Spectral Theorem 6

7/29/2019 Spectral Theorem 6

1/12

PDE LECTURE NOTES, M ATH 237A-B 157

11. Introduction to the Spectral Theorem

The following spectral theorem is a minor variant of the usual spectral theo-

rem for matrices. This reformulation has the virtue of carrying over to general(unbounded) self adjoint operators on infinite dimensional Hilbert spaces.

Theorem 11.1. Suppose A is an n n complex self adjoint matrix, i.e. A = Aor equivalently Aji = Aij and let be counting measure on {1, 2, . . . , n}. Thenthere exists a unitary map U : Cn L2({1, 2, . . . , n}, d) and a real function : {1, 2, . . . , n} R such that U A = U for all Cn. We summarize thisequation by writing U AU1 = M where

M : L2({1, 2, . . . , n}, d) L2({1, 2, . . . , n}, d)

is the linear operator, g L2({1, 2, . . . , n}, d) g L2({1, 2, . . . , n}, d).

Proof. By the usual form of the spectral theorem for self-adjoint matrices, thereexists an orthonormal basis {ei}

n

i=1 of eigenvectors ofA, say Aei = iei with i R.

Define U : Cn L2({1, 2, . . . , n}, d) to be the unique (unitary) map determinedby U ei = i where

i(j) =

1 if i = j0 if i 6= j

and let : {1, 2, . . . , n} R be defined by (i) := i.

Definition 11.2. Let A : H H be a possibly unbounded operator on H. We let

D(A) = {y H : z H 3 (Ax,y) = (x, z) x D(A)}

and for y D(A) set Ay = z.

Definition 11.3. If A = A the A is self adjoint.

Proposition 11.4. Let (X, ) be finite measure space, H = L2(X,d) and

f : X C be a measurable function. Set Ag = f g = Mfg for all

g D(Mf) = {g H : f g H}.

Then D(Mf) is a dense subspace of H and Mf = Mf.

Proof. For any g H = L2(X,d) and m N, let gm := g1|f|m. Since|f gm| m |g| it follows that fgm H and hence gm D(Mf). By the dominatedconvergence theorem, it follows that gm g in H as m , hence D(Mf) isdense in H.

Suppose h D(Mf) then there exists k L2 such that (Mfg, h) = (g, k) for all

g D(Mf), i.e. ZX

fgh d =

ZX

gk d for all g D(Mf)

or equivalently

(11.1)

ZX

g(f h k)d = 0 for all g D(Mf).

Choose Xn X such that Xn X and (Xn) < for all n. It is easily checkedthat

gn := 1Xnf h kf h k

1|f|n


2/12

158 BRUCE K. DRIVER

is in D(Mf) and putting this function into Eq. (11.1) shows

ZX f h k 1|f|nd = 0 for all n.Using the monotone convergence theorem, we may let n in this equation tofind

RX

f h k

d = 0 and hence that f h = k L2. This shows h D(Mf) and

Mfh = f h.

Theorem 11.5 (Spectral Theorem). Suppose A = A then there exists (X, ) a finite measure space, f : X R measurable, and U : H L2(x, ) unitarysuch that UAU1 = Mf. Note this is a statement about domains as well, i.e.U D(Mf) = D(A).

I would like to give some examples of computing A and Theorem 11.5 aswell. We will consider here the case of constant coefficient differential operatorson L2(Rn). First we need the following definition.

Definition 11.6. Let a C(U), L = P||m a a mth order linear differ-ential operator on D(U) and

L

=X

||m

(1)|| [a]

denote the formal adjoint of L as in Lemma 5.4 above. For f Lp(U) we sayLf Lp(U) or Lploc(U) if the generalized function Lf may be represented by anelement of Lp(U) or Lploc(U) respectively, i.e. Lf = g L

ploc(U) iff

(11.2)

ZU

f L dm =

ZU

gdm for all Cc (U).

In terms of the complex inner product,

(f, g) := ZU

f(x)g(x)dm(x)

Eq. (11.2) is equivalent tof L~

= (g, ) for all Cc (U)

where

L~ :=X

||m

(1)|| [a] .

Notice that L~ satisfies L~ = L. (We do not write L here since L~ is to beconsidered an operator on the space on D0 (U) .)

Remark 11.7. Recall that if f, h L2 (Rn) , then the following are equivalent

(1) f = h.(2) (h, g) =

f, F1g

for all g Cc (R

n) .

(3) (h, g) =

f, F1g

for all g S(Rn) .

(4) (h, g) =

f, F1g

for all g L2 (Rn) .

Indeed if f = h and g L2 (Rn) , the unitarity ofF implies

(h, g) =

f , g

= (Ff, g) =

f, F1g

.


3/12


Hence 1 = 4 and it is clear that 4 = 3 = 2. If 2 holds, then again since F isunitary we have

(h, g) = f, F1g = f , g for all g Cc (Rn)which implies h = f a.e., i.e. h = f in L2 (Rn) .

Proposition 11.8. Letp(x) =P

||m ax be a polynomial onCn,

(11.3) L := p () :=X

||m

a

and f L2 (Rn) . Then Lf L2 (Rn) iff p(i)f() L2 (Rn) and in which case

(11.4) (Lf) () = p(i)f().

Put more concisely, letting

D(B) =

f L2 (Rn) : Lf L2 (Rn)

with Bf = Lf for all f D(B), we haveFBF1 = Mp(i).

Proof. As above, let

(11.5) L :=X

||m

a () and L~ :=

X||m

a () .

For Cc (Rn) ,

L~(x) = L~Z

() eixd() =X

||m

a (x)

Z () eixd()

=

Zp (i) () eixd() = F1

hp (i) ()

i(x)

So if f L2 (Rn) such that Lf L2 (Rn) . Then by Remark 11.7,

(cLf, ) = (Lf,) = hf, L~i = hf(x),F1 hp (i) ()i (x)i= hf(),

hp (i) ()

ii = hp (i) f(), ()i for all Cc (R

n)

from which it follows that Eq. (11.4) holds and that p (i) f() L2 (Rn) .

Conversely, if f L2 (Rn) is such that p (i) f() L2 (Rn) then for Cc (R

n) ,

(11.6)

f, L~

=

f ,FL~

.

Since

FL~ () = ZL~ (x) eixd(x) = Z (x) Lxeixd(x)=

Z (x) a

x eixd(x) =

Z (x) a (i)

eixd(x)

= p (i)(),

Eq. (11.6) becomesf, L~

=

f(), p (i)()

=p (i) f(), ()

=F1

hp (i) f()

i(x), (x)

.


4/12

160 BRUCE K. DRIVER

This shows Lf = F1hp (i) f()

i L2 (Rn) .

Lemma 11.9. Suppose p(x) = P||m ax is a polynomial onRn and L = p()is the constant coefficient differential operator B = P||m a with D(B) :=S(Rn) L2 (Rn) . Then

FBF1 = Mp(i)|S(Rn).

Proof. This is result of the fact that F(S(Rn)) = S(Rn) and for f S(Rn)we have

f(x) =

ZRn

f()eixd()

so that

Bf(x) =

ZRn

f()Lxeixd() =

ZRn

f()p(i)eixd()

so that

(Bf) () = p(i)f() for all f S(Rn) .

Lemma 11.10. Suppose g : Rn C is a measurable function such that |g(x)|

C

1 + |x|M for some constants C and M. Let A be the unbounded operator on

L2 (Rn) defined by D(A) = S(Rn) and for f S(Rn) , Af = gf. Then A = Mg.

Proof. If h D (Mg) and f D(A), we have

(Af,h) =

ZRn

gfhdm =

ZRn

fghdm = (f, Mgh)

which shows Mg A, i.e. h D(A) and Ah = Mgh. Now suppose h D (A

)

and A

h = k, i.e.ZRn

gfhdm = (Af,h) = (f, k) =

ZRn

fkdm for all f S(Rn)

or equivalently that ZRn

gh k

f dm = 0 for all f S(Rn) .

Since the last equality (even just for f Cc (Rn)) implies gh k = 0 a.e. we may

conclude that h D(Mg) and k = Mgh, i.e. A Mg.

Theorem 11.11. Suppose p(x) =P

||m ax is a polynomial on Rn and

A = p() is the constant coefficient differential operator with D(A) := Cc (Rn)

L2

(Rn

) such that A = L = p() on D(A), see Eq. (11.3). Then A

is the operatordescribed by

D(A) =


=n

f L2 (Rn) : p(i)f() L2 (Rn)o

and Af = Lf for f D(A) where L is defined in Eq. (11.5) above. Moreoverwe haveFAF1 = M

p(i).


5/12


Proof. Let D(B) = S(Rn) and B := L on D(B) so that A B. We are firstgoing to show A = B. As is easily verified, in general ifA B then B A. Sowe need only show A B. Now by definition, if g D(A) with k = Ag, then

(Af,g) = (f, k) for all f D(A) := Cc (Rn) .

Suppose that f S(Rn) and Cc (Rn) such that = 1 in a neighborhood of

0. Then fn(x) := (x/n)f(x) is in S(Rn) and hence

(11.7) (fn, k) = (Lfn, g) .

An exercise in the product rule and the dominated convergence theorem showsfn f and Lfn Lf in L

2 (Rn) as n . Therefore we may pass to the limitin Eq. (11.7) to learn

(f, k) = (Bf,g) for all f S(Rn)

which shows g D(B) and Bg = k.By Lemma 11.10, we may conclude that A = B = M

p(i)and by Proposition

11.8 we then conclude thatD (A) =

nf L2 (Rn) : p(i)f() L2 (Rn)

o=


and for f D (A) we have Af = Lf.

Example 11.12. If we take L = with D(L) := Cc (Rn) , then

L = = FM||2F1

where D() =

f L2 (Rn) : f L2 (Rn)

and f = f.

Theorem 11.13. Suppose A = A andA 0. Then for allu0 D(A) there existsa unique solution u C1([0, )) such that u(t) D(A) for all t and

(11.8) u(t) = Au(t) with u(0) = u0.Writing u(t) = etAu0, the map u0 e

tAu0 is a linear contraction semi-group, i.e.

(11.9) ketAu0k ku0k for all t 0.

So etA extends uniquely to H by continuity. This extension satisfies:

(1) Strong Continuity: the map t [0, ) etAu0 is continuous for allu0 H.

(2) Smoothing property: t > 0

etAu0

\n=0

D(An) =: C(A)

and

(11.10) kAketAk ktk ek for all k N.

Proof. Uniqueness. Suppose u solves Eq. (11.8), then

d

dt(u(t), u(t)) = 2Re(u, u) = 2Re(Au,u) 0.

Hence ku(t)k is decreasing so that ku(t)k ku0k.This implies the uniqueness as-sertion in the theorem and the norm estimate in Eq. (11.9).


6/12

162 BRUCE K. DRIVER

Existence: By the spectral theorem we may assume A = Mf acting on L2(X, )for some finite measure space (X, ) and some measurable function f : X [0, ). We wish to show u(t) = etfu

0 L2 solves

u(t) = f u(t) with u(0) = u0 D(Mf) L2.

Let t > 0 and || < t. Then by the mean value inequalitye(t+4)f etf4 u0

= maxn|f e(t+4)fu0| : between 0 and o |fu0| L2.

This estimated along with the fact that

u(t +) u(t)

4=

e(t+4)f etf

4u0

point wise fetfu0 as 0

enables us to use the dominated convergence theorem to conclude

u(t) = L2 lim0

u(t + 4) u(t)

4

= etff u0 = f u(t)

as desired. i.e. u(t) = f u(t).The extension of etA to H is given by Metf. For g L

2,etfg

|g| L2 and

etfg egf pointwise as t , so the Dominated convergence theorem showst [0, ) etAg H is continuous. For the last two assertions, let t > 0 andf(x) = xketx. Then (ln f)0(x) = k

x+ t which is zero when x = k/t and therefore

maxx0

xketx

= |f(k/t)| =

k

t

kek.

Hence

kAketAkop maxx0

xketx

k

t

kek < .

Theorem 11.14. Take A = FM||2F1 so A|S = then

C(A) :=\n=1

D(An) CRd

i.e. for all f C(A) there exists a version f of f such that f C(Rd).

Proof. By assumption ||2nf() L2 for all n. Therefore f() = gn()1+||2n for

some gn L2 for all n. Therefore for n chosen so that 2n > m + d, we haveZRd

||m|f()|d kgnkL2

||m

1 + ||2n

2

<

which shows ||m|f()| L1 for all m = 0, 1, 2, . . . We may now differentiate the

inversion formula, f(x) =R

f()eixd to find

Df(x) =

Z(i)f()eixd for any

and thus conclude f C.

Exercise 11.1. Some Exercises: Section 2.5 4, 5, 6, 8, 9, 11, 12, 17.


7/12


11.1. Du Hammels principle again.

Lemma 11.15. Suppose A is an operator on H such that A is densely defined

then A is closed.

Proof. If fn D(A) f H and Afn g then for all h D(A)

(g, h) = limn

(Afn, h)

while

limn

(Afn, h) = limn

(fn, Ah) = (f,Ah),

i.e. (Ah,f) = (h, g) for all h D(A). Thus f D(A) and Af = g.

Corollary 11.16. If A = A then A is closed.

Corollary 11.17. Suppose A is closed and u(t) D(A) is a path such that u(t)

and Au(t) are continuous in t. Then

A

ZT0

u()d =

ZT0

Au()d.

Proof. Let n be a sequence of partitions of [0, T] such that mesh(n) 0 asn and set

fn =Xn

u(i)(i+1 i) D(A).

Then fn RT0

u()d and

A fn =

XnAu(i)(i+1 i)

ZT0

Au()d.

ThereforeRT0

u()d D(A) and ART0

u()d =RT0

Au()d.

Lemma 11.18. Suppose A = A, A 0, and h : [0, ] H is continuous. Then

(s, t) [0, ) [0, ) esAh(t)

(s, t) (0, ) [0, ) AkesAh(t)

are continuous maps into H.

Proof. Let k 0, then if s ,

Ak

esAh(t) eAh()

=

AkeA

e(s)Ah(t) h()

AkeAe(s)A [h(t) h()] + e(s)Ah() h()

k

kek

hkh(t) h()k +

e(s)Ah() h()i .So

lims an d t

Ak esAh(t) eAh() = 0


8/12

164 BRUCE K. DRIVER

and we may take = 0 if k = 0. Similarly, if s ,

Ak

esAh(t) eAh() = A

kesA h(t) e(s)Ah()

AkesA hkh(t) h()k + h() e(s)Ah()i

k

s

kek

hkh(t) h()k +

h() e(s)Ah()iand the latter expression tends to zero as s and t .

Lemma 11.19. Let h C([0, ), H) , D :=

(s, t) R2 : s > t 0

and

F(s, t) :=Rt0 e

(s)Ah()d for (s, t) D. Then

(1) F C1(D, H) (in fact F C(D, H)),

(11.11)

tF(s, t) = e(st)Ah(t)

and

(11.12)F(s, t)

s=

Zt0

Ae(s)Ah()d.

(2) Given > 0 let

u(t) := F(t + , t) =

Zt0

e(t+)Ah()d.

Then u C1 ((, ), H) , u(t) D(A) for all t > and

(11.13) u(t) = eAh(t) + Au(t).

Proof. We claim the function

(s, t) D F(s, t) := Zt

0

e(s)Ah()d

is continuous. Indeed if (s0, t0) D and (s, t) D is sufficiently close to (s0, t0) sothat s > t0, we have

F(s, t) F(s0, t0) =

Zt0

e(s)Ah()d

Zt00

e(s0)Ah()d

=

Zt0

e(s)Ah()d

Zt00

e(s)Ah()d

+

Zt00

he(s)A e(s

0)Ai

h()d

so that

kF(s, t) F(s0, t0)k Ztt0e(s)Ah() d + Zt

0

0he(s)A e(s0)Aih() d

Ztt0kh()k d +

Zt00

he(s)A e(s0)Aih() d.(11.14)By the dominated convergence theorem,

lim(s,t)(s0,t0)

Ztt0kh()k d = 0


9/12


and

lim

(s,t)(s0

,t0

)Zt0

0 he(s)A e(s

0)A

ih() d = 0which along with Eq. (11.14) shows F is continuous.

By the fundamental theorem of calculus,

tF(s, t) = e(st)Ah(t)

and as we have seen this expression is continuous on D. Moreover, since

se(s)Ah() = Ae(s)Ah()

is continuous and bounded for on s > t > , we may differentiate under the integralto find

F(s, t)

s=

Zt0

Ae(s)Ah()d for s > t.

A similar argument (making use of Eq. (11.10) with k = 1) shows F(s,t)s

is con-tinuous for (s, t) D.

By the chain rule, u(t) := F(t + , t) is C1 for t > and

u(t) =F(t + , t)

s+

F(t + , t)

t

= eAh(t) +

Zt0

Ae(s+)Ah()d = eAh(t) + u(t).

Theorem 11.20. Suppose A = A, A 0, u0 H and h : [0, ) H iscontinuous. Assume further that h(t) D(A) for all t [0, ) and t Ah(t) is

continuous, then

(11.15) u(t) := etAu0 +

Zt0

e(t)Ah()d

is the unique function u C1((0, ), H) C([0, ), H) such that u(t) D(A) forall t > 0 satisfying the differential equation

u(t) = Au(t) + h(t) for t > 0 and u(0+) = u0.

Proof. Uniqueness: If v(t) is another such solution then w(t) := u(t) v(t)satisfies,

w(t) = Aw(t) with w(0+) = 0

which we have already seen implies w = 0.Existence: By linearity and Theorem 11.13 we may assume with out loss of

generality that u0 = 0 in which case

u(t) =

Zt0

e(t)Ah()d.

By Lemma 11.18, we know [0, t] e(t)Ah() H is continuous, so theintegral in Eq. (11.15) is well defined. Similarly by Lemma 11.18,

[0, t] e(t)AAh() = Ae(t)Ah() H


10/12

166 BRUCE K. DRIVER

and so by Corollary 11.17, u(t) D(A) for all t 0 and

Au(t) = Zt

0

Ae(t)Ah()d = Zt

0

e(t)AAh()d.

Let

u(t) =

Zt0

e(t+)Ah()d

be defined as in Lemma 11.19. Then using the dominated convergence theorem,

suptT

ku(t) u(t)k suptT

Zt0

e(t+)A e(t)Ah() d

ZT0

eA Ih() d 0 as 0,suptT

kAu(t) Au(t)k

ZT0

eA I

Ah()

d 0 as 0

and Zt0

eAh()d

Zt0

h()d

Zt0

eA Ih() d 0 as 0.Integrating Eq. (11.13) shows

(11.16) u(t) =

Zt0

eAh()d +

Zt0

Au()d

and then passing to the limit as 0 in this equations shows

u(t) =

Zt0

h()d +

Zt0

Au()d.

This shows u is differentiable and u(t) = h(t) + Au(t) for all t > 0.

Theorem 11.21. Let > 0, h : [0, ) H be a locally Holder continuousfunction, A = A, A 0 and u0 H. The function

u(t) := etAu0 +

Zt0

e(t)Ah()d

is the unique function u C1((0, ), H) C([0, ), H) such that u(t) D(A) forall t > 0 satisfying the differential equation

u(t) = Au(t) + h(t) for t > 0 and u(0+) = u0.

(For more details see Pazy [2, 5.7].)

Proof. The proof of uniqueness is the same as in Theorem 11.20 and for existencewe may assume u0 = 0.

With out loss of generality we may assume u0 = 0 so that

u(t) =

Zt0

e(t)Ah()d.

By Lemma 11.18, we know [0, t] e(t)Ah() H is continuous, so theintegral defining u is well defined. For > 0, let

u(t) :=

Zt0

e(t+)Ah()d =

Zt0

e(t)AeAh()d.


11/12


Notice that v() := eAh() C(A) for all and moreover since AeA is abounded operator, it follows that Av() is continuous. So by Lemma 11.18, itfollows that [0, t] Ae(t)Av() H is continuous as well. Hence we knowu(t) D(A) and

Au(t) =

Zt0

Ae(t)AeAh()d.

Now

Au(t) =

Zt0

Ae(t+)Ah(t)d +

Zt0

Ae(t+)A [h() h(t)] d,Zt0

Ae(t+)Ah(t)d = e(t+)Ah(t)|=t=0 = e(t+)Ah(t) eAh(t)

and

Ae(t+)A [h() h(t)]

e11

(t + )kh() h(t)k

Ce1 1(t + ) |t | Ce1 |t |1 .

These results along with the dominated convergence theorem shows lim0 Au(t)exists and is given by

lim0

Au(t) = lim0

he(t+)Ah(t) eAh(t)

i+ lim

0

Zt0

Ae(t+)A [h() h(t)] d

= etAh(t) h(t) +

Zt0

Ae(t)A [h() h(t)] d.

Because A is a closed operator, it follows that u(t) D(A) and

Au(t) = etAh(t) h(t) +

Zt0

Ae(t)A [h() h(t)] d.

Claim: t Au(t) is continuous. To prove this it suffices to show

v(t) := A

Zt0

e(t)A(h() h(t))d

is continuous and for this we have

v(t + 4) v(t) =

Zt+40

Ae(t+4)A(h() h(t + 4))d

Zt0

Ae(t)A(h() h(t))d

= I+ II

where

I =

Zt+4t

Ae(t+4)A(h() h(t + 4))d and

II = Zt0hAe(t+4)A(h() h(t + 4)) Ae(t)A(h() h(t))i d

=

Zt0

hAe(t+4)A(h() h(t)) Ae(t)A(h() h(t))

id

+

Zt0

hAe(t+4)A(h(t) h(t + 4))

id

= II1 + II2


12/12

168 BRUCE K. DRIVER

and

II1 = Zt

0

A he(t+4)A e(t)Ai (h() h(t))d and

II2 =h

e(t+)A eAi

(h(t) h(t + 4)).

We estimate I as

kIk

Zt+4t

Ae(t+4)A(h() h(t + 4)) d

C

Zt+4t

1

t + 4 |t + 4 |

d

= CZ||

0

x1dx = C1 ||

0 as 0.

It is easily seen that kII2k 2C||

0 as 0 and

Ah

e(t+4)A e(t)Ai

(h() h(t))

C|t |1

which is integrable, so by the dominated convergence theorem,

kII1k

Zt0

A he(t+4)A e(t)Ai (h() h(t)) d 0 as 0.This completes the proof of the claim.

Moreover,

Au(t) Au(t) = e(t+)Ah(t) etAh(t) + h(t) eAh(t)

+

Zt0

A

e(t+)A e(t)A

[h() h(t)] d

so that

kAu(t) Au(t)k 2

h(t) eAh(t)

+

Zt0

Ae(t)A

eA I

[h() h(t)]

d

2h(t) eAh(t)+ e1 Zt

0

1

|t |

eA I [h() h(t)] dfrom which it follows Au(t) Au(t) boundedly. We may now pass to the limit inEq. (11.16) to find

u(t) = lim0

u(t) = lim0

Zt0

eAh()d +

Zt0

Au()d

=

Zt0

h()d +

Zt0

Au()d

from which it follows that u C1((0, ), H) and u(t) = h(t) + Au(t).

Documents

Spectral Theorem 6