Linear Transformations On Hilbert Spaces

LINEAR FUNCTIONAL ANALYSIS W W L CHEN c

W W L Chen, 2001, 2008.

This chapter is available free to all individuals, on the understanding that it is not to be used for financial gain, and may be downloaded and/or photocopied, with or without permission from the author. However, this document may not be kept on any information storage and retrieval system without permission from the author, unless such system is not accessible to any individuals other than its owners.

Chapter 8 LINEAR TRANSFORMATIONS ON HILBERT SPACES

8.1. Adjoint Transformations We begin with a result which is a consequence of the Riesz-Fr´echet theorem first studied in Section 6.3. THEOREM 8A. Suppose that V and W are Hilbert spaces over F. For every linear transformation T ∈ B(V, W ), there exists a unique linear transformation T ∗ ∈ B(W, V ) such that hT (x), yi = hx, T ∗ (y)i

for every x ∈ V and y ∈ W .

Remark. Note that in the above, hT (x), yi is an inner product in the Hilbert space W , while hx, T ∗ (y)i is an inner product in the Hilbert space V . Definition. Suppose that V and W are Hilbert spaces over F. The unique linear transformation T ∗ ∈ B(W, V ) satisfying the conclusion of Theorem 8A is called the adjoint transformation of the linear transformation T ∈ B(V, W ). Proof of Theorem 8A. Suppose that y ∈ W is fixed. It is easy to check that the mapping S : V → F, given for every x ∈ V by S(x) = hT (x), yi, is a linear functional on V . Furthermore, we have |S(x)| = |hT (x), yi| ≤ kT (x)kkyk ≤ kT kkxkkyk = (kT kkyk)kxk

for every x ∈ V ,

so that S : V → F is a bounded, and continuous, linear functional on V . It follows from the Riesz-Fr´echet theorem that there exists a unique u ∈ V such that S(x) = hx, ui for every x ∈ V . Write u = T ∗ (y). Then T ∗ : W → V is a mapping satisfying hT (x), yi = hx, T ∗ (y)i Chapter 8 : Linear Transformations on Hilbert Spaces

for every x ∈ V and y ∈ W . page 1 of 6

c

Linear Functional Analysis

W W L Chen, 2001, 2008

Next we show that T ∗ ∈ B(W, V ). Suppose first of all that y, z ∈ W and c ∈ F. Then for every x ∈ V , we have hx, T ∗ (y + z)i = hT (x), y + zi = hT (x), yi + hT (x), zi = hx, T ∗ (y)i + hx, T ∗ (z)i = hx, T ∗ (y) + T ∗ (z)i and hx, T ∗ (cy)i = hT (x), cyi = chT (x), yi = chx, T ∗ (y)i = hx, cT ∗ (y)i, so that T ∗ (y + z) = T ∗ (y) + T ∗ (z) and T ∗ (cy) = cT ∗ (y) by Theorem 4B. It follows that T ∗ : W → V is a linear transformation. Furthermore, for every y ∈ W , we have kT ∗ (y)k2 = hT ∗ (y), T ∗ (y)i = hT (T ∗ (y)), yi ≤ kT (T ∗ (y))kkyk ≤ kT kkT ∗ (y)kkyk. Suppose that kT ∗ (y)k > 0. Then dividing the above by kT ∗ (y)k, we obtain kT ∗ (y)k ≤ kT kkyk. Note that this last inequality is satisfied trivially if kT ∗ (y)k = 0. It follows that kT ∗ (y)k ≤ kT kkyk

for every y ∈ W ,

and so T ∗ : W → V is bounded, whence T ∗ ∈ B(W, V ). Finally, suppose that T1 , T2 ∈ B(W, V ) satisfy hT (x), yi = hx, T1 (y)i = hx, T2 (y)i

for every x ∈ V and y ∈ W .

Then it follows from Theorem 4B that T1 (y) = T2 (y) for every y ∈ W , so that T1 = T2 . The uniqueness of T ∗ ∈ B(W, V ) follows immediately. Example 8.1.1. Suppose that a, b ∈ R and a < b. Consider the vector space L2 [a, b] of all complex valued Lebesgue measurable functions that are square integrable on [a, b]. We know that the norm !1/2

b

Z

2

kf k =

|f (t)| dt

,

a

given in Example 7.1.3, is in fact induced by the inner product Z

!1/2

b

hf, gi =

f (t)g(t) dt

.

a

Let φ ∈ C[a, b] be chosen and fixed, and consider the bounded linear operator T : L2 [a, b] → L2 [a, b], where for every f ∈ L2 [a, b], the function T (f ) ∈ L2 [a, b] is defined by (T (f ))(t) = φ(t)f (t)

for every t ∈ [a, b],

as discussed in Example 7.1.3. It follows from Theorem 8A that the adjoint operator T ∗ satisfies hT (f ), gi = hf, T ∗ (g)i In other words, we must have Z b Z φ(t)f (t)g(t) dt = a

for every f, g ∈ L2 [a, b].

b

f (t)(T ∗ (g))(t) dt

for every f, g ∈ L2 [a, b].

a

Clearly (T ∗ (g))(t) = φ(t)g(t)

for every t ∈ [a, b]

would be sufficient. Hence by uniqueness, the adjoint operator T ∗ : L2 [a, b] → L2 [a, b] is given for every g ∈ L2 [a, b] by this. Chapter 8 : Linear Transformations on Hilbert Spaces

page 2 of 6

c

Linear Functional Analysis

W W L Chen, 2001, 2008

Example 8.1.2. Suppose that a, b, c, d ∈ R, with a < b and c < d. Consider the vector spaces L2 [a, b] and L2 [c, d]. We know that the respective norms !1/2

b

Z

2

|f (t)| dt

kf k =

2

|h(s)| ds

khk =

and

!1/2

d

Z

,

c

a

given in Example 7.1.4, are in fact induced by the respective inner products Z

!1/2

b

hf, gi =

f (t)g(t) dt

Z and

hh, ki =

a

!1/2

d

h(s)k(s) ds

.

c

Let φ : [c, d] × [a, b] → C be a fixed continuous function, and consider the bounded linear transformation T : L2 [a, b] → L2 [c, d], where for every f ∈ L2 [a, b], the function T (f ) ∈ L2 [c, d] is defined by b

Z (T (f ))(s) =

φ(s, t)f (t) dt

for every s ∈ [c, d],

a

as discussed in Example 7.1.4. It follows from Theorem 8A that the adjoint operator T ∗ satisfies hT (f ), ki = hf, T ∗ (k)i

for every f ∈ L2 [a, b] and k ∈ L2 [c, d].

In other words, we must have Z

d

Z

b

!

b

Z

f (t)(T ∗ (k))(t) dt

φ(s, t)f (t) dt k(s) ds = c

a

for every f ∈ L2 [a, b] and k ∈ L2 [c, d].

a

By Fubini’s theorem, clearly (T ∗ (k))(t) =

Z

d

φ(s, t)k(s) ds

for every t ∈ [a, b]

c

would be sufficient. Hence by uniqueness, the adjoint transformation T ∗ : L2 [c, d] → L2 [a, b] is given for every k ∈ L2 [c, d] by this.

8.2. Hermitian Operators We conclude our discussion by studying a special type of adjoint operators. Definition. Suppose that V is a Hilbert space over F. A linear operator T ∈ L(V ) is said to be self-adjoint or Hermitian if T ∗ = T . Example 8.2.1. Suppose that a, b ∈ R and a < b. Consider the Hilbert space L2 [a, b] of all complex valued Lebesgue measurable functions that are square integrable on [a, b], as discussed in Example 8.1.1. Let φ ∈ C[a, b] be chosen and fixed. For the bounded linear operator T : L2 [a, b] → L2 [a, b], where for every f ∈ L2 [a, b], the function T (f ) ∈ L2 [a, b] is defined by (T (f ))(t) = φ(t)f (t) for every t ∈ [a, b], we have shown earlier that the adjoint operator T ∗ : L2 [a, b] → L2 [a, b] is given for every g ∈ L2 [a, b] by (T ∗ (g))(t) = φ(t)g(t) for every t ∈ [a, b]. Hence T : L2 [a, b] → L2 [a, b] is Hermitian if φ ∈ C[a, b] is real valued. The following result gives a technique for finding the norm of a Hermitian operator. Chapter 8 : Linear Transformations on Hilbert Spaces

page 3 of 6

c

Linear Functional Analysis

W W L Chen, 2001, 2008

THEOREM 8B. Suppose that V is a Hilbert space over F. Suppose further that T ∈ B(V ) is an Hermitian operator. Then kT k =

|hT (x), xi|.

sup x∈V, kxk=1

Proof. For every x ∈ V satisfying kxk = 1, we have |hT (x), xi| ≤ kT (x)kkxk ≤ kT kkxk2 = kT k, so that kT k ≥

|hT (x), xi|.

sup x∈V, kxk=1

To prove the opposite inequality, let M=

sup

|hT (x), xi|.

x∈V, kxk=1

For any non-zero vector u ∈ V , the vector u/kuk has norm 1. It follows easily from linearity that |hT (u), ui| ≤ M kuk2

for every u ∈ V .

For every x, y ∈ V , noting that T ∗ = T , it is not difficult to check that hT (x + y), x + yi = hT (x) + T (y), x + yi = hT (x), xi + hT (x), yi + hT (y), xi + hT (y), yi = hT (x), xi + hT (x), yi + hy, T ∗ (x)i + hT (y), yi = hT (x), xi + hT (x), yi + hT (x), yi + hT (y), yi = hT (x), xi + 2RehT (x), yi + hT (y), yi, and similarly hT (x − y), x − yi = hT (x), xi − 2RehT (x), yi + hT (y), yi, and so 4RehT (x), yi = hT (x + y), x + yi − hT (x − y), x − yi ≤ M (kx + yk2 + kx − yk2 ) = 2M (kxk2 + kyk2 ), the last step as a consequence of the Parallelogram law. We can replace x by λx, where λ ∈ F satisfies |λ| = 1 and RehT (λx), yi = |hT (x), yi|. Then |hT (x), yi| ≤ 21 M (kxk2 + kyk2 )

for every x, y ∈ V .

Suppose first of all that T (x) 6= 0. Then taking y=

kxk T (x), kT (x)k

we have kyk = kxk and kxk |hT (x), T (x)i| ≤ M kxk2 , kT (x)k

so that

kT (x)k ≤ M kxk.

The last inequality holds trivially if T (x) = 0, and therefore holds for every x ∈ V . It follows that kT k ≤ M as required. Chapter 8 : Linear Transformations on Hilbert Spaces

page 4 of 6

c

Linear Functional Analysis

W W L Chen, 2001, 2008

Problems for Chapter 8 1. Suppose that V and W are Hilbert spaces over F, and that T ∈ B(V, W ). a) Show that hy, (T ∗ )∗ (x)i = hy, T (x)i for every x ∈ V and y ∈ W , using condition (IP1) twice. Explain why this implies that (T ∗ )∗ = T . b) Deduce that kT ∗ k = kT k. You may wish to study the proof of Theorem 8A for some useful information. c) Show that kT (x)k2 ≤ kT ∗ T kkxk2 for every x ∈ V . Explain why this implies the inequality kT k2 ≤ kT ∗ T k. d) Deduce that kT ∗ T k = kT k2 . 2. Suppose that V , W and U are Hilbert spaces over F, and that T ∈ B(V, W ) and S ∈ B(W, U ). Show that (ST )∗ = T ∗ S ∗ . 3. Suppose that V and W are Hilbert spaces over F. Show that for every c, a ∈ F and T, S ∈ B(V, W ), we have (cT + aS)∗ = cT ∗ + aS ∗ . 4. Suppose that V and W are Hilbert spaces over F. Show that the function f : B(V, W ) → B(W, V ), defined for every T ∈ B(V, W ) by f (T ) = T ∗ , is continuous in B(V, W ). [Hint: Show that kf (T ) − f (S)k = kT − Sk for every T, S ∈ B(V, W ).] 5. Suppose that V is a complex Hilbert space, and that x1 , x2 ∈ V are fixed. Consider the bounded linear operator T : V → V , where T (x) = hx, x1 ix2 for every x ∈ V . Show that the adjoint operator T ∗ : V → V is given by T ∗ (y) = hy, x2 ix1 for every y ∈ V . 6. Consider the vector space `2 of all square summable infinite sequences of complex numbers, with inner product hx, yi =

∞ X

!1/2 xi yi

.

i=1

For each of the given bounded linear operators T : `2 → `2 , find the adjoint operator T ∗ : `2 → `2 : a) T (x) = (0, x1 , x2 , x3 , . . .) for every x = (x1 , x2 , x3 , . . .), as discussed in Example 7.1.6. b) T (x) = (0, 2x1 , x2 , 2x3 , x4 , . . .) for every x = (x1 , x2 , x3 , x4 , . . .), as discussed in Problem 1 in Chapter 7. 7. Suppose that (xn )n∈N is an orthonormal basis in a Hilbert space V over C, and that (cn )n∈N is a fixed bounded sequence of complex numbers. Consider the bounded linear operator T : V → V such that T (xn ) = cn xn for every n ∈ N, discussed in Problem 3 in Chapter 7. Find the adjoint operator T ∗ : V → V . 8. Suppose that V and W are Hilbert spaces over F, and that T ∈ B(V, W ). Suppose also that R(T ) and R(T ∗ ) denote respectively the range of the linear transformations T : V → W and T ∗ : W → V . a) Show that hx, zi = 0 for every x ∈ ker(T ) and z ∈ R(T ∗ ). b) Deduce that ker(T ) ⊆ (R(T ∗ ))⊥ . c) Show that hT (u), T (u)i = 0 for every u ∈ (R(T ∗ ))⊥ . d) Deduce that (R(T ∗ ))⊥ ⊆ ker(T ). e) It follows from parts (b) and (d) that ker(T ) = (R(T ∗ ))⊥ . Use this and Problem 1 to show that ker(T ∗ ) = (R(T ))⊥ . 9. Suppose that V is a Hilbert space over F. Suppose further that T ∈ B(V ) is invertible, so that T T −1 = T −1 T = I, where I ∈ B(V ) is the identity linear operator. a) Show that I ∗ = I. b) By studying the adjoint of the equation T T −1 = T −1 T = I, show that T ∗ is invertible, with inverse (T ∗ )−1 = (T −1 )∗ . Chapter 8 : Linear Transformations on Hilbert Spaces

page 5 of 6

Linear Functional Analysis

c

W W L Chen, 2001, 2008

10. Suppose that V is a Hilbert space over F. Suppose further that H is the subset of all Hermitian operators in B(V ). a) Show that cT + aS ∈ H for every c, a ∈ R and T, S ∈ H. b) Show that H is a closed subset of B(V ). [Hint: Use Problem 4.] 11. Suppose that V is a Hilbert space over F, and that T ∈ B(V ). a) Show that T ∗ T and T T ∗ are both Hermitian. b) Show that there exist Hermitian R, S ∈ B(V ) such that T = R + iS.

Chapter 8 : Linear Transformations on Hilbert Spaces

page 6 of 6