MTL 411: Functional Analysis

Home , Operator (mathematics), Unitary operator

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 ∈ 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 , ∀x ∈ H1, ∀y ∈ 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 ∈ 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 ∈ B(H1,H2), there exists a unique operator T ∈ B(H2,H1) such that ∗ hT x, yi = hx, T yi , ∀x ∈ H1, ∀y ∈ H2 and ||T || = ||T ∗||.

Proof. Fix y ∈ H2. Deﬁne fy(x) = hT x, yi, x ∈ 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

|fy(x)| ≤ ||T || ||y|| ||x||, ∀x ∈ H1.

Then by Riesz representation theorem, there exists a unique z ∈ H1 such that fy(x) = hx, zi. Therefore for each y ∈ H2, there exists a unique z ∈ H1 such that fy(x) = hx, zi = hT x, yi. ∗ Now, we deﬁne T y = z, for each y ∈ H2. Thus ∗ hx, T yi = hT x, yi , ∀x ∈ H1, ∀y ∈ 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 ∈ H2, x ∈ H1 implies T1 − T2 = 0. Claim: T ∗ is linear.

Consider y, z ∈ H2 and α ∈ 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 , ∀x ∈ H1.

1 ∗ ∗ ∗ Hence T (αy + z) = αT y + T z, ∀y, z ∈ H2, ∀α ∈ K. Claim: T ∗ is bounded and ||T || = ||T ∗||. ∗ Since the equation (1.1) is satisﬁed for all elements in H1 and H2, choose x = T y and using CSI, we get ∗ 2 ∗ ||T y|| ≤ ||T || ||T y|| ||y||, ∀y ∈ H2. ∗ =⇒ ||T y|| ≤ ||T || ||y||, ∀y ∈ H2. Therefore, T ∗ is bounded and ||T ∗|| ≤ ||T ||. By a similar argument using (1.1), we can show that ||T || ≤ ||T ∗||.

Remark. If we have an operator S such that hx, Syi = hT x, yi , ∀x ∈ H1, ∀y ∈ 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 ∈ [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 |T f(x)|2 ≤ |k(x, t)|2dt |f(t)|2dt (using CSI in continuous variable) a a 1 1 1  b  2  b b  2  b  2 Z Z Z Z 2 2 2  |T f(x)| dx ≤  |k(x, t)| dtdx  |f(t)| dt . a a a a Now we calculate T ∗ : b Z hT f, gi = T f(x)g(x)dx a b  b  Z Z =  k(x, t)f(t)dt g(x)dx a a b  b  Z Z =  k(x, t)g(x)dx f(t)dt a a b  b  Z Z = f(t) k(x, t)g(x)dxdt. a a

b Z Therefore, T ∗g(y) = k(s, y)g(s)ds, y ∈ [a, b]. a

2 Exercises

1. Find the adjoint of right shift operator:

2 T (x1, x2,...) = (0, x1, x2,...), x = (x1, x2,...) ∈ ` .

2. Find the adjoint of multiplication operator:

Mf(x) = g(x)f(x), f ∈ 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 ∈ X and y ∈ Y.

(b) If T : X → X, where X is complex, and hT x, xi = 0 for all x ∈ 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, ∀α ∈ C∀x, y ∈ X. (1.3)

Choose α = 1 and α = i, we get hT x, yi = 0, ∀x, y ∈ 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. ||TT ∗|| = ||T ∗T || = ||T ||2

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 ∈ H1, y ∈ H2. ∗ ∗ Then h((T ) − T )x, yi = 0, for all x ∈ H1, y ∈ H2.

3 2. Recall that the norm of composition operators ||SH|| ≤ ||S|| ||H||. Henceforth, we get ||T ∗T || ≤ ||T ∗|| ||T || = ||T ||2. For the other inequality, consider ! ! ||T x|| ||T x|| ||T ||2 = sup sup 06=x∈H1 ||x|| 06=x∈H1 ||x|| ||T x||2 hT x, T xi = sup 2 = sup 2 06=x∈H1 ||x|| 06=x∈H1 ||x|| hT ∗T x, xi = sup 2 06=x∈H1 ||x|| ∗ ||T T ||||x||||x|| ∗ ≤ sup 2 = ||T T ||. 06=x∈H1 ||x||

∗ 3. Suppose x ∈ N (T ), then T x = 0. Thus hT y, xi = hy, T xi = 0, for all y ∈ H2. Therefore, x ∈ R(T ∗)⊥. Similarly, it is easy to show that R(T ∗)⊥ ⊂ N (T ). ∗ ∗ 4. For z ∈ R(T ), there exists a y ∈ H2 such that T y = z. Then hz, xi = hT ∗y, xi = hy, T xi = 0 for all x ∈ 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 ∈ N (T ) and x0 ⊥ R(T ). In particular,

∗ hx0,T yi = 0, ∀y ∈ H2

=⇒ hT x0, yi = 0, ∀y ∈ H2

=⇒ T x0 = 0, ⊥ =⇒ x0 ∈ 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 ∈ B(H1,H2),T ∈ B(H2,H3)

∗ ∗ ∗ 2.( αT + βS) = αT + βS for S, T ∈ B(H1,H2), and α, β ∈ 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 , √x2 , √x3 ,...), (x ) ∈ `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 ∈ 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. (Hint. T = 2iI, where I is the identity operator on a complex Hilbert space).

Let us discuss a simple criterion for self-adjointness.

Theorem 1.5. (Self-adjointness) Let T : H → H be a bounded linear operator on a Hilbert space H. Then:

(a) If T is self-adjoint, then hT x, xi is real for all x ∈ H.

(b) If H is complex Hilber space and hT x, xi is real for all x ∈ H, then T is self-adjoint.

Proof. (a) Given T = T ∗, we get hT x, xi = hx, T xi = hT x, xi, for all x ∈ H. Hence the imaginary part of hT x, xi = 0 for all x ∈ H. (b) Consider hT x, xi = hT x, xi = hx, T ∗xi = hT ∗x, xi for all x ∈ H. Thus h(T − T ∗)x, xi = 0, for all x ∈ H. Hence by zero operator lemma, we get T = T ∗.

Remarks.

1. Suppose S and T are self-adjoint operators. Then (ST )∗ = T ∗S∗ = ST . Therefore, ST is self-adjoint iﬀ S and T commutes.

2. The set of self-adjoint operators in B(H) is a closed set with respect to operator norm.

Now we discuss some basic properties of unitary operators.

Theorem 1.6. (Unitary operator) Let H be a Hilbert space and the operators U : H → H and V : H → H be unitary. Then:

(a) U is isometric, i.e., ||Ux|| = ||x||, for all x ∈ H.

(b) ||U|| = 1, provided H 6= {0}.

(d) A bounded linear operator T on a complex Hilbert space H is unitary if and only if T is isometric and surjective.

Proof. (a) Since UU ∗ = U ∗U = I, we get hU ∗Ux, xi = hIx, xi = ||x||2. Thus ||Ux|| = ||x||. x (b) If H 6= {0}, choose 0 6= x ∈ H, then ||U( ||x|| )|| = 1. (c) Since U −1 = U ∗, we have (U −1)−1 = (U ∗)−1 = (U −1)∗. Next, (UV )∗UV = V ∗U ∗UV = I. Then the result follows.

5 (d) Suppose T is a unitary operator on a complex Hilbert space H, then it is clear that T is isometric and surjective. Conversely, suppose T is isometric and surjective on a complex Hilbert space H. Then, consider

h(U ∗U − I)x, xi = 0 for all x ∈ H, then by zero operator lemma, we get U ∗U = I. Since U is injective and surjective, U is invertible. Therefore, U −1 = U ∗.

Exercises Problems 4, 5, 6, 14, 15 from E. Kreyzig (Pages 207-208).

Reference E. Kreyzig, Introductory Functional Analysis with Applications, John Wiley & Sons. Inc, 1978.