# 2. Let H 1 and H 2 be Hilbert spaces and let T : H 1 H 2 be a bounded linear operator. Prove that [T (H 1 )] = N (T ). (6p)

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1 Uppsala Universitet Matematiska Institutionen Andreas Strömbergsson Prov i matematik Funktionalanalys Kurs: F3B, F4Sy, NVP Skrivtid: 9 14 Tillåtna hjälpmedel: Manuella skrivdon, Kreyszigs bok Introductory Functional Analysis with Applications och Strömbergssons häfte Spectral theorem for compact, self-adjoint operators. 1. Let X and Y be normed spaces and fix some elements f 1, f 2 X and y 1, y 2 Y. For each x X we define T (x) = f 1 (x) y 1 + f 2 (x) y 2. Prove that T is a bounded linear operator from X to Y. 2. Let H 1 and H 2 be Hilbert spaces and let T : H 1 H 2 be a bounded linear operator. Prove that [T (H 1 )] = N (T ). 3. (a). Prove that the set M = {y 1, y 2, y 3,...} is not total in l 2 if y 1 = (1, 1, 0, 0, 0,...) y 2 = (1, 1, 1, 0, 0, 0,...) y 3 = (1, 1, 1, 1, 0, 0, 0,...) y 4 = (1, 1, 1, 1, 1, 0, 0, 0,...) (b). Prove that the set M = {x 1, x 2, x 3,...} is total in l 2 if x 1 = (1, 1, 0, 0, 0,...) x 2 = (1, 1, 1, 0, 0, 0,...) x 3 = (1, 1, 1, 1, 0, 0, 0,...) x 4 = (1, 1, 1, 1, 1, 0, 0, 0,...) (Hint: One may eg. use Theorem ) (4p)

2 2 4. Define T : l 1 l by T ( (ξ 1, ξ 2, ξ 3,...) ) ( = ξ j, j=2 ξ j, j=3 ξ j, ) j=4 ξ j,... Prove that T is a bounded linear operator T : l 1 l and compute the norm T. 5. Let α n,m be complex numbers with α n,m 1 for all n, m 1, and assume that the limit α m = lim n α n,m exists for all m 1. For each n 1 we let T n : l 1 l 1 be the bounded linear operator given by T n ((ξ 1, ξ 2, ξ 3,...)) = (α n,1 ξ 1, α n,2 ξ 2, α n,3 ξ 3,...). Prove that the sequence (T n ) is strongly operator convergent. Also give an example to show that (T n ) is not necessarily uniformly operator convergent. 6. Let X be the normed space given by X = {(ξ n ) ξ n C, and N Z + : n N : ξ n = 0}, (ξ n ) := ξ n 2. n=1 Prove that X is meager in itself. 7. Define T : l l by T ((ξ 1, ξ 2, ξ 3,...)) = (0, ξ 1, ξ 2, ξ 3,...). Prove that 1 σ 2 r(t ), i.e. prove that = 1 belongs to the residual 2 spectrum of T. GOOD LUCK!

3 3 Solutions 1. T is linear since for all x 1, x 2 X and all α 1, α 2 K we have T (α 1 x 1 + α 2 x 2 ) = f 1 (α 1 x 1 + α 2 x 2 ) y 1 + f 2 (α 1 x 1 + α 2 x 2 ) y 2 = (α 1 f 1 (x 1 ) + α 2 f 1 (x 2 )) y 1 + (α 1 f 2 (x 1 ) + α 2 f 2 (x 2 )) y 2 = α 1 (f 1 (x 1 ) y 1 + f 2 (x 1 ) y 2 ) + α 2 (f 1 (x 2 ) y 1 + f 2 (x 2 ) y 2 ) = α 1 T (x 1 ) + α 2 T (x 2 ). T is bounded since for all x X we have T (x) = f 1 (x) y 1 + f 2 (x) y 2 f 1 (x) y 1 + f 2 (x) y 2 = f 1 (x) y 1 + f 2 (x) y 2 f 1 x y 1 + f 2 x y 2 = ( f 1 y 1 + f 2 y 2 ) x. 2. [T (H 1 )] = 1 {y H 2 z T (H 1 ) : y, z = 0} = 2 {y H 2 x H 1 : y, T x = 0} = 3 {y H 2 x H 1 : T y, x = 0} = 4 {y H 2 T y = 0} = 5 N (T ). 1. By definition of orthogonal complement. 2. By definition of T (H 1). 3. By definition of T. 4. By Lemma and the trivial fact that 0, x = 0 for all x H By definition of N (T ) 3. Note that y n (1, 1, 0, 0, 0,...) for all n 1. Hence by Theorem 3.6-2(a), M = {y 1, y 2,...} is not total in l 2.

4 4 (b). Let x = (ξ n ) l 2 be an arbitrary vector which is orthogonal to M = {x 1, x 2,...}. Then etc. It follows that (ξ n ) x 1 = ξ 1 ξ 2 = 0; (ξ n ) x 2 = ξ 1 + ξ 2 ξ 3 = 0; (ξ n ) x 3 = ξ 1 + ξ 2 + ξ 3 ξ 4 = 0, ( j ) (ξ n ) x j = ξ n ξ j+1 = 0, ξ 2 = ξ 1 ; n=1 ξ 3 = ξ 1 + ξ 2 = 2ξ 1 ; ξ 4 = ξ 1 + ξ 2 + ξ 3 = 4ξ 1 ξ 5 = ξ 1 + ξ 2 + ξ 3 + ξ 4 = 8ξ 1 We get by induction: ξ n = 2 n 2 ξ 1 for n 2. Hence if ξ 1 0 then ξ n 2 = ξ 1 ( (n 2) ) =. n=1 This is impossible since (ξ n ) l 2. Hence ξ 1 = 0, and thus ξ n = 2 n 2 ξ 1 = 0 for all n 2. Hence there does not exist any nonzero vector x l 2 which is orthogonal to every element in M. By Theorem 3.6-2(b), this proves that M is total in l Note that if (ξ j ) l 1 then ξ j is absolutely convergent, and hence all series ξ j (n = 1, 2, 3,...) are also absolutely convergent. Hence also ξ j ξ j ξ j = (ξ j ), so that T ((ξ j )) is indeed a well-defined element in l. T is linear, for if (ξ n ), (η n ) l 1 and α, β K then where and thus ν n = n=2 T (α(ξ n ) + β(η n )) = (ν n ), (αξ j + βη j ) = α ξ j + β η j, T (α(ξ n ) + β(η n )) = (ν n ) = αt ((ξ n )) + βt ((η n )).

5 (The manipulations are permitted since all sums involved are absolutely convergent.) T is bounded since T ((ξ n )) = sup n 1 ξ j sup n 1 ξ j = ξ j = (ξ n ) for all (ξ n ) l 1. This also proves T 1. Let e 1 = (1, 0, 0, 0,...) (vector in l 1 or in l ). Then T (e 1 ) = e 1, and e 1 = 1 both in l 1 and in l. Hence T = Let T : l 1 l 1 be the bounded linear operator given by T ((ξ 1, ξ 2, ξ 3,...)) = (α 1 ξ 1, α 2 ξ 2, α 3 ξ 3,...), We claim that (T n ) is strongly operator convergent to T. Let x = (ξ m ) be an arbitrary vector in l 1. Then T n x T x = ( (α n,1 α 1 )ξ 1, (α n,2 α 2 )ξ 2, (α n,3 α 3 )ξ 3,... ) = α n,m α m ξ m. m=1 Let ε > 0. Then since (ξ m ) l 1 there is some M such that m=m+1 ξ m < ε. Furthermore, for each m there is some N 10 m 1 such that α n,m ε α m < for all n N 10M(1+ ξ m ) m, since lim n α n,m = α m. Hence, for all n max(n 1, N 2,..., N M ) we have: M T n x T x = α n,m α m ξ m + α n,m α m ξ m m=1 M m=1 M m=1 ε 10M(1 + ξ m ) ξ m + ε 10M + 2 ε 10 < ε. m=m+1 m=m+1 2 ξ m This proves that (T n ) is strongly operator convergent to T. We now give an example to show that (T n ) is not necessarily uniformly operator convergent to T. Let { 1 if n m α n,m = 0 if n > m. Then α m = lim n α n,m = 0 for all m 1. Hence T = 0 above. Hence if (T n ) is uniformly operator convergent then the limit must be T = 0, 5

6 6 since (T n ) is strongly operator convergent with limit T = 0. We would then have T n 0 0. However, T n ((ξ m )) = (0, 0,..., 0, ξ n, ξ n+1, ξ n+2,...), and in particular, for e n = (0,..., 0, 1, 0,...) (the 1 in the nth position), T n (e n ) = e n, and thus T n T n(e n ) e n = 1. This shows that T n 0 does not hold! Hence (T n ) is not uniformly operator convergent in this case. 6. Let U N = {(ξ n ) X n N : ξ n = 0}. Then by definition, X = N=1 U N. Hence it suffices to prove that each U N is rare in X. But U N is finite dimensional, hence U N is closed in X (Theorem 2.4-3). Hence it only remains to prove that U N has no interior points. Let x = (ξ n ) U N and r > 0. Then the vector v (ξ 1, ξ 2,..., ξ N, r/2, 0, 0, 0,...) X has distance r/2 from (ξ n ) (since ξ n = 0 for all n > N), and thus v B(x, r). We also have v / U N. Hence B(x, r) U N. This is true for every x U N and every r > 0. Hence U N has no interior points. 7. Let = 1. Note that 2 T ((ξ n )) = ( ξ 1, ξ 1 ξ 2, ξ 2 ξ 3,...) = ( 1 2 ξ 1, ξ ξ 2, ξ ξ 3,...). Hence if (η n ) = T ((ξ n )) for (ξ n ) l then ξ 1 = 2η 1 ξ 2 = 2η 2 4η 1 ξ 3 = 2η 3 4η 2 8η 1... ξ n = n 2j η n+1 j... This proves that T is injective, i.e. T 1 exists. It follows from the above computation that } ( ) {(η ) n ) l n (ξn ) l for ξ n = 2 j η n+1 j.

7 (In fact we have ) = {(η n ) l (ξn ) l for ξ n = } n 2 j η n+1 j, for if (η n ) l and ξ n = n 2j η n+1 j then one checks T ((ξ n )) = (η n ), and hence if (ξ n ) l then (η n ) ). However, we only need (*) for our discussion.) We prove that ) is not dense in l by proving e 1 = (1, 0, 0,...) / ). Assume to the contrary that e 1 = (1, 0, 0,...) ). ξ n = n 2j η n+1 j ; then since (η n ) by (*) above. But η 1 1 < 1 and η 10 n < Then there is some (η n ) ) such that (η n) e 1 < 1. Define 10 ) we have (ξ n) l, for all n 1 and hence, for each n 2, n ξ n = 2 j n 1 η n+1 j = 2n η j η n+1 j n 1 2 n η 1 2 j η n+1 j > 2 n (1 1 ) n j 1 10 = 2 n ( ) 1 10 (2n 2) > 2 n (1 1 5 ) > 2n 1. This proves that (ξ n ) / l, a contradiction. Alternative (inspired by the solution of Patrik Thunström): We prove that ) is not dense in l by proving (1, 1, 1,...) / 1 ). Assume to the contrary that (1, 1, 1,...) D(T ). Then ) such that (η n) (1, 1, 1,...) < 1, i.e. η n 1 < 1 for all n. Then Re η n > 0 for all n, and hence defining ξ n = n 2j η n+1 j we have n there is some (η n ) Re ξ n = 2 j Re η n+1 j < 2 n Re η 1. Since Re η 1 > 0 this implies that Re ξ n as n. It follows that (ξ n ) / l, a contradiction.

8 8 Alternative approach, also proving that ) is closed. (Inspired by the solution of Martin Linder.) Note that if (η n ) = T ((ξ n )), then for each k 2: 2 j η j+k = = 2 j (ξ j+k 1 1ξ 2 j+k) 2 j ξ j+k 1 2 (j+1) ξ j+k = ξ k 1. (All sums above are clearly absolutely convergent, since (η n ) l and (ξ n ) l. Hence the above manipulations are permitted.) In particular, using the above for k = 2 it follows that ( ) η 1 = 1 2 ξ 1 = j η j+2 = 2 1 j η j. j=2 Conversely, let (η n ) be any vector in l satisfying (*). Then define (ξ k ) through ξ k = 2 j η j+k+1. Then ξ k 2 j η j+k+1 (η n ) 2 j = 2 (η n ). Hence (ξ k ) l. We now look at T ((ξ k )). The first entry in T ((ξ k )) is 1ξ 2 1 = 1 2 j η 2 j+2 = 2 1 j η j = η 1, j=2 because of (*). The n:th entry in T ((ξ k )) is, for n 2: ξ n ξ n = 2 j η j+n 1 2 = η n + 2 j η j+n+1 2 j η j+n 2 (j+1) η (j+1)+n = η n.

9 Hence T ((ξ k )) = (η n ), and thus (η n ) R(T ). We have proved that (η n ) l belongs to R(T ) if and only if (*) holds, i.e. { } ) = R(T ) = (η n ) l η1 = 2 1 j η j. But note that f((η n )) := n=1 21 n η n is a bounded linear functional f (l ), and by the above formula, ) = {(η n ) l f((ηn )) = 0 j=2 } = N (f). Hence ) is closed in l (cf. Cor ), and ) = N (f) l, since (e.g.) f((1, 0, 0, 0,...)) = 1 0. Hence ) is not dense in l. 9

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