Functional Analysis HW #6

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1 Functional Analysis HW #6 Sangchul Lee November 13, 2015 Contents 1 Solutions Solutions Exercise.1. If T is a linear transformation defined on the Hilbert space H, then then T is bounded if and only if sup{ (T f, f ) : f H, f = 1} <. Proof. Let C = sup{ (T f, f ) : f H, f = 1}. (= ) If T is continuous, then by the Cauchy-Schwarz inequality (T f, f ) T f f T f 2. This implies that C T <. ( =) Now assume that C < and define Q( f ) = (T f, f ). Then we have the following polarization identity (T f, g) = 1 ωq( f + ωg). (1.1) ω:ω =1 Indeed, this easily follows from Q( f + ωg) = (T f + ωtg, f + ωg) = (T f, f ) + (Tg, g) + ω(tg, f ) + ω(t f, g). 1

2 Our goal is to prove that sup{ T f : f H, f = 1} is bounded. To this end, we may assume further that T f 0. Then with g = T f / T f, we have T f = (T f, g) = 1 ωq( f + ωg). ω:ω =1 Then using this identity together with the fact f = g = 1, we have T f 1 Q( f + ωg) 1 C f + ωg 2 C. ω:ω =1 ω:ω =1 Therefore T is continuous. Exercise.8. If T is an operator on H, then T is normal if and only if T f = T f for f H. Moreover, a complex number λ is an eigenvalue for a normal operator T if and only if λ is an eigenvalue for T. The latter statement is not valid for general operators on infinite-dimensional Hilbert spaces. Proof. (a) Let T L(H). From the polarization identity (1.1), we find that T f = T f f H (T T f, f ) = T f 2 = T f 2 = (TT f, f ) f H (T T f, g) = (TT f, g) f, g H T T = TT. (b) Let f 0 be such that T f = λ f. Then 0 = (T λ) f 2 = ((T λ)(t λ) f, f ) = ((T T λt λt + λ 2 ) f, f ) = ((TT λt λt + λ 2 ) f, f ) = ((T λ)(t λ) f, f ) = (T λ) f 2 and hence λ is an eigenvalue for T. The converse is straightforward. (c) Let H = l 2 (N 0 ) and define T as the left shift operator T (a 0, a 1, a 2, ) = (a 1, a 2, a 3, ). It is clear that T 1 and hence T L(H). Also for any λ D = {z C : z < 1} we know that 2

3 a = (1, λ, λ 2, ) l 2 (N 0 ) and Ta = λa. Thus any such λ is an eigenvalue for T. On the other hand, T is the right shift operator which is easily verified from the calculation T (a 0, a 1, a 2, ) = (0, a 0, a 1, ), (Ta, b) = a 1 b 0 + a 2 b 1 + = (a, T b) for a, b l 2 (N 0 ). We claim that T has no eigenvalue. To this end, assume otherwise that λ C is an eigenvalue for T and a 0 is a non-zero eigenvector corresponding to λ. From the relation TT = id l 2 (N 0 ), we know that 0 a = λta and hence λ 0. Then 0 = (1 λ 1 T )a = (a 0, a 1 λ 1 a 0, a 2 λ 1 a 1, ). So we have the recurrence relation a n+1 = λ 1 a n with a 0 = 0. This relation boils down to a = 0, which is a contradiction! Therefore T has no eigenvalue. Exercise.9. If S and T are self-adjoint operators on H, then ST is self-adjoint if and only if S and T commute. If P and Q are projections on H, then PQ is a projection if and only if P and Q commute. Determine the range of PQ in this case. Proof. (a) If S and T are self-adjoint, then ST is self-adjoint T S = T S = (ST) = ST. (b) Now assume that P and Q are projections. If PQ is also a projection, then it is self-adjoint and hence P and Q commute. Converse, suppose that P and Q commute. Since P and Q are self-adjoint, the previous part says that PQ is self-adjoint as well. Moreover, (PQ) 2 = PQPQ = PPQQ = PQ and hence PQ is idempotent. Therefore PQ is a projection. (c) In this case, we claim that Indeed, both im(pq) = im P im Q. im(pq) im P and im(pq) = im(qp) im Q imply that im(pq) im P im Q. In order to prove the other direction, let h im P im Q so that 3

4 h = P f = Qg for some f, g H. Then PQh = PQ 2 g = PQg = Ph = P 2 f = P f = h and hence h im(pq). This proves im P im Q im(pq) and hence completes the proof. Exercise.10. If H and K are Hilbert spaces and A is an operator on H K, then there exists unique operators A 11 A 12, A 21 and A 22 in L(H), L(K, H), L(H, K) and L(K), respectively, such that A h, k = A 11 h + A 12 k, A 21 h + A 22 k. In other words A is given by the matrix [ ] A11 A 12. A 21 A 22 Moreover show that such a matrix defines an operator on H K. Proof. (a) Let ι 1 : H H K and ι 2 : H H K be inclusions ι 1 h = h, 0 and ι 2 k = 0, k and ι 1 : H K H and ι 2 : H K K be projections ι 1 h, k = h and ι 2 h, k = k. (Here ι k and ι k are adjoint to each other as the notation suggests, though we do not need this observation.) It is clear that these maps are all continuous and ι 1 ι 1 = [projection onto H {0}], ι 2ι 2 = [projection onto {0} K]. Thus we obtain the following identity ι 1 ι 1 + ι 2ι 2 = id H K Finally, define A i j by A i j = ι i Aι j. Then for any h, k H K, A h, k = (ι 1 ι 1 + ι 2ι 2 )A(ι 1ι 1 + ι 2ι 2 ) h, k = (ι 1 ι 1 + ι 2ι 2 )A(ι 1h + ι 2 k) = ι 1 (A 11 h + A 12 k) + ι 2 (A 21 h + A 22 k) = A 11 h + A 12 k, A 21 h + A 22 k.

5 and hence (A i j ) is the desired decomposition. The uniqueness is straightforward since for any other representation A = à 11 h + à 12 k, à 21 h + à 22 k = ι i à i j ι j, 1 i, j 2 we can extract each entry by à i j = ι i Aι j, which is exactly A i j. (b) Conversely, let A be of the form A h, k = A 11 h + A 12 k, A 21 h + A 22 k for A 11 A 12, A 21 and A 22 in L(H), L(K, H), L(H, K) and L(K), respectively. Then A is a bounded linear operator on H K from the representation A = ι i A i j ι j. 1 i, j 2 Alternatively, the boundedness may also be derived from the crude estimate A h, k 2 = A 11 h + A 12 k 2 + A 21 h + A 22 k 2 ( A 11 h + A 12 k ) 2 + ( A 21 h + A 22 k ) 2 {( A 11 + A 12 ) 2 + ( A 21 + A 22 ) 2 } h, k 2. Exercise.11. If H and K are Hilbert spaces, A is an operator on L(K, H), and J is the operator on H K defined by the matrix [ ] I A, 0 0 then J is an idempotent. Moreover, J is a projection if and only if A = 0. Further, every idempotent on a Hilbert space L can be written in this form for some decomposition L = H K. Proof. (a) We have J 2 h, k = J h + Ak, 0 = h + Ak, 0 = J h, k for h, k H K and hence J is idempotent. Also (J h, k, f, g ) = ( h + Ak, 0, f, g ) = (h + Ak, f ) = (h, f ) + (Ak, f ) and likewise ( h, k, J f, g ) = ( h, k, f + Ag, 0 ) = (h, f + Ag) = (h, f ) + (h, Ag). 5

6 Consequently, J is self-adjoint of if and only if (Ak, f ) = (h, Ag) for any h, f H and k, g K, which is possible exactly when A = 0 (by setting h, g = 0 and f = Ak). (b) Let J be idempotent on L. Let H = im(j) and K = H. Then we claim that L = H K and J admits the desired matrix form. For the first claim, it suffices to show that H is closed. Suppose that (J f n ) H converges to some g L. Then g = lim n J f n = lim n J 2 f n = J( lim n J f n ) = Jg show that g H and hence H is closed. Then by the previous exercise, we know that J can be written as [ ] J11 J J = 12. J 21 J 22 But since im(j) = H, we have J 21 = 0 and J 22 = 0. Moreover, if h H then h = J f for some f L and hence Jh = J 2 f = J f = h. Thus J 11 = id H and hence [ ] I J12 J =. 0 0 This completes the proof. Exercise.13. If (X, S, µ) is a probability space and ϕ is a function in L (µ), then λ is an eigenvalue for M ϕ if and only if the set {x X : ϕ(x) = λ} has positive measure. Proof. If µ(ϕ 1 (λ)) > 0, then 1 ϕ 1 (λ) is a non-zero element of L (µ) such that Therefore λ is an eigenvalue for M ϕ. M ϕ 1 ϕ 1 (λ) = λ1 ϕ 1 (λ). Conversely, let λ be an eigenvalue for M ϕ and f L (µ) be a non-zero eigenvector of M ϕ corresponding to λ. Then (ϕ λ) f = (M ϕ λ) f = 0 = ϕ λ f = 0. Now for each r > 0, let ψ r = min{r, 1/ ϕ λ }. Them ψ r L (µ) and min{r ϕ λ, 1} f = ψ r ϕ λ f = 0. Now assume that µ(ϕ 1 (λ)) = 0. Then min{r ϕ λ, 1} 1 a.s. as r and hence 0 = min{r ϕ λ, 1} f r f a.s. This contradicts the assumption that f 0. Therefore we must have µ(ϕ 1 (λ)) > 0. 6