each On is dense in JR.

Q is. and so Cn = JR \ On is closed and nowhere dense. Since JR U (U n21 Cn) is a complete metric space (as well as a locally compact space). it is of second category. so Q or some since

Cn should have nonempty interior. But this is impossible. The notion of category can lead to interesting restrictions on a set. For example in a nondiscrete Hausdorff space. any countable set is first category since the one-point set is closed and nowhere dense. Hence in such a space every second category set is uncountable. In particular. nonfmite complete pseudometric and locally compact spaces are uncountable.

Exercises 1.7A Let X be a Baire space. Show that (i) X is a second category set; (ii) if U c: X is open. then U is Baire.

1.7B Let X be a topological space. A set is called an

1'0

if it is a countable union of closed sets.

and is called a (j& if it is a countable intersection of open sets. Prove that the following are equivalent: (i) X is a Baire space; (ii) any first category set in X has a dense complement; (iii) the complement of every first category

1'0

-set is a dense (j& -set;

(iv) for any countable family of closed sets {Cn ) satisfying X = Un 21 Cn' the open set Un 2Iint(Cn} is dense in X. (Hint: First show that (ii) is equivalent to (iv). For (ii) implies (iv). let Un = Cn \ int(Cn) so that Un 21 Un is a first category set and therefore X\ U n21 U n is dense and included in U n2I int(Cn}. For the converse. assume X is not Baire so that A = n n 2 1Un is not dense. even though all Un are open and dense. Then X = cl(A} U {X \ Un I n= 1.2 •... }. Put Fa = cl(A}. Fn = X \ Un and show that int(Fn} = int(cl(A}} which is not dense.}

1.7C Show that there is a residual set E in the metric space C([O. Il. JR} such that each fEE

is not differentiable at any point. Do this by following the steps below. (i) Let E£ denote the set of a\1 f E C([O. Il. lR) such that for every x E [0. Il.

Section 1.7 Baire Spaces

diam { f(x + h~ - f(x)

II

39

< Ih I <

e}

> 1

for e > O. Show that E£ is open and dense in C([O. 1J. JR). (Hint: For any polynomial p E C([O. 1]. JR). show that p + /) cos(kx) E E£ for /) small and /)k large.) (ii) Show that E = no~l E l/0 is dense in C([O. IJ. JR). (Hint: Use the Baire Category

Theorem.) (iii) Show that if fEE. then f has no derivative at any point.

1.7D Prove the following: In a complete metric space (M. d) with no isolated points. no countable dense set is a Gil-set. (Hint: Suppose E = {Xl' x2 •...

}

is dense in M and is also a Gil

set. i.e .• E = no>o Vo with Vo open. n = 1.2•.... Conclude that Vo is dense in M. Let Wo = Vo \ {xl' ...• Xo} . Show that Wo is dense in M and that no>oWo contradicts the Baire property.)

= 0. This

Chapter 2 BANACH SPACES AND DIF'F'ERENTIAL CALCULUS Manifolds have enough structure to allow differentiation of maps between them. To set the stage for these concepts requires a development of differential calculus in linear spaces from a geometric point of view. The goal of this chapter is to provide this perspective. Perhaps the most important theorem for later use is the Implicit Function Theorem. A fairly detailed exposition of this topic will be given with examples appropriate for use in manifold theory. The basic language oftangents. the derivative as a linear map. and the chain rule. while elementary. are important for developing geometric and analytic skills necessary for mastering manifold theory. The main goal is to develop the theory of finite-dimensional manifolds. However. it is instructive and efficient to do the infinite-dimensional theory simultaneously. To avoid being sidetracked by infinite-dimensional technicalities at this stage. some functional analysis background and other topics special to the infinite-dimensional case are presented in supplements. With this arrangement readers who wish to concentrate on the finite-dimensional theory can do so with a minimum of distraction.

§2.1 Banach Spaces It is assumed the reader is familiar with the concept of an abstract vector space over the rcal or complex numbers. Banach spaces are vector spaces with additional structure. While most of this book is concerned with finite-dimensional spaces. much of the theory is really no harder in the general case. and the infinite-dimensional case is actually essential for certain applications. Thus it makes sense to work in the setting of Banach spaces. In addition. although the primary concern is with real Banach spaces. the basic concepts for complex spaces are introduced with little extra effort. 2.1.1 Definition A norm on a real (complex) vector space

E is a mapping from E into the real

numbers. II· II ; E ~ 1R; e ~ II e II. such that N1 II e II ~ 0 for all e E E and II e II =0 iff e =0 (positive definiteness); N2 II Ae II =I AI II e IIfor all e E E and A E 1R (homogeneity); N3

II e 1 + e 2 II :5 II e 1 II + II ~ II for all e 1• e2

E

E (triangle inequality).

41

Section 2.1 Banach Spaces

The pair (E, II . II) is sometimes called a normed space. If there is no danger of confusion, we sometimes just say "E is a normed space." To distinguish different norms, different notations are sometimes used, e.g., II . liE' II . Ill' 111·111, etc.Jor the norm. The triangle inequality N3 has the following important consequence:

which is proved in the following way:

II e211

II e\ + (e2 - e\) II

:5;

II e\ II + II e\ - e211 ,

II e\ II = II e2 + (e\ - e2) II

:5;

II e211 + II e\ - e 2 II ,

=

so that both II e211-11 e\ II and II e\ 11- II ~ II are smaller than or equal to II e\ -

~

II.

If N1 in 2.1.1 is replaced by N1' the mapping

where x

II e II ~ 0 for all e E E and II e II =0 implies e =0, II . II : E --t IR is called a seminorm. For example, IRn with standard norm

= (xl' ... , xn ), is a normed space.

Actually, the standard norm on IRn comes from a

more special structure defined as follows. 2.1.2 Definition An inner product on a real vector space E is a mapping

<.,. ):E x E

--t

IR

which we denote (e\, e 2) ..... <e\, e2 ) such that 11 < e, e\ + e2 ) =< e, e\ ) + < e, e2 );

12 < e, ae\ ) =a< e, e\ ); 13 < el' e 2 ) =< e2, e\ ); 14 <e, e ) ~ 0 and <e, e) = 0 iff e = O.

The standard inner product on IRn is n

<.,.)

=

L

xiYi '

i=\

and 11 - 14 are readily checked. For vector spaces over the complex numbers, the definition is modified slightly as follows. 2.1.2' Definition A complex inner product (or a Hermitian inner product) on a complex vector

Chapter 2 Banach Spaces and DijJerenJiDl Calculus

42

space E is a mapping <-,·):ExE~C

such that the/allowing conditions hold:

CI1

(e, e l + e 2 ) = (e, e l ) + ( e, e 2 );

CI2 (a.e, e l ) =a(':':"':I); CI3 (e p e2 ) = ( e 2 , e l ) (so (e, e) is real); CI4 (e, e) ~ 0 and (e, e) = 0 iff e = O.

These properties are to hold for all e, e l , e2 E E and a E C; conjugation. Note that CI2 and CI3 imply that (el' ae2 )

= a( el' e2 ).

denotes complex

Properties CI1 - CI3

are also known in the literature under the name sesquilinearity. As is customary, for a complex number a we shall denote by Re a = (a + 0.)/2,

1m a = (a - 0.)/2,

1a 1 =

(a 0.)1/2

its real and imaginary parts and its absolute value. The standard inner product on the product space

cn = C

X ...

x C is defined by n

( x, w) =

~>iWi i=

,

1

and C11-C14 are readily checked. Also C' is a normed space with n

II Z 112

=

L1

Zi

12

i= 1 In lItn or

cn,

property N3 is a little harder to check directly. However, as we shall show in

Proposition 2.1.4, N3 follows from 11-14 or CI1-CI4. In a (real or complex) inner product space E, two vectors e p e2 E E are called orthogonal and we write e l -L e 2 provided (el' e2 ) = O. For a subset ACE, the set A.L defined by A.L = (e EEl (e, x) =0 for all x E A} is called the orthogonal complement of A. Two sets A, B C E are called orthogonal and we write A -L B if (A, B) = 0; that is, e l -L e 2 for all e l E A and e 2 E B. 2.1.3 Cauchy-Schwartz Inequality In a (real or complex) inner product space,

Equality holds iff e l , e2 are linearly dependent.

43

Section 2.1 Banach Spaces

Proof It suffices to prove the complex case. If a.

~

E

C we have

If we set a = ( e z, ez )' and ~ = - ( e 1, e2 ). then this becomes

and so

If ez = O. equality results in the statement of the proposition and there is nothing to prove. If ez "Ie0, the tenn (ez' ez ) in the preceding inequality can be cancelled since (e z• "Ie- 0 by C14. Taking square roots yields the statement of the proposition. Finally, equality results if and

ez)

only if ae l + ~ez =( ez' eZ)el - ( el' e2 )ez

= O.

•

2.1.4 Proposition Let (E, (', . » be a (real or complex) inner product space and set II e II = ( e, e )IIZ. Then (E, II . 10 is a normed space. Proof N1 and N2 are straightforward verifications. As for N3, the Cauchy-Schwartz inequality and the obvious inequality Re(( e 1, ez » ~ I ( e 1, I imply

ez )

II e1 + ezll z

»

= II e1 112 + 2 Re« ~

el' ez + II ezll z ~ II e1ll z + 21 (el' ez ) I + II ezll z

II e 1 liZ + 211 e 1II II ezll + II ezll z = (II e1 II + II ez 10z . •

Some other useful facts about inner products are given next.

2.1.5 Proposition nonn. Then

Let (E,(·,

.»

be an inner product space and II· II the corresponding

(i) (Polarization)

4( el' ez)

=

j

II el + ez liZ - II el - ez liZ + i II el + ez liZ - i II el - iez liZ if E is complex , II el + ez liZ - II el - Cz liZ if E is real .

(ii) (Parallelogram law)

2 II e 1 liZ + 2 II ez liZ

Proof (i) II e 1+ ez If - II e 1 -

= II e1+ ez liZ + II e1-

e2 UZ .

ez 112 + i II e 1 + iez 112 - i II e1 - iezll2

= II e 1 112 + 2Re« el' ez

»+ II e2112 -II C1 112 + 2Re«

el' e2 » -II

ez 112 +

Chapter 2 Banach Spaces and DifferentiJJl Calculus

44

+ ill e l 112 + 2iRc« cl' ic2 » + i)l c 2 112-i II e l 112 + 2iRe« cl' ie2

»- i II c2112

= 4Rc« el' ~» + 4iRc(-i (el' e 2 » = 4Re«

el' e2

»+ 4i Im«

el' c2

»=4 ( el' e2 ).

The real case is proved in a simnar way. (ii) II e l + e 2 112 + II e l

-

e2 112 = II e l 112 + 2Re« el' e2

»+ II c2 112 + II e

l

If - 2Rc« Cl' c2

»c2 112

= 2 II e l 112 + 2 II C2 If. • Not all norms come from an inner product. For cxamplc in JRR, R

II) x III =

L

I xii

i=1

is a norm, but there is no inner product with this as norm, since this norm fails to satisfy thc parallelogram law (see Exercise 2.IA for a discussion). 2.1.6 Proposition Let

(E,II·II) be a rwnned (or a semirwnned) space and define d(c l , c 2)

=

II e) - e211. Then (E, d) is a metric (pseudometric) space. Proof The only non-obvious verification is the triangle ineqUality. By N3 we have d(el' c 3 )

= II e l - e 3 II = lI(e l - e2) + (e2 = d(el' c 2) + d(e2, e3 ) . •

e3 ) II S II e) - e211 + II e 2 - e3 II

Thus we havc the following hicrarchy of gcncrality:

inner product spaces

c

normcd spaces

C

mctric spaces

C

topological spaccs.

Since inner product and normcd spaccs arc mctric spaces, we can use thc conccpts from Chaptcr 1. In a normcd spacc, N1 and N2 imply that the maps (el' e2) 1-+ c 1 + c2' (a, c) 1-+ ac of Ex E

~

E, and ex E

o fixed, thc mappings

e

~

1-+

E, respectively, are continuous. Hence for eo E E, a o E C, a o ~ eo + c, e 1-+ aoc are homeomorphisms. Thus U is a neighborhood

of thc origin iff c + U = {c + x I x E U} is a ncighborhood of c E E. In othcr words, all thc neighborhoods of e E E are sets that contain translates of disks ccntcred at the origin. This constitutes a complete description of the topology of a normed vector space (E, II . II).

Section 2.1 Banach Spaces

4S

Finally, note that the inequality III e) 11- II e2111 $ II e l - e2 II implies that the nonn is unifonnly continuous on E. In inner product spaces, the Cauchy-Schwartz inequality implies the continuity of the inner product as a function of two variables.

2.1.7 Definition Let (E, II . II) be a rwrmed space. If the corresponding metric d is complete, we say (E, II· II) is a Banach space. If (E, (.,.» is an inner product space whose corresponding metric is complete, we say (E, CO» is a Hilbert space. For example, it is proven in books on advanced calculus that JR." is complete. Thus, JR." with the standard nonn is a Banach space and with the standard inner product is a Hilbert space. Not only is the standard nonn on JR." complete, but so is the nonstandard one

"

IIl x lll=L IXil. i=l

However, it is equivalent to the standard one in the following sense.

2.1.8 Definition Two rwrms on a vector space E are equivalent if they induce the same topology on E.

2.1.9 Proposition Two rwrms II· II and III . ilion E are equivalent iff there is a constant M such that M-) III e III $ II e II $ Mill e III for all e E E.

Proof Let B:(x) = {y E E III y - x II $ r},

B~(x) = {y E E 1111 y - x III $ r}

denote the two closed disks of radius r centered at x E E in the two metrics defined by the nonns II· II and III . III, respectively. Since neighborhoods of an arbitrary point are translates of neighborhoods of the origin, the two topologies are the same iff for every R > 0, there are constants Ml' M2 > 0 such that

B~ 1(0) c Bk(O) c B~ 2(0). The first inclusion says that if III x III Thus, if e ~ 0, then

$

M I' then II x II

$

R, i.e., if III x III

$

1, then

II x II $ RlM l .

that is, II e II $ (R/M)1I1 e III for all e E E. Similarly, the second inclusion is equivalent to the assertion that (MzIR) III e III $ II e II for all e E E. Thus the two topologies are the same if there exist

46

Chapter 2 Banach Spaces and DiJJerentUd Calculus

constants N 1 > 0, N2 > 0 such that NI

III e III

II e II

~

~

N2111 e III

for all e E E. Taking M = max(N2, lIN I) gives the statement of the proposition. • If E and F are nonned vector spaces, the map II· II : E x F ~ lR defined by II (e, e') II = II e II + II e' II is a nonn on Ex F inducing the product topology. Equivalent nonns on Ex Fare (e, e') ...... max(1I e II, II e' II) and (e, e') ...... (II e 112 + II e' 112)1/2. The nonned vector space Ex F is

usually denoted by E ED F and called the direct sum of E and F. Note that E ED F is a Banach space iff both E and F are. These statements are readily checked.

2.1.10 Proposition. Let E be afinite-dimensional real or complex vector space. Then (i) there is a nonn on E; (ii) all nonns on E are equivalent; (iii) all nonns on E are complete. Proof Let e l ,

... ,

en denote a basis of E, where n is the dimension of E.

(i) A nonn on E is given, for example, by n

lie

II = Llail, i=1

where n

e =

L aiei i=1

(ii) Let

II· II' be any other nonn on E. If n

e=

L aiei

n

and f=

i=1

L biei' i=1

the inequality n

III e II' -II f II' I ~ II e - f II' ~

L I ai - bi III ei II' i=

shows that the map

1

Section 2.1 Banach Spaces

47

is continuous with respect to the III· III -nonn on en. Since the set S = {x E en 1111 x 111= 1} is closed and bounded, it is compact. The restriction of this map to S is a continuous, strictly positive function, so it attains its minimum MI and maximum M2 on S; that is,

for all (x I, ... , xn)

E

= 1.

en such that III (x I, ... , xn) III

Thus

i.e., MIll e II' $; II ell'$; M2 II e II, where n

e =

L

xiei

i=1

Taking M =max(M 2, 1IM I), Proposition 2.1.9 shows that II· II and II· II' are equivalent. (iii) It is enough to observe that n

(xl, ... , xn)

E

en

1-+

L

xic;

E

E

i=1

is a nonn-preserving map (Le., an isometry) between (en, III . III) and (E, II . II). • The unit spheres for the three common nonns on ]R2 are shown in Figure 2.1.1.

y

y

y

x

x

max(x. yl

\.'C

+ r

Figure 2.1.1

Ixl + l,vl

48

Chapter 2 Banach Spaces and Differential Calculus

The foregoing proof shows that compactness of the unit sphere in a finite-dimensional space is crucial. This fact is exploited in the foIIowing supplement.

~

SUPPLEMENT 2.IA

A Characterization of Finite-Dimensional Spaces 2.1.11 Proposition A normed vector space is finite dimensional iff it is locally compact. (Local

compactness is equivalent to the fact that the closed unit disk is compact.) Proof If E is finite dimensional. the previous proof of (iii) shows that E is 10caIIy compact.

Conversely. assume the closed unit disk B1(0)

C

E is compact and let {D 1/2(x i ) I i = 1•...• n}

be a finite cover with open disks of radii 1/2. Let F = span {xl' ...• xn }. Then F is finite

dimensional. hence homeomorphic to Ck (or

]Rk)

for some k ~ n. and thus complete. Being a

complete subspace of the metric space (E. II . II). it is closed. We shaII prove F

=E.

Suppose the contrary. that is. there exists vEE. v II' F. Since F = cl(F). the number d = inf{1I v - e II leE F} is strictly positive. Let r > 0 be such that B/v) set Br(v)

nF

n

F

* 0. Note that the

is closed and bounded in the finite-dimensional space F. and thus is compact.

Since inf{ II v - e III e E F} = inf{1I v - e III e E B/v) C F} and the continuous function defined by e E Br(v) n F ~ II v - e II E 1o. 00 [ attains its minimum. there is a point eo E Br (v) n F such that d =II v - eo II. But then there is a point

Xi

such that

so that

Since eo + II v - eo II

Xi

E F. we get II v - eo - (II v - eo lI)x i II

~

d. which is a contradiction .

• {l::rJ 2.1.12 Examples

sUPx E

A Let X be a set and F a normed vector space. Define the set B(X. F) = (f: X ~ F I X II f(x) II < 00 } . Then B(X. F) is easily seen to be a normed vector space with respect to

the so-caIled sup-norm. II f II~ = sUPx E X II f(x) II. We prove that if F is complete. then B(X. F) is a Banach space. Let {fn} be a Cauchy sequence in B(X. F). i.e.. II fn - fm II~ < E for n. m ~ N(E). Since for each x E X. II f(x) II ~ II f II~. it foIlows that {fn (x)} is a Cauchy sequence in F. whose limit we denote by f(x). In the inequality II fn(x) - fm(x) 11< E for all n. m ~ N(E). let m ~ 00 and get II fn (x) - f(x) II ~ E for n ~ N(E). i.e.• IIfn - f II~ ~ E for n ~ N(E). This shows that fn - f E B(X. F). i.e.• that f E B(X. F). and that II fn - f II~ ~ 0 as n ~ 00. As a particular case. we get the Banach space cb consisting of all bounded real sequences {an} with the

49

Section 2.1 Banach Spaces

norm II {an} II"" = sUPnl an I. B If X is a topological space, the space CB(X, F) = {f: X ~ F I f is continuous, f E B(X, F)} is closed in B(X, F). Thus if F is Banach, so is CB(X, F). In particular, C(X, F) = {f: X ~ F

I f continuous}, for

X compact and F Banach, is a Banach space. For example,

the vector space C([O, I], JR) = {f: [0, 1] ~ JR I f is continuous} is a Banach space with the norm IIfll",,=sup{lf(x)llxE [O,I]}. C (For readers with some knowledge of measure theory.) Consider the space of real valued square integrable functions defined on an interval [a, b]

C

JR, i.e., functions f that satisfy

The function II . II : f......

(

f. I

)1/2

b

f(x) 12 dx

is, strictly speaking, not a norm on this space; for example, if

f(x) = {

0 for x;to a , 1 for x = a ,

then II f II = 0, but f;to O. However, II· II does become a norm if we identify functions which differ only on a set of measure zero in [a, b], i.e., which are equal almost everywhere. The resulting vector space of equivalence classes [f] will be denoted L2 [a, b]. With the norm of the equivalence class [f] defined as b

II [f] II = ( 11 f(x) 12 dx

)1/2

,

L2 [a, b] is an (infinite dimensional) Banach space. The only nontrivial part of this assertion is the completeness; this is proved in books on measure theory, such as Royden [1968]. customary, [f] is denoted simply by f. In fact, L2 [a, b] is a Hilbert space with

( f, g) =

f

As is

f(x) g(x) dx .

If we use square integrable complex-valued functions we get a complex Hilbert space L2([a, bl, C) with (C, g) =

f

f(x) g(x) dx .

D The space LP([a, b)) may be defined for each real number p ~ 1 in an analogous

so

Chapter 2 Banach Spaces and Differential Calculus

fashion to L2 [a, b]. Functions f: [a, b] ~ JR. satisfying

are considered equivalent if they agree almost everywhere. LP([a, b]) is then defined to be the vector spac~ of equivalence classes of functions equal almost everywhere.

II' lip: LP [a, b]

~ JR. given by If] ~ (

f.

b

I f(x)I P dx

)I/P

defines a norm, called the LP -norm, which makes LP [a, b] into an (infmite dimensional) Banach space. E

Denote by C([a, b],JR.) the set of continuous real valued functions on [a, b]. With the

O-norm, C([a, bJ,JR.) is not a Banach space. For example, the sequence of continuous functions fn shown in Figure 2.1.2 is a Cauchy sequence in the L I-norm but does not have a continuous limit function. On the other hand, with the sup norm

II f II

sup

= x

E

I f(x) I ,

[0.11

C([O, 1]) is complete, i.e., it is a Banach space, as in Example B.

•

y

_ - - - - - - - - y = fn(x)

--4-----L---~------------------- X

Figure 2.1.2. fn converges in L I, but not in C. As in the case of vector spaces, quotient spaces of normed vector spaces playa fundamental role. 2.1.13 Proposition Let E be a normed vector space, F a closed subspace, ElF the quotient

vector space, and It: E ~ ElF the canonical projection defined by (i) The mapping II· II : ElF ~ JR.,

It

(e) = leJ = e + F

E

ElF.

51

Section 2.1 Banach Spaces

II [elll = inf{1I e + v III v E F} defines a norm on ElF. (ii) 1t is continuous and the topology on ElF defined by the norm coincides with the quotient topology. In particular, 1t is open. (iii) If E is a Banach space, so is ElF. Proof (i) Clearly II [elll ~ 0 for all [el E ElF and II [0111 = inf{ II v III v E F} = O. If II [elll = 0, then there is a sequence {V n } C F such that limn --> ~ II e + vn II = O. Thus limn --> ~ vn = - e and since F is closed, e E F; i.e., [el = O. Thus N1 is verified and the necessity of having F closed becomes apparent. N2 and N3 are straightforward verifications. (ii) Since

II [elll :511 e II, it is obvious that limn --> ~ en = e implies limn --> ~ 1t (en) =

limn --> ~[enl = [el and hence 1t is continuous. Translation by a fixed vector is a homeomorphism. Thus to show that the topology of ElF is the quotient topology, it suffices to show that if [01 E U and 1t -1(U) is a neighborhood of zero in E, then U is a neighborhood of [01 in ElF. Since 1t""1(U) is a neighborhood of zero in E, there exists a disk D/O)

C

1t""1(U).

But then 1t(Dr(O» C U and 1t (D/O» = {[ell e E Dr(O)} = {[ell II [elll < r }, so that U is a neighborhood of [01 in ElF. (iii) Let {[enl} be a Cauchy sequence in ElF. We may assume without loss of generality

that II [enl- [en+1lll :51/2n. Inductively, we find points e'n E [enl such that II e'n -e'n+l ll < 10.0. Thus {e'n} is Cauchy in E so it converges to, say, e E E. Continuity of 1t implies that limn-->~ [en 1 = [el . •

The codimension of F in E is defined to be the dimension of ElF. We say F is of jiniU codimension if ElF is finite dimensional. 2.1.14 Definition The closed subspace F of the Banach space E is said to be split, or complenrented, if there is a closed subspace GeE such that E = F E!:) G.

~

SUPPLEMENT 2.1B Split Subspaces

Definition 2.1.14 implicitly asks that the topology of E coincide with the product topology of FEEl G. We shall show in Supplement 2.2C that this topological condition can be dropped; i.e., F is split iff E is the algebraic direct sum of F and the closed subspace G. We note that if E = F EEl G then G is isomorphic to ElF. However, F need not split for

ElF

to be a Banach space, as we proved in 2.1.13. In finite dimensional spaces, any subspace is

closed and splits; however, in infinite dimensions this is false. For example, let E = LP(SI) and let F={fE Elf(n)=O forn
Chapter 2 Banach Spaces and DifferenJial Calculus

52

1 f(n) = -2 7t

r"

L"

f(9)e - ine de

is the nth Fourier coefficient of f. Then F is closed in E, splits in E for 1 < P < 00 by a theorem ofM. Riesz (Rudin [1966], Theorem 17.26) but does not split in E for p = 1 (Rudin [1973], Example 5.19). The same result holds if E = CO(SI, C) and F has the same definition. Another example worth mentioning is E and F

= co' the subspace of

= ~~, the Banach space of all bounded sequences,

~~ consisting of all sequences convergent to zero. The subspace F

= Co is closed in E = ~~. but does not split. However, Co splits in any separable Banach space which contains it isomorphic ally as a closed subspace by a theorem of Sobczyk; see Veech [1971]. If every subspace of a Banach space is complemented, the space must be isomorphic to a Hilbert space by a result of J. Lindenstrauss and L. Tzafriri [1971]. Supplement 2.2B gives some general criteria useful in nonlinear analysis for a subspace to be split. But the simplest situation occurs in Hilbert spaces. 2.1.15 Proposition

If E is a Hilbert space and

F a closed subspace, then E = F 9 Fl.. Thus

every closed subspace of a Hilbert space splits. The proof of this theorem is done in three steps, the first two being important resul ts in their own rights. 2.1.16 Existence and Uniqueness of Minimal Norm Elements in Closed Convex Sets

If C is a closed convex set (i.e., x, y E C and 0 ~ t ~ 1 implies tx + (1 there exists a unique eo E C such that II eo II =inf{1I e III e E C}. Proof Let

Vd =inf{1I e III e E

II (en + em )/2112 ~ d.

E

C) in E, then

C}. Then there exists a sequence {en} satisfying the inequality

d ~ II en If < d + l/n; hence II en 112 -t d. Since (en + em )/2 that

- t)y

E

C, C being convex, it follows

By the parallelogram law,

< ~ + ..!.. + ~ + _1__ d = 22n22m

1..2(n 1.. +m 2-).'

that is, {en} is a Cauchy sequence in E. Let limn ..... ~ en = eo' Continuity of the norm implies that

Vd = limn ..... ~ II en II = II eo II,

and so the existence of an element of minimum nonn in C is

proved. Finally, if fa is such that

II eo II = II fa II = Vd. the parallelogram law implies

Section 2.1 Banach Spaces

i.e., eo = fo·

53

•

2.1.17 Lemma Let FeE, F"* E be a closed subspace of E. Then there exists a nonzero element eo E E such that eo.l F. Proof

Let e E E, ell' P. The set e - P = {e - v 1 v E F} is convex and closed, so by the

previous lemma it contains a unique element eo = e - Vo E e - P of minimum norm. Since F is closed and ell' F, it follows that eo "* O. We shall prove that eo.l P. Since eo is of minimal norm in e - P, for any v E P and A E C (resp., JR), we have

II eo II = lIe-voll

~ lIe-vo +Avll = lIeo+Avll,

i.e., 2Re(A( v,e o » + I AI2 II v 112 ~ o. If A = a (eo' v), a E JR, a"* 0, this becomes a 1 (v, eo ) 12(2 + a II v 112) ~ 0 for all v E P, and a E JR, a"* O. This forces (v, eo) = 0 for all v E P, since if - 2/ II v 112 < a < 0, the preceding expression is negative. • Proof of Proposition 2.1.15 It is easy to see that F.L is closed (Exercise 2.1C). We now show that FEB p.L is a closed subspace. If {en + e'n} c FEB p.L, {en}

C

F, {e'n}

C

p.L, the relation

II (en + e' n) - (em + e'm)1I2 = II en - e'nll2 + II em - e'mll2 shows that {en + e'n} is Cauchy iff both {en} c: P and {e'n} c: F.L are Cauchy. Thus if {en + e'n} converges, then there exist eE P, e'E F.l such that Iimn---+~en=e, Iimn---+~e'n=e'. Thus limn ---+ ~ (en + c'n) = e + e' E FEB p.L. If FEB F.L "* E, then by the previous lemma there exists eo E E, eo II' P EB p.l, e o "* 0, eo.l (F EB p.L). Hencc eo E p.L and eo E P so that contradiction. • IbJ

(eo' co) = II eo 112 = 0; i.e., eo = 0, a

Exercises 2.1A Show that a normed space is an inner product space iff the norm satisfies the parallclogram law. (Hint: Use the polarization identities over JR and C to gucss the corresponding inner-products.) Concludc that if n ~ 2, III x III = L I xi I on JRn does not arise from an inner product. 2.18

Let Co be thc space of real sequenccs {an} such that an ~ 0 as n ~ 00. Show that Co is a closed subspace of the space c b of bounded sequcnces (see Example 2.1.12A) and conclude that Co is a Banach space.

54 2.1C

Chapter 2 Banach Spaces and Differential Calculus

Let EI be the set of all C l functions f: [0, 1] ~ JR with the norm

II f II

=

sup XE

(i)

(ii)

[0, II

Prove that EI is a Banach space. inclusion map EI

2.1E

XE

Let Eo be the space of CO maps f: [0, 1] ~ JR, as in 2.1.12. Show that the

Ea. 2.1D

I f(x) I + sup I f'(x) I

[0, II

~

Eo is compact; i.e., the unit ball in E J has compact closure

(Hint: Use the Arzela-Ascoli Theorem.)

Let (E, <. , .» be an inner product space and A, B subsets of E. Define the sum of A and B by A + B = {a + b I a E A, b E B}. Show that: (i) A C B implies B.l C A.l; (ii)

A.l is a closed subspace of E;

(iii) (iv)

A.l = (c1(span(A»~, (A.l).l = c1(span(A»; (A + B).l = A.l n B.l;

(v)

(c1(span(A»

A sequence {en}

n c1(span(B»).l =A.l + B.l (not C

necessarily a direct sum).

E, where E is an inner product space, is said to be weakly convergent

to e E E iff all the numerical sequences
and put

Show that: (i) (ii) (iii)

in any inner product space, convergence implies weak convergence; F(C) is an inner product space; the sequence (1,0, 0, ... ), (0, 1, 0, ... ), (0, 0, 1, ... ), ... is not convergent but is

weakly convergent to

°

in ~2(C).

Note. ~2(C) is in fact complete, so it is a Hilbert space. The ambitious reader can attempt

a direct proof or consult a book on real analysis such as Royden [1968]. 2.1F

Show that a normed vector space is a Banach space iff every absolutely convergent series is convergent. (A series L;;'=IX n is called absolutely convergent if L;;'=llIxnll converges.)

2.1G Let E be a Banach space and FI C F2 C E be closed subspaces such that F2 splits in E. Show that F J splits in E iff F J splits in F2. 2.1H Let F be closed in E of finite codimension. Show that if G is a subspace of E

ss

Section 2.1 Banach Spaces

containing F, then G is closed.

2.11

Let E be a Hilbert space. A set {e i }i E I is called orthonormal if (e i, ej

)

=

<\' the

Kronecker delta. An orthonormal set {eJ i EI is a Hilbert basis if cl(span {ei}i E I) = E. (i)

Let (ei}i E I be an orthonormal set and (ei(I)' ... , ei(n)} be any finite subset. Show that n

I. I ( e, %) )

12:::; II e 112

j= 1

for any e E E. (Hint: e' = e - L j = 1..... n ( e, eiU ) )e iU ») is orthogonal to all (eiUJ Ij = 1, ... , n}.)

(iii)

Deduce from (i) that for any positive integer n, the set {i E I I I ( e, e i ) I > l/n} has at most n II e 112 elements. Hence at most countably many i E I satisfy (e, ei ) 0, for any e E E. Show that any Hilbert space has a Hilbert basis. (Hint: Use Zorn's Lemma and

(iv)

2.1.17.) If {ei}iE

(ii)

*

I

is a Hilbert basis in E, e

such that (e, eiU ) )

* 0,

E

E, and (eiU )} is the (at most countable) set

show that

j= 1

(Hint: If e' = L j = 1..... ~ ( e, e iU ) )e iU »' show that (e i, e - e') =

°

for all i

E

I

and then use maximality of (eih E I') (v)

Show that E is separable iff any Hilbert basis is at most countable. (Hint: For the "if' part, show that the set

is dense in E. For the "only if' part, show that since

IN 2 (vi)

II e i - ej 112

= 2, the disks of radius

centered at ei are all disjoint.)

If E is a separable Hilbert space, it is algebraically isomorphic either with C n or F(C) (JR.n or ~2(JR.)), and the algebraic isomorphism can be chosen to be norm

preserving.

~2.2

Linear and Multilinear Mappings

This section deals with various aspects of linear and multilinear maps between Banach spaces. We begin with a study of continuity and go on to study spaces of continuous linear and multilinear maps and some related fundamental theorems of linear analysis. 2.1.2 Proposition Let A: E

~

F be a linear map of normed spaces. Then A is continuous if

and only if there is a constann M > 0 such that

II Ae IIF

:5;

Mile liE for all e E E .

Proof Continuity of A at eo E E means that for any r) O. there exists p > 0 such that

(0li. denotes the zero element in E and Bs(O)E denotes the closed disk of radius s centered at the origin in E). Since A is linear. this is equivalent to: if II e II :5; p. then II Ae II :5; r. If M = rip. continuity of A is thus equivalent to the following: II e liE :5; 1 implies II Ae IIF :5; M. which in turn is the same as: there exists M> 0 such that II Ae IIF :5; Mil e liE' which is seen by choosing the vector elll e liE in the preceding implication. • Because of this proposition one says that a continuous linear map is bounded. 2.2.2 Proposition If E is finite dimensional and A: E ~ F is linear, then A is continuous. Proof Let (e t •...• en} be a basis for E. Letting M t = aIel + ... + ane n• we see that

IlAeIl

=max( II Ae t II •...• II Aen II)

and setting e

II alAe l + ... + anAe n II :5;

I a l III Ae l II + ... + I an III Ae n II

:5;

M I( I at 1+ ... + I an I) .

Since E is finite dimensional, all norms on it are equivalent. Since III e III =L I ai I is a norm, it follows that III e III :5; Cil e II for a constant C. Let M = MI C and use 2.2.1. • 2.2.3 Definition If E and Fare normed spaces and A: E ~ F is a continuous linear map,let the operator norm of A be defined by

57

Section 2.2 Linear and Multilinear Mappings

II A II = su

- e ~ -IIAell1 lIell

E

E, e ;t 0}

(which is finite by 2.2.1). Let L(E, F) denote the space of all continuous linear mLlpS of E to F. If F = C (resp. lR), then L(E, C) (resp. L(E, JR.» is denoted by E* and is called the complex (resp. real) dual space of E. (It will always be clear from the context whether L(E, F) or E* means the real or complex linear maps or dual space; in most of the work later in this book it will mean the real case. A straightforward verification gives the following equivalent definitions of II A II : II A II = inf{M > 0 I II Ae II :5 Mile II for all e = sup{IIAelllllell:5l} = sup{1I Ae II

E

E}

I lIell=l}

1111 e II· L(E, F) and B E L(F, G), where E, F, and G are nonned spaces, then

In particular, II Ae II :5 II A

If A

E

II (B 0 A)(e) II = II B(A(e» II :5 II B 1111 Ae II :5 II B 1111 A 1111 ell, and so

II (B 0

A)

II :5 II B 1111 A II .

Equality does not hold in general. A simple example is obtained by choosing E = F = G = 1R2, A(x, y) = (x, 0), and B(x, y) = (0, y), so that BoA = 0 and II A

II = II B II =

1.

2.2.4 Proposition L(E, F) with the norm just defined is a normed space. It is a Banach space F is. Proof Clearly

if

II A II ~ 0 and II 0 II = O. If II A II = 0, then for any e E E, II Ae II :5 II A 1111 e II = 0,

so that A = 0 and thus N1 (see Definition 2.1.1) is verified. N2 and N3 are also straightforward to check. Now let F be a Banach space and {An} inequality II Ane - Ame II :5 II An - Am 1111 e

C

L(E, F) be a Cauchy sequence. Because of the

II for each e E E, the sequence {Ane} is Cauchy in· F

and hence is convergent. Let Ae = limn ..... ~ Ane. This defines a map A: E ~ F, which is evidently linear. It remains to be shown that A is continuous and II An - A II ~ O. If e> 0 is given, there exists a natural number N(e) such that for all m, n ~ N(e) we have

II ~ - Am II < e. If II e II :5 1, this implies II Ane - Ame II < e, and now letting m ~ it follows that II Ane - Ae II :;; e for all e with II ell:;; 1. Thus An - A E L(E, F), hence A E L(E, F) and II ~ - A II :;; e for all n ~ N(e); i.e., II An - A II ~ O. • 00,

If a sequence {An} converges to A in L(E, F) in the sense that II ~ -A II ~ 0, i.e., if ~ -+ A in the nonn topology, we say ~ ~ A in norm. This phrase is necessary since other

Chapter 2 Banach Spaces and DifferenJiol Calculus

58

topologies on L(E, F) are possible. For example, we say that An -+ A strongly if for each e E E. Since II Ane - Ae II

~

~e

-+ Ae

II An - A 1111 e II, norm convergence implies strong

convergence. The converse is false as the following example shows. Let

with inner product

({~}, {bn} )

=

L ~bn n=

1

Let en = (0, ... , 0, 1,0 ... )

E

E

F = IR , and

~

= (en") E L(E, F) ,

where the 1 in en is in the nth slot. The sequence {An! is not Cauchy in the operator nonn since II An - Am II = ";2, but if e = {am}, ~(e) = (en' e) = ~ -+ 0, i.e., An -+ 0 strongly. If both E and F are finite dimensional, strong convergence implies nonn convergence. (To see this, choose a basis el' ... , en of E and note that strong convergence is equivalent to Ake j ~ Ae j as k -+ 00 for i = 1, ... , n. Hence maxjll Aej II = III A III is a norm yielding strong convergence. But all norms are equivalent in finite dimensions.)

~

SUPPLEMENT 2.2A

Dual Spaces Recall from elementary linear algebra that the dual space of a finite dimensional vector space of dimension n also has dimension n and so the space and its dual are isomorphic. For general Banach spaces this is no longer true. However, it is true for Hilbert space. 2.2.5 Riesz Representation Theorem Let E be a real (resp. complex) Hilbert space. The map e .... ( " e) is a linear (resp. antilinear) norm-preserving isomorphism of E with E*;for short, E:; E*. (A map A: E -+ F between complex vector spaces is called antilinear ifwe have the identities A(e + e') = Ae + Ae' ,and A(ae) = ( f Ae.) Proof Let fe =(', e). Then IIfell=lIell and thus feE E*. Themap A:E-+E*, Ae=fe is clearly linear (resp. anti linear), norm preserving, and thus injective. It remains to prove surjectivity. Let f E E* and kerf = {e EEl f(e) = O}. kerf is a closed subspace in E. If kerf = E, then f = 0 and f = A(O) so there is nothing to prove. If ker f *- E, then by lemma 2.1.17 there exists e*-O such that e.l ker f. Then we claim that f = A(f(e)elll e 11 2 ). Indeed, any VEE can be written as f(v) f(v) f(v) v = v - f(e) e + f(e) e and v - f(e) e E ker f. •

Section 2.2 Linear and Multilinear Mappings

59

Thus. in a real Hilbert space E ev<;ry continuous linear function 2 : E ~ IR can be written 2(e) = ( e. eo> for some eo E E and II 2 II = II eo II· In a general Banach space E we do not have such a concrete realization of E*. However. one should not always attempt to identify E and E*. even in finite dimensions. In fact. distinguishing these spaces is fundamental in tensor analysis. We have a canonical map i: E ~ E ** defined by i(e)(2) = 2(e). (pause and look again at this strange but natural formula: i(e) E E** = (E*)*. so i(e) is applied to the element 2 E E*.) It is easy to check that i is norm preserving. One calls E reflexive if i is onto. Hilbert spaces are reflexive. by 2.2.5. For example. let V = L2( IRD) with inner product ( f. g > =

in

f(x) g(x) dx •

and let a: L2(IRD) ~ JR be a continuous linear functional. Then the Riesz representation theorem guarantees that there exists a unique g E L2(IRD) such that a(O =

r g(x)f(x) dx JRn

=

(g. f)

for all f E L2(JRD). In general. if E is not a Hilbert space and we wish to represent a linear functional a in the form of a(O = ( g. f >. we must regard g(x) as an element of the dual space E*. For example. let E = CO(n. lEt). where n C JRD. Each x E n defines a linear functional Ex: CO(n.lR) ~ IR; f ..... f(x). This linear functional cannot be represented in the form Ex(f) = (g. f> and. indeed. is not continuous in the L2 norm. Nevertheless. it is customary and useful to write such linear maps as if ( • > were the L2 inner product. Thus one writes. symbolically.

E"o(f) =

10

O(x - xo)f(x) dx.

which defines dIe Dirac delta/unction at xo; i.e .• g(x) = O(x - xo).

tl:n

Next we shall discuss integration of vector valued functions. We shall require the following.

2.2.6 Linear Extension Theorem Let E. F. and G be normed vector spaces where (i) FeE;

(ii) G is a Banach space; and (iii) T

E

L(F. G).

Then the closure cl(F)

0/ F is a normed vector subspace 0/ E and T can be uniquely extended

60

Chapter 2 Banach Spaces and Differential Calculus

to a rruzp 'Ie L(cl(F). G). Moreover. we have the equality II T II = II 'I II. Proof The fact that cl(F) is a linear subspace of E is easily checked. unique by continuity. Let us prove the existence of

rz:

N~te

that if 'I exists it is

If e e cl(F). we can write c '"

limn ..... ~ cn' whcre cn e F. so that

which shows that the sequcnce {Ten} is Cauchy in thc Banach space G. Let 'Ie =limn ..... ~ Tcn . This limit is indcpendent of thc sequcncc {cn}. for if c = lim e'n' then

which provcs that limn ..... ~ (Tcn) = limn ..... ~ (Te'n)' It is simple to chcck the linearity of

rz:

Since

Te = 'Ie for e e F (because e = limn ..... ~ e). 'I is an extension of T. Finally.

shows that 'Ie L(cl(F). G) and

II 'I II ~ II T II. The inequality II T II ~ II 'I II is obvious sincc 'I

extends T. • As an application of this lemma we define a Banach space valued integral that will be of use latcr on. Fix the closed interval [a. bJ c: 1R and the Banach space E. A map f: [a. bJ called a step function if there exists a partition a = to < t1 < '-2 ... < tn

=b

~

E is

such that f is constant

on each interval [ti• ti+1 [. Using the standard notion of a refinement of a partition. it is clear that the sum of two step functions and the scalar multiples of step functions are also step functions. Thus the set 8([a. bJ. E) of step functions is a vector subspace of B([a. bJ. E). the Banach spacc of all bounded functions (see Example 2.1.12). The integral of a step function f is defined by

f L b

n

f =

(1; + 1 - t) f(t i )

.

i= 0

a

It is easily verified that this definition is independent of the partition. Also note that

II where

that is.

f II ~ f f

II f II

~

(b -

a)1I f

II~

Section 2.2 Linear and Multilinear Mappings

r:

8([a. bJ. E)

~

61

E

is continuous and linear. By the linear extension theorem. it extends to a continuous linear map

r

E L(cl(8 ([a. b], E».E).

2.2.7 Definition The linear map

r

is called the Cauchy-Bochner integral. Note that

The usual properties of the integral such as

are easily verified since they clearly hold for step functions. The space cl(8([a.bJ. E) contains enough interesting functions for our purposes. namely cl([a. bJ. E)

C

cl(8([a. bJ. E»

C

B([a. bJ. E) .

The first inclusion is proved in the following way. Since [a.bJ is compact. each f E CO([a. bJ. E) is uniformly continuous. For E> O. let 0 > 0 be given by uniform continuity of f for £/2. Then take a partition a = to < ... < tn = b such that I ti+1 - tl I < 0 and define a step function g by g I[ti• ~+I[ = f(t;). Then the E-disk De(f) in B([a. bJ. E) contains g. Finally. note that if E and F are Banach spaces. A E L(E. F). and f E cl(8([a. bJ. E». we have AofE cl(8([a. bJ. F» since IIAofn-Aofll:5 IIAllllfn-fll~ where fn are step functions in E. Moreover.

since this relation is obtained as the limit of the same (easily verified) relation for step functions. The reader versed in Riemann integration should notice that this integral for E = IR is less general than the Riemann integral; i.e.• the Riemann integral exists also for functions outside of cl(8([a. bJ. lR». For purposes of this book. however. this integral will suffice. Next we turn to multilinear mappings. If E\, .... I\: and F are linear spaces. a map

Chapter 2 Banach Spaces and Differentiol Calculus

62

A : E) x·.. x

~ -t

F

is called k-multilinear if A(e), ... , <1,;) is linear in each argument separately. Linearity in the first argument means that

We shall study multilinear mappings in detail in our study of tensors. They also come up in the study of differentiation, and we shall require a few facts about them for that purpose.

2.2.8 Definition The space of continuous k-multilinear maps of E\, ... , ~ to F is denoted L(E), ... , ~; F). If E j = E, 1 :5; i :5; k, this space is denoted Lk(E, F). As in 2.2.1, a k-multilinear map A is continuous if and only if there is an M > 0 such that II A(e\, ... , <1,;) II :5; Mil e) II ... II <1,; II for all ej E Ej' 1:5; i :5; k. We set

which makes L(E) ,... , ~; F) into a nonned space that is complete if F is. Again II A II can also be defined as IIAII = inf(M>OIIlA(e\, ... ,en)II:5;Mlle)II···lIenll}

= sup(1I A(e\, ... , en) 11111 e) 11:5; 1, ... , II en II :5; I}

= sup ( II A (e\, ... , en) 11111 e) II =... =II en II = 1} 2.2.9 Proposition There are (natural) norm-preserving isomorphisms

where (i\, ... , ik ) is a permutation of (1, ... , k).

Proof For A

E

L(E\, L(E2 ,

••• , ~;

F», define A'

E

L(E\, ... , ~; F) by

The association A ..... A' is clearly linear and II A' II = II A II. The other isomorphisms are proved similarly. •

Section 2.2 Linear and Multilinear Mappings

63

In a similar way. we can identify LOR, F) (or L(e, F) if F is complex) with F: to A E L(1R, F) we associate A(l) E F; again II A II = II A(l) II. As a special case of 2.2.9 note that L(E,

E*) == L2(E, JR.) (or

L2(E; C), if E is complex). This isomorphism will be useful when we

consider second derivatives. We shall need a few facts about the permutation group on k elements. The information we cite is obtainable from virtually any elementary algebra book. The permutation group on k elements, denoted Sk' consists of all bijections 0: (1, ... , k)

~

(I, ... , k) together with the

structure of a group under composition. Clearly, Sk has order k!. Letting (lR., x) denote JR\(O} with the multiplicative group structure, we have a homomorphism sign: Sk

~

(JR., x).

That is, for O;t E Sk' sign (0 0 't) = (sign o)(sign 't). The image of "sign" is the subgroup (- 1, I), while its kernel consists of the subgroup of even permutations. Thus, a permutation 0 is even when sign 0 = + 1 and is odd when sign 0 = - 1. A transposition is a permutation that swaps two clements of (I, ... , k), leaving the remainder fixed. An even (odd) permutation can be written as the product of an even (odd) number of transpositions. The group Sk acts on the space Lk(E; F); i.e., each 0 E Sk defines a map 0: Lk(E; F) ~ Lk(E; F) by (oA)(el' ... , ~) = A(ecr(l)' ... , ecr(k)' Note that ('to)A = 't(oA) for all 't,0 E Sk' Accordingly, A E Lk(E, F) is called symmetric (antisymmetric) if for any permutation 0 E Sk' oA = A (resp. oA = (sign o)A.)

2.2.10 Definition Let E and F be normed vector spaces. Let L\(E; F) and LkaCE; F) denote

the subspaces of symmetric and antisymmetric elements of Lk(E; F). Write SO(E, F) = F and Sk(E, F) = {p: E ~ F I pee) = A(e, ... , e) for some A E Lk(E; F)}. We call Sk(E, F) the space

of homogeneous polynomials of degree k from E to F. Note that Lk,(E; F) and L\(E; F) are closed in Lk(E; F); thus if F is a Banach space, so are L\(E; F) and L\(E; F). The anti symmetric maps Lka(E; F) will be studied in detail in Chapter 7. For technical purposes later in this chapter we will need a few facts about Sk(E, F) which are given in the following supplement.

~ SUPPLEMENT 2.2B Homogeneous Polynomials

2.2.11 Proposition (i) Sk(E, F) is a normed vector space with respect to the following norm: IIfll = inf{M>Olllf(e)II$Mlleln = sup {lIf(e)lIll1ell$ I} = sup{llf(e)lIll1ell=i}

It is Complete

if F is.

(ii) IffE Sk(E, F) and gE sn(F, G), then gofE Skn(E,G) and IIgofll$lIglillfli. (iii) (Polarization). The mapping ': Lk(E, F) ~ Sk(E, F) defined by N(e) = A(e, ... , e)

restricted to L\(E; F) has an inverse ': Sk(E, F) ~ Lk,(E, F) given by

64

Chapter 2 Banach Spaces and DilJerentiJJJ Calculus

f(t l e l + ... + tke k) is a polynomial in tp ...• tk, so there is no problem in understanding what the derivatives on the right hand side mean).

(Note that

(iv) For A E Lk(E,F). II A' II ~ II A II ~ (\(k/k!) II A' II. which implies the mflPS

•

are

continuous.

Proof (i) and (ii) are proved exactly as for L(E. F) =SI(E. F). (iii) For A E Lks(E; F) we have

where each ei appears ai times. and

I

ak -'t

-'t

°1"'Ok

t=O

a

1

a

t I ... ti' =

j

1. if k = j

0, if k

*j

It follows that

and for j

*k.

This means that '(A') = A for any A E Lks(E. F). Conversely, if f E Sk(E. F). then 1 rO'(e) = , f(e •...• e) = -k' .

atl a... atk k

I t= 0

f(tle + ... + tke)

(iv) II A'(e) II = II A(e •... , e) II :0; II A 1111 e Ilk, so II A' II inequality, note that if A E Lk.<E; F), then

:0;

II A II.

To prove the other

Section 2.2 Linear and Multilinear Mappings

6S

where the sum is taken over all the 2k possibilities el = ± 1•...• ~ = ± 1. Put II e l II = ... = 1I~1I=1 andget

II A'(ele l + ... + ~~) II

elel + ... + ~~ II

$

II A'

$

II A' II (I el III e l II + ... + I ~ III ~ Il)k

1111

k

whence

Le .•

where each e i appears ai times. If e = tIel + ... +

f(e) = '[(e •...• e) =

~en'

L

the proof of (iii) shows that

cal ... ant~l ... ~n;

8 1+···+an=k

i.e.. f is a homogeneous polynomial of degree k in t l •...• tn in the usual algebraic sense. The constant kk/k! in (iv) is the best possible. as the following example shows. Write elements of JR.k as x = (xl •...• xk) and introduce the norm III (xl •...• xk) III = I xii + ... + I xk I. Define A

E

Lks(JR.k, JR.) by

I""

A(xl.···. xk) = -kl• where xi = (Xi I, ...• Xik)

E

£..J

X; I ••• X; k , 1

k

JR.k and the sum is taken over all permutations of {I ..... k}. It is

easily verified that II A II = l/k!, and II A' II = l/kk; i.e., II A II = (kk/k!) II A'

II. Thus. except for k

== 1, the isomorphism ' is not norm preserving. (This is a source of annoyance in the theory of

formal power series and infinite-dimensional holomorphic mappings.)

$:n

~ SUPPLEMENT 2.2C The Three Pillars of Linear Analysis

The three fundamental theorems of linear analysis are the Hahn-Banach theorem, the open mapping theorem, and the uniform boundedness principle. This supplement gives the classical proofs of these three fundamental theorems and derives some corollaries that will be used later. In

66

Chapter 2 Banach Spaces and Dif{erenJiol Calculus

finite dimensions these corollaries are all "obvious." 2.2.12 Hahn-Banach Theorem Let E be a real or complex vector space, 11·11 : E

~ lR

a

seminorm, and FeE a subspace. If f E F* satisfies 1f(e) 1:511 e II for all e E F, then there exists a linear map f': E ~ JR (or

C) such that f' I F = f and 1f'(e) 1:511 e II for all e E E.

Proof Real Case. First we show that f E F* can be extended with the given property to FEEl span{ eo), for a given eo i!: F. For el' e2 E F we have

so that and hence

Let a E JR be any number between the sup and inf in the preceding expression and define f': F EEl span {eo} ~ JR by f'(e + teo) = f(e) + tao It is clear that f' is linear and that f' I F = f. To show that 1 f'(e + teo) 1 :511 e + teo II, note that by the definition of a,

so that by multiplying the second inequality by t ~ 0 and the first by t < 0, we get the desired result. Second, one verifies that the set S = {(G, g) I F C GeE, G is a subspace of E, g E G*, g IF = f, and 1g(e) 1:511 e II for all e E G} is inductively ordered with respect to the ordering

Thus by Zorn's Lemma there exists a maximal element (Fo, fo) of S. Third, using the first step and the maximality of (Fo, fo)' one concludes that Fo = E. Complex Case. Let f = Re f + i 1m f and note that complex linearity implies that (1m f)(e) = - (Re f)(ie) for all e E F. By the real case, Re f extends to a real linear continuous map (Re f) : E ~ JR, such that 1 (Re f)'(e) 1 :5 II e II for all e E E. Define f': E ~ C by f'(e) = (Re f)'(e) - i(Re f)'(ie) and note that f is complex linear and f' I F =f. To show that 1f'(e) 1:5 II e II for all e E E, write f'(e) = 1f'(e) lexp(i9), so complex linearity of f' implies f'(e· exp(- i9» E JR, and hence

1f'(e) I

= f'(e· exp(- is)) = (Re f)'(e . exp(- i9»

:5 II e . exp (- i9) II

= II ell·

•

2.2.13 Corollary Let (E, II . II) be a normed space, F c: E a subspace, and f E F* (the topological dual). Then there exists f' E E* such that f' I F = f and II f' II = II f II.

67

Section 2.2 Linear and Multilinear Mappings

Proof We can assume f"'O. Then III e III = II filII e II is a norm on E and I f(e) I $11 fll·1I e I :: III e III for all e E F. Applying the preceding theorem we get a linear map f: E ~ lR (or C) with the properties f

IF

= f and I f '(e) I $ III e III for all e

since f extends f. it follows that II f II

E. This says that II f II $ II f II. and

E

II f II; i.e.. II f' II = II f II and f

$

E*. •

E

Applying the corollary to the linear function ae 1-+ a. for e E E a fixed element. we get the following. 2.1.14 Corollary Let E be a normed vector space and e '" 0. Then there exists f E E* such

that f(e) '" O. In other words if f(e) =

°

for all f E E*. then e = 0; i.e .• E* separates points of

E. 2.2.15 Open Mapping Theorem of Banach and Schauder Let E and F be Banach

spaces and suppose A E L(E. F) is onto. Then A is an open mapping. Proof To show A is an open mapping. it suffices to prove that A(cl(DI(O») contains a disk centered at zero in F. Let r> 0. Since E = Un~IDnr(O). it follows that F = Un~I(A(Dnr(O») and hence U n~lcl(A(Dnr(O») = F. Completeness of F implies that at least one of the sets cl(A(Dnr(O») has a nonempty interior by the Baire Category Theorem 1.7.2. Because the mapping e E E 1-+ ne C

E

E is a homeomorphism. we conclude that cl(A(Dr(O))) contains some open set V

F. We shall prove that the origin of F is in int{cl[A(D/O»]} for some r> O. Continuity of

(ei' e2 )

E

E x E 1-+ e l - e2

{e l - e2 1 el' e 2

E

U}

E

E assures the existence of an open set U

D/O). Choose r> 0 such that Dr«O»

C

C

C

E such that U - U =

U. Then cl(A(D/O»)::J

cl(A(U) - A(U» ::J cl(A(U» - cl(A(U» ::J V-V. But V - V = U e EV(V - e) is open and clearly contains 0 E F. It follows that there exists a disk D1(O)

C

F such that D1(O)

C

cl(A(Dr(O»).

Now let £(n) = 1/2n+l. n = 0.1.2•...• so that 1 = l:n~ £(n). By the foregoing result for each n there exists an 11(n) > 0 such that D'1(n)(O)

C

cl(A(DE(n)(O». Clearly 11(n) ~ 0. We

shall prove that D'1(O) C A(cl(DI(O»). For v E D'1(O)(O) C cl(A(DE(O)(O») there exists eo E DE(O)(O) such that II v - Aeo II < 11(1) and thus v - Aeo E cl(A(D E(1)(O»). so there exists e l E DE(I)(O) such that II v - Aeo - Ae l II < 11(2). etc. Inductively one constructs a sequence en E D'1(n) such that II v - Aeo - ... - Aen II < 11(n+ 1). The series l:n ~ 0 en is convergent because m

I

m

L ei II i=n+1 L 1d+ i=n+1

1•

$

and E is complete. Let e = l:n ~ 0 en

E

L- 1/2

n+ 1

n=O

E. Thus

Ae=LAen=v. n=O and

= 1,

68

Chapter 2 Banach Spaces and Differentiol Calculus

II e II i.e., v E O,,(olO) implies v

-

~ ~ II en II ~ ~o 2n1+ 1 =

=Ae,

II e II ~

1; that is, 0,,(0)(0) c: A(c1(OI(O»). •

An immediate consequence is the following.

2.2.16 Banach's Isomorphism Theorem A continuous linear isomorphism of Banach spaces is a homeomorphism.

Thus, if F and G are closed subspaces of the Banach space E and E is the algebraic direct sum of F and G, then the mapping (e, e') E F x G ...... e + e' E E is a continuous isomorphism, and hence a homeomorphism; i.e., E = F e G; this proves the comment at the beginning of Supplement 2.1B. 2.2.17 Closed Graph Theorem Let E, F be Banach spaces. A linear map A: E

~

F is

continuous iffits graph r A = {(e, Ae) E Ex Fie E E} is a closed subspace of E CD F.

Proof It is readily verified that r A is a linear subspace of E E9 F. If A E L(E, F), then r A is closed (see Exercise l.4B). Conversely, if r A is closed, then it is a Banach subspace of E e F, and since the mapping (e, Ae) ErA ...... e E E is a continuous isomorphism, its inverse e E E ...... (e, Ae) ErA is also continuous by 2.2.16. Since (e, Ae) ErA ...... Ae E F is clearly continuous, so is the composition e ...... (e, Ae) ...... Ae. • The Closed Graph Theorem is often used in the following way. To show that a linear map A : E ~ F is continous for E and F Banach spaces, it suffices to show that if en ~ 0 and ACn ~ e', then e' = O. 2.2.18 Corollary Let E be a Banach space and F a closed subspace of E. Then F is split iff there exists P E L(E, E) such that PoP =P and F = {e EEl Pe =e}.

Proof If such a P exists, then clearly ker P is a closed subspace of E that is an algebraic complement of F; any e E E is of the form e =e - Pe + Pe with e - Pe E ker P and Pe E F. Conversely, if E = Fe G, define P: E ~ E by P(e) =el' where c =e 1 + e 2, e 1 E F, c2 E G. P is clearly linear, p2 = P, and F = {e EEl Pc =e}, so all there is to show is that P is continuous. Let en =e 1n + e2n ~ 0 and P(en) =e 1n ~ e'; i.e., - e 2n ~ e', and since F and G are closed this implies that e' E F G = (OJ. By the closed graph theorem, P E L(E, E). •

n

2.2.19 Fundamental Isomorphism Theorem Let A E L(E, F) be surjective where E and F are Banach spaces. Then E/ker A and F are isomorphic Banach spaces.

Section 2.2 Linear and M ultiJinear Mappings

Proof The map tel

f-+

Ae is bijective and continuous (since its norm is $

69

II A II ), so it is a

homeomorphism. • A sequence of maps

of Banach spaces is said to be split exact if for all i, ker Ai + 1 =range Ai and both ker Ai and range Ai split. With this terminology, 2.2.18 can be reformulated in the following way: If 0 ~

G

E

~

F

~

~

0 is a split exact sequence of Banach spaces, then E/G is a Banach space

isomorphic to F (thus F;:; G EB F).

2.2.20 Uniform Boundedness Principle of Banach and Steinhaus Let E and F be normed vector spaces, with E complete, and let (A)i E

(II Aie Ill i E

I

is bounded in F, then

Proof Let q>(e) = sup{11 Ai e

Sn

(II Ai II) i E

1

I C

L(E, F). Iffor each e E E the set

is a bounded set of real numbers.

III i E I} and note that

= {e EEl

q>(e) $ n}

=

niEI

{e E E

III ~e II .,; n}

= E. Since E is a complete metric space, the Baire Category Theorem 1.7.2 says that some Sn has nonempty interior; i.e., there exist r > 0 and eo E E such that q>(e)

is closed and Un" 1 Sn $

M, for all e E cl(D/e o))' where M > 0 is come constant. For each i E I, and II e II = 1, we have II Ai(re + eo) II $ q>(re + eo) $ M, so that

i.e.,

II Ai II.,; (M + q>(eo))/r for all i E I. •

2.2.21 Corollary If (An) C L(E, F) is a strongly convergent sequence (i.e., limn ..... ~ Ane = Ae exists for every e E E), then A E L(E, F). Proof A is clearly a linear map. Since {Ane} is convergent, it is a bounded set for each e E E, that by 2.2.20, (II ~ II) is bounded by, say, M > O. But then

SO

i.e., A

E

L(E, F)..

b

70

Chapter 2 Banach Spaces and DjfferentiIJI Calculus

Exercises 2.2A

If E =]Rn and F =]Rm with the standard norms. and A: E --+ F is a linear map. show that (i) II A II is the square root of the absolute value of the largest eigenvalue of AA T. where AT is the transpose of A and (ii) if n. m ~ 2. this norm does not come from an inner product. (Hint: Use Exercise 2.IA.)

2.2B Let E = lItn and F =]Rn with the standard norms and A. B E L(E. F). Let ( A. B ) = trace(ABT). Show that this is an inner product on L(E. F). 2.2C Show that the map L(E. F) x L(F. E) --+ lit;

(A. B)

1-+

trace(AB)

gives a (natural) isomorphism L(E. F)*;: L(F. E). 2.2D Let E. F. G be Banach spaces. A: DeE --+ F. and B : DeE --+ G two closed operators with the same domain D. (See Supplement 7.4A for a full account of closed operators; all that is needed here is the definition). Show there exist constants MI' M2 > 0 such that II Ae II ~ MI(IIBell+llell) and

II Bell

~

MiliAell+llelD

forall eE E. (Hint: Norm EeG by II (e.g) 11= lie 1I+lIgll and define T:rB--+G by T(e. Be) =Ae. Use the Closed Graph Theorem to show that T E L(rB' G).) 2.2E

Linear transversality Let E. F be Banach spaces. Fo C F a closed subspace. and T E L(E. F). T is said to be transversal to Fo. if il(Fo) splits in E and T(E) + Fo = {Te + fie E E. f E F} = F. Prove the following. (i) T is transversal to Fo iff 1t 0 T E L(E. FIFo) is surjective with split kernel; here 1t: F --+ FIFo is the projection. (ii) If 1t 0 T E L(E. FIFo) is surjective and Fo has finite codimension. then ker(1t 0 T) has the same codimension and T is transversal to Fo. (Hint: Use the algebraic isomorphism T(E)/(Fo T(E» ;: (T(E) + Fo)lFo to show Elker(1t 0 T);: FIFo; now use 2.2.18.) (iii) If 1t 0 T E L(E, FIFo) is surjective and if ker T and Fo are finite dimensional. then ker(1t 0 T) is finite dimensional and T is transversal to Fo. (Hint: Use the exact sequence 0 --+ ker T --+ ker(1t 0 T) --+ Fo T(E) --+ 0.)

n

n

2.2F

Let E and F be Banach spaces. Prove the following. (i) If f E cl(8([a. bl. L(E. F») and e E E. then

Section 2.2 Linear and MulJiJinear Mappings

ff(t)edt

71

=( ff(t) dt J(e).

(Hint: T 1-+ Te is in L(L(E. F), F).) (ii)

If f E cl(8([a, b], JR) and v E F, then

(Hint: t

(iii)

1-+

multiplication by t in F is in L(JR, L(F, F»); apply (i).)

Let X be a topological space and f: [a, b] x X --t E be continuous. Then the mapping

g : X --t E, g(x) =

f

f(t, x)dt

is continuous. (Hint: for t E JR, x' E X and E> 0 given, II f(s, x) - f(t, x') II < E if (s, x) E U I X U X ',!; use compactness of [a, b] to find Ux' as a finite intersection and such that II f(t, x) - f(t, x') II < E for all t E [a. b]. x E U x") 2.2G Show that the Banach Isomorphism Theorem is false for normed incomplete vector spaces

in the following way. Let E be the space of all polynomials over JR normed as follows:

II (i) (ii)

(iii)

aa + a l x + ... + a"XR II

= max {I

aa I•...• 1a" I}.

Show that E is not complete. Define A: E --t E by

and show that A E L(E, E). Prove that A-I : E --t E exists. Show that A-I is not continuous.

2.2H Let E and F be Banach spaces and A E L(E. F). If A(E) has finite codimension. show

that it is closed. (Hint: If Fo is an algebraic complement to A(E) in F, show there is a continuous linear isomorphism E/ker A ;: FIFo; compose its inverse with E/ker A --t A(E).)

2.21

Symmetrization operator Define Symk : Lk(E. F)

--t

Lk(E. F), by

Chapter 2 Banach Spaces and DjfferenJiol Calculus

72

SymkA = k!

LoA,

ae!\

where (oA)(el' ... , ~) = A(ecr(l)' ... , ecr(k». Show that: (i) Symk(Lk(E, F» = Lk.(E, F). (ii) (Symk)2 = Symk.

2.2J

II Symk II ~ 1.

(iii) (iv)

If F is Banach, then Lks(E, F) splits in Lk(E, F). (Hint: Use 2.2.18.)

(v)

(SymkA)' = A'.

Show that a k-multilinear map continuous in each argument separately is continuous. (Hint: for k = 2: If II e l II ~ 1, then II A(el' e 2) II ~ II A(·, e 2) II, which by the uniform boundedness principle implies that II A(el' . ) II

2.2K

~

M for II e l II

~

1.)

(i) Prove the following theorem of Mazur and Ularn [1932] following the steps below (see Banach [1955], p. 166): Every isometric surjective mapping
that
Fix xl' x2 E E and define HI = {X III x - Xl II = II x - x211 = i II Xl - x211 } , Hn

={ XE ~-l IIIX- z

Show that

diarn(~) ~

Il15 idiarn(Hn-l), ZE

_1-1 diarn(H l ) 2n-

~-l}.

~ ~l II Xl n 2

x211.

-

'*

Conclude that if nn ~ lHn 0, then it consists of one point only. (b) Show by induction that if x E Hn, then Xl + x2 - X E ~. (c)

Show that (Xl + x 2)!2 = nn ~ lHn. (Hint: Show inductively that (Xl + x2)J2 E Hn using (b).)

(d) From (c) deduce that

(ii)

Use
Section 2.2 Linear and Multilinear Mappings

73

inequality only for colinear points. Show that if F is strictly convex. then

and

Show that

(b)

Show that. in general. the assumption on

(a. b • ..; ab). a. b > 0


(- a. b. - ..; ab). a. b > 0


(a. - b. - ..; ab). a. b > 0 ;


(- a. - b. - ..; ab). a. b > 0

Show that

= (0. O. 0). (Hint:

r

82 I).) 2.2L Let E be a complex n dimensional vector space. (i) Show that the set of all operators A E L(E. E) which have n distinct eigenvalues is open and dense in E. (Hint: Let p be the characteristic polynomial of A. i.e .• p(A) = det(A - AI). and let ~I' .... ~n _ 1 be the roots of p'. Then A has multiple eigenvalues iff P(~I) ... P(~n -I) = O. The last expression is a symmetric polynomial in ~I' .... ~n _ I' and so is a polynomial in the coefficients of p' and therefore is a polynomial q in the entries of the matrix of A in a basis. Show that q-I(O) is the set of complex n x n matrices which have multiple eigenvalues; q-I(O) has open dense complement by Exercise 1.1.L.) (ii) Prove the Cayley-Hamilton Theorem: If p is the characteristic polynomial of A E L(E. E). then p(A) = O. (Hint: If the eigenvalues of A are distinct. show that thc matrix of A in the basis of eigenvectors e p .... en is diagonal. Apply A. A2 ..... An -I. Then show that for any polynomial q the matrix of q(A) in the same basis is diagonal with entries q(A). where Ai are the eigenvalues of A. Finally. let q =

74

Chapter 2 Banach Spaces and Differentiol Calculus p. If A is general, apply (i).)

2.2M Let E be a normed real (resp. complex) vector space. (i) Show that A: E -t IR ( resp. C) is continuous if and only if ker A is closed. (Hint: Let e E E satisfy A(e) = 1 and choose a disk D of radius r centered at e such that D n (e + ker A) = 0. Then A,(x)"," 1 for all xED. Show that if xED then A(X) < 1. If not, let a = A(X), I a I > 1. Then II x/a II < r and A(x/a) = 1.) (ii)

Show that if F is a closed subspace of E and G is a finite dimensional subspace, then G + F is closed. (Hint: Assume G is one dimensional and generated by g. Write any x E G + F as x = A(X)g + f and use (i) to show A is continuous on G + F.)

2.2N Let F be a Banach space. (i) Show that if E is a finite dimensional subspace of F, then E is split. (Hint: Define

(ii)

(iii) (iv)

P : F -t F by P(x) = ~i = I, ... , n ei(x)e i, where {e l , ... , en} is a basis of E and {e l , ... , en} is a dual basis, i.e., ei (ej ) = Ow Then use Corollary 2.2.18.) Show that if E is closed and finite codimensionaI, then it is split. Show that if E is closed and contains a finite codimensional subspace G of F, then it is split. Let A: F -t IR be a linear discontinuous map and let E = ker A. Show that the codimension of E is 1 and that E is not closed. Thus finite codimensional subspaces of F are not necessarily closed. Compare this with (i) and (ii) and with 2.2H.

2.20 Let E and F be Banach spaces and T E L(E, F). Define T*: F* -t E* by (T*~, e) = ( ~, Te) for e E E, ~ E F*. Show that: (i) T* E L(F*, E*) and T** I E = T. (ii) (iii)

(iv)

ker T* = T(E)O = {~ E F* I ( ~, T e) = 0 for all e E E) and ker T = (T*(F*»O = feE EI(T*~,e)=O forall ~E F*}. If T(E) is closed, then T*(F*) = (ker T)o. (Hint: The induced map E/ker T ~ T(E) is a Banach space isomorphism; let S be its inverse. If A E (ker T)O, define the clement ~ E (E/ker T)* by ~([e]) = A(e). Let v E F* denote the extension of S*(~) E (T(E»* to v E F* with the same norm and show that T*(v) = A.) If T(E) is closed, then ker T * is isomorphic to (F/T(E»* and (ker T)* is isomorphic to E*/ T*(F*).

~2.3

For a differentiable function f: U point

Uo E

The Derivative C

JR

~

JR, the usual interpretation of the derivative at a

U is the slope of the line tangent to the graph of f at Uo. To generalize this, we

interpret Df(uo) = f'(u o) as a linear map acting on the vector (u - u o). Then we can say that Df(Uo) is the unique linear map from JR into JR such that the mapping g :U is tangent to f at

U

~

JR

given by

u

1-+

g(u) = f(u o) + Df(uo) . (u - u o)

o (see Figure 2.3.1). This version can be extended to vector valued mappings

of a vector variable as follows.

u R Uo

Figure 2.3.1 2.3.1 Definition Let E, F be rwrmed vectorspaces, with maps f, g : U open in E. We say f and g are tangent at the point Uo E U

C

E ~ F where U is

if

lim IIf(u) - g(u)1I = 0 , u--+uo lIu - Uoll where

\1.\1 represents the rwrm on the appropriate space.

2.3.2 Proposition For f: U C E ~ F and U o E U there is at most one L E L(E, F) such that the map gL: U C E ~ F given by gL(u) =f (uo) + L(u - u o) is tangent to f at uo.

76

Chapter 2 Banach Spaces and DifferenJiol Calculus

Proof Let LI and L2 E L(E, F) satisfy the conditions of the proposition. If e and u

=U o + A.e

for A. E lR (or C), then for A. small and u

II Lie - L 2e II = II LI(u - Uo) - L 2(u - Uo) II/II u Ilf(u) - f(Uo) - LI(u - uo)1I $

E

= 1; therefore,

IIf(u) - f(Uo) -

~(u

=

°

and thus LI

= L 2·

2.3.3 Definition If, in 2.3.2, there is such an L

and define the derivative of f at

Uo

E

E satisfying

- uo)1I

+ ------~--~----lIu - Uoll

lIu - Uoll

II LI - L211

E, II e II = I,

U o II

As A. ~ 0, the right hand side approaches zero so that II (L I - L 2)e II = II ell

E

U, we have

E

°

for all e

•

L(E, F). we say f is differentiable at uO'

to be Df(Uo) = L. The evaluation of Df(u o) on e E E

will be denoted Df(uo) . e. If f is differentiable at each Df: U

~

U, the map

Uo E

L(E, F); u 1-+ Df(u)

is called the derivative of f. Moreover. if Df is a continuous map (where L(E, F) has the norm topology). we say f is of class C l (or is continuously differentiable). Proceeding inductively we define D'"f: = D(D'"-If) : U

C

E ~ U(E, F)

if it exists. where we have identified L(E, U-I(E, F» with U(E, F) (see 2.2.9). If Drf exists and is norm continuous. we say f is of class

cr.

We shall reformulate the definition of the derivative with the aid of the somewhat imprecise but very convenient Landau symbol : o(d') will denote a continuous function of e defined in a

neighborhood of the origin of a normed vector space E, satisfying lime -> o(o(e k )/II e Ilk) = O. C E ~ F is differentiable at

The collection of these functions forms a vector space. Clearly f: U Uo E

U iff there exists a linear map Df(u o) E L(E, F) such that f(u o + e)

= f(u o) + Df(uo) . e + o(e)

Let us use this notation to show that if Df(u o) exists. then f is continuous at uo: lim f(uo + e) = lim (f(uo) + Df(uo)' e + o(e» = f(uo). e->O

e->O

2.3.4 Linearity of the Derivative Let f, g : U C E ~ F be r times differentiable mappings and a a real (or complex) constant. Then af and f + g: U C E ~ Fare r times differentiable

with D'(f + g)

= D'f + D'g

and D'(af)

= aD'"f

.

Section 2.3 The Derivative

77

Proof If u E U and e E E. then f(u + e) = f(u) + Df(u)·e + o(e)

and

g(u + e) = g(u) + Dg(u) . e + o(e) •

so that adding these two relations yields (f + g)(u + e) = (f + g)(u) + (Df(u) + Dg(u» . e + o(e) The case r> 1 follows by induction. Similarly af(u + e) = af(u) + aDf(u) . e + ao(e) = af(u) + aDf(u) . e + o(e). • 2.3.5 Derivative of a Cartesian Product Let fi : U

differentiable mappings.

Then f= fl x ... x fn : U c E (fl(u) •...• fn(u» is r times differentiable and

E

C

~

FI

~

Fi' 1:0::; i :0: ; n. be r times Fn defined by f(u) =

X ... X

Proof For u E U and e E E. we have f(u + e) = (fl(u + e) •...• fn(u + e» = (fl(u) + Dfl(u)· e + o(e) •...• fn(u) + Dfn(u) . e + o(e» = (fl(u) ....• fn(u» + (Dfl(u) •...• Dfn(u» . e + (o(e) •...• o(e» = f(u) + Df(u) . e + o(e) . the last equality follows using the sum norm in FI x ... x Fn :

II (o(e) •...• o(e» II = II o(e) II + ... + II o(e) II • so (o(e) •...• o(e»

=: o(e).

•

Notice from the definition that for L

E

L(E. F). DL(u) = L for any u E E. It is also clear

that the derivative of a constant map is zero. Usually all our spaces will be real and linearity will mean real-linearity. In the complex case. differentiable mappings are the subject of analytic function theory. a subject we shall not pursue in this book (see Exercise 2.3F for a hint of why there is a relationship with analytic function theory). In addition to the forcgoing approach. there is a more traditional way to differentiate a function f: U C JRn ~ JRm. We write out f in component form using the following notation: f(xl •...• xn) = (fl(x I •... xn) •...• F(x I •... xn» and compute partial derivaties. afi/ax i for j = 1•...• m and i = 1•...• n. where the symbol afi/ax i means that we compute the usual derivative of f.i with respect to xi while keeping the other variables x I....• xi-I. xi+l •...• xn fixed. For f: JR ~ JR. Df(x) is just the linear map "multiplication by df/dx." i.e.. df/dx = Df(x)·l. This fact. which is obvious from the definitions. can be generalized to the following theorem.

Chapter 2 Banach Spaces and Di/lerential Calculus

78

2.3.6 Proposition Suppose that U c: jRD is an open set and that f: U ~ jRm is differentiable. Then the partial derivatives ap/ax i exist. and the matrix of the linear map Df(x) with respect to the standard bases in jRD and jRm is given by afl axl ar axl

afl

afl

ax 2

ax D

ar

ar

ax 2

axD

where each partial derivative is evaluated at x = (xl •...• xD). This matrix is called the Jacobian matrix of f. Proof By the usual definition of the matrix of a linear mapping from linear algebra. the (j. i)th matrix element ai i of Df(x) is given by the jib component of the vector Df(x)· ei• where e l • ...• eD is the standard basis of JRD. Letting y = x + hei• we see that II f(y) - f(x) - Df(x)(y - x) II II Y_ x II

1

= liJ II f(x , •...• x'+h •...• x I '

D

I

) -

f(x •...• xD )

-

hDf(x)ei II

approaches zero as h ~ O. so the jib component of the numerator does as well; i.e.• lim h..... O

-Ih1-I

1 fj

. I (x I •...• xi +h •...• xD ) - f(x •...• xD ) - hal.i 1 = O.

which means that aii = ap/ax i. • In computations one can usually compute the Jacobian matrix easily. and this proposition then gives Df. In some books. Df is called the differential or the total derivaJive of f. 2.3.7 Example Let f:

jR2 ~ JR.3.

whose matrix in the standard basis is

f(x, y) = (x 2 • x3y. x4y 2). Then Df(x, y) is the linear map

79

Section 2.3 The Derivative

afl

afl

ax ay ar al ax ay ar

ax

2x

0

3x2y

x

2x4y

4x 3l

ar

ay

3

One should take special note when m = 1, in which case we have a real-valued function of n variables. Then Df has the matrix

[4

'00.'

ax

afn ] ax

and the derivative applied to a vector e =(aI, ... , an) is n

Df(x) . e =

L i=1

~ ai ax'

It should be emphasized that Df assigns a linear mapping to each x E U and the definition of Df(x) is independent of the basis used. If we change the basis from the standard basis to another one, the matrix elements will of course change. If one examines the definition of the matrix of a linear transformation, it can be seen that the columns of the matrix relative to the new basis will

be the derivative Df(x) applied to the new basis in JRn with this image vector expressed in the new basis in JRm. Of course, the linear map Df(x) itself does not change from basis to basis. In the ease m = 1, Df(x) is, in the standard basis, a 1 x n matrix. The veetor whose components are the same as those of Df(x) is called the gradient of f, and is denoted grad f or Vf. Thus for f: U C JRn -t JR, grad f=[4 ax

,00',

afn ] ax

(Sometimes it is said that grad f is just Df with commas inserted!) The formation of gradients makes sense in a general inner product spaee as follows.

2.3.8 Definition (i) Let E be a normed space and f: U C E -t JR be differentiable so that Df(u) E L(E, JR) = E*. In this case we sometimes write df(u) for Df(u) and call df the differential of f. Thus df: U-t E*. (ii) If E is a Hilbert space, the gradient of f is the map grad f = Vf: U -t E

defined by

( Vf(u), e

>=

df(u)· e

80

Chapter 2 Banach Spaces and DilJerentiBJ Calculus

where df(u)· e means the linear map df(u) applied to the vector e. Note that the existence of Vf(u) requires the Riesz Representation Theorem (see 2.2.5). The notation Of/au instead of (grad f)(u)

= Vf(u)

is also in wide use, especially in the case in

which E is a space of functions. See Supplement 2.4C below.

,» be a real inner product and f(u) = II u 112. Since II u 112 = II Uo 112 II u - Uo 11 2, we obtain df(Uo)' e = 2( u o' e) and thus Vf(u) = 2u. Hence f is

2.3.9 Example Let (E, ( + 2( u o' u - u o ) + of class

ct.

But since Df(u)

=2( u, .)

E

E* is a continuous linear map in u

E

E, it follows

that D2f(u) = Df E L(E, E*) and thus I)kf = 0 for k ~ 3. Thus f is of class C"". The mapping f considered here is a special case of a polynomial mapping (see 2.2.10). • We close this section with the Fundamental Theorem of Calculus in real Banach spaces. First a bit of notation. If cp : U c: JR.

~

F is differentiable, then Dcp(t)

L(lR, F) is isomorphic to F by A t-+ A(l), 1 E JR.; note that

~;

cp'(t) =

= Dcp(t)· I, 1

E

E

L(JR., F). The space

II A II = II A(1) II. We denote JR..

cp'(t) = lim cp(t + h) - cp(t) h ..... O h and cp is differentiable iff cp' exists. 2.3.10 Fundamental Theorem of Calculus (i) If g: [a, bl ~ F is continuous,for F a real

normed space, the map f: (a, b)

~F

defined by

f(t)

= fg(S)dS

is differentiable and we have f' =g. (ii) If f: [a, bl ~ F is continuous, is differentiable on la, b[ and f' extends to a continuous map on [a, bl, then f(b)-f(a) =

Proof (i) Let to

E

f

f'(s)ds.

la, b[. Since the integral is linear and continuous,

II f(to + h) - f(to) - hg(to) II

=

II

J'o'o+h (g(s) - g(to» ds

:;> Ihl sup {II g(s)-g(to) II The sup on the right-hand side

~

II

:;>

I to:;> s:;> to+h}

0 as I h I ~ 0 by continuity of g at to.

81

Section 2.3 The Derivative

(ii) Let

h(t) =

(f

f '(s) ds ) - f(t).

*

By (i). h'(t) =0 on lao b[ and h is continuous on [a. bl. If for some t E [a. bl. h(t) h(a). then by the Hahn-Banach Theorem there exists a E F* such that (a 0 h)(t) (a 0 h)(a).

*

Moreover. a 0 h is differentiable on lao b[ and its derivative is zero (Exercise 2.3D). Thus by elementary calculus. a 0 h is constant on [a. bl. a contradiction. Hence h(t) = h(a) for all t E [a. bl. In particular. h(a)

=h(b).

•

Exercises 2.3A Let B: E x F --t G be a continuous bilinear map of normed spaces. Show that B is COO and that DB(u. v)(e. f) = B(u. f) + B(e. v) . 2.3B Show that the derivative of a map is unaltered if the spaces are renormed with equivalent norms. 2.3C If f E Sk(E. F). show that

and IYf(O)

=0

for i

= 1•...• k -

I .

2.3D Let f: U C E --t F be a differentiable (resp. C,) map and A E L(F. G). Show that A 0 f: U C E --t G is differentiable (resp. 0) and D'(A 0 f)(u) = A 0 D'f(u). (Hint: Use induction.) 2.3E

Let f: U

C

E --t F be r times differentiable and A E L(G. E). Show that

exists for all i ~ r. where v is an affine map. 2.3F

(i) (ii)

E

A-I (U). and gl' .... gi E G. Generalize to the case where A

Show that a complex linear map A E L(C. C) is necessarily of the form A(z) = A.z. for some A. E C. Show that the matrix of A E L(C. C). when A is regarded as a real linear map in

82

Chapter 2 Banach Spaces and Differential Calculus

(Hint: A = a + ib.) (iii)

Show that a map f: U

C ~ C. f = g + ih. g. h : U

C

C

]R2 ~ JR is complex

differentiable iff the Cauchy-Riemann equations ag ah ax=ay'

ag ah ay=-ax

are satisfied. (Hint: Use (ii) and 2.3.6.) 2.3G Let (E. (

.»

be a complex inner product space. Show that the map f(u) = II u 112 is

not differentiable. Contrast this with 2.3.9. (Hint: Df(u). if it exists. should equal 2Re( ( u •. ) ).) 2.3H Show that the matrix of D2f(x)

E

L2(JRR. JR) for f: U

a2 f

a2f

axiaxi

ax iax2

a2f

a2f

axRaxi

axRax 2

C

JRR -+ JR. is given by

(Hint: Apply 2.3.6. Recall from linear algebra that the matrix of a bilinear mapping B E L(JRR.]Rm; JR) is given by the entries B(ej (first index = row index. second index =

.9

column index). where {el' .... eR} and {fl' .... fm } are ordered bases of JRR and JRm • respectively.)

§2.4 Properties of the Derivative In this section some of the fundamental properties of the derivative are developed. These properties are analogues of rules familiar from elementary calculus. Let us begin by strengthening the fact that differentiability implies continuity. 2.4.1 Proposition Suppose U

C

E is open and f: U ~ F is differentiable on U. Then f is

continuolLS. InfactJor each \10 E U there is a constant M > 0 and a

00 > 0 with the property

that II u - Uo II < 00 implies II feu) - f(u o) II :S M II u - Uo II. (This is called the Lipschitz property.)

Proof

III feu) - f(uo) II -II Df(uo) . (u - \10) III :S II feu) - f(uO> - Df(uo) . (u - uo) II

= II o(u - uo) II :S II u -

Uo

II

for II u - U o II :S 00 , where 00 is some positive constant depending on uo; this holds since lim u --> "0

o(u - no) --;;"'=0.

lIu-no II

Thus,

II feu) - f(u o) II :S II Df(ua> . (u - uo) II + II u - U o II :S (II Df(uo) II + 1)11 u - Uo II

Perhaps the most important rule of differential calculus is the chain rule. To facilitate its statement, the notion of the tangent of a map is introduced. The text will begin conceptually distinguishing points in U from vectors in E. At this point it is not so clear that the distinction is important, but it will help with the transition to manifolds in Chapter 3. 2.4.2 Definition Suppose f: U Tf: U x E

~

C

E ~ F is of class Cl. Define the tangent of f to be the map

FxF

given by

Tf(u, e)

= (f(u), Df(u) . e)

where we recall that Df(u) . e denotes Df(u) applied to e E E as a linear map. If f is of class cr, define Trf = T(T'-l f) inductively.

From a geometric point of view, T is more "natural" than D. If we think of (u, e) as a vector with base point u, and vector part e then (f(u), Df(u) . e) is the image vector with its

84

Chapter 2 Banach Spaces and Dilferentiol Calculus

base point f(u), as in Fig. 2.4.1. Another reason for this is its simple behavior under composition,

as given in the next theorem.

l

E

f

~

U

F

\~(U) ff(U)'e

Figure 2.4.1 2.4.3 C r Composite Mapping Theorem Suppose f: U

E -+ V

C

are differentiable (respectively Cr ) maps. Then the composite g

F and g: V

C 0

f :U

C

C

F -+ G

E -+ G is also

differentiable (respectively 0) and T(g 0 f) = Tg 0 Tf,

(respectively, T(g 0 f)

=Tg

0

Tf). Theformula T(g 0 f) = Tf 0 Tf is equivalent to the chain rule

in terms of D: D(g 0 f)(u) = Dg(f(u»

Proof Since f is differentiable at u

E

0

Df(u).

U and g is differentiable at f(u)

f(u + e) = f(u) + Df(u) . e + o(e) for e

E

E

V, we have

E

and for v = f(u) we have g(v + w) = g(v) + Dg(v) . w + o(w). Thus (g 0 f)(u + e) = g(f(u) + Df(u) . e + o(e»)

=(g

0

f)(u) + Dg(f(u» . (Df(u) . e) + Dg(f(u»(o(e» + o(Df(u) . e + o(e».

For e in a neighborhood of the origin, II Df(u) . e + o(e) II II ell

;5;

(II Df(u) II +

lIo(l~~I)1I ) M

for some constant M > 0, and IIDg(f(u» . o(e)1I

;5;

IIDg(f(u)1I lIo(e)lI.

;5;

M

Section 2.4 Properties of the Derivative

85

Thus

II o(lli(u)· e + o(e» II lie II

II (o(Df(u) . e + o(e» II II Df(u)· e + o(e) II

II Df(u) . e + o(e) II II e II

------~M

II (o(Df(u) . e + o(e» II . I Df(u) . e + o(e) II

Hence we conclude Dg(f(u» . (o(e» + o(Df(u) . e + o(e» = o(e) and thus D(g 0 f)(u) . e = Dg(f(u» . (Df(u) . e). Denote by cp: L(F. G) x L(E. F) --7 L(E. G) the bilinear mapping cp(B. A) = BoA and note that cp

E

L(L(F. G). L(E. F); L(E. G») since II BoA II ~ II B II II A II; i.e .• II cp II ~ 1. Let

(Dg 0 f) x Df: U --7 L(F. G) x L(E. F) be defined by [(Dg 0 f) x Df](u) = (Dg(f(u». Df(u»; notice that this map is continuous if f and g are of class CI. Therefore the composite function cp 0 (Dg 0 f) x Df) = D(g 0 f) : U --7 L(E. G) is continuous if f and g are CI. that is. go f is CI.

Inductively suppose f and g are

cr.

Then Dg is

cr- I •

so Dg 0 f is Cr- I and thus the

map (Dg 0 f) x Df is Cr - I (see 2.3.5). Since cp is COO (Exercise 2.3A). again the inductive hypothesis forces cp 0 «Dg 0 f) x Df) = D(g 0 f) to be Cr- I ; i.e .• go f is cr. The formula T'(g 0 f) = T'g 0 Ff is a direct verification for r = 1 using the chain rule. and the rest follows by induction. • If E=JRm. F=JRD. G=JRP

and f=(fl •...• fD). g=(gl •...• gP).where P:U--7JR

and gi : V --7 JR. by 2.3.6 the chain rule becomes 1

a(g 0 f) (x)

af(x)

which. when read componentwise. becomes the usual chain rule from calculus: a(gof)i(x) =

i

axi The chain rule applied to B fOllowing.

k=1

E

agi(f(>..» af(x) t

al

i= 1,... , m.

axi

L(Ft' F2; G) and fl x f2 : U

C

E --7 FI

X

F2 yields the

86

Chapter 2 Banach Spaces and DijJerentiol Calculus

2.4.4 The Leibniz Rule Let fj: U

C

E ~ Fj' i = 1,2, be differentiable (respectively Cr ) 0 (fl x f2) : U C E ~ G is

maps and B E L(FI' F 2; G). Then the mapping B(fl' f2) = B differentiable (respectively 0) and

In the case FI = F2 =JR and B is multiplication, 2.4.4 reduces to the usual product rule for derivatives. Leibniz' rule can easily be extended to multilinear mappings (Exercise 2.4C). The first of several consequences of the chain rule involves the directional derivative. 2.4.5 Definition Let f: U direction e E E at u

C

E ~ F and let u E U. We say that f has a derivative in the

if ddt f(u+te)1 1=0

exists. We call this element of F the directional derivative of f in the diurection e at u.

Sometimes a function all of whose directional derivatives exist is called Gaeaux differentiable, whereas a function differentiable in the sense we have defined is called Frichet differentiable. The latter is stronger, according to the following. (See also Exercise 2.4J.)

2.4.6 Proposition If f is differentiable at u, then the directional derivatives of f exist at u and are given by ddt feu + te)

I

1=0

=

Df(u) . e .

Proof A path in E is a map from I into E, where I is an open interval of lR. Thus, if c is differentiable, for tEl we have Dc(t) E L(lR, E), by definition. Recall that we identify L(JR. E) with E by associating Dc(t) with Dc(t) . I (1 E JR). Let dc dt(t) = Dc(t)· I . For f: U rule that

C

E ~ F of class C l we consider f 0 c, where c : I ~ U. It follows from the chain d

dt (f(c(t)))

=

D(f

0

c)(t) . I

=

~

Df(c(t» . dt .

The proposition follows by choosing c(t) = u + te, where u, e E E, I = 1 - A, A[, and A is sufficiently small. • For f: U C JRn ~ JR, the directional derivative is given in terms of the standard basis {el' .... en} by

Section 2.4 Properties o/the Derivative

Df(u) . e = -

df

x

dX I where e

= x lei + ...

I

87

df n + ... + - - x , dx n

+ xne n • This follows from 2.3.6 and 2.4.6.

The formula in 2.4.6 is sometimes a convenient method for computing Df(u)· e. For example, let us compute the differential of a homogeneous polynomial of degree 2 from E to F. Let f(e) = A(e, e), where A

E

L2(E; F). By the chain and Leibniz rules,

Df(u) . e = dd A(u + te, u + te) t

I

1=0

= A(u, e) + A(e, u) .

If A is symmetric, then Df(u)· e = 2A(u, e). One of the basic tools for finding estimates is the following.

2.4.7 Proposition Let E and F be real Banach spaces, f: U

C

E ~ F a Cl-map, x, Y E U,

and c a C l arc in U connecting x to y; i.e., c is a continuous map c: [0, 1] ~ U, which is Cion ]0, 1[, c(O)

= x, and

c(1)

= y.

Then 1

f(y) - f(x) = 50 Df(c(t» . c'(t) dt .

If U is convex and c(t) = (1 - t)x + ty, then f(y)-f(x)

Proof If g(t)

= 50IDf(O-t)x+ty).(y-x)dt =

= (f

0

(fDf(O-t)X+ty)dt).(y-x).

c)(t), the chain rule implies g'(t)

= Df(c(t»

. c'(t) and the fundamental

theorem of calculus gives 1

g(I) - g(O) = 50 g'(t) dt ,

which is the first equality. The second equality for U convex and c(t) 2.2F(i). •

2.4.8 Mean Value Inequality Suppose U for all x, y E U

C

=(1 -

E is convex and f: U

II f(y) - f(x) II ~ [suPO:SI:SI II Df«l - t)x + ty)

C

t)x + ty is Exercise

E ~ F is C l . Then

liJ II x - y II.

Thus, if II Df(u) II is uniformly bounded on U by a constant M > 0, then for all x, y II f(y) - f(x) II ~ Mil y - x II. If F = JR, then f(y) - f(x) = Df(c)· (y - x) for some c on the line joining x to y.

E

U

Chapter 2 Banach Spaces and Differentiol Calculus

88

Proof The inequality follows directly from 2.4.7. The last assertion follows from the intermediate value theorem as in elementary calculus. •

2.4.9 Corollary Let U C E be an open set; then the/ollowing are equivalent: (i) U is connected; (ii) every differentiable milP f: U C E ---+ F satisfying Df = 0 on U is constant. Proof If U = U 1 U U 2 , U 1 n U 2 = 0, where U 1 and U 2 are open, then the mapping

1

0, if u

f(u) =

E

U1

e, if u E U 2

where e E F, e"", 0 is a fixed vector, has Df = 0, yet is not constant. Conversely, assume that U is connected and Df = O. Then f is in fact C-. Let

Uo E U be fixed and consider the set S = {u E U I f(u) = f(u o)}' Then S"'" 0 (since Uo E S), S C U, and S is closed since f is continuous. We shall show that S is also open. If u E S, consider

v E Dr(u)

C

U and apply 2.4.8 to get

II f(u)-f(v) II

~ sup {

II Df«I-t)u+ tv) III t E [0, III II u-v II =0;

i.e., f(v) = f(u) = f(u o) and hence Diu)

C

S. Connectedness of U implies S = U. •

If f is Giiteaux differentiable and the Giiteaux derivative is in L(E, F); i.e., for each u there exists Gu E L(E, F) such that ddt f(u + te)

I

1=0

=

E

V

Gue ,

and if u ...... G u is continuous, we say f is Cl-Gateaux. The mean value inequality holds, replacing CI everywhere by "Cl-Giiteaux" and the identical proofs work. When studying differentiability the following is often useful. 2.4.10 Corollary If f: U

C

E ---+ F is CI-Gateaux then it is CI and the two derivatives coincide.

Proof Let u E U and work in a disk centered at u. Proposition 2.4.7 gives

and the sup converges to zero as, e ---+ 0, by uniform continuity of the map t E [0, 1] ...... Gu+1e L(E, F). This says that Df(u)· e exists and equals Gue. •

E

Section 2.4 Properties a/the Derivative

89

It will be convenient to consider partial derivatives in this context. We shall discuss only functions of two variables. the generalization to n variables being obvious.

2.4.11 Definition Let f: U -+ F be a mapping defined on the open set U C EI EEl Ez and let Uo = (uOI ' u02 ) E U. The derivatives of the mappings v I 1-+ f(v l' Uo2)' v 2 1-+ f(uol' v2). where vI E EI and v2 E Ez. if they exist, are called partial derivatives of f at Uo E U and are denoted by Dlf(u o) E L(EI' F). Di(u o) E L(E2, F). 2.4.12 Proposition Let U C EI EEl E2 be open and f: U -+ F. (i) If f is differentiable, then the partial derivatives exist and are given by

= Df(u) . (el' 0)

D/(u) . e l

and

D2f(u)· e 2 = Df(u) . (0.

~).

(ii) If f is differentiable, then

Df(u) . (el' e 2) = DI f(u) . e l + D2f(u) . e 2 . (iii) f is of class Cr

Proof (i) Let

j!

X:EI

iff Dl: U -+ L(Ei • F), i = 1,2 both exist and are of class Cr - I .

-+ EI EEl ~ be defined by .i.!(VI) = (vI' u2)' where u = (ul' U2). Then

is C~ and D.i.!(uI) =

]1 E

L(EI' EI EEl~) is given by ]I(el) = (el'O). By the chain rule, Dlf(u) = D(f ° j!)(uI) = Df(u) ° ]1'

which proves the first relation in (i). One similarly defines t']2 and proves the second relation. ]1°

(ii) Let Pi (el' e 2) = ei• i = 1.2 be the canonical projections. Then compose the relation PI +]2 ° P2 = identity on EI EEl E2 with Df(u) on the left and use (i).

(iii) Let «l>i E L(L(E I EEl Ez. F). L(Ei' F» and 'I'i E L(L(Ei, F). L(E I EEl E 2• F» be defined by «l>i(A) = A o]i and 'I'i (B i) = Bi ° Pi' i = 1.2. Then (i) and (ii) become

Dl=«l>ioDf.

Df='I'loDlf+'I'2oD2f.

This shows that if f is differentiable. then f is Cr iff Dlf and D2f are Cr - I. Thus to conclude the proof we need to show that if Dlf and D2f exist and are continuous, then Df exists. By 2.4.7 applied consecutively to the two arguments. we get f(u l + el' ~ + e2) - f(ul' u2) - Dlf(ul' u2) . e l - D2f(ul' u2) . e 2 = f(u l + el' u 2 + e 2) - f(ul'

~

+ e2) - Dlf(ul' u 2) . e l

+ f(ul' u2 + e2) - f(ul' u2) - D2f(ul' =

(f +

~)

. e2

(Dlf(ul + tel' U2 + e2) -Dlf(ul. u2» dt ) . el

(fal (D2f(ul' U2 + te2) - D2f(uI.u2)dt ) . e2.

90

Chapter 2 Banach Spaces and Differential Calculus

II e l II ~ II e l II + II e 2 II - II(e l • e2)11.

Taking nonns and using in each tenn the obvious inequality we see that

II f(u l + e p u2 + e2) - f(u p

~

Uz) - Dlf(u p u 2)·e l - D2f(u p u 2) . e 2 II

( sup

OStSI

II Dlf(ul + tel' u2 + e2) - Dlf(ul' u2 + e2) II

Both sups in the parentheses converge to zero as (e l • e 2)

~

(0. 0) by continuity of the partial

derivatives. • If EI

=E2 = \R.

and {el' e 2 } is the standard basis in \R.2 we see that

af (x. y) -_ lim ...:f(_x_+_h.:....:. y'-!-)_-_f.o....(x..;....:.. :. y) = D I f (x. y) . el E F. a x h-->O h Similarly. (af!ay)(x. y) = DzC(x. y) . e2 E F. Define inductively highcr dcrivatives

2.4.13 Example As an application of the fonnalism just introduced we shall prove that for f: U \R. 2 ~

C

F 2

D feu) . (v. w) = v

I

I W

a 2f I 2 a 2f 2 I a2f 2 2 a 2f (u) + V W ~.ax (u) + v w a ~. (u) + v W - 2 (u) • ax V] XV] ay

-2

if

(vi. v2 )

a 2f ayax (u)

-(u) ax 2 a 2f axay (u)

a 2f -(u) al

(::)

E U. V. WE \R. 2• V = vlel + v 2e 2 • w = wle l + w 2 e 2 • and {e l • e 2 } is the standard basis of \R.2. To prove this. note that by definition.

where u

D2f(u) . (v. w) = D(Df)(·) . w)(u) . v Applying the chain rule to DfO' w = Tw : A

E

L(\R. 2 • F)

f-+

A· w

E

F. the above equals

D(DfO . w)(u) . v (by 2.4.12(ii»

Section 2.4 Properties of the Derivative

91

D ( w Iaf - + w2af) (u)' v ax dy

ay

af ) (u)· v Iel + D2 ( ax af ) (u)· v2~ ] + W2 [ DI ( dy af) (u)· v Iel + D2 ( af) (u)· v2e2 ] WI[ DI ( ax I I a 2f 2 I a 2f I 2 a 2f 2 2 a 2f v w -2 (u) + v w a ::l,. (u) + v w ::l,.a (u) + v w -2 (u). • ax XV] V] x ay For computation of higher derivatives, let us note that by repeated application of 2.4.6,

In particular for f: U

C

Rm

~

RD the components of I>'"f(u) in terms of the standard basis are

Thus f is of class Cr iff all its r-th order partial derivatives exist and are continuous. 2.4.14 Proposition (L. Euler) If f: U

C

E

~

Cr, then Drf(u)

E

Us(E, F); i.e., Drf(u) is

symmetric. Proof First we prove the result for r =2. Let u E U, v, wEE be fixed; we want to show that D2f(u) . (v, w) = D2f(u) . (w, v). To this, define the linear map a: ~2 ~ E by a(e l ) = v, and a(e2) = w, where e l and e2 are the standard basis vectors of R2. For (x, y)

=xv + yw.

Now define the affine map A: R2 ~ E by A(x, y)

=u

E

]R2, then a(x,y)

+ a(x, y). Since D2(f 0

A)(x, y) . (el' e 2) = D2f(u) . (v, w) (Exercise 2.3E), it suffices to prove this formula: D2(f . (x,y) . (e l , e2) = D2(f 0 A)(x, y) . (e 2, e l ); i.e.,

(see 2.4.13). Let g

=f

0

A : V = A-I (U)

C

0

A)

R2 ~ F. Since for any A. E F*, a 2(A. 0 g)/axdy

=

A.<()2g!axdy), using the Hahn-Banach theorem 2.2.13, it suffices to prove that

Where

92

Chapter 2 Banach Spaces and Differenlilll Calculus Sh,k = [cp(x + h. y + k) - cp(x. y + k)]- [cp(x + h. y) - cp(x. y)] =

(~ (Ch,k' Y + k) - ~(Ch.k' Y»)k

a

2cp axay (Ch,k' dh,k)hk

for some ch,k' dh,k lying between x and x + h and y and y + k. respectively. By interchanging the two middle terms in Sh,k we can derive in the same way that

Equating these two formulas for Sh,k' cancelling h. k. and letting h D2cp gives the result.

~

O. k

~

O. the continuity of

For general r. proceed by induction: DTf(u)· (vI' v 2•···• vo ) = D2(DT-2f)(u) . (vi' v2)· (v 3 •...• v o ) = D2(DT-2f)(u) . (v 2• VI) . (v 3 •···• v o ) = DTf(u) . (v 2• vi' v3 •••.• vo). Let a be any permutation of {2 •...• n}. so by the inductive hypothesis

Take the derivative of this relation with respect to u (Exercise 2.4F):

E

U keeping v 2•...• Vo fixed and get

Since any permutation can be written as a product of the transposition {I. 2. 3 •...• n} 3 •...• n} (if necessary) and a permutation of {2 •...• n}. the result follows. • Suppose U

C

~

{2. 1.

E is an open set. Then as +: E x E ~ E is continuous. there exists an open

set DeE x E with these three properties: (i) U x {O} c D. (ii) u + I;h E U for all (u. h) E D and 0 $; I; $; 1. and (iii) (u. h) E D implies u E U. For example let D= {(+)-I(U)} () (UxE) Let us call such a set D. temporarily. a thickening of U. See Figure 2.4.2.

Section 2.4 Properties of the Derivative

£

93

U

(; ~ thickening of U

£

1 Figure 2.4.2 2.4.15 Taylor's Theorem A map f: U

C

E

~

F is of class Cr

iff there are continuous

mappings
p=I ..... r

-

R:D~Us(E.F).

and

where U is some thickening of U. such that for all (u. h)

E

-

U.


•

where hP = (h ..... h) (p times) and R (u. 0) = O. If f is Cr then necessarily
rl

R(u. h) = Jo

(I - t)r-I r r (r-I)! (D f(u + th) - D f(u» dt.

Proof We shall prove the "only if' part. The converse is proved in Supplement 2.4B. Leibniz' rule gives the foIl owing integration by parts formula. If [a. b] cue 1R and IJfj: U C 1R ~ E j, i

= 1,2 are C l mappings and B E L(EI' E2 ; F) is a bilinear map of EI x ~ to F, then

Assume f is a Cr mapping. If r = I, then by 2.4.7

f(u + h) = f(u) +

(f

Df(u + th) dt) . h = f(u) + Df(u) . h +

(L

1 (Df(u + th) - Df(u» dt) . h

and the formula is proved. For general k ~ r proceed by induction choosing in the integration by

Chapter 2 Banach Spaces and Dif/erentiIJJ Calculus

94

parts formula EI = JR, E2 = E, B(s, e) = se, 'l'2(t) = J>kf(u + th)·hk , and 'I',(t) = - (1 - tnk!, and taking into account that ( (I - t)k 1 Jo -k-!- dt = (k + 1)1" Since Dkf(u)

E

L\(E, F) by 2.4.14, Taylor's formula follows. •

Note that R(u, h) . hr =o(h') since R(u, h) ~ 0 as h ~ O. If f is Cr+1 then the mean value inequality and a bound on Dr+lf gives R(u, h) . hr = o(hr+I). See Exercise 2.4M for the differentiability of R. The proof also shows that Taylor's formula holds if f is (r - 1) times differentiable on U and r times differentiable at u. The estimate R(u, h) . hr =o(h') is proved directly by induction; for r = 1 it is the definition of the Frechet derivative. If f is COO (that is, is Cr for all r) then we may be able to extend Taylor's formula into a convergent power series. If we can, we say f is of class em, or analytic. A standard example of a COO function that is not analytic is the following function from lR to lR (Fig. 2.4.3)

l

ex P{-l/(1 - x2)} , I x 1<1

9(x) =

o

,Ixl~t.

y

-\

Figure 2.4.3 This function is COO, and all derivatives are 0 at x = it. (To see this note that for I x I < I, r
Section 2.4 Properties of the Derivative

and if the maps (t. u)

E

[a. b] xU 1-+ Di(g(t»(u)

defined by h(u) = is

cr

and nih(u) =

E

95

US (E. F) are continuous. then h : U ~ F.

f f f ~f(t. f(t. u) dt =

g(t)(u) dt

u)dt.

j = 1•...• r.

where Du means the partial derivative in u. For r = 1. write

II h(u + e) -

h(u) -

f

D (g(t»(u)'e dt

II

=

~ (b - a)1I e II

II

f({

(D(g(t))(u + se)-e - D(g(t))(u)-e)dS) dt I

a~~~~&t~'

II D(g(t»(u + se) - D(g(t»(u) 11= o(e) .

For r > lone can also use an argument like this. but the converse to Taylor's theorem also yields the result rather easily. Indeed. if R(t. u. e) denotes the remainder for the cr Taylor expansion of get). then with

(j)p

=

D~ =

f

oP[g(t)] dt. the remainder for h is clearly R(u. e) =

f

R(t. u. e) dt.

But R(t. u. e) dt 1-+ 0 as e 1-+ 0 uniformly in t. so R(u. e) is continuous and R(u. 0) = Thus h is cr.•

o.

SUPPLEMENT 2.4A The Leibniz and Chain Rules I@"

Here the explicit formulas are given for the k-th order derivatives of products and compositions. The proofs are straightforward but quite messy induction argunlents. which will be left to the interested reader.

1 The Higher Order Leibniz Rule Let E. F,. F2. and G be Banach spaces. U C E an open set. f:U~F" and g:U~F2 of class Ck and BE L(F" F2; G). Let fxg:U~F, x F2 denote the mapping (f x g)(e) = (f(e). gee»~ and let B(f. g) = B 0 (f x g). Thus B(f. g) is of class Ck and by Leibniz' rule. DB(f. g)(p) . e

= B(Df(p) . e. g(p» + B(f(p). Dg(p) . e).

Higher derivatives of f and g are maps

Where

96

Chapter 2 Banach Spaces and Differential Calculus

Denote by

the bilinear mapping defined by

is defined by

for p

E

U. Leibniz' rule for k-th derivatives is k

DkB(f. g) = Symk

0

~ (~}}.k-i(Dif. Dk-ig). i=O

where Sym k : Lk(E; G) ~ Lks(E; G) is the symmetrization operator. given by (see Exercise 2.21):

(SymkA)(el •...• ek)

:! ~

A(ecr(I).···.e(k».

crESk

where Sk is the group of permutations of {I •...• k}. Explicitly. taking advantage of the symmetry of higher order derivaties. this formula is

k

~

~ (~)B(Dif(p).(ecr(I)"'"

crE~

i=O

cr(I)<" '
ecr(i»' Dk-ig(p)(ecr(i+I).···. ecr(k») .

2 The Higher Order Chain Rule Let E. F. and G be Banach spaces and U

C

F and V

C

F

be open sets. Let f: U ~ V. and g: V ~ G be maps of class Ck. By the usual chain rule. go f: U ~ G is of class C k and D(g 0 f)(p) = Dg(f(p» for p

E

0

Df(p)

U. For every tuple (i. jl' ...• j). where i> 1. and jl + ... + j; = k. define the

Section 2.4 Properties of the Derivative

97

continuous multilinear map

by for A

E

Li(F; G), Bl

E

Lit(E; F), ~ = 1, ... , i, and e" ... , ek

E. Since J)itf: U ~ Lit(E; F),

E

we can define by

With these notations, the k-th order chain rule is

where Symk : Lk(E; G) ~ Lks(E; G) is the symmetrization opertor. Taking into account the symmetry of higher order derivatives, the explicit formula at p

E

U and e" ... , ek

E

E, is

where the third sum is taken for ~, < "'~J' 1, ... , ~J' 1 + "'+J' 1-1 +, < ... < ~k'

~ SUPPLEMENT 2.4B The Converse to Taylor's Theorem

This theorem goes back to Marcinkiewicz and Zygmund [1936], Whitney [1943a), and Glaeser [1958). The proof of the converse that we shall follow is due to Nelson [1969). Assume the formula in the theorem holds where

expression. If r = 1, the formula reduces to the definition of the derivative. Hence
Chapter 2 Banach Spaces and Differential Calculus

98

1 p-l p-l f(u + h + k) = f(u + h) + Df(u + h) . k + ... + - - - D f(u + h) . k (p - I)! 1 + , cpp(u + h) . kP + R1(u + k. k) . k P; p. 1 p-l p-l f(u + h + k) = f(u) + Df(u) . (h + k) + ... + (p _ I)! D f(u)' (h + k) 1

+ ,cpp(u)· (h + k)P + R 2 (u. h + k ) . (h + k)P. p. Subtracting them and collecting tenns homogeneous in kj we get: go(h) + gl(h) . k +.. + gp_l(h) . kP-l + ~(h) . kP = R1(u + h. k) . kP - R 2(u. h + k) . (h + k)P.

where g/h)

gj(h) =

E

V(E; F). g/O) = 0 is given by

1· . I)lf(u + h) -I)lf(u) -

~[ J.

I

p-l-j

.

1=1

1

1· . . 1 . 1)l+lf(u)' hI - -_( ')' cpp(u)· hP-] • P J.

~

O$j$p-2;

1.

and

Let 1\ kII satisfy ~ 1\ h 1\ $ 1\ k 1\ $ ~ 1\ h 1\. Since 1\ R1(u + h. k)·kP - R 2 (u. h + k)·(h + k)P - gp(h).kP 1\

$ (1\ R1(u + h. k) 1\ + 1\ ~(h) 1\) II k I\P + 1\ $ {I\ R1(u + h. k) 1\ + 1\ gp(h) 1\ + 1\ and the quantity in braces { } ~ 0 as h Hence

~

Rz
Rz
O. it follows that

R1(u + h. k) . kP - R 2(u. h + k) . (h + k)P -

~(h)

. kP = o(hP).

go(h) + gl (h)·k + ... + gk-l (h) . kP-l = o(h P). We claim that subject to the condition ~I\ h 1\ $ 1\ k 1\ $ ~ II h 1\. each tenn of this sum is o(h P). If A. 1..... A. p are distinct numbers. replace k by A.jk in the foregoing. and get a p x p linear system in the unknowns go(h) ..... ~_t
= ~-l (h)

. kP-l

= o(hP).

Section 2.4 Properties o/the Derivative

99

Using polarization (see Supplement 2.2B) we get

II DP-If(u + h) - DP-If(u) -
l

II (DP-If(u + h) -

(~-_I~;I p

.

sup

II

(DP-If(u + h) - DP-If(u) -
II

II ellS I

(p - I r l

( _ 1)' p .

DP-If(u) -
sup

(D

p-I

f(u + h)

-

D

p-I

2k ) . . ( __ f(u) -
P-I

IIkllSllhi1/2

l

(2(p - I)r I sup (p - 1)!II h II P- IIkllSllhll/2 p-I (2(p o(hP). (p - I)!II h lip-I

I

(DP-If(u + h) - DP-If(u) -
II

I»

Since o(hP)/11 h liP ~ 0 as h ~ 0, this relation proves that Dp-If is differentiable and DPf(u)

=

<j)p(u). Thus f is of class CP, 0) of 1 into E is a Banach space with respect to the norm

II fll k = max sup II Dif(t) II ISiSk

tEl

(see Exercise 2.4H). If U is open in E, then the set C'(I; U) = (f E C'(I; E) I f(l) checked to be open in C'(I; E). 2.4.17 Proposition The evaluation map defined by

ev : C'(I; U) x ]0, 1[ ev(f, t) = f(t)

~

U

C

U} is

100 is

Chapter 2 Banach Spaces and Differentinl Calculus

cr and its kth derivative is given by k

Dkev(f, t)'«g\, s\), ... , (gk, sk»

= Dkf(t)·(s\, ... , sk) +

LD~\Mt)·(S\, ... ,Si-\' 8;+\, ... , sk) i=\

i = 1, ... , k.

where Proof For (g, s)

E

CT(I; E) x lR, define the norm I/(g, s)1/ = max(ll g Ilk' I s I). Note that the

right-hand side of the formula in the statement is symmetric in the arguments

(~,

s), i = 1, ... , k.

We shall let this right-hand side be denoted
11 0 ,

lim

E

by uniform continuity of f on I since

Since

(g, s) .... (0,0)

for all t

=0

DTg(t)'ST 1/ (g, s) liT

=

0

]0, 1[, by Taylor's theorem for g we get ev(f + g, t + s) = f(t + s) + g(t + s) T

=

L~

(Dif(t) , Si + Dig(t) . si) + R(t, s) . sr

I.

i=O T

= f(t) +

~

1 LJ 'T
i

r

t) . (g, s) + R«f, t), (g, s» . (g, s) ,

i=\

where r

R«f, t), (g, s»' «g\, s\), .. " (gr' ST» = R(t, s), (s\, ... , ST) +

L

Drgi(t)· (S\, ... , Sr),

i=\

which is symmetric in its arguments and R«f, t), (0, 0»

= O.

By the converse to Taylor's

theorem, the proposition is proved if we show that every
by the mean value theorem, the inequality II (
+

I, 1/ Dk-\gi(t) - Dk-\gi(S) III sl I ... I si-I I I si+\1 ... I Sk I i=\

implies 1/
II okf(t) - okg(s) 1/ + kit - s I

Section 2.4 Properties of the Derivative

101

II Dkf(t) - Dkf(s) II + II Dkf(s) - Dkg(s) II + kit - s I

$

II Dkf(t) - Dkf(s) II + 2k lI(f. t) - (g. s)ll.

$

Thus the uniform continuity of Dkf on I implies the continuity of


Let us consider now the omega lemma. (This is terminology of Abraham [19631. Various results of this type can be traced back to earlier works of Sobolev [1939] and Eells [1958].) Let M be a compact topological space and E. F be Banach spaces. With respect to the norm

II fll

II f(m) II.

sup

=

mEM

the vector space CO(M. E) of continuous E-valued maps on M. is a Banach space. If U is open in E. it is easy to see that CO(M. U)

= (f E

2.4.18 Omega Lemma Let g: U

is also of class

cr.

-?

CO(M. E)

F be a

cr

I f(M)

C

U} is open.

map. r> O. The map

The derivative of Og is DOg(f) . h = [(Dg) [DOg(f) . h](x)

i.e ..

0

f] . h

= Dg(f(x)) . h(x).

The formula for DOg is quite plausible. Indeed. we have [DOg(f)'h](x) =

~~

Og(f + Eh)(x)

ut

I£=0

= dd g(f(x) + Eh(x)) E

I . £=0

By the chain rule this is Dg(f(x))· h(x). This shows that if Og is differentiable. then DOg must be as stated in the proposition. Proof Let f E CO(M. U). By continuity of g and compactness of M.

II Og(f) - Og(f') II

=

sup

II g(f(m)) - g(f'(m)) II

mEM

is small as soon as

be given by

II f - f II is small; i.e .• Og is continuous at each point

f. Let

102

Chapter 2 Banach Spaces and DifferentWlCalculus

for HE CaCM, U(E; F)), hi"'" hi E CaCM, E) and mE M. The maps Ai are clearly linear and are continuous with II Ai II .,:; 1. Since Dig: U ~ UseE; F) is continuous, the preceding argument shows that the maps

are continuous and hence

is continuous. The Taylor theorem applied to g yields r

I. ~ Dig(f(m))'h(m)i + R(f(m), h(m))'h(mi

g(f(m) + hem)) = g(f(m)) +

1.

i=1

so that defining and [R(f, h)·(h l ,... , hr)](m) = R(f(m), h(m))·(hl(m), ... , h/m)) we see that R is continous, R(f, 0)

=0, and r

ng(f +

kv = g

0

(f + h)

=g f + 0

I. ~ (Dig i=1

1.

Thus by the converse of Taylor's theorem Ding = Ai

0

0

f) . hi

+ R(f, h) . hi

nDig and ng is of class

C .•

This proposition can be generalized to the Banach space C(I, E), I = [a, b], equipped with the norm II· IIr given by the maximum of the norms of the first r derivatives; i.e., II f IIr = max OSi ';'- suptEIIi f{i)(t) II. If g is cr+q, then ng: C(I, E) ~ cr-k(l, F) is Cq+k. Readers are invited to convince themselves that the foregoing proof works with only trivial modifications in this case. This version of the omega lemma will be used in Supplement 4.1C. For applications to partial differential equations, the most important generalizations of the two previous propositions is to the case of Sobolev maps of class HS; see for example Palais [1968], Marsden [1974b] and Marsden and Hughes [1982] for proofs and applications.

Section 2.4 Properties of the Derivative

103

~

Supplement 2.4C The Functional Derivative and the Calculus of Variations Differential calculus in infinite dimensions has many applications, one of which is to the calculus of variations. We give some of the elementary aspects here. We shall begin with some notation and a generalization of the notion of the dual space. Let E and F be Banach sp~ces. A continuous bilinear functional <,): E x F ~ JR. is called E-non-degenerate if <x, y)

=0

for all y E F implies x

= O.

Similarly, it is F-

non-degenerate if <x, y) = 0 for all x E E implies y = O. If it is both, we just say <,) is non-degenerate. Equivalently, the two linear maps of E to F* and F to E* defined by x ...... <x, -) and y ...... <-, y ), respectively, are one to one. If they are isomorphisms, <,) is called Eor F-strongly non-degenerate. A non-degenerate bilinear form <,) thus represents certain linear functionals on F in terms of elements in E. We say E and F are in duality if there is a non-degenerate bilinear functional <,): E x F ~ JR., also called a pairing of E with F. If the functional is strongly non-degenerate, we say the duality is strong. 2.4.19 Examples

A

Let E

= F*.

Let < ,) : F* x F ~ JR. be given by <
=
so the map E ~ F*

is the identity. Thus, <,) is E-strongly non-degenerate by the Hahn-Banach theorem. It is easily checked that <,) is F-non-degenerate. (If it is F* strongly non-degenerate, F is called

reflexive. ) B

Let E = F and <,): E x E ~ JR.

non-degenerate since

be an inner product on E.

Then <,) is

<,) is positive definite. If E is a Hilbert space, then <,) is a strongly

non-degenerate pairing by the Riesz representation theorem.

+

2.4.20 Definition Let E and F be normed spaces and < , ) be an E-weakly nOli-degenerate

pairing. Let f: F ~ JR. be differentiable at the point ex E F. The functional derivative 8f/8ex of f with respect to ex is the unique element in E, if it exists, such that

Df( ex) .

Ukewise, if g: E

~ JR.

~

=

<:~ ,~)

for all

~

E

and <,) is F-weakly degenerate, we define

(I)

F.

~~

E

F

if it exists, by

'<'

Dg(v) . v =

v, 8v 8f) for all VEE. ,

(1 )'

Often E and F are spaces of mappings, as in the following example. 2.4.21 Example

Let

n

C

JR.n be an open 00unded set and consider the space E

= CO(D), of

Chapter 2 Banach Spaces and Differential Calculus

104

continuous real valued functions on 0 where 0

=cl(fl).

Take F =CO(D) =E. The L2-pairing

on E x F is the bilinear map given by

Let r be a positive integer and define f: E

~

lR. by f(q» =

"2I Irn [q>(x)) r dn x.

Then using

the calculus rules from this section. we find Df(q» . 'l' =

8f

Th us. &p

r {~r-I 'Y

Irn r[q>(x)) r-I 'l'(s) dn x.

.•

Suppose. more generally. that f is defined on a Banach space E of functions q> on a region fl in lR. n . The functional derivative

8f oq>

E

~

of f with respect to q> is the unique element

E • I'f"It eXists. suc hth at

Df(q» . 'l' =

< ~ . 'l' > =

fn (~ )<X)'l'(X) dnx

for all

E

E.

The functional derivative may be determined in examples by

5.n

8f d -r(x)'l'(x) dn x = -d uq> E

I £=0

f(q> + E'l') .

(2)

A basic result in the calculus of variations is the following.

2.4.22 Proposition Let E be a space of functions, as above. A necessary condition for a

differentiable function f: E ~ lR. to have an extremwn at q> is that

o. Proof If f has an extremum at q>. then for each 'l'. the function h(E) = f(cp + E'V) has an extremum at E = O. Thus. by elementary calculus. h'(O) = O. Since 'l' is arbitrary. the result follows from (2). • Sufficient conditions for extrema in the calculus of variations are more delicate. See. for example. Bolza (1904) and Morrey (1966).

105

Section 2.4 Properties of the Derivative

2.4.23 Examples A Suppose that Q

C

JR. is an interval and that f, as a functional of
E

Ck(Q), k ~ I, is

of the fonn f(
fn F( x,
for some smooth function F: Q x JR. x JR. call F the density

~

(3)

JR., so that the right hand side of (3) is defined. We

associated with f. It can be shown by using the results of the preceding

supplement that f is smooth. Using the chain rule,

fn ~ (x)\jI(x) dx :€ 1£=0 fn ~ x,

€\jI, d(
fn D2~ x,
second and third arguments. Integrating by parts, this becomes

fn D2F( x,
)\jI(X) dx

fan D3 F ( x,
Let us now restrict our attention to the space of \jI's which vanish on the boundary

aQ

of

Q. In that case we get

Rewriting this according to the designation of the second and third arguments of F as

d
Ch '

respectively, we obtain (4)

By a similar argument, if Q

C

JR.n, (4) generalizes to

(4)

(Here, a sum on repeated indices is assumed.) Thus, f has an extremum at

106

Chapter 2 Banach Spaces and Differential Calculus

aF

a

d dx k

aF

a(:; )

o.

This is called the Euler-Lagrange equation in the calculus of variations. B Assume that in example A, the density F associated with f depends also on higher d

In F(x,

By an analogous argument, fommla (4) generalizes to

(6)

C Consider a closed curve y in JR3 such that y lies above the boundary an of a region

n

in the xy-plane, as in Figure 2.4.4.

z

y

Figure 2.4.4 Consider differentiable surfaces in JR3 (i.e., two dimensional manifolds of JR3) which are gr~phs of Ck functions
Section 2.4 Properties of the Derivative

107

From equation (5). a necessary condition for

2


for (x. Y)

E

(7)

O.

Q. We relate this to the classical theory of surfaces as follows. A surface has two

principal curvatures KI and

Kz;

the mean curvature K is defined to be their average: that is. K =

(K I + Kz)/2. An elementary theorem of geometry asserts that K is given by the formula

(8)

If the surface represents a sheet of rubber. the mean curvature represents the net force due to internal stretching. Comparing (7) and (8) we find the well-known result that a minimal surface. i.e .. a surface with minimal area. has zero mean curvature. • Now consider the case in which f is a differentiable function of n variables. i.e .. f is defined on a product of n function spaces F i• i = 1•...• n; f: FI x ... x Fn have pairings (')i: Ei x Fi ~ R 2.4.24 Definition

O~i

The i-th partial functional derivative

~

lR and we

of f with respect to
E

Fi

is defined by

The totalfunctional derivative is given by

n

L D/(
0)

i=1

2.4.25 Examples A Suppose that f is a function of n functions
E

Ck(Q). where Q

C

lRn. and their

Chapter 2 Banach Spaces and Differential Calculus

108

It follows that (10)

(sum on k).

B Classical Field Theory As discussed in Goldstein (1980), §12, Lagrange's equations for a field T\ = T\(x, t) with components T\a follow from Hamilton's variational principle. When the Lagrangian L is given by a Lagrangian density L, i.e., L is of the form

(11)

the variational principle states that T\ should be a critical point of L. Assuming appropriate boundary conditions, this results in the equations of motion

(12)

(sum on k is understood). Regarding L as a function of T\a and ~a =

itrta at'

the equations of

motion take the form:

d oL dt o~a C Let

ncR"

and let ~(n) stand for the

(13)

C'

functions vanishing on

-+ R be given by the Dirichlet integral

Using the standard inner product (,) in R", we may write

Differentiating with respect to cp: Df(cphV

I

~ fn (V(cp + £'1'), V(cp + £'1'»

=

:£

=

fn (Vcp,V'I') dUx

[:0()

dUx

an.

Let f: c~(n)

Section 2.4 Properties of the Derivative

= -

Thus

~

In V2
109

(integrating by parts) .

= _V2
D The Stretched String Consider a string of length ~ and mass density a, stretched horizontally under a tension 't, with ends fastened at x = a and x =~. Let u(x, t) denote the vertical displacement of the string at x, at time t. We have u(a, t) = u(~, t) = a. The potential energy V due to small vertical displacements is shown in elementary mechanics texts to be

V =

I '2 t 1

a

't

(dU)2 dX dx,

and the kinetic energy T is

T

Ia '2 t 1

=

a

(du)2 at dx.

From the definitions, we get

oT

and

=

au.

1)u Then with the Lagrangian L = T - V, the equations of motion (13) become the wave equation

Next we formulate a chain rule for functional derivatives. Let ( , ) : Ex F

~

lR. be a weakly

nondegenerate pairing between E and F. If A E L(F, F), its adjoint A * E L(E, E), if it exists, is defined by (A *v, a) = (v, Aa) for all vEE and a E F. Let
~

F be a differentiable map and f: F

~

lR. be differentiable at a E F. From the

chain rule, D(f 0
for

~

E F.

Hence assuming that all functional derivatives and adjoints exist, the preceding relation implies 1)(fo
<~' ~)

1)[

=

< oy' D
1)[ =


oy'

~

)

where y=
of

D
(14)

110

Chapter 2 Banach Spaces and DijJerentWl Calculus

D(\jf

f)(a) . ~ = D\jf(f(a» . (Df(a) . ~)

0

where the first dot on the right hand side is ordinary multiplication by D\jf(f(a» E llt Hence

i.e. , ,f ()f ~ = \jf «a)) Sa'

(15)

We conclude with a discussion of extrema for real valued functions on Banach spaces. 2.4.26 Definition Let f: U

C

~

E

JR be a continuous function, U open in E. We say f has

a local minimum (resp. maximum) at U o E U, if there is a neighborhood V of uo' V C U such that f(u o) ~ f(u) (f(u o) ~ f(u» for all u E V. If the inequality is strict, U o is called a strict local minimum (resp. maximum). The point Uo is called a global minimum (resp. maximum) if f(u o)

~

f(u) (resp. f(u o)

~

for all u E U. Local maxima and minima are called local

f(u»

extrema. 2.4.27 Proposition Let f: U

E

C

~

JR be a continuous function differentiable at U o E U. If f

has a local extremum at uo' then Df(Uo) =

o.

Proof If U o is a local minimum, then there is a neighborhood V of U such that f(u o + th) f(u o) ~ 0 for all hE V. Therefore, the limit of [f(uo + th) - f(Uo)]!t as t ~ 0, t ~ 0 is ~ 0 and as t

~

0, t ~ 0 is ~ O. Since both limits equal Df(u o)' it must vanish. • This criterion is not sufficient as the elementary calculus example f: JR ~ JR, f(x) = x3

shows. Also, if U is not open, the values of f on the boundary of U must be examined separately. 2.4.28 Proposition Let f: U

C

E

~

JR be twice differentiable at Uo E U.

is a local minimum (maximum), then D2f(u o)' (e, e) ~ 0 (~O) for all e E E. (ii) If U o is a non-degenerate critical point f, i.e., Df(u o) = 0 and D2f(u o) defines an (i) If

Uo

isomorphism of E with E*, and

if

D2f( uo) . (e, e) > 0

«

0) for all e

to

0, e E E, then

Uo

is a strict local minimum (maximum) of f. Proof

TDf(uo)(h, h)

(i) By Taylor's formula, in a neighborhood V of U o, 0 ~ f(uo + h) - f(uo) =

+ 0(h2) for all hE V. If e E E is arbitrary, for small t E JR, te E V, so that 0 ~ ..!..D 2f(Uo)(te, te) 2

+ 0(t 2e 2) implies D 2f(u o)(e, e) + 20(t2e 2)!t2 ~ O. Now let t ~ O. (ii) Denote by T: E ~ E* the isomorphism defined by e t--+ D2f(u o) . (e, .), so that there exists a > 0 such that

Section 2.4 Properties of the Derivative

a II e II ~ II Te II

=

111

sup I (Te, e') I = sup I D2f(Uo) . (e, e')I.

lIe'II=1

Ile'II=1

By hypothesis and symmetry of the second derivative,

which is a quadratic form in s. Therefore its discriminant must be negative, i.e.,

and we get a II e II ~ sup I D2f(uo) . (e, e')1 ~ II D2f(uo) 111/2[D2f(uo) . (e, e)] 112. 110'11=1

Therefore, letting m = a2~1 D2f(u o) II, the following inequality holds for any e

E

E:

Thus, by Taylor's theorem we have f(uo + h) - f(uo)

=

'21 D2f(uo)

2 2 2 . (h, h) + o(h ) ~ m II h II (2 + o(h ).

Let E> 0 be such that if II h II < E, then I 0(h2) I ~ m II h 112/4, which implies f(uo + h) - f(uo) ~ m II h 112/4> 0 for h

* 0, and thus

Uo

is a strict local minimum of f. •

The condition in (i) is not sufficient for f to have a local minimum at uo' For example, = 0, Df(O, 0) = 0, D 2f(O, 0) . (x, y)2 = 2x2 ~ 0 and

f: JR.2 ~ JR., f(x, y) = x2 - y4 has f(O, 0)

in any neighborhood of the origin, f changes sign. The conditions in (ii) are not necessary for f to have a strict local minimum at u o' For example f: JR. ~ IR., f(x) = x4 has f(O)

= £'''(0) = 0,

= £'(0) = f'(O)

[<4)(0) > 0 and 0 is a strict global minimum for f. On the other hand, if the

conditions in (ii) hold and

Uo

is the only critical point of a differentiable function f:

U~

JR., then

Uo is a strict global minimum of f. For if there was another point u l E U with f(u l ) ~ f(uo) on the straight line segment (1 - t)u o + tu l , t E [0, 1] there exists a point u2 such that f(u 2) > f(uo)

;::. f(u l ) since by (ii) ~ uO'

Uo is a strict local minimum. Therefore, there exists u 3 on this segmenI, u3 u l such that f(u 3 ) = f(uo)' But then by the mean value theorem 2.4.8 there exists u 4 uO'

*

u3 such that Df(u 4) = 0 which contradicts uniqueness of the critical point. Finally, care has to be taken with the statemenI in (ii): non-degeneracy holds in the topology of E. If E is continuously embedded in another Banach space F and D2f(u o) is non-degenerate in F only, even be a minumum. For example, consider the smooth map

U

o need not

Chapter 2 Banach Spaces and Differential Calculus

112

4

f: L ([0, 1)) ~ JR ,f(u) =

1 rl 2 '2J o (u(x) -

4

u(x) ) dx

and note f(O) = 0, Df(O) = 0, and D2f(0)(v, v) =

fol v(d dx

> 0 for vi' 0,

and that D2f(0) defmes an isomorphism of L4([0, 1)) with L 4/3([0, 1]). Alternatively, D 2f(0) is non-degenerate on L2([O, 1]) not on L 4([0,1]). Also note that in any neighborhood of 0 in L 4([0, 1)), f changes sign: f(lIn) = (n 2 - 1)!2n4 ~ 0 for n ~ 2, but f(u n) = -12/n < 0 for n ~ 1 if

1

2, on [0, lin)

~=

0, elsewhere

and both lin, un converge to 0 in L4([0, 1)). Thus, even though D2f(0) is positive, 0 is not a minimum of f. (See Ball and Marsden [1984) for more sophisticated examples of this sort.)

fl:n

Exercises 2.4A Show that if g : U

E

C

~

(g(u»(v), u E U, V E F is also the evaluation map.) 2.4B Show that if f: U h :U

C

C

E

~

L(F, G) is C r , then f: U x F

cr.

(Hint:

L(F, G), g: U

~

G, defined by feu, v) =

Apply the Leibniz rule with L(F, G) x F

C

E

~

~

G

L(G, H) are Cr mappings then so is

E ~ L(F, H), defined by h(u) = g(u) 0 feu).

2.4C Extend Leibniz' rule to multilinear mappings and find a formula for the derivative. 2.4D Define a map f: U C E ~ F to be of class TI if it is differentiable, its tangent map Tf : U x E ~ F x F is continuous and II Df(x) II is locally bounded. (i) For E and F finite dimensional, show that this is equivalent to C I. (ii) (project) Investigate the validity of the chain rule and Taylor's theorem for T' maps. (iii) (Project) Show that the function developed in Smale [1964) is T2 but is not C2. 2.4E Suppose that f: E ~ F (where E, F are real Banach spaces) is homogeneous uf degree k; i.e., fete) = tkf(e) for all t E JR, and e E E. (i) Show that if f is differentiable, then Df(u)· u = kf(u). (Hint: Let get) = [(tu) and compute dg/dt.)

Section 2.4 Properties of the Derivative

113

(ii) If E = JRn and F = JR. show that this relation is equivalent to n

"£... xi a;.df

=

kf(x).

I

i='

Show that maps multilinear in k variables are homogeneous of degree k. Give other examples. (iii) If f is Ck show that f(e) = (Ilk!) Dkf(O)· ek • i.e .• f E Sk(E. F) and thus it is C~. (Hint:

f(O)

=0;

inductively applying Taylor's theorem and replacing at each step h by

tho show that

2.4F Let e, •...• en_,

E

E be fixed and f: U c: E ~ F be n times differentiable. Show that the

map g: U c: E ~ F defined by g(u) = Dn-'f(u) . (e" ...• en_i) is differentiable and Dg(u) . e = Dnf(u) . (c. e" ...• en_i)' 2.4G (i) Prove the following refinement of 2.4.14. If f is C' and D, D 2f(u) exists and is

continuous in u. then D2D,f(u) exists and these are equal. (ii) The hypothesis in (i) cannot be weakened: show that the function

I

xy(x2 - / )

f(x. y) =

2

x

2

+Y

. • If

O.

is

ct. has

(x. y) '" (0. 0)

if (x. y) = (0. 0)

d 2 f!dxdy. d2 f!dydx continuous on JR2\ (O. 0»). but that d2 f(0. O)/dxdy '"

d 2 f(0. O)ldydx. 2.4H

For f: U c E

~

F. show that the second tangent map is given as follows:

T2f: (U x E) x (E x E) ~ (F x F) x (F x F) (u. e" e2• e 3 )

t-+

(f(u). Df(u) . e" Df(u) . e 2• D2f(u) . (e,. e2) + Df(u) . e 3 ).

2.41 Let f: JR2 ~ JR be defined by f(x. y) = 2x 2 y/(x 4 + y2) if (x. y) '" (0. 0) and 0 if (x. y) =

(0. 0). Show that (i) f is discontinuous at (0.0). hence is not differentiable at (0.0); (ii) all directional derivatives exist at (0.0); i.e .• f is Gateaux differentiable. 2.4J Differentiating sequences Let fn : U c: E ~ F be a sequence of cr maps. where E and F are Banach spaces. If {fnl converges pointwise to f: U ~ F and if {]).ifn). O:::;j:::; r.

Chapter 2 Banach Spaces and Differential Calculus

114

converges locally uniformly to a map gi: U ~

0 seE, F), then show that

f is C r , Dif = gi

and {fn } converges locally uniformly to f. (Hint: For r = I use the mean value inequality and continuity of gl to conclude that II feu + h) - feu) - gl(U) . h II ~ II feu + h) - fn(u + h) - [feu) - fn(u)]II

+ II fn(u + h) - fn(u) - Dfn(u) . h II + II Dfn(u) . h - gl(u) . h II ~

£ II h II.

For general r use the converse to Taylor's theorem.)

2.4K

a

lemma In the context of 2.4.18 let a(g) = g

0

f. Show that a is continuous linear and

hence is COO.

2.4L Consider the map : CI([O, I]) ~ CO([O, 1]) given by (f)(x) = exp[f'(x)]. Show that is COO and compute D.

2.4M (H. Whitney [1943a]) Let f: U

C

E ~ F be of class C k+p with Taylor expansion

1 k k feb) = f(a) + Df(a) . (b - a) + ... + k! D f(a) . (b - a)

+

{I

I 0

tl-

I (l } (k-_ I)! [Dkf«(l - t)a + tb) - Dkf(a)] dt . (b - at

(i) Show that the remainder Rk(a,b) is C k+p for b"" a and CP for a, bEE. If E

=F

=JR, Rk(a, a) = 0, and lim (I b - a Ii Di+PRk(a, b» = 0, 1 ~ i ~ k.

b......

(For generalizations to Banach spaces, see Tuan and Ang [1979].) (ii) Show that the conclusion in (i) cannot be improved by considering f(x) = I X Ik+p+1!2 2.4N (H. Whitney [1943b]). Let f: JR (resp. f(x» =

~

JR be an even (resp. odd) function; i.e., f(x)

= f( -x)

-fe-x»~.

(i) Show that f(x) = g(x2) (resp. f(x) = xg(x 2» for some g. (ii) Show that if f is C 2k (C 2k+1) then g is C k (Hint: Use the converse to Taylor's theorem). (iii) Show that (ii) is still true if k = 00. (iv) Let f(x) = I X 12k +I+I!2 to show that the conclusion in (ii) cannot be sharpened. 2.40 (M. Buchner, 1. Marsden, and S. Schecter [1983]). Let E = L 4([0, 1]) and let q>: JR ~ JR be a

C~

function such that q>'(A)

= 1,

if -1:'> A:'> 1 and q>'(A)

=

°

if I A I ~ 2. Assume

q> is monotone increasing with q> = -M for A:'> -2 and q> = M for A ~ 2. Define the map h: E ~ JR by

115

Section 2.4 Properties of the Derivative

1

fI

'3 Jo

h(u) =


(i) Show that h is C3 using the converse to Taylor's theorem. (Hint: Let 'V(A) =
V'h(u) =

where 'V(A) =
1

'3 'V'(u)

V'h: E

~

E is Co but is not Cl.

derivative would be v ~ 'V "(u)v/3. Let a E [0, 1] be such that
(Hint:

*0

Its

and let

un = a on [0, l/n], un = 0 elsewhere; vn = n 1/4 on [0, l/n], vn = 0 elsewhere. Show that in L 4([0, 1]), un ~ 0, II Vn II = 1, 'V "(un) . vn does not converge to 0, but 'V "(0) = 0.) Using the same method, show h is not C4 on L4([0, 1]). (iii) Show: if q is a positive integer and E = U ([0, 1]), then h is Cq-l but is not cq. (iv) Let feu)

I fl =2" Jo I u(x) 12 dx + h(u).

Show that on L 4([0, 1]), f has a formally non-degenerate critical point at 0 (i.e., D2 f(0) defines an isomorphism of L2([0, 1]), yet this critical point is not isolated.

(Hint:

Consider the function un = -Ion [O,I/n]; 0 on ]1/n, 1].) This exercise is

continued in 5.4H.

2.4P Let E be the space of maps A: lit3 ~ ITt3 with A(x) ~ 0 as x ~ 0 sufficiently rapidly. Let f: E ~ ITt and show 8 8f oA f(curI A) = curI oA . (Hint: Specify whatever smoothness and fall-off hypotheses you need; use A· curI B B • curI A = div(B x A), the divergence theorem, and the chain rule.)

2.4Q (i) Let E = {B I B is a vector field on ITt3 vanishing at and such that div B = O} and pair E with itself via (B, B') = B(x) • B'(x) dx. Compute 8F/oB, where F is defined by F = (1/2) II B 112 d 3x. (ii) Let E = {B I B is a vector field on lit3 vanishing at such that B = V' x A for some A} 00

f

f

00

and let

F = {A' lA' is a vector field on ITt3, divA'=O}

fA.

with the pairing (B, A') = A' d 3x. Show that this pairing is well defined. Compute of/oB, where F is as in (i). Why is your answer different?

§2.5 The Inverse and Implicit Function Theorems The Inverse and Implicit Function Theorems are pillars of nonlinear analysis and geometry. so we give them special attention in this section. Throughout. E. F ..... are assumed to be Banach spaces. In the finite-dimensional case these theorems have a long and complex history; the infinite-dimensional version is apparently due to Hildebrandt and Graves [1927]. We first consider the Inverse Function Theorem. This states that if the linearization of the equation f(x) = y is uniquely invertible. then locally so is f; i.e .• we can uniquely solve f(x) = y for x as a function of y. To formulate the theorem. the following terminology is useful. 2.5.1 Definition A map f: U

class

C

E -t V

C

F (U. V open) is a C r diffeomorphism

cr. is a bijection (that is. one-to-one and onto

2.5.2 Inverse Mapping Theorem Let f: U

C

V). and r- I is also of class

E -t F be of class

cr.

if

f is of

E

U. and

cr.

r 2': 1. Xu

suppose Df(x o) is a linear isomorphism. Then f is a C diffeomorphism of some neighborhood of Xo onto some neighborhood of f(x o) and, moreover, Dr-I(y) = [Df(f-I(y»]-I for y in this neighborhood of f(x o)' Although our immediate interest is the finite-dimensional case. for Banach spaces it is good to keep in mind the Banach Isomorphism Theorem: If T: E -t F is linear. bijective. and continuous. then T-I is continuous. (See 2.2.19.) To prove the theorem. we assemble a few lemmas. First recall the contraction mapping principle from § 1.2. 2.5.3 Lemma Let M be a complete metric space with distance function d: M x M -t R

Let F : M -t M and assume there is a constant A. 0 $ A< 1, such that for all x. y d(F(x). F(y»

$

E

M.

Ad(x, y) .

Then F has a unique fixed point Xo E M; that is, F(x o) = xo' This result is the basis of many important existence theorems in analysis. The other fundamental fixed point theorem in analysis is the Schauder Fixed Point Theorem. which states that a continuous map of a compact convex set (in a Banach space. say) to itself, has a fixed point -- not necessarily unique. however. 2.5.4 Lemma The set G.L(E, F) of linear isomorphisms from E to F is open in L(E. F).

117

Section 2.5 The Inverse and Implicit Function Theorems

Proof We can assume E = F. Indeed, if
E

GL(E, F), the linear map", 1-+ q>0-1 0'" from

L(E, F) to L(E, E) is continuous and GL(E, F) is the inverse image of GL(E, E). Let

II a II = sup II a(e) II eEE

lIell=1

be the norm on L(E, F) relative to given norms on E and F. For
E

GL(E, E), we need to

prove that '" sufficiently near

II", _ q> II < II
implies '" E GL(E, E). The key is that II· II is an algebra norm. That is, II ~ 0 a II :5 II ~ for a

E

L(E, E) and ~

E

"'»,

L(E, E) (see Section 2.2). Since '" = q> 0 (1- q>-I 0 (q> -

",)11 < 1, it is sufficient to show that II < 1. (I is the identity operator.) Consider the following

invertible, and our norm assumption shows that II q>-I 0 (q> I - ~ is invertible whenever II ~

sequence called the Neumann series): ~o

I,

~I

I+~

~2

I +~ +~

~n

0

~

= I + ~ + ~ 0 ~ + ... + (~ 0 ~ 0 ... 0 ~) .

Using the triangle inequality and the foregoing norm inequality, we can compare this sequence to the sequence of real numbers, 1, 1 + II ~ II, 1 + II ~ II + II ~ 11 2, ... , which we know is a Cauchy sequence since

II ~ II < 1. Because L(E, E) is complete,

~n

must converge. The limit, say p, is

the inverse of I -~. Indeed (I - ~)~n = I - (~ 0 ~ 0 ... 0 ~), so the result follows.

•

2.5.5 Lemma Let I: GL(E, F) ~ GL(F, E) be given by q> 1-+ q>-I. Then I is of class C~ and

D I(-I 0 '" 0 -I 0 '" 0 q>-I, then it will follow from Leibniz' rule that

I is of class

C~.

Indeed D I

= B(J,

J) where B

E

L2(L(F,E); L(L(E, F), L(F, E») is defined by B(",\, "'2)(A) = - "'loA 0 "'2' where "'\, "'2 E L(F, E) and A E L(E, F), which shows inductively that if I is Ck then it is Ck+I. Since the mapping '" 1-+

-

q>-I

0'" 0
E

L(E, F», we must show that

Note that '1'1 - (q>-I _ q>-I 0'" 0 q>-I + q>-I

0

'1'1 _ 2q>-1 +
0

0



118

Again, using

Chapter 2 Banach Spaces and Differential Calculus

II ~ a II $ II a 1111 ~ II for a E L(E, F) and 0

~ E

L(F, G), we get

With this inequality, the limit is clearly zero. _ To prove the Inverse Mapping Theorem, we first note that it is enough to prove it under the

=0, f(x o) =0, E =F, and Df(O) is the identity. Indeed, replace f [f(x + xo) - f(x o»)' Now let g(x) = x - f(x) so Dg(O) = 0. Choose r> so that II x II $ r implies II Dg(x) II $ 1(2, which is possible by continuity of Dg. Thus by the Mean Value Inequality, II x II $ r implies

simplifying assumptions Xo by h(x)

=Df(xo)-I

0

°

II g(x) II $ r(2. Let B£(O) = {x E E III x II $ e}. For y E Br /2(O), let g/x) = y + x. If y II g(x) II $ r(2, so

E

B,12(O) and xl' X2 E B,(O), then II y II $ r(2 and II ~(x) II $ II y II + II g(x) II $ r, (i)

and, by the Mean Value Inequality, (ii) II ~(xI) - gi x2) II $11 xl - x211/2. Thus by Lemma 2.5.3, ~ has a unique fixed point x in BiO). This point x is the unique solution of f(x)

=y.

Thus f has an inverse

From (ii) with y = 0, we have

II (xl - f(x I» - (x 2 - f(x 2» II $11 Xl - x211(2, and so

i.e., Thus we have (iii) II f-I(YI) - f- I(Y2) I so f- is continuous.

II $ 211 YI - YzlI,

From Lemma 2.5.4 we can choose r small enough so that Df(x)-I exists for x Moreover, by continuity, well. If y\, Yz

E

II Df(x)-I II

M for some M and all x 0,12(0), Xl = f-I(y\), and X z = l I(Y2)' then $

II f-I(YI) - f- I(Y2) - Df(x 2)-I. (YI - Y2) II

E

E

0iO).

0,(0) can be assumed as

= II x l - x2 - Df(x2)-1 . [f(x I) - f(x z») II = II Df(x 2)-I. (Df(x 2)· (xl - x2) - f(x l ) + f(x 2)} $ Mil f(x I ) - [(x 2) - Df(x 2) . (xl - x2)

II

II .

This, together with (iii), shows that II is differentiable with derivative Df(x)-I at f(x); i.e.,

Section 2.5 The Inverse and Implicit Function Theorems

119

D(f- I ) = 10 Df 0 f- I on Vo = Dr12 (O). This formula, the chain rule, and Lemma 2.5.5 show inductively that if f- I is Ck - I then f- I is Ck for 1:<:; k:<:; r. • This argument also proves the following: if f: U E and V

C

C

F are open sets, and Df(u)

E

~

V is a Cr fwmeomorphism where U

aL(E, F) for u

E

U, then f is a

cr

diffeomorphism. For a Lipschitz Inverse Function Theorem see Exercise 2.5K.

~

SUPPLEMENT 2.SA The Size of the Neighborhoods in the Inverse Mapping Theorem An analysis of the preceeding proof also gives explicit estimates on the size of the ball on which f(x) = y is solvable. Such estimates are sometimes useful in applications. The easiest one to use in examples involves estimates on the second derivative (we thank M. Buchner for suggestions about this). 2.5.6 Proposition Suppose f: U isomorphism. Let

C

E

~

F is of class Cr, r

~

2, Xo

E

U and Df(x o) is an

Assume

Let p=

minC~M ' R).

Q= minC~, ~, P).

S = min(

2~M' ~, Q).

Here N = 8M 3 K Then f maps an open set G C Dp(x o) dijfeomorphicallyonto D p12M (Yo) and r l maps an open set He DQ(yo) diffeomorphicallyonto DQf2L(x O)' Moreover, BQf2L(xO) C G C

Dp(x o) and BSf2M (yo)

Proof We can assume Xo

Df(x) = Df(O) +

f

C

H

=0

C

DQ(yo)

C

D pf2M (Yo)' See Figure 2.5 .1.

and f(x o) = O. From

D(Df(tx» . x dt = Df(O)' { 1+ [Df(O)r l . fol D 2f(tx) . x dt}

and the fact that 11(1 + Arl II :<:; 1 + II A II + II A 112 + '" = - - -

l-IIA II

for II A II < 1 (see the proof of 2.5.5), we get

120

Chapter 2 Banach Spaces and Differential Calculus - I .

II Dfi( x) II i.e., if

$

2M

If

II x II

$

R

II x II

and

$

1 2MK

II x II $ P.

Figure 2.5.1 As in the proof of the Inverse Function Theorem, let g/x) = [Df(O»)-1 . (y + Df(O)x - f(x». Write q>(x) = Df(O)' x - f(x) =

fo

I

to obtain ~(x) = [Df(O)-I) . (y + q>(x»,

-1 J

I rl

Dq>(sx)' x ds =

2 o D f(tsx)· (sx, x) dt ds

II q>(x) II $ KII x 112 if II x II $ P, and

II ~(x) II $ M(II y II + KII x 112) . Hence for II y II $ P/2M, gy maps Bp(O) to B5 (0). Similarly we get II g/x l ) - ~(x2) II $ II xI -

Xz 1112 from the Mean Value Inequality and the estimate

if II x II $ P. Thus, as in the previous proof, r-I : B p!2M(O) ~ Bp(O) is defined and there exists an open set G

C Bp(O) diffeomorphic via f to the open ball D Pf2M (O). Taking the second derivative of the relation r- I 0 f= identity on G, we get

D 2f- l (f(x»(Df(x) . ul' Df(x) . u2) + Df-I(f(x») . D 2f(x)(ul' u 2) = 0 for any ul' u 2

E

E. Let Vj = Df(x) . uj, i = 1,2, so that

D 2r- I(f(x» . (v!' v 2 ) = - Dr-I (f(x» . D 2 f(x)(Df(x)-1 . VI' Df(x)-l . v 2) and hence

121

Section 2.5 The Inverse and Implicit Function Theorems

since x

E

G

C

Dp(O) and on Bp(O) we have the inequality

II Df(x)-l II :-:; 2M. Thus on

Bp12M (O) the following estimate holds:

By the previous argument with f replaced by f- I, R by P/2M, L by M, and K by N = 8M3 K, it follows that there is an open set He DQ(O), Q = mini I/2KM, Q/2L, Q} such that II : H ~ D Q/2L(O) is a diffeomorphism. Since II is a diffeomorphism on BQ(O) and H is one of its open subsets, it follows that BQ12L(O) C G. Finally, replacing R by Q/2L, we conclude the existence of a ball B S12M (O), where S = mini I/2KM, Q/2L, Q). on which II is a diffeomorphism. Therefore B S/2M (O)

C

H. •

fl:n

In the study of manifolds and submanifolds, the argument used in the following is of central importance. 2.5.7 Implicit Function Theorem Let veE, V

C

F be open and f: V x V

~

G be

cr,

r

~

V, Yo E V assume Di(x o' Yo) : F ~ G is an isomorphism. Then there are neighborhoods Vo of Xo and Wo of f(x o' Yo) and a unique cr map g: Vo x Wo ~ V such 1. For some Xo

E

thatJor all (x, w)

E

Vo x W o' f(x, g(x, w)) = w .

Proof Define the map q,: V x V

~

E x G by (x, y) ...... (x, f(x, y». Then Dq,(x o' Yo) is given

by

which is an isomorphism of Ex F with Ex G. Thus q, has a unique cr local inverse, say q,-I : Vo x Wo ~ V x V, (x, w) ...... (x, g(x, w». The g so defined is the desired map. • Applying the chain rule to the relation f(x, g(x, w», one can compute the derivatives of g: DIg(x, w) = - [D 2f(x, g(x, w))]-I

0

Dlf(x, g(x, w» ,

D 2g(x, w) = [D2f(x, g(x, w))]-I . 2.5.8 Corollary Let veE be open and f: V ~ F be

cr,

r ~ 1. Suppose Df(x o) is surjective

and ker Df(x o) is complemented. Then f(V) contains a neighborhood of f(x o)' Proof Let EI = ker Df(x o) and E = EI EEl E 2. Then D/(x o): E2 ~ F is an isomorphism, so the hypotheses of 2.5.7 are satisfied and thus f(V) contains Wo provided by that theorem. •

122

Chapter 2 Banach Spaces and Differentiol Calculus

Since in finite-dimensional spaces every subspace splits, the foregoing corollary implies that if f: U

C

]R.n --t ]R.m, n

~

m, and the Jacobian of f at every point of U has rank m, then f is

an open mapping. This statement generalizes directly to Banach spaces, but it is not a consequence of the Implicit Function Theorem anymore, since not every subspace is split. This result goes back to Graves [1950]. The proof given in Supplement 2.58 follows Luenberger [1969]. E --t F is C l and Df(uo) is onto for some U o U, then f is locally onto; i.e., there exist open neighborhoods U I of U o and VI of f(uo)

2.5.9 Local Surjectivity Theorem If f: U E

C

such that fl U I : U I --t VI is onto. In particular, open mapping. ~

if

Df(u) is onto for all u

E

U, then f is an

SUPPLEMENT 2.SB

Proof of the Local S urjectivity Theorem Proof Recall from Section 2.1 that E/ker Df(uo) =Eo is a Banach space with nom1 lI[x]11 = inf{ II x + u III u E ker Df(u o)}, where [x] is the equivalence class of x. To solve f(x) = y we set up an iteration scheme in Eo and E simultaneously. Now Df(u o) induces an isomorphism T: Eo --t F, so j l E L(F, Eo) exists by the Banach Isomorphism Theorem. Let x =Uo + h and write f(x)

=y

as jl(y - f(uo + h»

=0

.

To solve this equation, define a sequence Ln

» and ~

of ker Df(uo

E

Ln

C

E E/ker Df(u o) (so the element Ln is a coset E inductively by Lo =ker Df(uo)' ho E Lo small, and

(1)

and selecting hn E Ln such that

(2) The latter is possible since II Ln - L n_1 II = jl(Df(uo) . hn_I), so

= inf{ II h -

hn_1 II

I

h

E

Ln)' Since hn_1 E Ln_l' Ln_1

Subtracting this from the expression for Ln_1 gives

For £ > 0 given, there is a neighborhood U of U o such that II Df(u) - Df(u o) II < £ for u E U, since f is C I. Assume inductively that U o + ~_I E U and U o + hn_2 E U. Then from the Mean Value Inequality,

(4)

Section 2.S The Inverse and Implicit Function Theorems

123

By (2).

Thus if

I':

is small,

TII ho II. Uo + ~ remains inductively in U since

Starting with ho small and II hI - ho II <

II ~ II

~

II ho II + II hI - ho II + II h2 - hI II + ... + II

~

(1 +

~ + ... +

2n 1_ I

II ho II

)

~

~

- hn _ 1 II

2 II ho II .

It follows that h n is a Cauchy sequence, so it converges to some point. say h. Correspondingly, Ln converges to L and hE L. Thus from (1). 0 = il(y - f(u o + h» and so

y = f(u o + h). • This proves that for y near Yo = f(u o)' f(x) = y has a solution. If therc is a solution g(y) which is CI, then Df(x o) 0 Dg(yo) = I and so range Dg(yo) is an algebraic complement to ker Df(x o)' It follows that if range Dg(yo) is closed, then ker Df(x o) is split.

=x

In many applications to nonlinear partial differential equations, methods of functional analysis and elliptic operators can be used to show that ker Df(x o) does split, even in Banach spaces. Such a splitting theorem is called the Fredholm alternative. For illustrations of this idea in geometry and relativity, see Fischer and Marsden [1975], [1979], and in elasticity, see Marsden and Hughes [1983, Chapter 6]. For such applications, 2.S.8 usually sufficies.

/l:;J

The locally injective counterpart of this theorem is the following. 2.5.10 Locallnjectivity Theorem Let f: U

C

E ~ F be a CI map. Df(uo)(E) be closed in

F. and Df(u o) E G.L(E, Df(uo)(E». Then there exists a neighborhood V of U o. V which f is injective. The inverse r- I : f(V) ~ U is Lipschitz continuous.

C

U, on

Proof Since (Df(uo»-I E L(Df(uo)(E), E). there is a constant M > 0 such that II Df(u o) . e II ;:;Mil e II for all e E E. By continuity of Df, there exists r > 0 such that II Df(u) - Df(u o) II < M/2 whenever II u -

Uo

II < 3r. By the Mean Value Inequality. for el' e 2 E D/uo)

II f(el) - f(e2) - Df(Uo)(el - e2) II

sup II Df(el + t(e2 - el» - Df(uo) 1111 el - e21!

~ t E

since II Uo - e l

-

t(e2

-

e l ) II < 3r. Thus

[0. Il

Chapter 2 Banach Spaces and DUJerentUd Calculus

124

i.e. ,

which proves that f is injective on Dr(uO) and that II : f(D/Uo» -t U is Lipschitz continuous .

•

We now give an example of the use of the Implicit Function Theorem to prove an existence theorem for differential equations. For this and related examples, we choose the spaces to be infinite dimensional. In fact, E, F, G, ... will be suitable spaces of functions. The map f will often be a nonlinear differential operator. The linear map Df(x o) is called the linearization of f about xo' (Phrases like "first variation," "first-order deformation," and so forth are also used.) 2.5.11 Example Let E be the space of all CI-functions f: [0, 1] -t lR with the norm

II fli!

=

sUP I f(x) I + sup

XE

[0,1]

XE

[6,1]

df(x) I I -dx

and F the space of all CO-functions with the norm II f 110 = sUPx E

[0.1]

I f(x) I. These are Banach

spaces (see Exercise 2.1C). Let : E -t F be defined by (f) = df/dx + f3. It is easy to check that is

C~

and D(O) =d/dx : E -t F. Clearly D(0) is surjective (Fundamental Theorem of

Calculus). Also ker D(O) consists of EI

= all

constant functions. This is complemented

because it is fmite dimensional; explicitly, a complement consists of functions with zero integral. Thus Corollary 2.5.8 yields the following: There is an e >

°

such that if g : [0, 1] -t lR is a continuous function with

I g(x) I < e, then

there is a C I function f: [0, 1] -t lR such that

df ..3 dx + r(x) = g(x) . •

~ SUPPLEMENT 2.5e An Application of the Inverse Function Theorem to a Nonlinear Partial Differential Equation

Let

nC

lRn be a bounded open set with smooth boundary. Consider the problem

Section 2.5 The Inverse and Implicit Function Theorems

125

for given f and g. We claim that for f and g small, this problem has a unique small solution. For partial differential equations of this sort one ean use the Sobolev spaces HS(n, 1R) consisting of maps
n

~ IR whose first s distributional derivatives lie in L2.

(One uses Fourier

transforms to define this space if s is not an integer.) In the Sobolev spaces E H S- 2(n, 1R) x HS-1I2(an, lit), if s > nl2 the map

is

c~

= H'(n, lit),

F

=

(use Supplement 2.4B) and the linear operator

is an isomorphism. The fact that D(O) is an isomorphism is a result on the solvability of the Dirichlet problem from the theory of elliptic linear partial differential equations. See, for example, Friedman [1969]. (In the Ck spaces, D(O) is not an isomorphism.) The result claimed above

;lJJ

now follows from the Inverse Function Theorem.

The following series of consequences of the Inverse Function Theorem are important technical tools in the study of manifolds. The first two results give, roughly speaking, sufficient conditions to "straighten out" the range (respectively, the domain) of f in a neighborhood of a point, thus making f look like an inclusion (respectively, a projection). 2.5.12 Local Immersion Theorem Let f: U

C

E

~

F be of class Cr , r

~

1,

110 E

U and

suppose that Df(u o) is one to one and has a closed split image Fl with closed complement F2. (If E = IRm and F = 1R", assume only that Df(u o) has trivial kernel.) Then there are two open sets U' C F and veE EEl F2 where f(u o) E U' and a C diffeomorphism
= Fl = 1R2,

F2

= IR

(i.e., m

= 2,

n

~

V such

= 3) is given in Figure 2.5.2.

The

function

and B E L(E', P) is defined by (A, B)(e, e') = (Ae, Be').) By the Inverse Function Theorem there exist open sets U' and V such that (uo' 0) Eve E EEl F 2, and g(uo' 0) = f(u o) E U'

C

F

126

Chapter 2 Banach Spaces and DifJerentUzl Calculus

and a Cr diffeomorphism q>: U' ~ V such that q>-' (q>

0

g)(e, 0)

= (e, 0).

=g I V.

Hence for (e, 0)

E

V, (q>

0

f)(e) =

•

Figure 2.5.2 2.5.13 Local Submersion Theorem Let f: U C E ~ F be of class cr, r 2" 1, U o E U and suppose Df(u o) is surjective and has split kernel E2 with closed complement E,. (If E = jRm and F =JRD, assume only that rank (Df(uo» = n.) Then there are open sets U' and V such that U o E U' cue E and V C F (JJ ~ and a Cr diffeomorphism '1': V ~ U' with the property that (f 0 'I')(u, v) =u for all (u, v) E V. The intuition for E,

= E2 = F = jR

is given in Figure 2.5.3, which should be compared to

Figure 2.5.2. Note that this theorem implies the results of 2.5.9, the Local Surjectivity Theorem, but the hypotheses are more stringent. Proof By the Banach Isomorphism Theorem (Section 2.2), D,f(u o) E aL(E" F). Define the map g: U C E, (JJ E2 ~ F (JJ E2 by g(u!' u2) = (f(u!, u 2), u2 ) and note that

so that Dg(u o) E aL(E, F (JJ E 2). By the Inverse Function Theorem there are open sets U' and V such that Un E U' cue E, V C F (JJ ~ and a Cr diffeomorphism '1': V ~ U' such !hat 'I' -, = g I U'. Hence if (u, v) E V, (u, v) = (g 0 'I')(u, v) = (f('I'(u, v», 'I'/u, v», where

Section 2.5 The Inverse and Implicit Function Theorems

\jf = \jfl

X

\jf2: i.e., \jf2(u, v) = v and (f 0 \jf)(u, v) = u.

E,

f

0

~ constant

'"

E,

1

f

127

•

~ constant

~~~ F

------'---F

E,

Figure 2.5.3

2.5.14 Local Representation Theorem Let f: veE ~ F be of class cr, r > 1, U o E V and suppose Df(u o) has closed split image FI with closed complement F2 and split kernel Ez with closed complement E I. (If E = jRffi, F= jRn, assume that rank (Df(uo)) = k, k ~ n, k ~ m,

so that F2 = jRn-k, FI = jRk, EI = the property that

Uo E

E2 =

jRk,

V' eve E, V

C

that (f 0 \jf)(u, v) = (u, TJ(u, v)), where TJ : V Proof Write f = fl

X

and thus there exists a fl

0

f 2, where f i : V

cr

~

jRffi-k.)

Then there are open sets V' and V with

FI E9 E2 and a ~

Fz is a

cr Cr

diffeomorphism \jf: V ~ V' such

map satisfying DTJ(uo) = o.

Fi , i = 1, 2. Then f satisfies the conditions of 2.5.13,

diffeomorphism \jf: V

\jf is given by (flo \jf)(u, v) = u. Let TJ = f2

C 0

FI E9 E2

~

V'

C

E such that the composition

\jf. •

To use 2.5.12 (or 2.5.13) in finite dimensions, we must have the rank of Df equal to the dimension of its domain space (or the range space). However, we can also use the Inverse Function Theorem to tell us that if Df(x) has constant rank k in a neighborhood of xo' then we can straighten out the domain of f with some invertible function \jf such that f 0 \jf depends only on k variables. Then we can apply Proposition 2.5.12. This is the essence of the following theorem. Roughly speaking, the theorem says that if Df has rank k on jRffi, then m - k variables are redundant and can be eliminated. As a trivial example, if f:

jRz

~

jR

is defined by

setting f(x, y) = x - y, Df has rank 1, and so we can express f using just one variable, namely, let \jf(x, y) = (x + y, y) so that (f 0 \jf)(x, y) = x, which depends only on x.

2.5.15 Rank Theorem Let f: veE ~ F be of class Cr , r;:' 1, U o E V and suppose Df(u o) has closed split image FI with closed complement F2 and split kernel E z with closed Complement E I. In addition, assume that for all u in a neighborhood of U o E V, Df(u)(E) is a closed subspace of F and Df(u) I EI : EI ~ Df(u)(E) is a Banach space isomorphism. (In case E:: jRffi and F = JRR, assume only that rank (Df(u)) = k for u in a neighborhood of uo') Then there exist open sets VI C FI E9 E2, V 2 c: E, VI C F, and V 2 C F and there are c r

Chapter 2 Banach Spaces and Differential Calculus

128

diffeomorphisms q>:

VI ~ V 2

and'll: U I

~

U 2 such that (q>

0

f

0

'V)(x, e) = (x, 0).

The intuition is given by Figure 2.5.4 for E = ]R2, F =]R2 and k = 1.

F,

F,

£,

Figure 2.5.4 Remark It is clear that the theorem implies EI EEl ker(Df(u» = E and Df(u)(E) EEl F2 = F for u in U o in U, because q> 0 f 0'11 has these properties. These seemingly stronger conditions can in fact be shown directly to be equivalent to the hypotheses in the theorem by the use

a neighborhood of

of the openness of aL(E, E) in L(E, E). Proof By the Local Representation Theorem there is a Cr diffeomorphism'll: U I

C

FI EEl E2

~

'V)(x, y) = (x, ll(x, y». Let PI: F ~ FI be the projection. Since Df(x, y) . (w, e) = (w, Dll(x, y) . (w, e», it follows that PI 0 Df(x, y»(w, e) = (w, 0), U2

C

E such that f(x, y) := (f

0

for WE FI and e E E2. In particular PI 0 Df(x, y) I FI x (OJ = I\, the identity on FI' which shows that Df(x, y) I FI x {OJ: FI x (OJ ~ Df(x, y)(FI EEl Ez) is injective. In finite dimensions this implies that it is an isomorphism, since dim(F I ) = dim(Df(x, y)(FI EEl E 2». In infinite dimensions this is our hypothesis. Thus we get Df(x, y) 0 PI I Df(x, y)(FI EEl E 2) = identity. Let (w, Dll(x, y)(w, e» (Df(x, y)

0

E

Df(x, y)(FI EEl

Ez).

PI)(w, Dll(x, y). (w, e»

Since Df(x, y). (w, 0) (w, Dll(x, y) . (w, 0»

(w, D\ll(x, y) . w)

Section 2.5 The Inverse and ImpliciJ Function Theorems

we must have DTl(x. y) . e = 0 for all e

E

129

E2 ; i.e .• D 2Tl(x. y) = O. However. D 2 f(x. y) . e =

(0. D 2Tl(x. y) . e). which says that D 2 f(x. y) = 0; i.e .. f does not depend on the variable y E

E2. Define fix)

=f(x. y) =(f

0

'V)(x. y). so f: P,'(V)

C

F, ~ F where P,': F, EB E2 ~ F, is

the projection. Now f satisfies the conditions of 2.5.12 at P ,'('I"(u o)) and hence there exists a C r diffeomorphism
0

f

0

'V)(x. y) = (x. 0). •

2.5.16 Example Functional functions fl' .... fn : U

~

Dependence Let U

C

JR.n

be an open set and let the

JR. be smooth. The functions f" ...• fn are said to be functionally

»

U if there is a neighborhood V of the point (f, (x o)•...• fn(x O E JR.n and a smooth function F: V ~ JR. such that DF"# 0 on a neighborhood of (f,(xo)' ...• fn(x O and

dependent at Xo

E

»'

for all x in some neighborhood of xO' Show: (i) If f" .... fn are functionally dependent at xo. then the determinant of Df. denoted

vanishes at Xo' (ii) If

on a neighborhood of xo. then fl' .... fn are functionally dependent, and fn = G(fl' "',[n-') for some G. Solution Let f

= (fl' .... fn)'

0 f = O. so DF(f(x» 0 Df(x) = o. Now if Jf(x o)"# O. Df(x) would be invertible in a neighborhood of xo. implying DF(f(x» = O. By the Inverse Function Theorem.

(i) We have F

this implies DF(y) = 0 on a whole neighborhood of f(x o)' (ii) The conditions of (ii) imply that Df has rank n - 1. Hence by the Rank Theorem. therc are mappings