Functional Analysis (2006)

Homework assignment 3

All students should solve the following problems:

1. (4.8: Problem 4.) Show that if the sequence (xn) in a normed spaceX is weakly convergent to x0 X, then lim infn ||xn|| = ||x0||.(Hint: You may find Theorem 4.3-3 useful.)

2. Let T1, T2, T3, be the following bounded linear operators `1 `:

T1((1, 2, 3, )) = (1, 1, 1, 1, 1, );

T2((1, 2, 3, )) = (1, 2, 2, 2, 2, );

T3((1, 2, 3, )) = (1, 2, 3, 3, 3, );

etc.

Prove that the sequence (Tn) is strongly operator convergent. Alsoprove that (Tn) is not uniformly operator convergent.

3. Define T : `1 `1 by

T ((1, 2, 3, ...)) = (2, 3, ...).

Determine the four sets (T ), p(T ), c(T ), r(T ).

4. Let T : H H be a compact and self-adjoint operator on a Hilbertspace H. We say that T is positive if Tx, x = 0 holds for everyx H. Prove that T is positive if and only if each eigenvalue of Tsatisfies = 0.

Students taking Functional Analysis as a 6 point course should alsosolve the following problems:

5. Let S and T be bounded self-adjoint operators on a Hilbert spaceH. Let E = (E) be the spectral family associated with T andlet F = (F) be the spectral family associated with S. Prove thatTS = ST holds if and only if EF = FE for all , R.

1

26. Let T be a (possibly unbounded) symmetric operator on a Hilbertspace H. Prove that the following three statements are equivalent:(a) T is self-adjoint.(b) T is closed and N (T i) = N (T + i) = {0}.(c) R(T i) = R(T + i) = H.

[Hints. (Partial score is given for proof of any fact mentioned inthe following hints.) To prove (b)(c) one may first prove thatN (T i) = {0} implies R(T i) = {0} so that R(T i) isdense in H. To prove R(T i) = H it then suffices to prove thatR(T i) is closed; to do this it may be useful to note ||(T + i)w||2 =||Tw||2 + ||w||2, w D(T ) (proof?), and use the fact ((b)) that Tis closed.To prove (c)(a) one may first prove that R(T i) = H impliesN (T + i) = {0}. One then uses this (plus other observations!) toprove D(T ) D(T ), which implies (a).]

Solutions to problems 1-4 should be handed in by Monday,March 13. Solutions to problems 5-6 should be handed in byTuesday, April 18. (Either give the solutions to me directly or putthem in my mailbox, third floor, House 3, Polacksbacken.)

3Functional Analysis

Solutions to homework assignment 3

1. If x0 = 0 then ||x0|| = 0 and the statement is obviously true. Nowassume x0 6= 0. Then by Theorem 4.3-3 there is some f X suchthat ||f || = 1 and f(x0) = ||x0||. Since (xn) is weakly convergent tox0 we have limn f(xn) = f(x0) = ||x0||. But f(xn) 5 |f(xn)| 5||f || ||xn|| = ||xn||. Hence lim infn ||xn|| = limn f(xn) = ||x0||,Q.E.D.

2. Let T : `1 ` be the bounded linear operator given by

T ((1, 2, 3, )) = (1, 2, 3, ).

(T is obviously linear, and it is bounded with norm ||T || 5 1, for if(j) `1 then

k=1 |k| = ||(j)||`1 < and hence |k| 5 ||(j)||`1for all k; hence ||(j)||` 5 ||(j)||`1.) We claim that (Tn) is stronglyoperator convergent to T .

Let x = (j) be an arbitrary vector in `1. Then

||Tnx Tx||` =(0, 0, , 0, n n+1, n n+2, n n+3, )

`

= supj=1

|n n+j|.

But since x = (j) `1 the sum

k=1 |k| is convergent. In particularthe individual terms tend to 0, i.e. limk |k| = 0. Hence, given any > 0 there is some K Z+ such that |k| 5 for all k = K. Then ifn = K we have for all j = 1: n+ j = K, hence |n| 5 and |n+j| 5 ,and thus

|n n+j| 5 |n|+ |n+j| 5 + = 2.

This is true for all j = 1. Hence

||Tnx Tx||` = supj=1

|n n+j| 5 2.

This is true for all n = K. Also, we have shown above that such a Kcan be found for any given > 0. Hence:

limn

||Tnx Tx||` = 0. (That is, Tnx Tx in `.)

This is true for every vector x `1. Hence (Tn) is strongly operatorconvergent to T .

4It follows from this that if (Tn) would be uniformly operator con-vergent, then the limit must be equal to T ! Hence to prove that (Tn)is not uniformly operator convergent, it suffices to prove that (Tn) isnot uniformly operator convergent to T . Now note that Tn T is thefollowing operator:

(Tn T )((1, 2, 3, )) = (0, 0, , 0, n n+1, n n+2, n n+3, ).

In particular if x = (0, 0, , 0, 1, 0, 0, 0, ) (with the 1 in the nthposition) then

(Tn T )(x) = (0, 0, , 0, 1, 1, 1, ).

Here ||x||`1 = 1 and ||(0, 0, , 0, 1, 1, 1, )||` = 1. Hence||Tn T || = 1. This is true for all n, and hence we do not havelimn ||Tn T || = 0. Hence (Tn) is not uniformly operator conver-gent to T , and hence, by our remarks above, (Tn) is not uniformlyoperator convergent.

3. Solution: Note that ||T || = 1, and hence by Theorem 7.3-4 thespectrum (T ) is contained in the disk given by || 5 1. In otherwords we now know that every C with || > 1 belongs to theresolvent set, (T ). Hence it only remains to analyze arbitary C with || 5 1.

It is easy to determine the point spectrum p(T ): Suppose that Cis an eigenvalue of T ; then Tx = x for some vector x = (j) `1, thus

(2, 3, 4, ) = (1, 2, 3, ).

This implies 2 = 1, 3 = 2 = 21, etc, thus n =

n11 for alln = 2, 3, , i.e.

x = 1(1, , 2, 3, ).

But if || = 1 then (1, , 2, 3, ) / `1, hence the above equa-tion forces 1 = 0 and x = 0. Hence no with || = 1 belongs top(T ). On the other hand, if || < 1 then (1, , 2, 3, ) / `1, since

j=1 |j| < , and this vector (1, , 2, 3, ) is an eigenvector of T

with eigenvalue . Hence

p(T ) = { C : || < 1}.

It now remains to analyze the case || = 1. Let us fix an arbitrary C with || = 1. We then know that T1 exists, since / p(T ).Note that

T((n)) = (2 1, 3 2, 4 3, ...).

5Hence if (n) = T((n)) for some (n) `1, (n) `1 then

()

2 = 1 + 1

3 = 2 + 1 + 21

4 = 3 + 2 + 21 +

31

...

n =n1

j=1 n1jj +

n11

...

Here it follows from (n) `1 that limn n = 0, hence also limn 1nn =0 (since || = 1). Using the above equations this gives:

limn

(n1j=1

jj + 1

)= 0,

i.e.

1 = limn

n1j=1

jj =

j=1

jj.

(Note that this sum is absolutely convergent, since

j=1 |jj| =

j=1 |j| < .) Using this together with (*) we now see, for eachn = 2:

n =

n1j=1

n1jj + n1

(

j=1

jj

)=

n1j=1

n1jj

j=1

n1jj

=

j=n

n1jj.

(Clearly this is also true for n = 1.) Conversely, let us note that if(n) `

1, (n) `1 and n =

j=n n1jj holds for all n = 1, then

(n) = T((n)). [Proof: Under the stated assumptions, the nth entryin T((n)) is: n+1 n =

j=n+1 njj +

j=n n1jj =

j=n+1 njj +

j=n njj = n.]

Let M be the set of those (n) `1 which have only finitely manynonzero entries. We then have M D(T1 ). Proof: If (n) M thenthere is some N Z+ such that n = 0 for all n = N , and then if wedefine n by n =

j=n n1jj, we get n = 0 for all n = N . Hence

(n) `1, and by assumption (n) M `1; hence T((n)) = (n)and (n) D(T

1 ). This is true for all (n) M , hence we have proved

the claim, M D(T1 ).

6But M is dense in `1! (Proof: Given x = (j) `1 we may define thesequence v1, v2, v3, M by vn = (1, 2, , n, 0, 0, 0 ). Then||vn x|| =

j=n+1 |j| 0 as n . Hence x is a limit point of M .

This is true for all x `1. Hence M = `1, as claimed.) From the factsM D(T1 ) and M dense in `

1 it follows that D(T1 ) is dense in `1.

We can now complete the solution using only general principles: Wehave seen above that the spectrum (T ) contains the set p(T ) = { C : || < 1}, i.e. the open unit disk. But we know from Theorem 7.3-4that (T ) is compact ; hence (T ) must also contain every boundarypoint of the unit disk, i.e. every with || = 1 belongs to (T ). Wehave also seen that for these we have / p(T ) and / r(T )(since T1 exists and D(T

1 ) is dense in `

1). Hence the only remainingpossibility is c(T ).

Answer: (T ) = { C : || > 1}, p(T ) = { C : || < 1},c(T ) = { C : || = 1}, r(T ) = .

Alternative proof that every with || = 1 belongs to c(T ).Fix an arbitrary C with || = 1. We have proved that T1exists and that D(T1 ) is dense in `

1. Hence it remains to provethat T1 is unbounded. To do this we let vn = (n

0, (n 1)1, (n 2)2, , 1 n1, 0, 0, 0, ) `1, for each n = 1. We then computewn = T(vn) = (1,2,3, ,n, 0, 0, 0, ). Here ||vn|| =n

j=1 |n + 1 j||j1| =

nj=1(n + 1 j) =

n1k=0 k =

n(n1)2

, and

||wn|| =n

j=1 | j| =

nj=1 1 = n. But vn = T

1 (wn); hence if

T1 were bounded then we would have ||vn|| 5 ||T1 || ||wn||, i.e.

n(n1)2

5 ||T1 || n, i.e. ||T1 || =

n12

. This would be true for everyn Z+. This is a contradiction, since no real number can be largerthan n1

2for all n Z+. Hence T1 is unbounded, Q.E.D.

4. Since T is self-adjoint we know that each eigenvalue is real. Sup-pose there is some negative eigenvalue; Tv = v with < 0, v 6= 0.Then Tv, v = v, v = ||v||2 < 0, i.e. T is not positive.

Conversely, assume that all eigenvalues of T are = 0. Let {n}Nn=1

be the nonzero eigenvalues of T , and {en}Nn=1 the corresponding or-thonormal sequence of eigenvectors, as in Theorem 1 in the text aboutcompact self-adjoint operators. (Here N Z+ or N = .) Thenour assumption says that n > 0 for all n. We will prove that Tis positive. Let x H be an arbitrary vector. We may then write

x =(N

n=1 nen

)+ z for some n C with

Nn=1 |n|

2 < and

7z N (T ). Now

Tx, x = N

n=1

nnen,

Nn=1

nen + z =N

n=1

n|n|2 = 0,

since all n = 0. Hence we have proved Tx, x = 0 for all x H.Hence T is positive.

Alternative proof using Theorems from 9.2 in the book.(Mainly of interest for students taking the 6p course.)

We will use the following fact, which we prove below: If T is anarbitrary compact self-adjoint operator T with spectral decompositionas in Theorem 1 in the text about compact self-adjoint operators, thenwe have

() (T ) = E where E =

{{n}Nn=1 if N (T ) = {0}

{0} {n}Nn=1 if N (T ) 6= {0}.

(The splitting in two cases is rather natural, for note that N (T ) 6= {0}holds if and only if 0 is an eigenvalue of T .)

Using (*), we may now solve the given problem using Theorem 9.2-1and Theorem 9.2-3 in the book. First, if T is positive then m = 0 inTheorem 9.2-1, and hence that theorem implies that (T ) [0,),and in particular p(T ) [0,), i.e. each eigenvalue of T is = 0.Conversely, if T is not positive then m < 0 in Theorem 9.2-1, andTheorem 9.2-3 says that m (T ). Then by (*) there exists some nwhich is < 0, i.e. T has a negative eigenvalue.

Finally, we give a proof of (*) (see Theorem 8.4-4 in the book for analternative proof in a more general case): Note that each E is aneigenvalue of T , i.e. E p(T ). Hence E (T ), and since (T ) is aclosed set (Theorem 7.3-4) it follows that E (T ).

Conversely, let be any complex number which does not belong toE. Then there is some r > 0 such that | | > r for all E.Recall from the text about compact self-adjoint operators that every

vector v H can be uniquely written v =(N

n=1 nen

)+ z, where

n C,N

n=1 |n|2 < and z N (T ). We now define a linear

operator A : H H by defining [we here assume N (T ) 6= {0}, andthus in particular || > r]:

A(v) = A

((N

n=1

nen

)+ z

):=

(N

n=1

(n )1nen

) 1z.

8This is in fact a bounded linear operator on H, since

||A(v)||2 =

(N

n=1

|n |2|n|

2

)+ ||2||z||2

5 r2

((N

n=1

|n|2

)+ ||z||2

)= r2||v||2.

(Thus ||A|| 5 r2.) We compute that for all v H represented asabove we have

AT(v) = A

((N

n=1

(n )nen

) z

)=

(N

n=1

nen

)+ z = v

and

TA(v) = T

((N

n=1

(n )1nen

) 1z

)=

(N

n=1

nen

)+ z = v.

Hence AT = TA = I, the identity operator, and thus A = T1 .

Hence T1 exists and is defined on all H, and is bounded. This proves (T ), i.e. / (T ). [In the case N (T ) = {0} the proof is simplymodified by removing the z-term in all sums above.] This is true forall / E. Hence (T ) E, and in total we have proved (T ) = E,i.e. (*)!

5. First assume TS = ST . Then by Lemma 9.8-2 (carrying overLemma 9.8-1(b)) we have ES = SE for all R. Fixing any R and again applying Lemma 9.8-2 (carrying over Lemma 9.8-1(b)), but this time applied to the operator S in place of T , we obtainEF = FE for all R.

Conversely, assume that EF = FE holds for all , R. Weknow that T =

dE and S =

dF, and in fact there is

a number A > 0 such that T = AA

dE and S = AA

dF (forexample we may take A = 1 + max(||T ||, ||S||)). This means that Tis obtained as the uniform limit of a sequence of operator sums of theform (for a given partition A = t0 < t1 < < tn = A)

T =n

j=1

tj(Etj Etj1),

9and similarly for S. Hence, given > 0 there exists a partition A =t0 < t1 < < tn = A such that if

T =n

j=1

tj(Etj Etj1)

and S =

nj=1

tj(Ftj Ftj1),

then

||T T || < and ||S S|| < .

The point now is that it follows from our assumption EF = FE,, R, that T S = S T ! Proof:

T S =

(n

j=1

tj(Etj Etj1)

)(n

k=1

tk(Ftk Ftk1)

)

=n

j=1

nk=1

tj(Etj Etj1)tk(Ftk Ftk1)

=

nj=1

nk=1

tk(Ftk Ftk1)tj(Etj Etj1)

=

(n

k=1

tk(Ftk Ftk1)

)(n

j=1

tj(Etj Etj1)

)= S T .

We also have

||TS T S || 5 ||T (S S )||+ ||(T T )S ||

5 ||T || ||S S ||+ ||T T || ||S ||

< ||T || + ||S || 5 ||T || + (||S||+ ||S S||)

< ||T || + ||S|| + 2,

and similarly

||S T ST || < ||T || + ||S|| + 2.

Hence

||TS ST || = ||TS T S + S T ST ||

5 ||TS T S ||+ ||S T ST ||

< 2(||T || + ||S|| + 2

).

10

But here > 0 is arbitrary, and by taking small we can make theright hand side arbitrarily small. Hence ||TSST || = 0, i.e. TS = ST ,Q.E.D.

6.(a) = (b). Assume (a), i.e. that T is self-adjoint. Then T = Tand T is closed (Theorem 10.3-3). Assume v N (T i). Thenv D(T ) = D(T ) and (T i)v = 0, hence (T i)v = 0, henceTv = iv, and this implies:

iv, v = iv, v = Tv, v = v, T v = v, Tv = v, iv = iv, v.

Hence v, v = 0, i.e. v = 0. This prove that N (T i) = {0}.Similarly one proves N (T + i) = {0}. Hence (b) holds.

(b) = (c). Assume (b), i.e. that T is closed and N (T i) =N (T + i) = {0}. We first prove R(T + i) = {0}: Let v be anarbitrary vector in R(T +i). Then w, v = 0 for all w R(T +i), i.e.(T + i)x, v = 0 for all x D(T ). Hence Tx, v = ix, v = x, ivfor all x D(T ). This implies that v D(T ) and T v = iv, i.e.(T i)v = 0, i.e. v N (T i). By our assumption (b), this impliesv = 0. Hence we have proved R(T + i) = {0}.

We next prove that R(T + i) is closed in H: Let v1, v2, v3, be anarbitrary sequence of points in R(T +i) such that v = limn vn existsin H. We have to prove v R(T + i). By definition there are vectorsw1, w2, w3, D(T ) such that vn = (T + i)wn. Since T is symmetricwe have for every w D(T ):

||(T + i)w||2 = (T + i)w, (T + i)w = ||Tw||2 + iw, Tw iTw, w+ ||iw||2

= ||Tw||2 + i(T w, w Tw, w

)+ ||w||2 = ||Tw||2 + ||w||2.

In particular

||vn vm||2 = ||(T + i)(wn wm)||

2 = ||T (wn wm)||2 + ||wn wm||

2.

Now (vn) is a Cauchy sequence, i.e. ||vn vm|| 0 as n, m ,and the above equality shows ||wn wm||

2 5 ||vn vm||2; hence also

||wnwm|| 0 as n, m , i.e. (wn) is a Cauchy sequence. Since His complete it follows that w = limnwn H exists. Similarly, theabove equality also shows that u = limn Twn H exists. Since Tis closed (by assumption (b)) it follows that w D(T ) and Tw = u,

11

and thus

(T + i)w = u + iw =(

limn

Twn

)+(

limn

iwn

)= lim

n(Twn + iwn)

= limn

(T + i)wn = limn

vn = v

Hence v R(T + i). This proves that R(T + i) is closed in H.Since R(T+i) is both dense and closed in H it follows thatR(T+i) =

H. Similarly one proves R(T i) = H. Hence (c) holds.(c) = (a). Assume (c), i.e. that R(T i) = R(T + i) = H. Let

us first prove N (T + i) = {0}: Assume v N (T + i), i.e. v D(T )and (T + i)v = 0. Then for all w D(T ) we have

0 = (T + i)v, w = T v, w+ iv, w = v, Tw v, iw = v, (T i)w.

But every vector in H can be expressed as (Ti)w, since R(Ti) = H(assumption (c)). Hence v is orthogonal to all H, and hence v = 0.This completes the proof that N (T + i) = {0}.

Now let v be an arbitrary vector in D(T ). Since R(T + i) = Hthere exists a vector w D(T ) such that (T + i)w = (T + i)v. Nowsince T is symmetric we have w D(T ) and T w = Tw. Hence alsov w D(T ), and

(T + i)(v w) = (T + i)v (T + i)w = 0.

Hence vw N (T +i). But we have seen above thatN (T +i) = {0}.Hence v w = 0, i.e. v = w D(T ). But v was an arbitrary vector inD(T ); hence we have proved D(T ) D(T ). On the other hand wehave T T since T is symmetric. Hence we actually have T = T ,i.e. T is self-adjoint. Hence (a) holds.