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MTL 411: Functional Analysis MTL 411: Functional Analysis Lecture A: Adjoint operators Recall, Riesz representation theorem: Suppose f is a continuous linear functional on a Hilbert space H. Then there exists a unique z in H such that f(x) = hx; zi for all x 2 H: As a consequence, for a given bounded linear operator, we can construct an associated bounded linear operator, which is called Hilbert-adjoint operator or simply adjoint operator. 1 Adjoint operator Let H1 and H2 be (Real/complex) Hilbert spaces and T : H1 ! H2 be a bounded linear ∗ ∗ operator. Then an adjoint operator T of T is an operator T : H2 ! H1 satisfies ∗ hT x; yi = hx; T yi ; 8x 2 H1; 8y 2 H2: Remark. In the above equation, the left hand side inner product is between elements in H2, that is, the inner product is from H2 and in the right hand side the inner product is from H1. ∗ Questions. Given T 2 B(H1;H2); does an adjoint operator T exist? If it exists, will it be unique and bounded operator? The answers are affirmative. ∗ Theorem 1.1. Given T 2 B(H1;H2), there exists a unique operator T 2 B(H2;H1) such that ∗ hT x; yi = hx; T yi ; 8x 2 H1; 8y 2 H2 and jjT jj = jjT ∗jj: Proof. Fix y 2 H2: Define fy(x) = hT x; yi, x 2 H1: Since T is bounded linear operator and using Cauchy-Schwarz inequality (CSI), we can show that fy is a continuous linear functional on H1, that is, fy is linear and jfy(x)j ≤ jjT jj jjyjj jjxjj; 8x 2 H1: Then by Riesz representation theorem, there exists a unique z 2 H1 such that fy(x) = hx; zi. Therefore for each y 2 H2; there exists a unique z 2 H1 such that fy(x) = hx; zi = hT x; yi. ∗ Now, we define T y = z; for each y 2 H2: Thus ∗ hx; T yi = hT x; yi ; 8x 2 H1; 8y 2 H2: (1.1) The uniqueness of T ∗ follows from the Riesz representation theorem or from a simple relation that h(T1 − T2)y; xi = 0 for all y 2 H2; x 2 H1 implies T1 − T2 = 0: Claim: T ∗ is linear. Consider y; z 2 H2 and α 2 K(R or C), we get hx; T ∗(αy + z)i = hT x; αy + zi = α hT x; yi + hT x; zi = α hx; T ∗yi + hx; T ∗zi ∗ ∗ = hx; αT y + T zi ; 8x 2 H1: 1 ∗ ∗ ∗ Hence T (αy + z) = αT y + T z; 8y; z 2 H2; 8α 2 K: Claim: T ∗ is bounded and jjT jj = jjT ∗jj: ∗ Since the equation (1.1) is satisfied for all elements in H1 and H2, choose x = T y and using CSI, we get ∗ 2 ∗ jjT yjj ≤ jjT jj jjT yjj jjyjj; 8y 2 H2: ∗ =) jjT yjj ≤ jjT jj jjyjj; 8y 2 H2: Therefore, T ∗ is bounded and jjT ∗jj ≤ jjT jj. By a similar argument using (1.1), we can show that jjT jj ≤ jjT ∗jj: Remark. If we have an operator S such that hx; Syi = hT x; yi ; 8x 2 H1; 8y 2 H2; then by uniqueness T ∗ = S: Examples. n n T T T T T T • A : R ! R , x 7! Ax. Then hAx; yi = (Ax) y = (x A )y = x (A y) = x; A y : Therefore, the adjoint A∗ is the transpose AT of A. n n T T T T T D E • B : C ! C , x 7! Bx. Then hBx; yi = (Bx) y = (x B )y = x (B y) = x; BT y : Therefore, the adjoint B∗ is the conjugate-transpose BT of B. • The integral operator T : L2[a; b] ! L2[a; b] defined by b Z T f(x) = k(x; t)f(t)dt; x 2 [a; b] a where k(x; t) is a real-valued continuous function on [a; b] × [a; b]: Then it is easy to show that T is a bounded linear operator. Indeed, b b Z Z jT f(x)j2 ≤ jk(x; t)j2dt jf(t)j2dt (using CSI in continuous variable) a a 1 1 1 0 b 1 2 0 b b 1 2 0 b 1 2 Z Z Z Z 2 2 2 @ jT f(x)j dxA ≤ @ jk(x; t)j dtdxA @ jf(t)j dtA : a a a a Now we calculate T ∗ : b Z hT f; gi = T f(x)g(x)dx a b 0 b 1 Z Z = @ k(x; t)f(t)dtA g(x)dx a a b 0 b 1 Z Z = @ k(x; t)g(x)dxA f(t)dt a a b 0 b 1 Z Z = f(t)@ k(x; t)g(x)dxAdt: a a b Z Therefore, T ∗g(y) = k(s; y)g(s)ds; y 2 [a; b]: a 2 Exercises 1. Find the adjoint of right shift operator: 2 T (x1; x2;:::) = (0; x1; x2;:::); x = (x1; x2;:::) 2 ` : 2. Find the adjoint of multiplication operator: Mf(x) = g(x)f(x); f 2 L2[a; b] where g(x) is a bounded function on [a; b]. Next we will discuss a simple lemma which will be useful in several occasions. Lemma 1.2. (Zero operator) Let X and Y be inner product spaces and T : X ! Y be a bounded linear operator. Then (a) T = 0 () hT x; yi = 0 for all x 2 X and y 2 Y: (b) If T : X ! X, where X is complex, and hT x; xi = 0 for all x 2 X; then T = 0: Proof. (a) The proof is trivial. (b) Consider hT (αx + y); αx + yi = 0 (1.2) α hT y; xi + α hT x; yi = 0; 8α 2 C8x; y 2 X: (1.3) Choose α = 1 and α = i, we get hT x; yi = 0; 8x; y 2 X: Then T = 0: Remark. In the above lemma, if X is real inner product space then the statment (b) is not 2 true. (Hint. Use rotation operator in R .) Properties of the adjoint operator Theorem 1.3. Let T : H1 ! H2 be a bounded linear operator. Then 1. (T ∗)∗ = T 2. jjTT ∗jj = jjT ∗T jj = jjT jj2 3. N (T ) = R(T ∗)? 4. N (T )? = R(T ∗) 5. N (T ∗) = R(T )? 6. N (T ∗)? = R(T ) Here N (T ) and R(T ) denote the null space and range space of T . ∗ ∗ ∗ Proof. 1. Consider h(T ) x; yi = hx; T yi = hT x; yi ; for all x 2 H1, y 2 H2. ∗ ∗ Then h((T ) − T )x; yi = 0, for all x 2 H1, y 2 H2. 3 2. Recall that the norm of composition operators jjSHjj ≤ jjSjj jjHjj: Henceforth, we get jjT ∗T jj ≤ jjT ∗jj jjT jj = jjT jj2: For the other inequality, consider ! ! jjT xjj jjT xjj jjT jj2 = sup sup 06=x2H1 jjxjj 06=x2H1 jjxjj jjT xjj2 hT x; T xi = sup 2 = sup 2 06=x2H1 jjxjj 06=x2H1 jjxjj hT ∗T x; xi = sup 2 06=x2H1 jjxjj ∗ jjT T jjjjxjjjjxjj ∗ ≤ sup 2 = jjT T jj: 06=x2H1 jjxjj ∗ 3. Suppose x 2 N (T ); then T x = 0: Thus hT y; xi = hy; T xi = 0; for all y 2 H2: Therefore, x 2 R(T ∗)?. Similarly, it is easy to show that R(T ∗)? ⊂ N (T ). ∗ ∗ 4. For z 2 R(T ); there exists a y 2 H2 such that T y = z. Then hz; xi = hT ∗y; xi = hy; T xi = 0 for all x 2 N (T ): Therefore, z ?N (T ): This implies that R(T ∗) ⊂ N (T )?, and hence R(T ∗) ⊂ N (T )? (why?). Claim. R(T ∗) = N (T )?: Suppose this is not true, that is, R(T ∗) is a proper closed subspace of N (T )?: Then by ? ∗ projection theorem, there exists a 0 6= x0 2 N (T ) and x0 ? R(T ): In particular, ∗ hx0;T yi = 0; 8y 2 H2 =) hT x0; yi = 0; 8y 2 H2 =) T x0 = 0; ? =) x0 2 N (T ) \N (T ) Therefore, x0 = 0 which is contradiction to x0 6= 0, hence the claim is proved. The other parts are exercise. Exercises Let H1;H2 and H3 be Hilbert spaces. Then show that ∗ ∗ ∗ 1.( TS) = S T for S 2 B(H1;H2);T 2 B(H2;H3) ∗ ∗ ∗ 2.( αT + βS) = αT + βS for S; T 2 B(H1;H2); and α; β 2 K: 3. T ∗T = 0 () T = 0: 4. Construct a bounded linear operator on `2 whose range is not closed. Hint: T (x ; x ;:::) = (x ; px2 ; px3 ;:::), (x ) 2 `2 . 1 2 1 2 3 n 5. Let H be a Hilbert space and T : H ! H be a bijective bounded linear operator whose inverse is bounded. Show that (T ∗)−1 exists and (T ∗)−1 = [T −1]∗: 6. Let T1 and T2 be bounded linear operators on a complex Hilbert space H into itself. If hT1x; xi = hT2x; xi for all x 2 H, show that T1 = T2. 7. Let S = I +T ∗T : H ! H, where T is linear and bounded. Show that S−1 : S(H) ! H exists. 4 Self-adjoint, Normal and Unitary operators Definition 1.4. A bounded linear operator T : H ! H on a Hilbert space H is said to be self-adjoint if T = T ∗; unitary if T is bijective and T ∗ = T −1; normal if T ∗T = TT ∗: Remarks. • T is unitary iff T ∗T = TT ∗ = I: • Unitary, Self-adjoint =) Normal. The converse is not true.
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