LECTURE 7. k=1 (, v k)u k. Moreover r

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1 LECTURE 7 Finite rank operators Definition. T is said to be of rank r (r < ) if dim T(H) = r. The class of operators of rank r is denoted by K r and K := r K r. Theorem 1. T K r iff T K r. Proof. Let T K r and let u 1, u 2,..., u r be an orthonormal basis in T(H). Then for any x H we have r r Tx = (Tx, u k )u k = (x, T u k )u k. k=1 Denote v k = T u k, then T = r k=1 (, v k)u k. Moreover r r (Tx, y) = ((x, v k )u k, y) = (x, (y, u k )v k ) = (x, T y). k=1 Therefore T = r k=1 (, u k)v k and thus T K r. Theorem 2. The uniform closure of the class of finite rank operators K coincides with S. Proof. Let T S. Then T maps the set B = {x : x 1} onto a relatively compact set. For any ε > 0 there exists a finite set of elements {y k } r k=1 such that for any y T(B) we have min y y k ε. Let P be the projection on the subspace spanned by y k. Clearly rank P r. Thus for any x s.t. x 1 we obtain k=1 k=1 Tx PTx min k Tx y k ε. Remark 1. Uniform closure cannot be replaced by the strong closure. Theorem 3. The strong closure of K(H) coincides with the class of all bounded operators. 1

2 2 Proof. Let {u k } k=1 be an orthonormal basis in H and let P n be the projectors on the subspaces spanned by {u k } n k=1. Then for any x H, P nx x 0 which means that s-lim P n = I. Thus s-lim P n T = T for any bounded operator T. Integral Operators Theorem 4. Let K : L 2 () L 2 (), R, be an integral operator Kf(x) = K(x, y)f(y) dy, such that Then K is compact. K(x, y) 2 dxdy <. Proof. Let {u j } be an orthonormal basis in L2 (). Then K(x, y) = K j (y)u j (x), where K j (y) = K(x, y)u j (x) dx for almost all y. Due to the Parseval identity we have for almost all y K(x, y) 2 dx = K j (y) 2 and (1) K(x, y) 2 dxdy = We now define the following operator of rank N K N f(x) = K N (x, y)f(y) dy, K j (y) 2 dy. where K N (x, y) = N K j(y)u j (x). By Cauchy-Schwartz inequality we obtain ( ) (K K N )f 2 = K(x, y) K N (x, y) f(y) dy 2 dx K(x, y) K N (x, y) 2 dxdy f 2

3 Thus by using that the right hand side in (1) is absolutely convergent, we find (K K N ) 2 K(x, y) K N (x, y) 2 dxdy = K(x, y) 2 dxdy K(x, y) = N K(x, y) K j (y)u j (x) dxdy N K j (y)u j (x) dxdy + K(x, y) 2 dxdy N N K j (y) 2 dy K j (y) 2 dy 0, as N. 3 Bounded Self-adjoint Operators Definition. A bounded operator T : H H is said to be self-adjoint if x, y H (Tx, y) = (x, Ty), (A = A ). Theorem 5 (Av.Fr ). Let T : be a bounded self-adjoint operator in a Hilbert space H. Then Proof. Clearly if x = 1, then T = sup (Tx, x). x =1 (Tx, x) Tx x = Tx T and therefore sup x =1 (Tx, x) T. In order to proof the inverse inequality we consider z H, z = 1, Tz 0 and u = Tz/λ, where λ = Tz 1/2. If we denote by α := sup x =1 (Tx, x), then ( ) Tz 2 = T(λz), u = 1 [( ) ( )] T(λz + u), λz + u T(λz u), λz u 4 α ] [ λz + u 2 + λz u 2 = α ] [ λz 2 + u α [ ] λ 2 + Tz = α Tz. 2

4 4 This implies that for any z H, z = 1 we have Tz α and hence T α = sup x =1 (Tx, x). Definition. If Tx = λx, x H, x 0, then x is called an eigenvector and λ is called an eigenvalue for the operator T. Theorem 6 (Av.Fr ). Let {λ j } and {x j } be eigenvalues and eigenvectors for a self-adjoint operator T in H. Then All λ j are real. If λ j λ k then the corresponding eigenvectors x j and x k are orthogonal. Proof. Let Tx = λx. Then λ x 2 = (Tx, x) = (x, Tx) = (Tx, x) = λ x 2. Let now Ty = µy, λ µ. Then since T is self-adjoint we have 0 = (Tx, y) (x, Ty) = λ(x, y) µ(x, y) = (λ µ)(x, y). Theorem 7. Let T 1 and T 2 be two bounded self-adjoint operators. The product T := T 1 T 2 is self-adjoint iff the operators T 1 and T 2 commute (T 1 T 2 = T 2 T 1 ). Proof. T = T 2 T 1 = T 2 T 1 = T 1 T 2 = T. Definition. A bounded operator T in H is called positive (T 0) if for any x H, (Tx, x) 0. A bounded operator T in H is called strictly positive (T > 0) if for any x H, (Tx, x) > 0. A bounded operator T in H is called positive definite if there exist a positive constant m > 0 such that (Tx, x) m x 2, x H.

5 If T is positive then we can define an inner product < x, y >= (Tx, y). The the Schwartz inequality gives us (2) (Tx, y) 2 =< x, y > 2 < x, x >< y, y >= (Tx, x)(ty, y). Lemma 1. If T 0 then Tx 2 T (Tx, x). 5 Proof. (Tx, y) 2 (Tx, x)(ty, y) (Tx, x) T y 2. The proof is complete if we substitute into the latter inequality y = Tx. Definition. We say that B A if B A 0. For any bounded operator T there exist two unique self-adjoint operators A and B such that T = A + ib. Then T = A ib and the operators A and B could be found via A = T + T, B = T T. 2 2i It is clearly that T T 0 and TT 0. The operators T and T do not always commute. Definition. If T T = TT then T is called normal. Positive operators & Unbounded operators Theorem 8. Let A n be a bounded monotone sequence of operators. Then {A n } is strongly convergent. Proof. Assume, for example, that A n. Since sup n A n < we obtain that for any x H the sequence (A n x, x) is convergent. Therefore due to polarisation (A n x, y) = 1 [( ) )] A n (x + y), x + y (A n (x y), x y 4 we have that lim n (A n x, y) exists for any x, y H. Such a limit defines a self-adjoint bounded operator A in H. Denote by c = sup n A n A. Then by using Lemma 1, Lecture 12, we find A n x Ax c[(a n x, x) (Ax, x)] 0 x H.

6 6 Definition. A square root of a positive operator A is a self-adjoint operator B such that B 2 = A. Theorem 9 (Av.Fr ). Every positive operator A has a unique positive square root B such that B commutes with any linear operator that commutes with A. Proof. See Av.Fr. page 223. Definition. Let A be a bounded operator. We define A = A A. Remark 2. λa = λ A if λ C. In general AB = A B or A = A is false. In general A + B A + B is not true. Definition. A bounded operator U in H is called unitary if UH = H and U U = I, where I is the identity, Ix = x. Example. Let A be a shift operator in l 2 such that A(x 1, x 2, x 3,... ) = (0, x 1, x 2, x 3,... ). Then A could be defined by A (x 1, x 2, x 3,... ) = (x 2, x 3,... ). Clearly A = A A = I. If the formula A = U A would hold true for some unitary operator U, then U should be equal to A. However, A is not unitary since (1, 0, 0,... ) H. Definition. A bounded operator U is called isometry if Ux = x for all x H. U is called a partial isometry if U is an isometry when restricted to (Ker U). Theorem 10 (Polar decomposition). Let A be a bounded operator in H. There exists a partial isometry U such that A = U A. The operator U is uniquely determined by the condition Ker U = Ker A. Proof. Let U : Ran A Ran A such that U( A x) = Ax. Then A x 2 = (x, A 2 x) = (x, A Ax) = Ax 2. Clearly A x = 0 is equivalent to Ax = 0 and therefore Ker A = Ker A and thus Ker U = Ker A.

7 7 1. UNBOUNDED OPERATORS Let A : H H be an operator in H with domain D(A) and range R(A) := A(H). By N(A) we denote the kernel of the operator A, N(A) = {x D(A) : Ax = 0}. The operator A is invertible if and only if N(A) = 0. In this case D(A 1 ) = R(A) and R(A 1 ) = D(A). The sesqui-linear form (x, y) A = (x, y) + (Ax, Ay), x, y D(A), define pre-hilbert space (a space with a scalar product which is not necessary complete). Theorem 11. The operator A has a bounded inverse if and only if there is a constant c > 0 such that Ax > c x. The biggest possible constant c in the latter inequality is equal to A 1 1. Proof. Let us point out that Ax > c x implies N(A) = 0 and therefore A 1 exists. Let y = Ax. Then y c A 1 y and thus c 1 y A 1 y, or A 1 c 1. If A 1 exists and bounded then A 1 y c 1 Ay, where c = A 1 1. Definition. We say that two unbounded operators A 1 and A 2 are equal A 1 = A 2 if D(A 1 ) = D(A 2 ) and A 1 x = A 2 x for any x D(A 1 ) = D(A 2 ). If D(A 1 ) D(A 2 ) and A 1 x = A 2 x for any x D(A 1 ), then A 2 is called an extension of A 1. Let G A = {(x, Ax) : x D(A)} H H. The scalar product ( ) (x, Ax), (y, Ay) := (x, y) + (Ax, Ay) define a pre-hilbert structure in G A. Theorem 12. An operator A is closed iff G A is closed in H and iff for any sequence x n D(A) lim n x n = x, lim n Ax n = y = x D(A) & y = Ax. Proof. See Th Av.Fr.

8 8 2. ADJOINT OPERATORS Definition. Let A be a linear operator in H and let D(A) = H. An element y H is said to be in D(A ) if there exists h such that (Ax, y) = (x, h), x D(A). In this case we define the adjoint operator A as A y = h and thus (Ax, y) = (xa y). If D(A) H then the element h is not unique and we are not able to define A. Home exercises. { 1. Let l 2 = x = {x j }, x } j C : j x 2 <. Define T : l 2 l 2 such that T(x 1, x 2, x 3,... ) = (0, x 1, x 2,... ). Show that T T TT. 2. Let P L and P M be two projectors on the the subspaces L, M H. Show that P L P M iff L M. 3. Show that an operator U is a partial isometry if and only if the operators P 1 = U U and P 2 = UU are projections. 4. (Av.Fr. &.6.1) Find the positive square root of the operator Ax(t) = a(t)x(t) in L 2 (0, 1) assuming that a(t) Prove that if A B 0, then A B (Heinz s inequality). (Hint: (A + ti) 1 (B + ti) 1 for any t > 0).

Functional Analysis Exercise Class

Functional Analysis Exercise Class Week: January 18 Deadline to hand in the homework: your exercise class on week January 5 9. Exercises with solutions (1) a) Show that for every unitary operators U, V,