Hilbert-Spaces MATH231B Appendix1of1

8/13/2019 Hilbert-Spaces MATH231B Appendix1of1

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222 BRUCE K. D RIVER

12. Hilbert Spaces

12.1. Hilbert Spaces Basics.

Definition 12.1. Let Hbe a complex vector space. An inner product onH is afunction,h, i: H H C, such that

(1) hax + by,zi= ahx, zi + bhy, zi i.e. x hx, zi is linear.(2) hx, yi= hy, xi.(3) kxk2 hx, xi 0 with equality kxk2 = 0 iffx = 0.

Notice that combining properties (1) and (2) that x hz, xi is anti-linear forfixedz H, i.e.

hz,ax + byi= ahz, xi + bhz, yi.

We will often find the following formula useful:

kx + yk2 =hx + y, x + yi= kxk2 + kyk2 + hx, yi + hy, xi

=kxk2 + kyk2 + 2Rehx, yi(12.1)

Theorem 12.2 (Schwarz Inequality). Let(H, h, i)be an inner product space, thenfor allx, y H

|hx, yi| kxkkykand equality holds iffx andy are linearly dependent.

Proof. Ify = 0,the result holds trivially. So assume that y 6= 0.First offnoticethat ifx = y for some C, thenhx, yi= kyk2 and hence

|hx, yi|= || kyk2 =kxkkyk.

Moreover, in this case := hx,yikyk2

.

Now suppose that x H is arbitrary, let z x kyk2hx, yiy. (So z is theorthogonal projection ofx onto y, see Figure 28.) Then

Figure 28. The picture behind the proof.

0 kzk2 =x hx, yikyk2 y

2

=kxk2 +|hx, yi|2

kyk4 kyk2 2Rehx,hx, yi

kyk2yi

=kxk2 |hx, yi|2

kyk2

from which it follows that 0 kyk2kxk2 |hx, yi|2 with equality iff z = 0 orequivalently iffx = kyk2hx, yiy.


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ANALYSIS TOOLS W ITH APPLICATIONS 223

Corollary 12.3. Let(H, h, i)be an inner product space andkxk:=p

hx, xi.Thenk kis a norm onH. Moreoverh, iis continuous onH H, whereH is viewed asthe normed space(H, kk).

Proof. The only non-trivial thing to verify that kk is a norm is the triangleinequality:

kx + yk2 =kxk2 + kyk2 + 2Rehx, yi kxk2 + kyk2 + 2kxk kyk= (kxk + kyk)2

where we have made use of Schwarzs inequality. Taking the square root of thisinequality showskx + yk kxk + kyk. For the continuity assertion:

|hx, yi hx0, y0i|= |hx x0, yi + hx0, y y0i| kykkx x0k + kx0kky y0k

kykkx

x0k + (kxk + kx

x0k) ky

y0k

=kykkx x0k + kxkky y0k + kx x0kky y0kfrom which it follows thath, iis continuous.

Definition 12.4. Let (H, h, i) be an inner product space, we say x, y H areorthogonal and write x y iff hx, yi = 0. More generally if A H is a set,x H is orthogonal to A and write x A iff hx, yi = 0 for all y A. LetA = {xH : xA} be the set of vectors orthogonal to A. We also say that aset S H is orthogonal ifx y for all x, y Ssuch that x 6= y. IfS furthersatisfies, kxk= 1 for all x S, then Sis said to be orthonormal.Proposition 12.5. Let(H, h, i) be an inner product space then

(1) (Parallelogram Law)

(12.2) kx + yk2

+ kx yk2

= 2kxk2

+ 2kyk2

for allx, y H.(2) (Pythagorean Theorem) IfS H is afinite orthonormal set, then

(12.3) kXxS

xk2 =XxS

kxk2.

(3) IfA His a set, thenA is aclosed linear subspace ofH.Remark 12.6. See Proposition 12.37 in the appendix below for the converse ofthe parallelogram law.

Proof. I will assume that H is a complex Hilbert space, the real case beingeasier. Items 1. and 2. are proved by the following elementary computations:

kx + yk2 + kx

yk2 =kxk2 + kyk2 + 2Rehx, yi + kxk2 + kyk2

2Rehx, yi

= 2kxk2 + 2kyk2,

and

kXxS

xk2 =hXxS

x,XyS

yi=Xx,yS

hx, yi

=XxS

hx, xi=XxS

kxk2.


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Item 3. is a consequence of the continuity ofh, iand the fact that

A=

x

Aker(h, xi)

whereker(h, xi) = {y H: hy, xi= 0} a closed subspace ofH.Definition 12.7. A Hilbert space is an inner product space (H, h, i) such thatthe induced Hilbertian norm is complete.

Example 12.8. Let (X, M, ) be a measure space then H := L2(X, M, ) withinner product

(f, g) =

ZX

f gd

is a Hilbert space. In Exercise 12.6 you will show every Hilbert space H is equiv-alent to a Hilbert space of this form.

Definition 12.9. A subset C of a vector space X is said to be convex if for allx, y C the line segment [x, y] := {tx + (1 t)y: 0 t 1} joining x to y iscontained in Cas well. (Notice that any vector subspace ofX is convex.)

Theorem 12.10. Suppose thatHis a Hilbert space andM Hbe a closed convexsubset ofH. Then for anyx Hthere exists a uniquey Msuch that

kx yk= d(x, M) = infzM

kx zk.

Moreover, ifMis a vector subspace ofH, then the pointy may also be characterizedas the unique point inMsuch that(x y) M.

Proof. By replacing M byM x:= {m x: m M} we may assumex = 0.Let:= d(0, M) = infmMkmk and y, z M, see Figure 29.

Figure 29.The geometry of convex sets.

By the parallelogram law and the convexity ofM ,

(12.4) 2kyk2+2kzk2 =ky+zk2+kyzk2 = 4k y+ z2

||2+kyzk2 42+kyzk2.

Hence ifkyk = kzk = , then 22 + 22 42 +ky zk2, so that ky zk2 = 0.Therefore, if a minimizer ford(0, )|Mexists, it is unique.


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Existence. Letyn Mbe chosen such that kynk= n d(0, M). Takingy= ym andz = yn in Eq. (12.4) shows 2

2m+ 2

2n 42 + kyn ymk2. Passing to

the limitm, n

in this equation implies,

22 + 22 42 + lim supm,n

kyn ymk2.

Therefore {yn}n=1 is Cauchy and hence convergent. Because M is closed, y :=

limn

yn M and because kk is continuous,

kyk = limn

kynk= = d(0, M).

Soy is the desired point in Mwhich is closest to 0.Now for the second assertion we further assume that M is a closed subspace of

H and x H. Let y Mbe the closest point in M to x. Then for w M, thefunction

g(t) kx (y+ tw)k2 =kx yk2 2tRehx y, wi + t2kwk2

has a minimum at t = 0. Therefore 0 = g0(0) =2Rehx y, wi. Since w M isarbitrary, this implies that (x y) M. Finally suppose y Mis any point suchthat(x y) M. Then for z M, by Pythagoreans theorem,

kx zk2 =kx y+ y zk2 =kx yk2 + ky zk2 kx yk2

which shows d(x, M)2 kx yk2. That is to say y is the point in M closest to x.

Definition 12.11. Suppose that A : H H is a bounded operator. The adjointofA, denote A, is the unique operator A : H Hsuch that hAx,yi= hx, Ayi.(The proof that A exists and is unique will be given in Proposition 12.16 below.)A bounded operatorA : H H is self - adjoint or Hermitian ifA = A.Definition 12.12. Let Hbe a Hilbert space and M H be a closed subspace.The orthogonal projection ofHontoMis the function PM :H Hsuch that forx H, PM(x) is the unique element in M such that(x PM(x)) M.Proposition 12.13. LetHbe a Hilbert space andM H be a closed subspace.The orthogonal projectionPM satisfies:

(1) PMis linear (and hence we will writePMx rather thanPM(x).(2) P2M=PM (PM is a projection).(3) PM=PM, (PM is self-adjoint).(4) Ran(PM) = M andker(PM) = M

.

Proof.

(1) Letx1, x2 Hand F, thenPMx1+ PMx2 M andPMx1+ PMx2 (x1+ x2) = [PMx1 x1+ (PMx2 x2)] M

showingPMx1+ PMx2 = PM(x1+ x2), i.e. PMis linear.(2) ObviouslyRan(PM) = M and PMx = x for all x M. ThereforeP2M =

PM.


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(3) Letx, y H, then since (x PMx) and (y PMy)are inM,hPMx, yi= hPMx, PMy+ y

PMyi

=hPMx, PMyi

=hPMx + (x PM), PMyi=hx, PMyi.

(4) It is clear thatRan(PM) M. Moreover, ifx M, then PMx= x impliesthatRan(PM) = M. Now x ker(PM)iffPMx= 0 iffx = x 0 M.

Corollary 12.14. Suppose that M H is a proper closed subspace of a HilbertspaceH, thenH= M M.

Proof. Givenx H, let y = PMxso that x y M. Thenx = y+ (x y) M+ M. Ifx

M

M, then x

x, i.e. kxk2 =hx, xi= 0. So M

M = {0} .

Proposition 12.15 (Riesz Theorem). Let H be the dual space of H (Notation3.63). The map

(12.5) z H j h, zi His a conjugate linear isometric isomorphism.

Proof. The map j is conjugate linear by the axioms of the inner products.Moreover, for x, z H,

|hx, zi| kxk kzk for all x Hwith equality whenx = z. This implies that kjz kH =kh, zikH =kzk . Therefore

j is isometric and this shows that j is injective. To finish the proof we must showthat j is surjective. So let f H which we assume with out loss of generality isnon-zero. ThenM= ker(f) a closed proper subspace ofH. Since, by Corollary12.14, H = M M, f : H/M= M F is a linear isomorphism. Thisshows that dim(M) = 1 and hence H = M Fx0 where x0 M \ {0} .28Choose z = x0 M such that f(x0) =hx0, zi. (So = f(x0)/ kx0k2 .) Then forx= m + x0 with m M and F,

f(x) = f(x0) = hx0, zi= hx0, zi= hm + x0, zi= hx, zi

which shows that f=jz.

Proposition 12.16 (Adjoints). LetH andK be Hilbert spaces andA : H Kbe a bounded operator. Then there exists a unique bounded operatorA : K Hsuch that

(12.6) hAx,yiK=hx, AyiH for allx H andy K.

Moreover(A + B) = A+B, A := (A) = A, kAk= kAk andkAAk=kAk2 for allA, B L(H, K) and C.

28 Alternatively, choosex0 M\{0}such thatf(x0) = 1.Forx M we havef(xx0) = 0provided that := f(x). Therefore x x0 MM = {0} , i.e. x = x0. This again showsthat M is spanned by x0.


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Proof. For each y K, then map xhAx,yiK is in H and therefore thereexists by Proposition 12.15 a unique vector z Hsuch that

hAx,yiK=hx, ziH for all x H.This shows there is a unique map A :K H such that hAx,yiK= hx, A(y)iHfor all xH and y K. To finish the proof, we need only show A is linear andbounded. To see A is linear, lety1, y2 Kand C, then for any x H,

hAx,y1+ y2iK=hAx,y1iK+hAx,y2iK

=hx, A(y1)iK+hx, A(y2)iK=hx, A(y1) + A(y2)iK

and by the uniqueness ofA(y1+ y2) we find

A(y1+ y2) = A(y1) + A(y2).

This shows A is linear and so we will now write Ay instead ofA(y). Since

hAy, xiH=hx, AyiH=hAx,yiK=hy,AxiK

it follows that A = A. he assertion that (A + B) = A + B is left to thereader, see Exercise 12.1.

The following arguments prove the assertions about norms ofA and A :

kAk= supkK:kkk=1

kAkk= supkK:kkk=1

suphH:khk=1

|hAk, hi|

= suphH:khk=1

supkK:kkk=1

|hk,Ahi|= suphH:khk=1

kAhk= kAk ,

kAAk kAk kAk= kAk2 andkAk2 = sup

hH:khk=1|hAh,Ahi|= sup

hH:khk=1|hh, AAhi|

suphH:khk=1

kAAhk= kAAk ,

wherein these arguments we have repeatedly made use of the Inequality.

Exercise 12.1. LetH,K,M be Hilbert space,A, B L(H, K), C L(K, M)and C. Show(A + B) = A+B and(CA)= AC L(M, H).Exercise 12.2. LetH= Cn andK= Cm equipped with the usual inner products,i.e. hz, wiH=z w for z , w H. Let A be an m n matrix thought of as a linearoperator fromHto K. Show the matrix associated toA : K His the conjugatetranspose ofA.

Exercise 12.3. Let K : L2()L2() be the operator defined in Exercise 9.12.Show K : L2() L2() is the operator given by

Kg(y) =ZX

k(x, y)g(x)d(x).

Definition 12.17. {u}A H is an orthonormal set ifu u for all 6= andkuk= 1.


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Proposition 12.18(Bessels Inequality). Let{u}A be an orthonormal set, then

(12.7) XA

|hx, ui|2

kxk2 for allx

H.

In particular the set{ A: hx, ui 6= 0} is at most countable for allx H.Proof. Let Abe any finite set. Then

0 kx X

hx, uiuk2 =kxk2 2Re

X

hx, ui hu, xi +X

|hx, ui|2

=kxk2 X

|hx, ui|2

showing that X

|hx, ui|2 kxk2.

Taking the supremum of this equation of A then proves Eq. (12.7).Proposition 12.19. Suppose A H is an orthogonal set. Then s =PvA vexists inH iff

PvA kvk

2 < . (In particularA must be at most a countable set.)Moreover, if

PvA kvk

2 < , then(1) ksk2 =

PvA kvk

2 and(2) hs, xi=

PvAhv, xi for allx H.

Similarly if {vn}n=1 is an orthogonal set, then s =

Pn=1

vn exists in H iff

Pn=1

kvnk2 0 there exists A

such that if A \ ,

(12.8)

Xv

v

2

= Xv

kvk2 < 2.

Hence by Lemma 3.72,P

vA v exists.For item 1, let be as above and set s:=

Pv v. Then

|ksk ksk| ks sk< and by Eq. (12.8),

0 XvA

kvk2 ksk2 =Xv/

kvk2 2.


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Letting 0 we deduce from the previous two equations that ksk ksk andksk

2 PvA kvk2 as 0and thereforeksk2 =PvA kvk

2.

Item 2. is a special case of Lemma 3.72.For the final assertion, let sN

NPn=1

vn and suppose that limN sN=s exists

in Hand in particular {sN}N=1 is Cauchy. So for N > M.

NXn=M+1

kvnk2 =ksN sMk2 0 as M , N

which shows thatPn=1

kvnk2 is convergent, i.e.

Pn=1

kvnk2 < .

Remark: We could use the last result to prove Item 1. Indeed, ifP

vA kvk2 0 there exists a polynomialp(y)such thatZ 1

1

f(y1/3) p(y)

2d(y)< 2.

However, by Eq. (12.13) we have

2 >

Z 11

f(y1/3) p(y)

2d(y) =

Z 11

f(x) p(x3)2 dx.

Alternatively, iff C([1, 1]),theng(y) = f(y1/3)is back inC([1, 1]).There-fore for any >0, there exists a polynomial p(y) such that

>kg pku = sup {|g(y) p(y)|: y [1, 1]}= supg(x

3) p(x3): x [1, 1]= sup f(x) p(x3): x [1, 1] .

This gives another proof the polynomials in x3 are dense in C([1, 1]) and hencein L2([1, 1]).12.3. Weak Convergence. Suppose H is an infinite dimensional Hilbert spaceand{xn}

n=1 is an orthonormal subset ofH. Then, by Eq. (12.1), kxn xmk2 = 2

for all m 6= n and in particular, {xn}n=1 has no convergent subsequences. From

this we conclude that C := {x H :kxk 1} , the closed unit ball in H, is notcompact. To overcome this problems it is sometimes useful to introduce a weakertopology on Xhaving the property that C is compact.

Definition 12.30. Let(X, kk)be a Banach space andX be its continuous dual.The weak topology, w, on X is the topology generated by X. If{xn}

n=1 X

is a sequence we will write xnw x as n to mean that xn x in the weak

topology.Because w = (X

) kk := ({kx k: x X} , it is harder for a functionf : X F to be continuous in the w topology than in the norm topology, kk.In particular if : X Fis a linear functional which isw continuous, then iskk continuous and hence X.Proposition 12.31. Let{xn}

n=1 Xbe a sequence, thenxn

w x Xasn iff(x) = limn (xn) for all X.


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Proof. By definition ofw, we have xnw x X ifffor all X and >0

there exists an N N such that |(x) (xn)| < for all n N and .This later condition is easily seen to be equivalent to (x) = lim

n(x

n) for all

X.The topological space (X, w)is still Hausdorff, however to prove this one needs

to make use of the Hahn Banach Theorem 18.16 below. For the moment we willconcentrate on the special case where X = H is a Hilbert space in which caseH = {z :=h, zi: z H} ,see Propositions 12.15. Ifx, y Hand z := y x6= 0,then

0< := kzk2 =z(z) = z(y) z(x).Thus Vx:= {w H: |z(x) z(w)|< /2} and Vy :={w H :|z(y) z(w)|< /2}are disjoint sets fromwwhich containxandyrespectively. This shows that(H, w)is a Hausdorffspace. In particular, this shows that weak limits are unique if theyexist.

Remark12.32. Suppose thatHis an infi

nite dimensional Hilbert space{xn}n=1 isan orthonormal subset ofH. Then Bessels inequality (Proposition 12.18) impliesxn

w 0 H asn . This points out the fact that ifxn w x H asn , itis no longer necessarily true that kxk= limn kxnk .However we do always havekxk liminfn kxnkbecause,

kxk2 = limn

hxn, xi liminfn

[kxnk kxk] =kxk lim infn

kxnk .

Proposition 12.33. LetHbe a Hilbert space, Hbe an orthonormal basis forH and{xn}

n=1 Hbe a bounded sequence, then the following are equivalent:

(1) xnw x H asn .

(2) hx, yi= limnhxn, yi for ally H.(3) hx, yi= limnhxn, yi for ally .

Moreover, ifcy := limnhxn, yi exists for ally , thenPy|cy|2 < andxn

w x:=Pycyy H asn .Proof. 1. = 2. This is a consequence of Propositions 12.15 and 12.31. 2. =

3. is trivial.3. =1. Let M:= supn kxnkand H0 denote the algebraic span of . Then for

y H andz H0,|hx xn, yi| |hx xn, zi| + |hx xn, y zi| |hx xn, zi| + 2Mky zk .

Passing to the limit in this equation implies lim supn |hx xn, yi| 2Mky zkwhich shows lim supn |hx xn, yi|= 0 since H0 is dense in H.

To prove the last assertion, let . Then by Bessels inequality (Proposition12.18), X

y|cy|2 = limn

Xy

|hxn, yi|2 liminfn kxnk2 M2.

Since was arbitrary, we conclude thatPy|cy|2 M < and hence wemay definex :=

Pycyy. By construction we have

hx, yi= cy = limn

hxn, yi for all y

and hencexnw x H asn by what we have just proved.


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Theorem 12.34. Suppose that{xn}n=1 H is a bounded sequence. Then there

exists a subsequenceyk :=xnk of{xn}n=1 andx Xsuch thatyk

w x ask .Proof. This is a consequence of Proposition 12.33 and a Cantors diagonalization

argument which is left to the reader, see Exercise 12.14.

Theorem 12.35 (Alaoglus Theorem for Hilbert Spaces). Suppose that H is aseparable Hilbert space, C := {x H: kxk 1} is the closed unit ball in H and{en}

n=1 is an orthonormal basis forH. Then

(12.14) (x, y) :=Xn=1

1

2n|hx y, eni|

defines a metric on C which is compatible with the weak topology on C, C :=(w)C={V C: V w} . Moreover(C, ) is a compact metric space.

Proof. The routine check that is a metric is left to the reader. Let bethe topology on C induced by . For any y H and n N, the map x Hhx y, eni= hx, eni hy, eni is w continuous and since the sum in Eq. (12.14) isuniformly convergent for x, y C, it follows that x (x, y) is C continuous.This implies the open balls relative to are contained in C and therefore C. For the converse inclusion, let z H, x z(x) = hz, xi be an element ofH, and for N N let zN :=

PNn=1hz, enien. Then zN =

PNn=1hz, enien is

continuous, being a finite linear combination of the en which are easily seen to be continuous. Because zN z as N it follows that

supxC

|z(x) zN(x)|= kz zNk 0 as N .

Thereforez|C is continuous as well and henceC=(z|C :z H) .The last assertion follows directly from Theorem 12.34 and the fact that sequen-

tial compactness is equivalent to compactness for metric spaces.

Theorem 12.36 (Weak and Strong Differentiability). Suppose that f L2(Rn)andv Rn \ {0} . Then the following are equivalent:

(1) There exists{tn}n=1 R\ {0} such that limn tn = 0 and

supn

f( + tnv) f()tn2

< .

(2) There existsg L2(Rn) such thathf, vi= hg, i for all Cc (Rn).(3) There existsg L2(Rn)andfn Cc (Rn)such thatfn L

2

f andvfn L2

gasn

.

(4) There existsg L2 such thatf( + tv) f()

tL2 g as t 0.

(See Theorem 19.7 for theLp generalization of this theorem.)

Proof. 1. = 2. We may assume, using Theorem 12.34 and passing to asubsequence if necessary, that f(+tnv)f()tn

w g for some g L2(Rn). Now for


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Cc (Rn),

hg, i= limn

hf( + tnv) f()

tn, i= lim

nhf,

( tnv) ()tn

i

=hf, limn

( tnv) ()tn

i= hf, vi,wherein we have used the translation invariance of Lebesgue measure and the dom-inated convergence theorem.

2. = 3. Let Cc (Rn,R) such thatRRn(x)dx = 1 and let m(x) =

mn(mx), then by Proposition 11.24, hm:= m f C(Rn) for all m andvhm(x) = vm f(x) =

ZRn

vm(x y)f(y)dy = hf, v[m(x )]i=hg, m(x )i= m g(x).

By Theorem 11.21, hm f L2(Rn) and vhm = m g g in L2(Rn) asm

. This shows 3. holds except for the fact thath

mneed not have compact

support. To fix this let Cc (Rn, [0, 1]) such that = 1in a neighborhood of0and let(x) = (x) and (v)(x) := (v) (x). Then

v(hm) = vhm+ vhm= (v) hm+ vhm

so that hm hm in L2 andv(hm) vhm in L2 as 0. Let fm= mhmwherem is chosen to be greater than zero but small enough so that

kmhm hmk2+ kv(mhm) vhmk2 < 1/m.Thenfm Cc (Rn), fm f andvfm g in L2 asm .

3. =4. By the fundamental theorem of calculustvfm(x) fm(x)

t =

fm(x + tv) fm(x)t

=1tZ 10

dds fm(x + stv)ds=

Z 10

(vfm) (x + stv)ds.(12.15)

Let

Gt(x) :=

Z 10

stvg(x)ds=Z 10

g(x + stv)ds

which is defined for almost everyx and is in L2(Rn)by Minkowskis inequality forintegrals, Theorem 9.27. Therefore

tvfm(x) fm(x)t

Gt(x) =Z 10

[(vfm) (x + stv) g(x + stv)] dsand hence again by Minkowskis inequality for integrals,

tvfm fm

t Gt

2

Z 1

0

kstv(vfm) stvgk2 ds= Z 1

0

kvfm gk2 ds.

Lettingm in this equation implies (tvf f) /t= Gt a.e. Finally one moreapplication of Minkowskis inequality for integrals implies,tvf ft g

2

=kGt gk2 =Z 10

(stvg g) ds2

Z 10

kstvg gk2 ds.


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By the dominated convergence theorem and Proposition 11.13, the latter term tendsto0 ast 0and this proves 4. The proof is now complete since 4. = 1. is trivial.

12.4. Supplement 1: Converse of the Parallelogram Law.

Proposition 12.37 (Parallelogram Law Converse). If(X, kk) is a normed spacesuch that Eq. (12.2) holds for allx, y X, then there exists a unique inner producton h, i such thatkxk :=

phx, xi for all x X. In this case we say thatkk is a

Hilbertian norm.

Proof. Ifkk is going to come from an inner product h, i, it follows from Eq.(12.1) that

2Rehx, yi= kx + yk2 kxk2 kyk2and

2Rehx, yi= kx yk2 kxk2 kyk2.Subtracting these two equations gives the polarization identity,

4Rehx, yi= kx + yk2 kx yk2.Replacingy byiy in this equation then implies that

4Imhx, yi= kx + iyk2 kx iyk2

from which we find

(12.16) hx, yi=1

4

XG

kx + yk2

whereG = {1, i} a cyclic subgroup ofS1

C. Hence ifh, iis going to exists

we must define it by Eq. (12.16).Notice that

hx, xi=1

4

XG

kx + xk2 =kxk2 + ikx + ixk2 ikx ixk2

=kxk2 + i1 + i|2

kxk2 i 1 i|2 kxk2 =kxk2 .

So to finish the proof of (4) we must show that hx, yi in Eq. (12.16) is an innerproduct. Since

4hy, xi=XG

ky+ xk2 =XG

k (y+ x) k2

= XG

ky+ 2xk2

=ky+ xk2 + k y+ xk2 + ikiy xk2 ik iy xk2=kx + yk2 + kx yk2 + ikx iyk2 ikx + iyk2= 4hx, yi

it suffices to show x hx, yi is linear for all y H. (The rest of this proof maysafely be skipped by the reader.) For this we will need to derive an identity from


16/28


Eq. (12.2). To do this we make use of Eq. (12.2) three times to find

kx + y+ zk2 =

kx + y

zk2 + 2kx + yk2 + 2kzk2

=kx y zk2 2kx zk2 2kyk2 + 2kx + yk2 + 2kzk2=ky+ z xk2 2kx zk2 2kyk2 + 2kx + yk2 + 2kzk2= ky+ z+ xk2 + 2ky+ zk2 + 2kxk2 2kx zk2 2kyk2 + 2kx + yk2 + 2kzk2.

Solving this equation for kx + y+ zk2 gives

(12.17) kx + y+ zk2 =ky+ zk2 + kx + yk2 kx zk2 + kxk2 + kzk2 kyk2.Using Eq. (12.17), for x, y,z H,

4 Rehx + z, yi= kx + z+ yk2 kx + z yk2=ky+ zk2 + kx + yk2 kx zk2 + kxk2 + kzk2 kyk2

kz yk2 + kx yk2 kx zk2 + kxk2 + kzk2 kyk2=kz+ yk2 kz yk2 + kx + yk2 kx yk2= 4Rehx, yi + 4Rehz, yi.(12.18)

Now suppose that G, then since ||= 1,

4hx,yi=1

4

XG

kx + yk2 =1

4

XG

kx + 1yk2

=1

4

XG

kx + yk2 = 4hx, yi(12.19)

where in the third inequality, the substitution was made in the sum. So Eq.(12.19) says hix,yi= ihix,yi and h

x, yi=

hx, yi. Therefore

Imhx, yi= Re (ihx, yi) = Rehix,yiwhich combined with Eq. (12.18) shows

Imhx + z, yi= Rehix iz,yi= Rehix,yi + Rehiz,yi= Imhx, yi + Imhz, yi

and therefore (again in combination with Eq. (12.18)),

hx + z, yi= hx, yi + hz, yifor all x, y H.Because of this equation and Eq. (12.19) to finish the proof that x hx, yi islinear, it suffices to show hx,yi= hx, yi for all >0. Now if = m N, then

hmx,yi= hx + (m 1)x, yi= hx, yi + h(m 1)x, yiso that by induction hmx,yi = mhx, yi. Replacing x by x/m then shows thathx, yi= mhm1x, yiso thathm1x, yi= m1hx, yi and so ifm, n N, we find

hn

mx, yi= nh

1

mx, yi=

n

mhx, yi

so thathx,yi= hx, yifor all >0 and Q.By continuity, it now follows thathx,yi= hx, yifor all >0.


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12.5. Supplement 2. Non-complete inner product spaces. Part of Theorem12.24 goes through when His a not necessarily complete inner product space. Wehave the following proposition.

Proposition 12.38. Let(H, h, i)be a not necessarily complete inner product spaceand Hbe an orthonormal set. Then the following two conditions are equivalent:

(1) x=Pu

hx, uiu for allx H.(2) kxk2 =

Pu

|hx, ui|2 for allx H.Moreover, either of these two conditions implies that His a maximal ortho-

normal set. However Hbeing a maximal orthonormal set is not sufficient toconditions for 1) and 2) hold!

Proof. As in the proof of Theorem 12.24, 1) implies 2). For 2) implies 1) let and consider

x Xu

hx, uiu2

=kxk2 2Xu

|hx, ui|2 +Xu

|hx, ui|2

=kxk2 Xu

|hx, ui|2 .

Sincekxk2 =Pu

|hx, ui|2, it follows that for every >0 there exists suchthat for all such that ,x

Xu

hx, uiu

2

=kxk2 Xu

|hx, ui|2 <

showing thatx =

Puhx, uiu.

Supposex = (x1, x2, . . . , xn, . . . ) .If 2) is valid thenkxk2 = 0,i.e. x= 0.Sois maximal. Let us now construct a counter example to prove the last assertion.

TakeH= Span{ei}i=1 2 and letun= e1(n+1)en+1for n= 1, 2 . . . .Apply-ing Gramn-Schmidt to {un}

n=1we construct an orthonormal set= {un}

n=1 H.

I now claim that H is maximal. Indeed ifx = (x1, x2, . . . , xn, . . . ) thenx un for all n, i.e.

0 = (x, un) = x1 (n + 1)xn+1.Therefore xn+1 = (n + 1)

1 x1 for all n. Since xSpan{ei}i=1, xN = 0 for someN sufficiently large and therefore x1 = 0 which in turn implies that xn = 0 for alln. So x = 0 and hence is maximal inH. On the other hand, is not maximalin 2. In fact the above argument shows that in 2 is given by the span ofv =(1, 12 ,

13 ,

14 ,

15 , . . . ).Let P be the orthogonal projection of

2 onto theSpan() = v.

Then Xi=1

hx, uniun= P x= x hx, vikvk2

v ,

so thatPi=1

hx, uniun = x iff x Span() = v 2. For example if x =(1, 0, 0, . . . ) H (or more generally for x = ei for any i), x / v and hencePi=1

hx, uniun6=x.


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12.6. Supplement 3: Conditional Expectation. In this section let (, F, P)be a probability space, i.e. (, F, P) is a measure space and P() = 1. Let G Fbe a sub sigma algebra ofF and write f

Gb

iff : C

is bounded and f is(G, BC) measurable. In this section we will write

Ef :=

Z

fdP.

Definition 12.39(Conditional Expectation). LetEG :L2(, F, P) L2(, G, P)

denote orthogonal projection ofL2(, F, P) onto the closed subspace L2(, G, P).Forf L2(, G, P), we say thatEGf L2(, F, P) is the conditional expecta-tion off .

Theorem 12.40. Let(, F, P) andG Fbe as above andf , g L2(, F, P).(1) Iff 0, P a.e. thenEGf 0, P a.e.(2) Iff g, P a.e. thereEGf EGg, P a.e.(3) |EGf| EG |f|, P a.e.(4) kEGfkL1 kfkL1 for allf L2.So by the B.L.T. Theorem 4.1, EG extends

uniquely to a bounded linear map fromL1(, F, P)to L1(, G, P)which wewill still denote byEG .

(5) Iff L1(, F, P) thenF =EGf L1(, G, P) iffE(F h) = E(f h) for allh Gb.

(6) Ifg Gb andf L1(, F, P), thenEG(gf) = g EGf, P a.e.Proof. By the definition of orthogonal projection forh Gb,

E(f h) = E(f EGh) = E(EGf h).

So iff, h

0then0

E(f h)

E(EGf h)and since this holds for all h

0in Gb,

EGf 0, P a.e. This proves (1). Item (2) follows by applying item (1). to f g.Iff is real, f |f| and so by Item (2), EGf EG |f|, i.e. |EGf| EG |f|, P a.e. For complexf , let h 0be a bounded and G measurable function. Then

E[|EGf| h] =Eh

EGf sgn(EGf)hi

= Eh

f sgn(EGf)hi

E[|f| h] =E[EG|f| h] .Since h is arbitrary, it follows that |EGf| EG|f| , P a.e. Integrating thisinequality implies

kEGfkL1 E|EGf| E[EG|f| 1] = E[|f|] =kfkL1 .Item (5). Suppose f L1(, F, P) and h Gb. Let fn L2(, F, P) be a

sequence of functions such that fn

f in L1(, F, P). Then

E(EGf h) = E( limn

EGfn h) = limn

E(EGfn h)

= limn

E(fn h) = E(f h).(12.20)

This equation uniquely determines EG , for ifF L1(, G, P)also satisfies E(Fh) =E(f h) for all h Gb, then takingh = sgn(F EGf) in Eq. (12.20) gives

0 = E((F EGf) h) = E(|F EGf|).


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This shows F =EGf, P a.e. Item (6) is now an easy consequence of this charac-terization, since ifh Gb,

E[(gEGf) h] =E[EGf hg] =E[f hg] = E[gf h] = E[EG(gf) h] .Thus EG(gf) = g EGf, P a.e.

Proposition 12.41. IfG0 G1 F. Then(12.21) EG0EG1 =EG1EG0 =EG0 .

Proof. Equation (12.21) holds on L2(, F, P) by the basic properties of or-thogonal projections. It then hold onL1(, F, P)by continuity and the density ofL2(, F, P)in L1(, F, P).

Example 12.42. Suppose that(X, M, )and (Y,N, )are two finite measurespaces. Let = X Y, F = M N and P(dx,dy) = (x, y)(dx)(dy) where L1(, F, ) is a positive function such that RXYd ( ) = 1. LetX : Xbe the projection map,X(x, y) = x, and

G := (X) = 1X (M) = {A Y :A M} .

Then f : R is G measurable ifff = F X for some function F : X Rwhich isN measurable, see Lemma 6.62. Forf L1(, F, P), we will now showEGf=F X where

F(x) = 1

(x)1(0,)((x))

ZY

f(x, y)(x, y)(dy),

(x) :=RY

(x, y)(dy). (By convention,RY

f(x, y)(x, y)(dy) := 0ifRY

|f(x, y)| (x, y)(dy) =.)

By Tonellis theorem, the set

E:={x X : (x) = } x X : ZY |f(x, y)| (x, y)(dy) = is a null set. Since

E[|F X |] =ZX

d(x)

ZY

d(y) |F(x)| (x, y) =

ZX

d(x) |F(x)|(x)

=

ZX

d(x)

ZY

(dy)f(x, y)(x, y)

ZX

d(x)

ZY

(dy) |f(x, y)| (x, y)< ,

F X L1(, G, P).Let h = H X be a bounded G measurable function, then

E[F

X h] = Z

X

d(x) ZY

d(y)F(x)H(x)(x, y)

=

ZX

d(x)F(x)H(x)(x)

=

ZX

d(x)H(x)

ZY

(dy)f(x, y)(x, y)

=E[hf]

and henceEGf=F X as claimed.


20/28


This example shows that conditional expectation is a generalization of the notionof performing integration over a partial subset of the variables in the integrand.Whereas to compute the expectation, one should integrate over all of the variables.See also Exercise 12.8 to gain more intuition about conditional expectations.

Theorem 12.43 (Jensens inequality). Let (, F, P) be a probability space and : R R be a convex function. Assume f L1(, F, P;R) is a function suchthat (for simplicity)(f) L1(, F, P;R), then(EGf) EG[(f)] , P a.e.

Proof. Let us first assume that is C1 andf is bounded. In this case

(12.22) (x) (x0) 0(x0)(x x0)for all x0, x R.Takingx0 = EGf andx = fin this inequality implies

(f) (EGf) 0(EGf)(f EGf)and then applyingEG to this inequality gives

EG[(f)]

(EGf) = EG [(f)

(EGf)]

0(EGf)(EGf

EGEGf) = 0

The same proof works for general , one need only use Proposition 9.7 to replaceEq. (12.22) by

(x) (x0) 0(x0)(x x0)for all x0, x Rwhere0(x0) is the left hand derivative of at x0.

Iffis not bounded, apply what we have just proved to fM =f1|f|M, to find

(12.23) EG

(fM) (EGfM).

Since EG : L1(, F, P;R) L1(, F, P;R) is a bounded operator and fM fand (fM) (f) in L1(, F, P;R) as M , there exists {Mk}k=1 such thatMk andfMk f and(fMk) (f), P a.e. So passing to the limit in Eq.(12.23) shows EG[(f)] (EGf), P a.e.12.7. Exercises.

Exercise 12.7. Let (X, M, ) be a measure space and H :=L2(X, M, ). Givenf L() let Mf :H Hbe the multiplication operator defined by Mfg = f g.Show M2f =Mf iffthere exists A Msuch thatf= 1A a.e.Exercise 12.8. Suppose (, F, P) is a probability space and A := {Ai}

i=1 F

is a partition of . (Recall this means =

i=1 Ai.) Let G be the algebragenerated by A. Show:

(1) B G iffB = iAi for some N.(2) g : R is G measurable iffg =Pi=1 i1Ai for some i R.(3) For f L1(, F, P), let E(f|Ai) := E[1Aif] /P(Ai) if P(Ai) 6= 0 and

E(f|Ai) = 0 otherwise. Show

EGf=Xi=1

E(f|Ai)1Ai .

Exercise 12.9. Folland 5.60 on p. 177.

Exercise 12.10. Folland 5.61 on p. 178 about orthonormal basis on productspaces.

Exercise 12.11. Folland 5.67 on p. 178 regarding the mean ergodic theorem.


21/28


Exercise 12.12(Haar Basis). In this problem, letL2 denoteL2([0, 1], m)with thestandard inner product,

(x) = 1[0,1/2)(x) 1[1/2,1)(x)and for k, j N0 := N{0} with 0 j


22/28


(4) Forf L2, let

Hnf :=hf, 1i1 +

n1Xk=0

2k1Xj=0

hf, kjikj .

Show (compare with Exercise 12.8)

Hnf=2n1Xj=0

2nZ (j+1)2nj2n

f(x)dx

!1[j2n,(j+1)2n)

and use this to show kf Hnfku 0 as n for all f C([0, 1]).Exercise 12.13. Let O(n) be the orthogonal groups consisting of n n realorthogonal matrices O, i.e. OtrO = I. For O O(n) and f L2(Rn) letUOf(x) = f(O

1x). Show(1) UOf is well defined, namely iff=g a.e. then UOf=UOg a.e.

(2) UO : L2

(Rn

) L2

(Rn

) is unitary and satisfies UO1UO2 = UO1O2 for allO1, O2 O(n). That is to say the map O O(n) U(L2(Rn)) theunitary operators on L2(Rn) is a group homomorphism, i.e. a unitaryrepresentation ofO(n).

(3) For eachf L2(Rn), the map OO(n)UOf L2(Rn) is continuous.Take the topology onO(n)to be that inherited from the Euclidean topologyon the vector space of alln nmatrices. Hint: see the proof of Proposition11.13.

Exercise 12.14. Prove Theorem 12.34. Hint: Let H0 := span {xn: n N} a separable Hilbert subspace ofH. Let {m}

m=1 H0 be an orthonormal basis

and use Cantors diagonalization argument to find a subsequence yk := xnk suchthatcm := limkhyk, mi exists for all m N. Finish the proof by appealing toProposition 12.33.

Exercise 12.15. Suppose that {xn}n=1 H and xn

w xH as n . Showxn x as n (i.e. limn kx xnk= 0)ifflimn kxnk= kxk .Exercise 12.16. Show the vector space operations ofXare continuous in the weaktopology. More explicitly show

(1) (x, y) X X x + y X is (w w, w) continuous and(2) (, x) F X x X is (F w, w) continuous.

Exercise 12.17. Euclidean group representation and its infinitesimal generatorsincluding momentum and angular momentum operators.

Exercise 12.18. Spherical Harmonics.

Exercise 12.19. The gradient and the Laplacian in spherical coordinates.Exercise 12.20. Legendre polynomials.

Exercise 12.21. In this problem you are asked to show there is no reasonablenotion of Lebesgue measure on an infinite dimensional Hilbert space. To be moreprecise, suppose H is an infinite dimensional Hilbert space and m is a measure onBH which is invariant under translations and satisfies, m(B0())> 0 for all >0.Show m(V) = for all open subsets ofH.


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12.8. Fourier Series Exercises.

Notation 12.44. LetCkper(Rd) denote the 2 periodic functions in Ck(Rd),

Ckper(Rd) :=

f Ck(Rd) : f(x + 2ei) = f(x)for all x Rd andi = 1, 2, . . . , d

.

Also leth, idenote the inner product on the Hilbert space H:=L2([, ]d)givenby

hf, gi:=

1

2

d Z[,]d

f(x)g(x)dx.

Recall that

k(x) := eikx :k Zd

is an orthonormal basis for H in particularfor f H,(12.24) f=

XkZd

hf, kik

where the convergence takes place in L2([, ]d). For f L1([, ]d), we willwrite f(k)for the Fourier coefficient,

(12.25) f(k) := hf, ki=

1

2

d Z[,]d

f(x)eikxdx.

Lemma 12.45. Lets > 0, then the following are equivalent,

(12.26)XkZd

1

(1 + |k|)s < ,

XkZd

1

(1 + |k|2)s/2< ands > d.

Proof. LetQ := (0, 1]d andk Zd. Forx = k + y (k+ Q),2 + |k|= 2 + |x y| 2 + |x| + |y| 3 + |x| and2 + |k|= 2 + |x y| 2 + |x| |y| |x| + 1

and therefore for s >0, 1

(3 + |x|)s 1

(2 + |k|)s 1

(1 + |x|)s.

Thus we have shown1

(3 + |x|)s

XkZd

1

(2 + |k|)s1Q+k(x) 1

(1 + |x|)s for all x Rd.

Integrating this equation then showsZRd

1

(3 + |x|)sdx

XkZd

1

(2 + |k|)s

ZRd

1

(1 + |x|)sdx

from which we conclude that

(12.27)XkZd

1

(2 + |k|)s < iffs > d.

Because the functions1 + t,2 + t,and

1 + t2 all behave liket as t ,the sumsin Eq. (12.26) may be compared with the one in Eq. (12.27) to finish the proof.

Exercise 12.22 (Riemann Lebesgue Lemma for Fourier Series). Show for fL1([, ]d) that f c0(Zd), i.e. f : Zd C and limk f(k) = 0. Hint: Iff H, this follows form Bessels inequality. Now use a density argument.


24/28


Exercise 12.23. Supposef L1([, ]d)is a function such that f 1(Zd)andset

g(x) :=XkZd

f(k)e

ikx

(pointwise).

(1) Show g Cper(Rd).(2) Show g (x) =f(x) for m a.e. x in [, ]d. Hint: Show g(k) = f(k)and

then use approximation arguments to showZ[,]d

f(x)h(x)dx=

Z[,]d

g(x)h(x)dxh C([, ]d).

(3) Conclude that f L1([, ]d) L([, ]d) and in particular f Lp([, ]d)for all p [1, ].

Exercise 12.24. Suppose m N0, is a multi-index such that || 2m andf C2mper(Rd)29 .

(1) Using integration by parts, show(ik)f(k) = hf, ki.

Note: This equality impliesf(k)

1

kkfkH

1

kkfku .

(2) Now let f=Pd

i=1 2f/x2i , Working as in part 1) show

(12.28) h(1 )mf, ki= (1 + |k|2)mf(k).Remark 12.46. Suppose that m is an even integer, is a multi-index and fCm+||per (Rd), then

XkZd

|k|f(k)

2

=XkZd

|hf, ki| (1 + |k|2)m/2(1 + |k|2)m/2

2

=

XkZd

h(1 )m/2f, ki

(1 + |k|2)m/2

2

XkZd

h(1 )m/2f, ki

2XkZd

(1 + |k|2)m

=Cm

(1 )m/2f2H

whereCm:=PkZd(1 + |k|2)m < iffm > d/2. So the smoother fis the faster

fdecays at infinity. The next problem is the converse of this assertion and hencesmoothness offcorresponds to decay offat infinity and visa-versa.

Exercise 12.25. Supposes R and ck C :k Zd are coefficients such thatXkZd

|ck|2 (1 + |k|2)s < .

29 We view Cper(R) as a subspace ofH by identifying f Cper(R) with f|[,] H.


25/28


Show ifs > d2 + m, the function f defined by

f(x) = XkZd

ckeikx

is in Cmper(Rd). Hint: Work as in the above remark to showX

kZd|ck| |k

|< for all || m.

Exercise 12.26 (Poisson Summation Formula). LetF L1(Rd),

E:=

x R

d :XkZd

|F(x + 2k)|=

and set

F(k) := (2)d/2 ZRd

F(x)eikxdx.

Further assume F 1(Zd).(1) Show m(E) = 0 and E+ 2k = E for all k Zd. Hint: ComputeR

[,]dP

kZd|F(x + 2k)| dx.(2) Let

f(x) :=

PkZdF(x + 2k) for x / E

0 if x E.Show f L1([, ]d)and f(k) = (2)d/2 F(k).

(3) Using item 2) and the assumptions on F, show f L1([, ]d)L([, ]d)and

f(x) = XkZd

f(k)eikx = XkZd

(2)d/2 F(k)eikx form a.e. x,

i.e.

(12.29)XkZd

F(x + 2k) = (2)d/2XkZd

F(k)eikx for m a.e. x.

(4) Suppose we now assume thatF C(Rd)and F satisfies 1)|F(x)| C(1 +|x|)s for somes > d andC < and 2) F 1(Zd),then show Eq. (12.29)holds for all x Rd and in particularX

kZdF(2k) = (2)d/2

XkZd

F(k).

For simplicity, in the remaining problems we will assume thatd = 1.

Exercise 12.27 (Heat Equation 1.). Let(t, x) [0, ) R u(t, x)be a contin-uous function such that u(t, )Cper(R) for all t0, u:= ut, ux, and uxx existsand are continuous when t > 0. Further assume that u satisfies the heat equationu = 1

2uxx. Let u(t, k) := hu(t, ), ki for k Z. Show for t > 0 and k Z that

u(t, k) is differentiable in t and ddt u(t, k) = k2u(t, k)/2. Use this result to show(12.30) u(t, x) =

XkZ

et

2k2 f(k)eikx


26/28


wheref(x) := u(0, x) and as above

f(k) =hf, ki= 1

2Z

f(y)eikydy.

Notice from Eq. (12.30) that(t, x) u(t, x)is C fort >0.Exercise 12.28 (Heat Equation 2.). Let qt(x) :=

12

PkZ e

t2k2eikx. Show that

Eq. (12.30) may be rewritten as

u(t, x) =

Z

qt(x y)f(y)dy

andqt(x) =

XkZ

pt(x + k2)

wherept(x) := 12t

e 1

2tx2 . Also show u(t, x)may be written as

u(t, x) = pt f(x) := ZRdpt(x y)f(y)dy.Hint: To show qt(x) =

PkZpt(x+ k2), use the Poisson summation formula

along with the Gaussian integration formula

pt() = 1

2

ZR

pt(x)eixdx=

12

et

22 .

Exercise 12.29(Wave Equation). Letu C2(RR)be such thatu(t, ) Cper(R)for all t R. Further assume that u solves the wave equation, utt = uxx. Letf(x) := u(0, x) and g(x) = u(0, x). Show u(t, k) := hu(t, ), ki for k Z is twicecontinuously differentiable int and d

2

dt2u(t, k) = k2u(t, k). Use this result to show

(12.31) u(t, x) = XkZ

f(k) cos(kt) + g(k)sin kt

k eikx

with the sum converging absolutely. Also show thatu(t, x)may be written as

(12.32) u(t, x) =1

2[f(x + t) + f(x t)] +1

2

Z tt

g(x + )d.

Hint: To show Eq. (12.31) implies (12.32) use

cos kt =eikt + eikt

2 , and sin kt =

eikt eikt2i

andeik(x+t) eik(xt)

ik =

Z tt

eik(x+)d.

Exercise 12.30. (Worked Example.) Let D := {z C : |z| < 1} be the openunit disk in C=R

2

, where we write z = x +iy = rei

in the usual way. Also let =

2

x2 + 2

y2 and recall that may be computed in polar coordinates by the

formula,

u= r1r

r1ru

+ 1

r22u.

Suppose thatu C(D) C2(D)and u(z) = 0 for z D. Let g = u|D and

g(k) := 1

2

Z

g(eik)eikd.


27/28


(We are identifying S1 = D :=

z D: |z|= 1 with [, ]/ ( ) by themap [, ] ei S1.)Let

(12.33) u(r, k) := 12

Z

u(rei)eikd

then:

(1) u(r, k) satisfies the ordinary differential equation

r1r(rru(r, k)) = 1

r2k2u(r, k)for r (0, 1).

(2) Recall the general solution to

(12.34) rr(rry(r)) = k2y(r)

may be found by trying solutions of the form y(r) = r which then implies2 = k2 or = k. From this one sees that u(r, k) may be written as

u(r, k) = Akr|k|

+Bkr|k|

for some constants Ak and Bk when k 6= 0. Ifk = 0, the solution to Eq. (12.34) is gotten by simple integration and theresult is u(r, 0) =A0+ B0ln r.Since u(r, k) is bounded near the origin foreach k, it follows that Bk = 0 for all k Z.

(3) So we have shown

Akr|k| = u(r, k) =

1

2

Z

u(rei)eikd

and letting r 1 in this equation implies

Ak = 1

2

Z

u(ei)eikd= g(k).

Therefore,

(12.35) u(rei) =XkZ

g(k)r|k|eik

for r


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Putting this altogether we have shown

u(rei) = 1

2Z

Pr(

)u(ei)d

where

Pr() := 1 r2

1 2r cos + r2is the so called Poisson kernel.

Exercise 12.31. ShowP

k=1 k2 = 2/6, by taking f(x) = x on [, ] and

computingkfk22 directly and then in terms of the Fourier Coefficientsf off .

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Hilbert-Spaces MATH231B Appendix1of1