Prove that this gives a bounded linear operator T : X l 1. (6p) Prove that T is a bounded linear operator T : l l and compute (5p)

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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 häften Spectral theorem for compact, self-adjoint operators and Mathematical statements and proofs. 1. Let X be a normed space and let g 1, g 2, g 3, be elements in X with g n n 2 for all n = 1, 2, 3,. We define: ) T x) = g 1 x), g 2 x), g 3 x),. Prove that this gives a bounded linear operator T : X l 1. 6p) 2. Let H 1 and H 2 be Hilbert spaces and let T : H 1 H 2 be a bounded linear operator. Prove that 3. Define T : l l by T ξ 1, ξ 2, ξ 3, )) = ξ 1, 1 2 T H 2 ) N T ). 2 ξ j, ξ j, ξ j, ). 5p) Prove that T is a bounded linear operator T : l l and compute the norm T. 5p) 4. Let Y = ξ j ) l at most finitely many ξ j 0. Show that Y is not complete. Also prove that the closure of Y in l equals Y = ξ j ) l r > 0 : N Z + : j N : ξ j < r. In words: Y consists of all sequences ξ j ) l such that for every r > 0, there are at most finitely many elements ξ j with ξ j r.) 6p)

2 2 5. Let e 1, e 2, e 3, be a total and orthonormal sequence in a Hilbert space H and let α 1, α 2, α 3, be a sequence of real numbers. Define the operator T : H H by the formula T x = α n x, e n e n. n=1 Prove that if T is a compact operator then lim α n = 0. 6p) 6. Let C[0, 1] be the Banach space of all complex valued continuous functions on [0, 1], with the norm x := sup t [0,1] xt). Let T : C[0, 1] C[0, 1] be the bounded linear operator given by [T x]t) = 1 + t 2 )xt). Determine the four sets ρt ), σ p T ), σ c T ), σ r T ). 6p) 7. Let X be a Banach space and let T 1, T 2, T 3, and S 1, S 2, S 3, be bounded linear operators from X to X such that S n T m = T m S n for all n, m 1. Assume that T n ) is strongly operator convergent with limit T and S n ) is strongly operator convergent with limit S. Prove that T S = ST. 6p) GOOD LUCK!

3 3 Solutions 1. We know that n=1 n 2 is a convergent sum; define C = n=1 n 2 in fact we know C = π 2 /6, but we won t need this). For any x X we have T x) l 1 = g n x) n=1 g n x n=1 n 2 x = C x. n=1 This proves that T x) l 1 for all x X. Also for any a, b C and x, y X we have ) T ax + by) = g 1 ax + by), g 2 ax + by), g 3 ax + by), ) = ag 1 x) + bg 1 y), ag 2 x) + bg 2 y), ag 3 x) + bg 3 y), ) ) = a g 1 x), g 2 x), g 3 x), + b g 1 y), g 2 y), g 3 y), = at x) + bt y). Hence T is a linear operator. Finally, our first computation shows that T x) C x for all x X; hence T is bounded. 2. Let v be an arbitrary vector in H 2. Then for every w N T ) we have T v), w = v, T w) = v, 0 = 0. Hence T v) N T ). This is true for every v H 2. Hence T H 2 ) N T ). 3. Take an arbitrary vector ξ j ) l, and let η k ) = T ξ j )). Then, for each k = 1, 2, 3,, η k = 1 k ξ j 1 k k k ξ j 1 k k ξ j ) = ξ j ). Since this is true for all k = 1, 2, 3, we have T ξ j )) = η k ) l. This is true for all ξ j ) l, hence T is a map from l to l. Also

4 4 for any a, b C and ξ j ), ν j ) l we have T aξ j ) + bν j ) ) = T aξ 1 + bν 1, aξ 2 + bν 2, aξ 3 + bν 3, ) = aξ 1 + bν 1, 1 2 = a ξ 1, aξ k + bν k ), ξ k, aξ k + bν k ), ) 3 ξ k, ) + b ν 1, 1 2 = at ξ j ) ) + bt ν j ) ). 2 ν k, ν k, ) Hence T is a linear operator. It follows from our inequality above that, for all ξ j ) l, T ξ j )) = η k ) = sup η k ξ j ). k Since this is true for all ξ j ) l, T is a bounded, and T 1. On the other hand, taking ξ j ) = 1, 1, 1, ) we obtain η k ) = T ξ j )) = 1, 1, 1, ), and T ξ j )) = ξ j ) = 1; hence T 1. Answer: T = Let x n = 2 1, 2 2, 2 3,, 2 n, 0, 0, 0, ). Then x 1, x 2,... Y. We also define x = 2 1, 2 2, 2 3, ) l. We now claim that x n x as n ; this follows from x x n = 0, 0,, 0, 2 n 1, 2 n 2, ) = sup 2 k = 2 n 1 0 as n. k n+1 Note that x / Y, by the definition of Y. Hence by Theorem 1.4-6b), Y is not closed as a subspace of l ). Hence by Theorem 1.4-7, Y is not complete. Next, we prove ) Y = ξ j ) l r > 0 : N Z + : j N : ξ j < r. First take an arbitrary vector x = ξ j ) in the right-hand set. Thus we assume ξ j ) l and that for every r > 0 there is an N Z + such that ξ j < r holds for all j N. Now form the sequence x 1, x 2,... where x n = ξ 1, ξ 2,, ξ n, 0, 0, 0, ).

5 Then x 1, x 2, Y. Note that x x n = sup ξ j ; j n+1 hence for any given r > 0, if we take N Z + as above then x x n r holds for all n N. This can be achieved for every r > 0, and hence x n x in l. It follows that x Y. Hence we have proved: ) Y ξ j ) l r > 0 : N Z + : j N : ξ j < r. Conversely, take an arbitrary vector x = ξ j ) Y. Then there is a sequence x 1, x 2, x 3, of vectors in Y such that x n x in l. Let r > 0 be given. Then there is n Z + such that x n x < r. Let us write x n = η j ). Since x n Y there is some N Z + such that η j = 0 for all j N. Also, x n x < r implies that η j ξ j < r for all j. Hence for all j N we have 0 ξ j < r, i.e. ξ j < r. Hence we have proved that for every r > 0 there is some N Z + such that ξ j < r holds for all j N. But here x was an arbitrary element in Y. Hence: ) Y ξ j ) l r > 0 : N Z + : j N : ξ j < r. Together, **) and ***) imply *), Q.E.D. 5. Assume that T is compact and lim α n 0. We will show that this leads to a contradiction. Since lim α n 0 there is some ε > 0 and a subsequence α nk ) with n 1 < n 2 < n 3 < such that α nk > ε for all k = 1, 2, 3,. Now e n1, e n2, e n3, is a bounded sequence in H, hence by Theorem in Kreyszig s book the sequence T e n1 ), T e n2 ), T e n3 ), must have a convergent subsequence, say T e nkj )) with k 1 < k 2 < k 3 <. Every convergent sequence is a Cauchy sequence, hence ) T e nkj ) T e ) nkj 0 as j, j. On the other hand, we have T e n ) = α n e n for each n, and hence for all j j : T e nkj ) T e ) αnkj nkj = e nkj α e nkj n kj = α nkj 2 + α nkj 2 ) > ε 2 + ε 2 > ε. We used Pythagoras theorem; this is ok since j j implies n kj n kj so that e nkj and e nkj are orthogonal to each other.) But ) and ) cannot both be true. This is a contradiction. 5

6 6 6. First let λ be an arbitrary complex number which does not belong to the real interval [1, 2]. Note that the operator T λ = T λ is given by [T λ x]t) = 1 + t 2 λ)xt). We know that the range of the function 1 + t 2 on [0, 1] is the interval [1, 2]. Hence 1 + t 2 λ 0 for all t [0, 1]; hence since [0, 1] is a compact set and 1+t 2 λ is continuous) there is a constant r > 0 such that 1 + t 2 λ r for all t [0, 1]. It follows that we can define a new bounded linear operator A : C[0, 1] C[0, 1] by [Ax]t) = 1 + t 2 λ) 1 xt). We see that A r 1.) It follows immediately from the definition that T λ A = AT λ = I, the identity operator on C[0, 1]. Hence T 1 λ = A exists and is a bounded operator. Hence λ belongs to the resolvent set ρt ). We have thus proved: C [1, 2]) ρt ). Next let λ be an arbitrary number in the interval [1, 2]. We still have [T λ x]t) = 1 + t 2 λ)xt). Let t 0 [1, 2] be the unique number which gives 1 + t 2 0 λ = 0 thus, explicitly: t 0 = λ 1). Assume that x C[0, 1] is a vector with T λ x = 0. Then 1 + t 2 λ)xt) = 0 for all t [0, 1], and this implies xt) = 0 for all t t 0 in [0, 1]. Hence since xt) is a continuous function we also have xt 0 ) = lim t t0 xt) = lim t t0 0 = 0 if t 0 = 0 then this limit should be interpreted as right limit; if t 0 = 1 it should be interpreted as left limit). Hence xt) = 0 for all t [0, 1], exists Theorem ). Note that if y = T λ x for some x C[0, 1] then yt 0 ) = 1 + t 2 0 λ)xt 0 ) = 0 xt 0 ) = 0; hence every y RT λ ) satisfies yt 0 ) = 0. It follows that RT λ ) is not dense in C[0, 1]. [Proof: Let z C[0, 1] be the constant function zt) = 1. Then for every y RT λ ) we have y z = sup t [0,1] yt) zt) yt 0 ) zt 0 ) = 0 1 = 1; i.e. x = 0 in C[0, 1]. Hence T λ is injective and T 1 λ hence z is not in the closure of RT λ ).] But RT λ ) = DT 1 λ ), hence we have proved that DT 1 λ ) is not dense in C[0, 1]. It follows that λ belongs to the residual spectrum, λ σ r T ). We have thus proved: [1, 2] σ r T ). Since the four sets ρt ), σ p T ), σ c T ), σ r T ) are known to be disjoint it follows that the answer is as follows. Answer: ρt ) = C [1, 2], σ p T ) =, σ c T ) =, σ r T ) = [1, 2].

7 7. One can give a fairly short proof by a computation using limits inside limits, and being very careful about the order of these limits in each step. Note that we make crucial use below of Lemma 4.9.5, and hence of the uniform boundedness theorem! The fact that T n ) is strongly operator convergent with limit T means that ) y X : lim T n y) = T y). The fact that S n ) is strongly operator convergent with limit S means that ) y X : lim S n y) = Sy). Now, for every x X: T Sx) = lim T n Sx)) Use *) with y = Sx). Use **) with y = x. = lim T n lim S mx) ) Each T n is continuous. m = lim lim T ns m x)) ) Use assumption in problem. m = lim lim S mt n x)) ) Use **) with y = T n x). m = lim STn x)) ) = S lim T n x) ) = ST x)). S is continuous by Lemma 4.9.5). Use *) with y = x. Since this holds for every x X, we have proved T S = ST. Alternative solution, with some comments: Note that we never claimed T Sx) = lim n,m T n S m x) in the above solution! We did see that T Sx) = lim limm T n S m x)) ), but this is not the same thing. The statement T Sx) = lim n,m T n S m x) means in precise terms: ε > 0 : N, M Z + : n N, m M : T Sx) T n S m x) < ε. This statement is in fact true for any fixed x X), but it is not obvious. In the solution below we give a proof of that statement, along the way to a complete solution. By Lemma we have T, S BX, X). Furthermore, since the sequence T n x) is bounded for every x X and X is complete), it 7

8 8 follows from the uniform boundedness theorem that there is a constant C > 0 such that T n C for all n this was also seen in the proof of Lemma 4.9-5). Similarly, there is a constant D > 0 such that S n D for all n. Now let x be an arbitrary vector in X. Let ε > 0 be given. Since T n ) is strongly operator convergent with limit T there is N Z + such that both T n x T x < ε and T n Sx T Sx < ε hold for all n N. Similarly, there is M Z + such that both S m x Sx < ε and S m T x ST x < ε hold for all m M. Hence we have for all n N, m M: and 1 T Sx T n S m x = T Sx T n Sx + T n Sx T n S m x T Sx T n Sx + T n Sx S m x) < ε + T n Sx S m x < 1 + C)ε, ST x S m T n x = ST x S m T x + S m T x S m T n x ST x S m T x + S m T x T n x) < ε + S m T x T n x < 1 + D)ε. However, by the assumptions of the problem we have S m T n = T n S m for all n, m; hence by choosing any n N, m M we obtain: T Sx ST x = T Sx T n S m x + S m T n x ST x T Sx T n S m x + ST x S m T n x < 1 + C D)ε. But we are free to choose ε > 0 arbitrarily; hence we can make 1 + C D)ε arbitrarily small. Hence T Sx ST x = 0, i.e. T Sx = ST x. This holds for all x H. Hence T S = ST, Q.E.D. 1 Note that by these two computations, if we note that such M, N can be found for any given ε > 0, we have now proved T Sx) = lim n,m T n S m x) and ST x) = lim n,m S m T n x)! However, we have chosen to complete the solution without using these limits explicitly, instead working with our precise bounds.