Nonstandard Analysis

(±


I

2

+cos~) x

is defined for x> O.

y

Fig. 9.4

113 Thus, for positive x, there are three regions of definition off

R + defined by y ~ +(x) where the slopes are positive, R _ defined by y ~ _(x) where the slopes are negative, R transition region between the graphs of + and -. The transition region is situated between the graphs of the two functions y=

±ix 3 •

9.3.2

Families of solutions

Here are two continuous families of solutions of the differential equation y' =f(x,y). For a

~

0, we have the family a for x ~ 0 { y= a+2x 2 forxE[10,1]

and for a

~

0, the family

°

y- { a for2 x ~ - a - 2x for x E [0, 1]. We discover the two solutions satisfying y(0) = 0 (I do not know if these are the only ones ...) y=

9.3.3

o for x ~ 0 { ± 2x 2 for XE [0,1].

Dependence of choice of infinitesimal

Take now an infinitesimal h >0 ofthe form h = 11k where k is an illimited odd integer. The approximate solution v goes through the points Po = (0,0),

PI = (h,O)ER+,

hence also through P 2 = (2h,4h 2 ), •••• Let us show that all subsequent points remain in R +. We even see that P 2 is situated above the graph of y = (3/2)x 3 hence also in R +: h < 1/3 => 4h 2 > 12h 3 = (3/2)(2h)3. Let us write

PlI PlI + I

(nh, all h2 ), = ((n+ I)h, (a,,+4n)h 2 •

=

Then we can show that if P is in R +, the same happens for P 11

11

+ l'

at least if

114

DIFFERBNTIAL BQUATIONS

x(PII) = nh < 2/3:

> (3/2)n 3 h => all + 1 = an + 4n > (3/2)n 3 h + 4n a,,+ 1 > (3/2)h(n 3 + 8n/3h). all

Only remains to check

n3 + 8n/3h > (n + 1)3

=

n3 + 3n 2 + 3n + 1

or

8n/3h> 3n 2 + 3n + 1,

h < 8n/(9n 2 + 9n + 3)

which is obviously true! Consequently, the solution constructed in the proof 9.2.2 would agree with 2x 2 on [0,2/3] hence also on [0, 1] (there is only one solution of our system going through the point Xo = 2/3 and Yo = 2(2/3)2 = 8/9). Starting from an infinitesimal h = 11k where k is an illimited even integer, we would similarly see that PIE R _ and all subsequent points would also remain in

R_. Thus, according to the nature of the infinitesimal h > 0, the solution with u(O) = 0 could be either 2x 2 or - 2x 2 (for x ~ 1).

CHAPTER

10

PERTURBATION OF A GREEN FUNCTION 10.1

10.1.1

GREEN FUNCTION

Review

We shall only consider the particular differential equation

y"-Aly=O

(with a real parameter A>O)

to which we refer by the initials ED. We are looking for solutions satisfying the conditions

y(O)=O

and

yEL 2(O, (0).

Let us recall that the Green function K of ED is the family of continuous functions K p (or the function with two variables K(r, p)= Kp(r» defined for p > 0 by the conditions (a) K p is a solution of ED except at most at r = p, (b) K p(O) = 0 and KpEL 2 (0, 00), (K p satisfies the boundary conditions), (c) K~ is discontinuous at p, with a unit jump at this point. Moreover, the preceding conditions imply the following symmetry (d) Kp(r) = K(r, p) = K(p, r)= K,(p).

The usefulness of this Green function is exhibited in the solution of the inhomogeneous equation

116

PBRTUlBATION OF A GRBEN FUNCTION

satisfying y(0) = 0 and y E L 2 (0,

00)

in the form

y(p)= [Kp(r)f(r)

dr.

10.1.2 Computation of the Green function As the solutions of ED are the linear combinations of exponentials exp( ± Ar) (or of Sh Ar and Ch A.r), we must have

Kp(r)=AShAr

for r
since Kp(O)=O.

Similarly,

Kp(r)=Be-;.r

since K p EL 2 (0, 00).

for r>p

Continuity of K p at r= p gives the condition

ASh Ap=Be-;'P and the unit jump of K p at r= p leads to

- lB e - J.p - A.A Ch lp = 1. These two equations can be solved and lead to the values

B= - l -1 Sh lp. Thus, we have found

K p(r)=

10.2 10.2.1

{ - A. - 1 e - J.p Sh lv' for -,t -1 Sh I.p e-J.r for

r

p.

NONSTANDARD PERTURBATION

Nonstandard differential equation

Let us now consider a perturbed differential equation ED' yO +(V _).2)y=0

with the same limit conditions y(0) = 0 and y E L 2 (0, (0). More precisely, we shall consider the case of a perturbation V = Vt of the following type a2 /f,2 Vt(r) = {

o

for

r <E:

for

r

(a >0)

~ f;

where c > 0 is infinitesimal: the support of the perturbation V is small, but the height ofthe perturbation itselfis large. The Green function ofthis equation is still defined: in particular, the derivative K ~ is required to be continuous at r = e.

117

NONSTANDARD PERTU1tIAnON

10.2.2 Computation of the Green function The form of the Green function for positive values of r is given in the following figure.

o

p

t

r

I

I

I

\:

\

I

4

¥

A sin ~ r E

Fig. JO.l

We have introduced a constant J1 satisfying J12 / [;2 = a2/ e2 - A. 2 i.e. Il = J(a 2 -£2 A. 2).

As before, the constants A, B, C, D depend on A, e and p. The continuity conditions can be written (1) A sin Il=B eAt+C e-:U, (2) B eAP + C e- AP =D e-;'P, (3) AilE - lCOS Ji. = BA eAr; - C..l

e - At.

We still have to write the condition giving a unit jump of (4) -

K~

at r=p.

D..l e-),P - BA. e AP + C;.. e-;'P = 1.

These four equations can be solved (the assiduous reader should check our computations ...) and furnish A= -

e),(t -

pI /

(Ji.£ - 1 cOSJi. + AsinJi.),

B = - e -Ap /(2..l), A

C = e (2t: - p) . Ji. cOSJl- ..le sinJi. 2..l J1 cOSJ1 + At sin Jl' AP D = C _e /(2A).

10.2.3 Discussion For this computation, we shall assume that the parameters a, ..l and pare appreciable (Le. limited and not infinitesimal), while E > 0 is infinitesimal. Since Jl=J(a 2 -e 2).2), we see that Jl~a. For comparison, we recall that the Green function for r

e -;'p

,_ e -).p

.

___ . eAr +--' e - Jor

2A

2A

'

118

PERTURBATION OF A GREEN FUNCTION

and in the perturbed case by (for e
e

2£).

.

Jl COSJl- ,i.e sin Jl

Jl cosJ.1 + ,i.e sin Jl

p, cOSJl- ).e sin Jl p, cos Jl + ,i.e sin Jl

"'------

This term describes the visible effect of the perturbation for r near p (or r < p). If Jl cos p, is not infinitesimal, this term is still:::::: 1and the perturbation has little effect. However, if p, COSJ,1::::::0,

i.e. J,1::::::(n+ 1/2)n,

the fraction Jl cos p, - Ae sin J,1 Jl cos Jl + Ae sin Jl

can be far from I! Indeed, this happens if we take the parameter a in the perturbation VI: of the form a=(n+ 1/2)n, with nelL limited (i.e. standard). In this case Jl* = st(,u) = a, and more precisely Jl = a(l- e2 ).2Ja'2)1/2 = a(1-e 2 ,i.'2/2a'2 +...) =a-e 2,i.'2/2a+ ... =(n+ 1/2)n-e'2A'2/2a+ ... Jll; - 1 cos Jl:::::: ± Jle - lsin e2,i. '2 /2a:::::: ±Jlel'2 2a infinitesimal.

The fraction Jl cos Jl- ;.e sin p, __ -,i. sin Jl Jl cos Jl + Ae sin p,

_ 1

+,i. sin Jl

produces an observable perturbation at an appreciable distance from the origin. The origin ofthis perturbation lies in the behaviour of the Green function in the first interval [0, e]. In this part, we have Kp(r) = A sin p,r/e

with A = -e eA(t- P ) !(p, cos Jl + ).e sin Jl)

so that A::::::O

when Jl cos Jl is not infinitesimal.

In this case, the perturbation is small since its effect is to replace the condition Kp(O)=O by Kp(e)::::::O: this difference cannot be detected experimentally. If we assume that a=(n+ 1/2)n so that p.e-l cos Jl::::::O, we shall have A= -

e).(t.- 1') /(jJ.f.

- 1

cos jJ. +). sin Jl) ~ - e -).1'!(), sin Jl)

'119

OllOIN OF THE PROBLBM

appreciable and thus Kp(e) = A sin J.l~ -e-J.P/).

appreciable.

The perturbation has now the visible effect of replacing the condition Kp(O)=O (unperturbed case) by Kp(e)~ -e-·'.P/l not infinitesimal. 10.3 10.3.1

ORIGIN OF THE PROBLEM

Schroedinger's equation

In physics, the wave function of a particle is a solution of the Schroedinger equation: this is a partial differential equation of the type

Jr'P = _ ~ a'P = ih 0'1' i

at

at

where .Yf denotes the Hamiltonian operator of the system (and h is Planck's constant divided by 2n). Let us assume that this Hamiltonian does not depend on time t and is simply given by h2

.1f= - 2m~+

V,

with a certain potential V, only depending on the space coordinates. The stationary solutions of Schroedinger's equation are those having the form 'I' = e - iwt I/J whence 0'1' ih - = ih ( - iw)' 'I' = hw· 'II = E'P

at

since hw = E is the energy of the particle. After a simplification by e - iw" we find the equation for the space part'" of 'II ~"'+2mh-2(E

- V)I/J=O.

With an attractive potential, bound states occur and we shall focus our attention to this case, namely v= -h 2 VJ2m<0, E= -h 2).2j2m
10.3.2 Spherical coordinates

It is well known that the Laplace operator in spherical coordinates is given by ~"'=t/J"+2r-l""+r-2·nt/J

(where

t/J'= ot/J/or ...)

120

PERTURBATION OF A aREEN FUNCT10N

with a partial differential operator n only acting on the angular variables oft/!. Let us consider radial functions only, so that the Laplace operator is simply given by

&l/J =11/' +2r- 1 t/J'. In this case, the above Schroedinger equation reduces to t/J" + 2,-1 ""

+(Vt -).2)l/J =0

or to rl/J" +2"" +(Vt

).2)r'" =0.

-

Since rl/J" + 2t/J'+(r",r, we shall put r"'=y and finally get the equation y" +(Vt -).2)y=0

precisely of the type considered in 10.2.1. It is also interesting to translate the limit conditions for'" in terms of y. Since the function t/J should be continuous at the origin, we have to impose y(0) = rt/J(r),.=o =0

(more precisely, the continuity of t/J at the origin implies that y(r) = (9 (r) is of the order ofr for r-+O: this condition is automatically satisfied for the solutions ofthe differential equation satisfying y(0) = 0). Finally, the condition t/J E L 2 (~3) gives in spherical coordinates 2

ao> 11"'11 =

II

2

I:

2

II"'1 • r dr dcr=4n

2

Ir"'1 dr=4n

f.

2

IYI dr.

This shows that y E L 2 (0, (0) and up to the factor 47[, t/J~rt/J= y:L 2 (~3)-+L2(0, (0) is an isometry.

10.3.3

NOI~tationary

problem

It is interesting to observe that the resolution of the nonstationary problem also uses Green's function since 0'11jot = -(ijh)Jf\f'

is solved by the evolution operator U,=exp( -(ijh).tt't) 'II(r, t)= U,' 'P(r, 0). This evolution operator can be computed by means of the functional calculus

f(JIf)= lim (27[i)-1 t-O+

f R

(RA+lt-RA-it)!().)d)..

ORIOIN OF THE PROBLEM

121

wherefdenotes any analytic function on R (the exponential in our case) and R is the resolvant of the autoadjoint operator .1f R). =(.1f _).1)-1.

This resolvant is precisely an integral operator having the Green function as kernel!

10.3.4 Final comment Recall that the perturbation was produced by a potential of the form V£(r} = {

a2 /e 2 forr
o

forr

~ 8

It is really interesting to observe that the perturbation produced by the short range potential V£ is observable in spite of the fact that the measure V£ dx dy dz is infinitesimal. The potential V£ has its support in the baD of radius 8 (with volume 2 2 47[8 /3) and its intensity is only ofthe order of l/e • In particular V£ is not a Dirac function: for any standard continuous function defined in a neighbourhood of 0 in ~, we have

8tfffF dxdydz =0. However, the perturbation is not negligible for certain limited values of the parameter a.

Ce qui a ete compris n'existe plus! (Paul Eluard) [What has once been understood ceases 10 exist!]

CHAPTER

11

INVARIANT SUBSPACES 11.1 11.1.1

SITUATION OF THE RESULT

Historical comments

Continuous linear operators (we shall simply say 'operator') in Hilbert spaces have been extensively studied since the beginning of the century. In particular, attempts have been made to generalize results obtained in the finite dimension. However. it is still unknown if these operators always possess nontrivial closed invariant subspaces. For special classes of operators, such results can be established. In particular, Hermitian or compact operators have many such spaces. This is clear by Riesz theory for compact operators having a spectrum different from {O} since any nonzero spectral value is an eigenvalue in this case (the corresponding eigenspace being finite dimensional). Compact operators having spectrum {O} are also called quasi-nilpotent and also have nontrivial invariant subspaces: this has been shown in an article [1] by Aronszajn and Smith. The problem for operators having a square which is a compact operator was mentioned by Halmos in a series of open problems which remained open until the Bernstein-Robinson article [2] appeared. We shall give an internal translation of this paper here (i.e. using Nelson's approach of NSA). Historically, the appearance of [2] was a first success of NSA: it showed that this theory is capable of solving unsettled conjectures. Specialists were of course puzzled by the method and Halmos himself translated the proof (with some help from Robinson) into classical terms. This lead him to publish [4] (both [2] and [4] appeared in the same issue of the Pacific J. of Mathematics!). Although Robinson's proofis simpler and more conceptual than Halmos' (this is particularly true of the internal version that we are going to explain), it lost some interest when Lomonosov gave a simpler proof [9] using the

124

rNV ARIANT SUBSPACES

Leray-Schauder fixed point theorem. Hilden was even able to avoid recourse to this fixed point theorem and present a completely elementary classical proof of the same result (cf. [5]). 11.1.2 Bernstein-Robinson Theorem Let H be an infinite dimensional Hilbert space and TEL(H) be an operator. If there exists a polynomial P:F 0 such that the operator p( 1) is compact, then T has a nontrivial closed invariant subspace. 11.1.3 An example Perhaps the simplest (trivial) example of a noncompact operator satisfying the assumptions of the Bernstein-Robinson Theorem is a projector. Indeed, projectors P are characterized by the property of being equal to their square: p 2 = P. The nonzero polynomial p(X) = X 2 - X thus vanishes on all projectors, and thus p(P)=O is a compact operator! Obviously, projectors in spaces of dimension greater than I always have nontrivial invariant subspaces.

11.2 11.2.1

PRELIMINARY RESULTS

Review of a definition

Recall that if A is a subset of a standard metric space X, we can define a standard part of A by A*=S{XEX: there is an yEX with dist (x, y)~O}.

Instead of dist (x, y)~O, we shall sometimes write simply We shall have to use the basic facts

x~y.

A * is always closed, if A is standard, then A * = A contains A

(they were established in 3.3.5-3.3.6). Moreover, it is clear that if H is a standard nonned vector space, and A = V is a vector subspace, then A * = V* is also a vector subspace of H.

11.2.2 Lemma For two subspaces We V of H dim

V/W~

I=>dim

V*/W*~

1.

11.2.3 Proof Assume that the codimension of W in V is ~ I and take arbitrary standard elements Xl' X2EV*. Thus, there are YlEV with Xl ~Yi' From the condimension

t'2S

PRELIMINARY RESULTS

hypothesis, there is a nontrivial linear relation a.y. +a2Y2EW. Dividing the coefficients of the preceding relation by a=ai if lail = max (a., a2»O, we can assume that one Q i = 1 and both are limited (more precisely, lail ~ 1). Taking standard parts, we can write

a!x t +Q!x2~a!Yl +a!Y2~alYl +Q2Y2 eW This proves indeed

a!x t +a!x 2eW*. Since this last relation is nontrivial (one of its coefficients is precisely 1), we see that the codimension of W* in V* is ~ 1. 11.2.4

Proposition

Let H be a standard Hilbert space and (e i ) c H any standard orthonormal basis. Then any near standard element y of H has infinitesimal components yi=(ejIY)~O for iIlimited i. 11.2.5

Proof

Consider first a standard element xeH. The sequence of components (x n ) of x is a standard sequence (in any standard orthonormal basis). Moreover, this sequence I iHX = (e1Ix) tends to 0 for i- 00 by the Bessel inequality, or by the Parseval relation giving IlxIl 2 =r.lx iI2 • Thus, XI~O for nonstandard i. Recall now that by Definition 3.3.1, near standard elements yeH satisfy there exists a standard x = y*eH

with y ~ x.

The Cauchy-Schwarz inequality gives IYi-xil~lIy-xll~O

for all i.

Hence Yi~Xi

11.2.6

for all i

and

Yj~Xj~O

for nonstandard i.

Application

The preceding proposition will be used as follows. Let us take a standard compact operator Tin H. Then, the image of the unit ball B 1 = B 1(0) c H must be contained in a standard compact subset of H. It follows that every element Tx (where Ilxll ~ 1 or more generally Ilxlllimited) is near standard. Thus, we have x limited => Tx near standard =>(Tx)i=(eiITx)~O for nonstandard i.

126

INVARIANT SUBSPACES

Taking in particular the unit basis vector x =ei' we see that the matrix coefficients a}=(eil Te j ) of T in the basis (eJ) satisfy a~~O

for i nonstandard and all j.

In other words, lines with nonstandard index i (in the matrix of T) only have infinitesimal elements (observe that this property really depends on the fact that the operator Tis compact: the identity operator has identity matrix with one 1 in each line, hence fails to have this property!). 11.3 11.3.1

PROOF OF THE BERNSTEIN-ROBINSON THEOREM

Preliminary reductions

Since the statement of the theorem is classical, it is enough to prove it for standard data (transfer taking care of the generalization). Thus, we shall assume that the operator T and the polynomial p:#=O (for which p(T) is compact) are standard. This also implies that the Hilbert space H in which T is defined is standard and if we write

Pn(X) = a/lX" +

... + ao

the degree n and all coefficients ai of p are standard. Let us choose a nonzero standard eeH. The smallest closed subspace of H containing e and all its transforms Tie is obviously standard, invariant under T, and separable. Thus we are reduced to the case of a separable Hilbert space H. Since we are assuming p:#= 0, its degree n is well defined and we can assume that an ~O. Dividing throughout by this coefficient we can assume that p is unitary, still standard, hence with limited coefficients. If p(T)=O, the finite dimensional space generated by the Tie for o~ i ~ n = deg p is obviously invariant under T and we are reduced to the finite dimensional case. Thus we could assume that p(T):#=O. In any case, let us choose and fix a standard nonned eeH with {Tie: ieN} total in H. The sequence is thus standard. 11.3.2

First part of the proof of 11.1.2

We shall construct a finite dimensional (hence closed) subspace Vin H which is nearly invariant under T in the following sense. Denote by P the orthogonal projector from H to Vand let Py be the restriction of PTto V(we consider that Py is an operator in V). The construction will be such that Tyx~ Tx

for aU limited xeV.

PROOF OF TRB .ERNSTBJN~ROBINSON THEOREM

127

Let us first use the Gram-Schmidt orthonormalization process to deduce from the sequence a standard orthonormal basis (e,,) of H with

eo = e,

e" = linear combination of the Tie for i ~ k

(recall that we assume that H is infinite dimensional, topologically generated by the linearly independent family ( Tie». The coefficient of T"e in e" is nonzero. By construction ofthe basis (el), the matrix A of Tin this basis will nearly be in upper triangular form a~=O

for i> j+ 1 (instead of i> j!).

By induction, it is easy to show that the powers TN will have matrices

A"=(aHn» with aHn)=O for i>j+n. Line by column matrix multiplication also shows that the coefficients in this first nonzero diagonal are simply given by

n

aj+n(n) =

a~+ 1

jsi~.j+("-l)

The matrix B=(b~) ofthe operator p(n will inherit the preceding property in the following form

pen

But the operator is compact, hence must have infinitesimal coefficients in nonstandard rows by 11.2.6. In particular, bj+"~O

for j illimited.

Since all is standard and nonzero, it implies that

n

a~+ 1 ~O

for illimited j.

jsisj+(n-l)

In this product of n (standard) terms, at least one must be infinitesimal. We can thus find an iIlimited v( ~j + 1) with a: _ 1 ~ 0 and the finite dimensional space V generated by the ej for i < v has the expected property, as we now show. For x limited and in V, the last component x v - 1 of this vector must also be limited (by the Cauchy-Schwarz inequality) and

TX=T(

I.

I
xiei+XV-lev_l)

128

INVAJtIANT SUBSPACES

By nonstandard induction, we see that TkX~(Tv)"x for all standard integers

k,

and in particular, we infer that p(T)x~p(Tv)x

for limited XEV.

The preceding construction can be illustrated by a drawing of the matrices A of T and resp. A v of T. The last is the square block defined by i and j < v in the first. It contains all nonzero elements of the first v columns of A except the coefficient a~-l just situated below the far end corner of A v .

ag a? ab a~ 0 af

a~_ I

ae at

Y-I aY-l

a Y-I y

ae-I

Y- I ••. a Y-2

0

0

0 0

0 0

0 0

0

0

0

...............................

a:- 1 a: 0 0

Fig. JI.J

11.3.3 Second part of the proof By the Jordan theorem for the finite dimensional operator Tv we can find a filtration of V by an increasing sequence of subspaces {O}=VOcV1c ... cJ-:=V having the properties dim(Vi)=i, Vi

is invariant under Tv.

Still call Pi the orthogonal projector from H onto Vi (hence P v = P). The standard subspaces H j = vr=S{xEH:3yeVj

with x~y}

of H will now play a crucial role. Obviously {O}=HocH 1c ... cHv=H.

PaOOF OF THE BERNSTB1N -ROBINSON THEOREM

129

I claim that these closed subspaces Hi of H are invariant under T. Take indeed a standard xeH j , say x ~ ye Vi' By the best approximation property of orthogonal projections, we shall a fortiori have x ~ PjX and S-continuity of T at the origin gives Tx~TPjX

. ~ TVPix

since PiX is limited,

and TyPjxeV j since Vj is invariant under Tv. This has shown that for each standard xeH i, Tx also belongs to Hi' My claim is proved by transfer since H j is standard.

11.3.4 Tbird part of tbe proof Let us now introduce the positive real numbers dj = IIp(Tv)Pjell. From PoeeVo = {O}, we see that do=O and at the other extreme, Pye=eeV~

p(T)e is standard and a linear combination of eo, ... en=p(Ty)eeV

hence dv =lIp(Ty)ell

is standard and> O.

Question: When can Hi be {O}? Answer: Only when d i ~O. Indeed, Pie is limited in V hence p(TY)Pje~p(T)Pje is near standard (because p( T) is compact!), say p(Ty)Pje ~x standard in HI = {O}. Thus P(TV)Pie~O

and

dj::::O.

Question: When can Hi be equal to H? Answer: Only when di :::: d. Indeed, if eeH l ' we can find ye V j with e:::: y and again by the best approximation property of orthogonal projections e~Pie. This implies p(Ty)e~p(TY)Pie, whence d=dy::::d j • The proof of the Bernstein-Robinson theorem can now be concluded by exhibiting a nontrivial closed invariant subspace among the Hi' Certainly Hi would be nontrivial if we could prove that Od/2: dj - 1 ~d,l2~dj-l

not ~d~Hj-l =l=H.

130

INVARIANT SUBSPAceS

Thus, • either Hj_ 1 i= {O} (and the proof is finished!) • or H j -

1

= {O} (and dj - 1 ::::0).

In the second case, dj>d/2~dj

not ::::O~Hji={O}.

But by Lemma 11.2.2, dim H j = 1 in this case, and we have again found a nontrivial closed invariant subspace of T among the Hi'

11.4 11.4.1

COMMENTS

Back to the proof of the Bernstein-Robinson Theorem

In the second part 11.3.3 of the preceding proof, the spaces Hi are standard for all indices i. But we have not defined a map it-+H j since the non classical relation (i, H j ) is not classical (and not set forming). It cannot be used to define a graph and the situation should be compared with that of defining the standard part x* of a real number XE[O, 1]. The space H j is well defined for each i, but there is no map it-+H j ! However, it-+d j is a map since it is classically defined, with certain (nonstandard) parameters Tv and Vj. It would have been possible to work with the distances di = IIp(T)Pjell, but even here Pi is not necessarily standard when i is!

11.4.2

An example

The construction of 11.3.3 can be illustrated in the example of operator T defined in [2

(N)=space of square summable sequences N~C

by 1{e j )=ej-j 1/0+ 1)

for i?O

(ej denotes the sequence having only one nonzero element equal to 1 at place i:ej is the characteristic function of the point i). This operator is compact with spectrum {O}: it is quasi-nilpotent. Since Tis already compact, we can take the polynomial p( X) = X here. The truncation

v = [2( [0, v[)

(with dim V = \I)

is adequate for T as soon as v is illimited, and

Tvej=Te j ifi
13t

COMMENTS

The Jordan filtration of V is given by the Vj =12([1'-i, v[).

Consequently, in this case, l ' - i illimited~Hj=

{OJ,

l ' - ; limited~Hj=12([v-i,oo[).

Since e=eo (as in the proof), P j eEVj =J2([v-i,

vD.

Pje=O for \1-;>0 (i.e. i
INDICATIONS FOR THE EXERCISES CHAPTER

1.9.1

1

Forget that n is nonstandard and try to prove the result for all integers n.

1.9.2 It is easy to construct set forming relations corresponding to the empty set. 1.9.3 If you have the impression that one of the (a), (b) or (c) parts is true, you are making a mistake .... For (a), consider the set of even numbers, for (b), consider the set of prime numbers, for (c), consider the numbers 2"'.

CHAPTER 2

2.8.1

Compare the standard sets E and 0.

2.8.2 Take a finite part F containing all standard elements of E ... (and use 2.8.1 to answer the second question). 2.8.3

If E (or F) is empty, the Cartesian product is empty. Prove E x F standard and not empty => E and F standard.

2.8.4 Consider the standard set

A=S{nEN: pen)}. 2.8.5 The sets Band C are standard, B and A have the same standard elements ... and &/2 E A but &/2 ~ C.

134

INDICAnoNS FOR THB BXBllCISBS

2.8.6 Compare the standard subsets of N x N

S{(x,y):R(x,y)} 2.8.7

and

tJ.

= {(x,x):xEN} (diagonal).

For any functionf: E --+ E, we can form the subset

= x}

{xEE:f(x)

c E

(this is just set theory, nothing to do with NSA). 2.8.8 (a) If E has only one element, this element is either standard or nonstandard. Proceed by (nonstandard) induction using (2.8.4). (b) Consider the set S{xEE:P(x)}. (c) If x is nonstandard, E and E - {x} have the same standard elements. 2.8.9

Use transfer.

CHAPTER 3

3.5.1

Take ·n E N nonstandard and consider e = 1/n!.

3.5.2 Even if B > 0 is rational, it is not always possible to find n E N with ne standard and > O. 3.5.3 (a) Consider a suitable linear function f(x) = kx. (b) Consider the set of integers n EN for which

If(x)l s; lin

for all Ixl

~

n

and use principle 2.4.4. 3.5.4 If A E N is nonstandard, we obviously have If(x) I < A (for all x).

3.5.5 Use transfer to prove (i) => (ii) and the preceding exercise for (iii) => (i). 3.5.6

Is there a set containing precisely the real numbers x with x*

= O?

3.5.7 Forget first that e is infinitesimal. Consider the setf-I(O) to answer the second question.

3.5.8 Let x be the number in question. Can you compute the square of x*? 3.5.9 Consider a set of the form [e, 1]. 3.5.10 When is the sequence m-+(1 + x/ny' standard? 3.5.11 Consider constant sequences all = e (with a = 0). Also consider sequences of the form all = Aln where A is illimited.

3.5.12 Consider only the case of standard sequences and use the characterization of convergence given in Theorem 3.4.1.

INDICATIONS FOR THE EXBRCISBS

135

3.5.14 (b) For each standard e > 0, we can write (in the real case) (l-e)aj ~ b l ~ (1 +e)aj

and one can sum these inequalities.

3.5.20 The set U A is the complement or (X - A)* (both sets are standard so that the equality X - U A = (X - A)* is checked on standard elements).

SOLUTIONS OF THE EXERCISES CHAPTER 1

1.9.1 The indicated result is classical when formulated for all initial intervals I,. c !\oJ. It is proved by (classical) induction on n. When proved for all ne!\oJ, can anyone doubt its validity for nonstandard n? (When a result P(n) is proved for aU integers ne N, should we have any doubts saying that pep) is true for a particular prime p?) 1.9.2 The property P(x)='x is standard and x is nonstandard' is nonclassical and is set forming (in any set E): it defines the empty set 0. 1.9.3 (a) The set of even integers is infinite (this is known!) hence contains a nonstandard element: let n = 2m be a nonstandard even number. This integer admits the standard prime divisor 2. (b) The set of primes is infinite (known since Euclid!) hence contains a nonstandard element. Let p be a nonstandard prime. This integer has only one prime factor and k = 1 is standard in this case. (c) Take a nonstandard meN and consider n = 2m > m. Thus n is also nonstandard. The only prime factor of n is the standard prime 2 (k = 1 in this case, whence another example for (b) I). One could also start with the infinite set consisting of the powers of 2 and take a nonstandard element in this set ....

CHAPTER 2

2.8.1 Recall that two standard sets are equal as soon as they have the same standard elements (transferred extensionality 2.2.4). In particular: a standard set having no standard element is empty (the empty set is standard).

138

SOLunONS OF THE EXERCISES

2.8.2 The complement C of a finite subset F of the infinite set E is infinite. When F contains all standard elements, C can only contain nonstandard elements. By 2.8.1 above, C cannot be standard. 2.8.3 The two projections PE: Ex F -. E and PI'": Ex F -. F are surjective as soon as Ex F is not empty, i.e. as soon as both E and F are not empty. If Ex F is standard and not empty, take a standard element (e,f) E Ex F. Then e and [are standard and e.g. E = {x: (x,[) E E x F} must be standard since it is uniquely characterized by a classical formula containing two parameters f and G = Ex F having standard values. 2.8.4 The standard set A = S { n EN: P( n)} coincides with N as we see by the following argument. The classical fonnula YSn(n E A

~

n + 1E A)

gives by transfer (the parameter A has a standard value) Yn(nEA

~n+ 1EA).

Since we are also assuming that OE A, a classical induction shows that A = N. Finally, only the standard elements of A are characterised by the property P: we can only conclude that P(n) is true for standard n. 2.8.5 We have B = C c A with c :F A (thanks to e/2). The set A is not standard since its upper bound is not standard (A and B have the same standard elements and are distinct, hence cannot be both standard). 2.8.6 The two sets given in the indication to this exercise are standard and have tbe same standard elements, hence coincide. 2.8.7 We know that there is no subset of [0, 1] consisting precisety of the standard elements. Consequently, there is no function f with [(x) = x* (the condition x = x* precisely characterizes the standard elements in the given interval). Let us repeat several arguments for the nonexistence of a set containing precisely the standard elements of I = [0, 1]. If a subset A consists only of standard elements, it must be standard and finite; but if A also contains all standard elements, it must coincide with I (since it is standard). Variant: if there was a subset A of I consisting precisely of the standard elements, we could define a subset

{nEN: l/nEA} c N consisting precisely of the standard integers, but this is impossible since there is no smallest nonstandard integer. 2.8.8 (a) We can use nonstandard induction as mentioned in the indication for this exercise (one could even start with the empty set, in which any relation is set fonning!). To prove the induction step for a standard cardinal n (hence n + 1 also

139

SOLUTIONS OF THE EXERCISES

standard) onejust uses the fact that an element has to be standard or nonstandard ('excluded middle third principle'). (b) In a standard finite set, all elements are standard and in S{x E E: P(x)}, all· elements are distinguished by the property P. (c) The sets E and E- {x} are distinct although they have the same standard elements. Hence one of them at least must be nonstandard. Since E is standard, E - {x} must be nonstandard.

2.8.9 If a and b :f:. 0 are standard, so is alb (this number is uniquely defined by a standard formula having parameters a and b, assuming standard values in our case). Conversely, if x E C there is a unique reduced representation as given in the exercise: b is uniquely defined classically from x and will be standard if x is (then a = bx will also be standard).

CHAPTER 3

If n is an illimited integer, n! is divisible by all standard integers. Take any standard rational alb =F 0 written in reduced form (as in Exercise 2.8.9) with b standard. Thus alb is an integral multiple of 1In!.

3.5.1

3.5.2 Take a nonstandard prime integer p (the set of primes is infinite) and consider 8 = P- 1 + P- 2 > O. This number is infinitesimal (e.g. since 8 < 2p-l = 2Ip). Let us find the integers n EN for which ne is standard. In this case nip

<

nip + nlp2

= ne ~ nip

is standard limited

=>nlp 2 = p-l(nlp) =:

o.

But n(p + l)j p 2 = ne

standard

and this quotient representation cannot be the reduced one since here the denominator pl is nonstandard: n(p + 1) and p2 are not relatively prime => p

must divide n(p + 1)

P must divide n(p and p + 1 are relatively prime) => n = n' p for some n' E N ~ n'(p + 1)1 p = ne standard ~

=> n'(p+ ~

1) and p are not relatively prime (as before)

p must divide n' (as before), say n' = n"p

~ n"(p + 1) = ne standard =>n" = 0 (and thus n' = 0, n

= 0).

This shows that the only standard multiple of 8 is O.

140

SOLunONS OF THE BXBRClSII

3.5.3 (a) Take an infinitesimal e > 0 and consider the function f(x) = ex. (b) The set A c N defined in the indication for this exercise contains all standard integers n. Since there is no set containing precisely the standard integers, this set contains a nonstandard integer v and we must hence havef(v) infinitesimal since If(v)1 S; 1Iv ~ o. 3.5.4 and 3.5.5 Obvious with the indications. 3.5.6 There is no set containing precisely the infinitesimal real numbers (otherwise there would be a largest infinitesimal of the form 1I n, corresponding to a minimal nonstandard n EN). Consequently, there is no function on [ - 1, 1] satisfying f(x) = x* (no standard or nonstandard f would do). 3.5.7 The functions/,. are well defined for all e > O. In particular, they are well defined when e is infinitesimal. If/,. is a standard function, the setf; 1(0) = [0, e[ is standard and thus e = sup [0, e[ is standard. If e > 0 is infinitesimal,/,. cannot be standard.

3.5.8 Using the rule (xy)* = x*y* for x = y, we infer (X*)2

= (x 2)* = (2 + e)* = 2.

Hence x* = J2 (note that x ~ 0 ~ x* ~ 0). Another way of proceeding would consist in guessing the answer and proving it by estimating the difference + e) in a more classical way

J(2

J2

J(2+8)-J2

= (J(2+e)-J2)(J(2+e) + J2) J(2+8)+J2

=

(2+e)-2 J(2+8)+J2

whence for example

J(2+8)-J2 ~ tlJ2 ~ e ~ o. 3.5.9 If 8 > 0 is infinitesimal, 8* = more simply, take the set A = {e}.

o~

[e, 1] although this set is closed. Even

3.5.10 The sequence nt-+(1 + xln)n is standard when x is standard. Use Theorem 3.4.1 since the limit is known to be c. 3.5.11 The condition (i) suggests a movement of the a,. towards a whereas condition (ii) is purely static! This interpretation shows how to fulfill each ofthem separately (if we are not required to exhibit standard sequences). 3.5.12 By transfer, it is enough to prove the result for standard sequences, and then it is possible to use condition (ii) of Theorem 3.4.1. To simplify the problem, subtract the constant standard value a from each all' Thus we can assume an -+ 0 and we have to prove the corresponding statement for the averages.

141

SOLUTIONS OF Tltl BXDCIIBS

Take any illimited n EN. Using nonstandard induction on m, we see that (a l + ... +am)/n ~ 0 for aU standard me N. By Robinson's lemma, there is a nonstandard m with the same property. The~ aUla,1 for m < i:s; n are infinitesimal and so must be their maximum. Afortiori

lam + l + ... +alll/n:s; lam + l + ... +a"I/(n-m) ~ O. 3.5.13 If x is limited with x* :F 0, afortiori x :F 0 and we can write for example Ixl ~ Ix*1 > Ix*I/2

whence Ixl > Ix*I/2 standard

and finally II/xl < 2/lx*l, i.e. l/x is limited. If x and I/x are limited, say for example Ix I and II/xl :s; a standard, then -log a

~

loglxl

~

log a

and thus log Ixl is limited. Finally, if log Ixl is limited, so is Ixl

3.5.14 Assume x appreciable and x 1- y/x

=

eO/x)

~

Conversely, if x is limited and x

I - y/ x =

E ~

~

y. Thus x - y =

E~

0 (since l/x is limited): x

~

= exp log Ixl.

0 and ~ y.

y,

0~x- y

= XE ~ 0: x

~

y.

Let us write the second part of the proof in the complex case. If ai then for each standard e > 0, we have

~

bi for all i,

II-ai/bil < e and thus

IIbi -

Llb,-ail < GLlbil, Iad/IIbd < e.

This shows

Ibi~Iai'

3.5.15 Let the galaxy.s;/ be defined by the functionf(resp. fJI by g). Since they are disjoint, we have f(x) limited ~ g(x) illimited, g(x) limited ~ f(x) illimited. Define A = {xeE:f(x) < g(x)} and B = {xeE:f(x»g(x)}. These subsets are obviously disjoint and x e.s;/ means f(x) limited, hence g(x) illimited and f(x) < g(x): x e A. Similarly fJI c B.

3.5.16 Simple exercise of style .... 3.5.17 When b is appreciable (Exercise 3.5.13), E* is the ellipse of equation x 2/ a 2 + y2/(b*)2 = 1. When b is illimited, E* is the union of the two horizontal

142 lines x axis.

SOLUTIONS OF TH!! EXUCISES

= ± a. Finally, when b is infinitesimal E* is the interval [-a, a] of the real

3.5.18 (a) Let x be a standard extremity of C, Le. x is an end point of an interval of a C m for a standard m. It is obvious that x E C m for all m' > m and in particular x E C n (n illimited). This proves that x E By transfer, the set of all extremities of C (this set is dense in C) is contained in the (standard) set C:. Since this last set is closed, we conclude C c (b) Conversely, if x is standard, the sequence m t-+ d(x, Cm) is standard too. But if x is a standard element of C:, we have d(x, C~ ~ O. This proves that d(x, C m) -+ 0 and xEnCm=C. This set C is the Cantor set. It is the simplest nontrivial example of a fractal. The interested reader will also be able to construct the 'von Koch curve' with NSA as follows. The first approximation is given by installing an equilateral triangle with base equal to the middle third of the segment [0, 1] and deleting this middle third as in the Cantor set construction. This first approximation consists thus of four connected segments (cf. next figure) of equal length 1/3. Repeat the construction above each of them to obtain a curve consisting of 16 = 4 2 segments of equal length 1/9. The process is easily continued by induction, and the von Koch curve (already considered in 1904) is the limit, or the standard part of an illimited step.

C:.

C:.

Fig. 3.3

143

SOLUTIONS OF THE EXERCISES

3.5.19 Since A is standard in X (also standard), the set of sequences (an) c A which converge to x E X is standard whenever x is standard. This set is not empty if XE A, hence contains some standard elements (Exercise 2.8.1).

CHAPTER 4 4.6.1 The function x", sin nx, ~ are continuous at all points x (and for all integers n). (a) If x is standard and Ixl < I, the sequence m--+x n is standard and tends to O. Hence x" ~O for illimited n (and x as above). This groves that X 1-+ x" (n illimited) is S-continuous at all limited points x with Ix* I < 1. But this function is not S-continuous at points x with Ixl ~ 1. For example, x=I-I/n ~ 1 but xn = (1- l/n)" ~ lie is not infinitely close to (x*)" = 1. More generally, if h is infinitesimal and y = x + h ~ x, we have f(y)-f(x) = (x+h)"- X " ~

nhx" -

1

~

nh

= nx"-lh+

... +h"

if x ~ 1 and

h > O.

J

This difference can be appreciable (h = 1/ n makes it iIIimited). (b) At the origin, the function sin nx behaves like nx and it cannot be S-continuous (we leave the reader to find trigonometrical formulas permitting to prove this without having recourse to differentiability). The same is true for all integral multiples ofn/n. Finally, let x be arbitrary: there is an integral multiple of n/n which is infinitely close to x. Consequently, the function xl-+sin nx (n illimited) is S-continuous nowhere. (c) The exponential function ex is not S-continuous at illimited points x > O. Take indeed such a point x and consider h ~ 0 with y = x + In(1

+ h) ~

x

(the In function is continuous and S-continuous at the standard point 1). We then have e' = eX . e1n(1 + h) = eX' (1 + h), eY - eX = heX

not always infinitesimal

x

(for example, h = e- makes this difference equal to 1). The reader can still show that the functionf(x) = l/x defined for x-::j:. 0 is not S-continuous at points x ~ 0, x -::j:. O. 4.6.2 Fix a < b and prove f(a) ~ f(b). Take the integer n sufficiently large to have (b - a)/n ~ 0 (if a and b are standard, it is enough to take n illimited, but in general, use the Archimedian property of ~ to show that such an n exists). Introduce the subdivision points Xl

= a + k(b - a)ln (k

~ 1).

144

SOLUTIONS OF THE 8X!lasU

Then f(a) S!(Xk) can be proved by induction on k, (f(s) s!(t) is a classical formula). 4.6.3 We can take a finite set F containing all standard real numbers and put n = Card(F) (illimited integer). There exists a polynomial p of degree < n having the required properties. Polynomial functions are continuous functions (still true ...). Let us show that if a continuous function! satisfies

f(r)

=1

for all standard rationals r,

!(s)

=0

for all standard irrationals s,

then f is not S-continuous at limited points. Let us examine first the case of a standard rational r. The sequence nt--+r+J2/n is standard, consists of irrational numbers and tends to r. The set of integers n such thatf(r + .)2/n) = 0 contains all standard n, hence contains some nonstandard n as well (principle 2.4.4). For such an iIIimited integer n,

f(r) = 1 but f(r + J2/n)

= 0

in spite of r + J21n ~ r.

At a standard irrational s,fis not S-continuous either. This is shown for example by consideration of the sequence m-.. [ns]/n which is standard if s is, and consists of rational numbers tending to s. Finally, if x is limited x* = s is standard and S-continuity at x is equivalent to S-continuity at s. 4.6.4 When I is compact, the functionfis uniformly continuous on I, however, we cannot infer thatf(x) ~!(y) as soon as x ~ y (counterexamples are given by f(x) = sin nx where n is illimited, or by the functions of the preceding exercise). Thus the condition Ix - y I < lin is not sufficient. But it is enough to add the hypothesis that f is standard. In this case, uniform continuity indeed gives

x For k = [xle], ke

~

~

y ~ f(x) ~f(y).

x and g(x) = f(ke)

~f(x)

as desired.

4.6.5 Take a finite part F containing all standard numbers. If N = Card F (illimited), the elements of F can be parametrized by the interval [0, N [: say F = {all: 0 s n < N}. Define then f(x) =

L 1/2"+ 1

finite sum extended over the n with all < x.

This function 'jumps' at all all E F hence is discontinuous at all standard points. This function is positive, increasing and bounded by

L

1/2"+ 1 = 1/2+ 1/4+

... =

1.

II€N

4.6.6 Since the sum of two infinitesimals is also infinitesimal, the sum of two S-continuous functions is S-continuous. The same is true for a finite sum having a limited number of terms (as is seen by nonstandard induction). Be more careful about products, since at points where one factor is illimited, infinitesimal jumps of

145

&OLtmONS OF THE BXERCJS!S

the other are magnified (take lex) = n illimited constant and g(x) = 0 for x ~ 0 whereas g(x) = lin for x > 0: these functions are S-continuous at the origin but their product is not S-continuous at the same point).

4.6.7 (a) For each e > 0, choose

Iy-xl < Take an arbitrary z and put c

~e

> 0 with

e ~ If(Y)-.f(x)1 <

= If(z) -

Iy-zl < e~ly-xl

f(x)l, d

br..

= Iz -

xl. Then

~

Iy-zl+lz-xl < c+d ~ \f(y)-f(x)l < ~r.+d => If(y) - /(z) 1 < c + ~£+d·

This proves that/is suounitnoc at z as soon as it is at x. We shall simply say thatf is suounitnoc at one point (hence at all points also). (b) When ~ > 0 has the property of the definition of suounitnoc with respect to some G > 0, it is obvious that the same ~ will also work for smaller c' ~ B. The point is thus to consider big values for B. The intuitive fonnulation of suounitnoc is thus easily seen to be

f

bounded on all bounded intervals.

For example, the function f(x) = x is suounitnoc. Constant functions are also suounitnoc. When c is illimited,f(x) = cis suounitnoc butf(O) = c is not limited and the criterium (c) can only be applied to standard functions. (c) Let us show that iffis standard

f

suounitnoc <::- f(x)

limited for all x limited.

By (b), we see that f is suounitnoc precisely when for each interval I, I bounded => f(l) bounded. When f is standard, transfer can be applied and the preceding condition is equivalent to for each standard interval I, 1 bounded => f(1) bounded. But by definition,

.

a limited <::- a belongs to a standard bounded interval.

Taking successively a = x and then a = f(x), we obtain the desired equivalence.

CHAPTER 5

5.6.1

Take an illimited n and a limited x. Then xln ~ 0 and log(l +xln)=xln+r.xln

where

e~O.

146

SOLunONS OF THB BXBRCISI8

Hence (1

+ x/n)" = en 10g(1 +X/r1) = e rl(x/n+u/r1) =ex+ tx =ex eEX ~ex

since ex ~ 0 and tHe' is S-continuous at

t

= O.

5.6.2 Iff is S-differentiable at the point a, there is a standard m with f(x)-f(a) x-a

m+e

with

f;=e(x)~O for all x~a.

Hence f(x)-f(a)=(m+e)(x-a)~O

for

x~a.

5.6.3 It is classical thatf:: is differentiable with derivative 0 at all non multiples of c (this is true for all e). At multiples of f.,!r. is not continuous and aforaori not differentiable.

CHAPTER 6

6.4.1

By definition

f

eX dx=st

i

L

eihh=st(l +eh+ ... +e(n+l)h).

Osjh< 1

o

It is easy to sum this geometric progression with ratio h, finding erlh_l

st h eh _ 1 =(e-l) st h/(e h - 1) (since nh = 1). But we have seen that the exponential is differentiable with derivative eX. In particular, at the origin, the derivative is 1 and the desired standard part is 1. We have found

r

eX dx=e-1.

o

The second integral can be computed in a similar fashion, observing that the sum

can be computed by derivation of the sum of the geometric series I

is the derivative of r(r"-I)/(r-l).

147

SOLUTIONS OP THE BXDClIBS

6.4.2 Proceeding as above with h = 1t/n (n illimited), the proposed integral is the standard part of

L

10g(I-2a cosjn/n+a1)·n/n

O5o}
211 =;;nlog [a-I a+ 1 (a -I) ].

(The product corresponds to 2nth-roots of 1, a = 1 appearing twice and a = - 1 being absent.) We see then that the standard part is n loga 1

if a > 1, 0 otherwise.

Indeed,

I

n

{7110g al 10g(I- 2a cos x + a1)dx = 0

iflal> 1 iflal < 1·

o

6.4.3 Take x ~ 1 and x < 1 so that 8t log(l +x)=log(1 +stx)=log 2 (the standard function log is continuous, hence also S-continuous at the standard point x = 1). On the other hand, choose an ilIimited integer N so that st(x-x 2 /2+ . ..)=st

L (_1)I-l x"/n.

"
We have to prove that

L

(_l)"-1x"/n~

II
L

(_I)"-1/n

II
since then we shall have ~

stL( _1)"-

00

1

x"/n=st

L (-I)"-l/ n = I( _1)"-1 In. II
1

1

Let us estimate the difference

L (- 1r - 1 x"/n - L (- 1r - 1In =

II
II
-

L (- 1)"- 1 (1 -

x")/n.

n
For this, let us write M,,(x)=(l +x+ . . .+xn-l)/n (average of powers of x)

148

SOLunONS OF THE EXBRClSU

so that the above difference can be written

L (-1)"-1 Mn(x).

(I-x)

n
The numbers M n(x) tend to zero monotonously so that their alternating sum is majorized by the first term 1. The conclusion results from the assumption x-I ~o. More generally, one can show (*)

L an~ ~ L an

n
if x < 1 and

x~1

n
as soon as

I

L

~

nanl

ls.n
limited (independent from pl.

(1

Indeed, the difference is as above (I-x)

L nanMn(x).

n
Abel's transformation shows that

L

nanM,,(x) is limited.

n
When (*) is satisfied, we conclude ao

L an=st 1 L t

L

an=st

anx"

s,,
t sn
a"xn~f(x)=

L a"xn

with 00

L lsn
1

if this last series converges. In particular, when the sequence nl-+a,. is standard (i.e. fstandard) 00

Lan = st/(x)=!(st x)=!(l). 1

CHAPTER 7 7.5.1 The part A = [ - n, n] is finite for each n, hence m(A) =0 but when n is iIlimited, A is not standard and m(A) is not equal to the standard part of m,,(A). 7.5.2 The first assertion is obvious. But the equality m(N)= 1/2 is only valid for the special invariant means defined by the procedure used in the text. For example, one could define m~(A)=(I/n)

Card (An[O, n[)

SOLUTlONS OF THB EXERCISES

149

and then, fixing an illimited n m'(A)=st m~(A)

for all standard A.

The reader will prove that these m' are also invariant means on lL and that for example m' (N)=m'([k, oo[)= 1, m'(] - 00, k])=O

for all kElL. There are-of course-still other invariant means on lL ...

7.5.3 The part 3A consists of the integers 3m where 3 21 + 1 <3Iml~32i+2

for some integer iEN.

let A + = AnN. It is obvious that the three parts 3A +, 3A + - 1

and

3A + - 2

constitute a partition of A'+ =A'nN. Hence 3m(3A+)=m(A'+) and similarly for the negative part of A. This proves that m(3A)=m(A')/3.

In the case m=st m,. with n=3 2", we have m(A)=

1/4, hence m(A')=3/4 and m(3A)= 1/4 (=m(A)!).

If, on the contrary, n is an odd power of 3, we have m(A)=3/4,

hence m(A')= 1/4

and

m(3A)= 1/12 (:;l:m(A)/3!).

First, m is not countably additive on parts of lL 2 (lL 2 is itself a countable union of points, each of which has zero invariant mean). Second, the inequality m(dA)=m(A)jd 2 is not valid for all parts A clL 2 • ••• However, the conclusion is correct as the following argument (due to Cesaro) shows. Let us work with the mean on N 2defined as in the preceding exercise (and defining an invariant mean on lL 2 ) Q is the square [1, n]2 where n is a fixed iIlimited integer,

7.5.4

mll(A)=n- 2 Card (AnQ), m(A)=st mll(A)

for all standard parts A c N 2 •

(This choice will have little importance-as we shall see-but is convenient since summations can then be extended over positive values for the integers k: this will be tacitly assumed.) Let us compute the density of QI = QnA 1 by estimating the number of couples of relatively prime integers in Q as follows. To obtain Card QI' it is necessary to subtract from Card Q the number of couples multiples of2, 3, ... (only taking into

150

SOLUTJONS OF THBSXBaClUI

account primes p ~ n). As there are [nlp]2 couples which are multiples of pin Q, a first approximation for the desired number is Card Ql :n 2-

I

[nlp]2.

p~rt

Effecting this difference, we have subtracted twice the couples which are a multiple of 6, ... more precisely, we have subtracted twice the couples having a gcd multiple of a product pq ~ n of two distinct primes p and q. We have to add a corrective term for the number of these couples. This term is

[nlpq] 2

for primes p
with pq~n.

Finally, we obtain a correct formula by writing the alternating sum Card

Ql

=n 2 - L [njp] 2 + psn

L

[nlpq]2_ ....

p
The density of Q1 in Q is n- 2 Card Ql = 1- In- 2[nlp]2+ In- 2[n/pq]2-. . .

(1)

(with the same restrictions on primes in these sums). We shall compute this density by comparing it to the alternating sum 1-

I

l/p2 + L I/ p2 q2 -

•••

(same p, q, ... as in (1)).

(2)

This second sum is indeed easy to estimate. Forgetting the restricting pq ~ FIt pqr ~ FI, ••• and keeping only in mind that p ~ n in the first sum, p < q ~ n in the second, ... we add terms (up to sign) taken from the tail of the standard convergent series I:.k - 2:

I

llk 2 ~ 0 (since n is iIlimited).

II>n

This shows that (2) is infinitely close to

l-LP-2+ L p~n

p
p-2 q -2_···=0(1_p-2) pSrt

and also infinitely close to the convergent infinite product

n (1-p-2)= 1j(2)=6In

2

•

(3)

p prime

It only remains to show that the difference between (1) and (2) is infinitesimal: this will give m(A d = 61n 2 independently from the choice of the illimited integer 11 chosen to define m. For this we have to estimate the differences

SOLUTIONS OF THB

lSI

BXE~C1SES

and take their sum for k:s n. By definition of the integral part, we have

n=[n/k]k+r with

O~r=rk
Hence

11k = [nlk]/n + rink, l/k 2- 2r/(nk 2 ) + r 2 j(n 2k 2) = [njk]2jn 2, and thus

O:s Ijk 2 -[njk]2jn 2 :S2r/(nk 2 ) -

•••

< 2r/(nk 2 )<2/nk.

The sum of these differences is majorized by

I

21nk = 2n - 1 I I jk.

/(Sll

(4)

Itsn

But it is clear that the averages M,,=n- 1

I

Ijk of the sequence 11k

lsks"

(monotonously decreasing to 0) are infinitesimal for n illimited (cf. Exercise 3.5.12). CHAPTER 8

8.5.1 (a) Here is an example of a continuous Dirac function which does not satisfy D(x) ~ 0 for all noninfinitesimal XElR: D(x) =

peak of surface 1- e, height Ije, supported in Ixl:s e peak of surface c, height 1, supported in lx-II ~e { vanishes otherwise.

(Make a picture!) (b) Look at Proof (8.1.3) and insert absolute values where needed! Let E = [0, 1]:::> A = {O, I/n,2/n, ... , njn = I} where n is an illimited integer. Then A is finite hence closed in E. But each XEE lies between two consecutive multiples of tin

8.5.2

k=[nx]=>k:snx l)jn=>lx-klnl ~ lln~O.

kjn~x«k+

Thus, (ii) is true but (i) fails. Observe that (i)=>(ii) in general: if A is dense in E and xEE, for each illimited integer nEN there exists aEA with d(x, a) ~ Iln::= O. When A is standard, (ii)=>(i) by transfer.

One further warning, my son: the use of books is endless. and much study is wearisome. (Ecclesiastes)

BIBLIOGRAPHY BOOKS E. Borel, Les Nombres Inaccessibles, Gauthier-Villars, Paris (1952). P. 1. Cohen, Set Theory and the Continuum Hypothesis, W. A. Benjamin Inc. New York (1966). [0] J. Dieudonne, Foundations ofModern Analysis, Academic Press, New York (1960). [Di] F. Dienes, Cours tfanalyse non standard, PubI. Univ. d'Oran (1983), (epuise). [E] L. Euler, Opera Omnia, lntroductio in Analysin lnjinitorum, Tomi Primi, Lausanne (1748). [H] P. R. Halmos, Naive Set Theory, van Nostrand Co. (1960). [HL] A. E. Hurd, and P. A. Loeb, An Introduction to Non Standard Real Analysis, Academic Press (1985). [K] J. H. Keisler, Elementary Calculus, Prindle, Weber & Schmidt, Boston (1976). [K'] J. H. Keisler, Foundations of Infinitesimal Calculus, Prindle, Weber & Schmidt, Boston (1976). [L] R. Lutz, and M. Goze, Non Standard Analysis. Springer Lect. Notes in Mathematics, # 881 (1981). [La] D. Laugwitz, Zahlen und Kontinuum (Eine Einfuhrung in die Infinitesimalmathematik), B. I.-Wissenschaftsverlag Mannheim{Wienj ZUrich (1986). [N] E. Nelson, Radically Elementary Probability Theory (in preparation). [R] A. Robinson, Non Standard Analysis, North-Holland Publ. Co. (1966).

[B] [C]

ARTICLES [I] Aronszajn, N., and Smith, K. T., Invariant subspaces of completely continuous operators, Ann. Math., 60 (1964), 345-350. [2] Bernstein, A. R.• and Robinson, A., Solution of an invariant subspace problem of K. T. Smith and P. R. Halmos, Pacific Journal of Math., 16 (3) (1966), 421-431.

154

BIBLIOGRAPHV

[3J Cartier, P., Perturbations singulieres des equations differentielles ordinaires et analyse non standard, Sem. Bourbaki, Nov. 1981, Asterisque 92-93 (1982), 21-44. [4J Halmos, P. R., Invariant subspaces of polynomially compact operators, Pacific Journal of Math., 16(3) (1966),433-437. [5J Hilden, H. M., Proof of Lomonosov's theorem, Math. Intelligencer, 4(1) (1982),34. [6J Hurd, A. E., ed., Nonstandard analysis, recent developments, Springer Lect. Notes in Mathematics, Nb. 983 (1983). [7] Lakatos, I., The Significance of Non Standard Analysis for the History and Philosophy of Mathematics, Math. Intelligencer, 1(3) (1978), 151-161. [8J Laugwitz, D., ed., Nichtstandard analysis, Math. Unterricht, 29(4) (1983). [9J Lomonosov, V. J., Invariant subspaces for operators commuting with compact operators. Funct. Anal. Appl., 7 (1973), 213-215. [IOJ Nelson, E., Internal set theory, a new approach to NSA, Bull. Amer. Math. Soc., 83 (1977), t t 65-1198. [11 J Richter, M., Ideale Punkte, Monaden und Nichtstandard Methoden, Braunschweig, Wiesbaden: Vieweg (1982). [12] Robert, A., Une approache naive de I'ANS, Dialectica 38/asc. 4 (1984),287-296. [13] Robert, A., L'Analyse Non Standard, L'Encyclopedie Philosophique, Vol. I, Presses Univ. France, Paris (1987). [14] van den Berg, I., Nonstandard asymptotic analysis, Springer Leet. Notes in Mathematics, Nb. 1249 (1987).

INDEX A

Analytic (function) 5.4.2 Appreciable number 3.5.13 Archimedian field 3.1.1 Asymptotic numbers 3.5.14

B Bernoulli (formulas) 6.1.2 Bernstein polynomials 8.3.1 Bernstein-Robinson (theorem) 11.1.2 Best approximation Lemma 8.4.2

C Cantor set, Exercise 3.5.18 Cardinal (of a set) 2.4.1 Cartesian product 2.5.2 Cesaro, sol. of Exercise 7.5.4 Classical (formula, statement) 1.2.2 Classical mathematics 1.1.3 Compact operator 11.2.6 Continuity (relation with S-continuity) 4.1.2 uniform -, 4.4 Continuous shade theorem 4.3.5 Couple 2.5.1, 2.5.2

D Definitions (implicit -) 2.2.5 Derivatives (higher -) 5.4 Difference operator V: 5.5.1

Differentiability 5.1.1 Strict - 5.3.1 Dirac function 8.1.1 E

Extension (axiom of-) 1.1.1 Extensionality 2.2.4 Transferred - 2.2.4 External set 1.8 F

Fejer's Theorem 8.2.2 Finite (cardinal) 2.4.1 Finite differences 5.5.1 Fourier coefficients 8.2.1 Fractal, sol. of Exercise 3.5.18 G

Galaxy 3.5.15 Graph of a map 2.5.2 Green function 10.1.1 H

Halo 3.5.15, 3.5.16 Higher derivatives 5.4

Idealization axiom (I) 1.2.4

156

INDEX

Illimited integer 1.6.1 - numbers 3.1.2 - vectors 3.2.5 Implicit definition of a subset 2.2.5 a standard map 2.5.4 continuity 4.3.1 differentiability 5. t.5 uniform continuity 4.4.4 Induction principle 1.5.5 nonstandard -, Exercise 2.8.4 Infinitesimal 3.1.2 Infinitely near 3.1.2 Integral (of a function) 6.2.1 Integral part [x] 3.1.6 Invariant mean (on 1) 7.1.1

K von Koch (curve), sol. of Exercise 3.5.18 Kuratowski (axiom of couples) 2.5.2 L

Limited integers 1.6.1 - numbers l.1.2 - vectors 3.2.5 M

Mean (invariant-) 7.1.1 Measure 7.1.2 N

Near standard 3.3.1

Q Quasi-nilpotent operator 11.1. t R

Relativization (of a statement) 2.6 Rolle's Theorem 5.2.1 Robinson's Lemma 3.4.9

S Safety rectangle 9.1.3 Schroedinger equation 10.3.1 S-continuity 4.1.1 S-differentiability 5.1.1 Separation principle (Diener-van den Berg) 3.5.15 Set forming relation 1.5.1 Shade of a set 3.3.4 Specification axiom 1.5.2 Standard (predicate) 1.1.2 Standard finite set 2.4.2 Standard part x* =st(x) 3.2.1, 3.3.1 - of a function 3.2.9 -of a vector 3.2.5 strict-of a function 6.3.5 Standardization axiom 2.1.1 Strict differentiability 5.3.1 - standard part of a function 6.3.5

T Transfer axiom 2.1.1

W Wallis (formulas) 6.1.2

p Parseval equality 8.4.1 Power set (j' (E) I. 7

Z Zermelo-Fraenkel axiomatic (ZF), (ZFC) 1.1.1

BASIC PRINCIPLES OF NSA 1. In every infinite set, there are some nonstandard elements (1.4.4).

Equivalently: A set which contains only standard elements is finite.

2. In any set, there is a finite part which contains all standard elements of E. 3. Let E be a standard infinite set. Then: (a) each subset A of E which contains all standard elements of E also contains some nonstandard elements (2.4.4). (b) each subset A of E which contains all nonstandard elements of E also contains some standard elements. 4. Let P be any property (classical or not) defined for all natural integers. Assume • P(O) is true, • 'VSn (p(n)==-p(n + 1».

Then pen) is true for all standard n E N (2.8.4).

5. Robinson's Lemma. Let (a/l)/IEN be a setquence of real (or complex) numbers. If a" ~O for every standard n EN, then there exists an illimited integer v EN with a,,~O

for all

n~v.

6. Let E and F be two standard sets. If a construction (classical or not) furnishes a standard element f(x) E F for each standard element x E E, then there is a unique standard mapf:E~F taking the prescribed values for standard XEE. In particular, if a standard element all E F is constructed for each standard n EN, then there is a unique standard sequence (a/l)/IEN in F extending the preceding definition.

1ST AXIOMS FOR NSA (I) IDEALIZATION Let R = R(x, y) be a classical relation. In order to be able to find an x with R(x, y) for all standard y, a necessary and sufficient condition is for each standard finite part F, it is possible to find an x = x F such that R(x, y) holds for all y E F. (S) STANDARDIZATION Let E be a standard set and P any property. There exists a (unique) standard subset A of E, denoted by S{xEE:P(x)} having the following property: the standard elements of A are precisely the standard elements of E satisfying P. (T) TRANSFER

Let F be a classical formula in which all parameters A, B, ... L have some fixed standard values. F(x, A, B, ... L) is true for all x as soon as it is true for all standard x.

Equivakntly If there exists an x such that F(x, A, B, ... L) is true, there also exists a standard x such that F(x, A, B, ... L) is true.

In particular If there is a unique x such that F(x, A, B, ... L) is true, then this x must be standard.