Solutions OPERATOR THEORY Mathematic87.Blogfa

8/2/2019 Solutions OPERATOR THEORY Mathematic87.Blogfa

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OPERATOR THEORY

Solution I

1. Prove that B(F, Y ) is not a Banach space if Y is not complete. [Hint:

take a Cauchy sequence (yn) in Y which does not converge and consider thesequence of operators (Bn),

Bnλ := λyn, ∀λ ∈ F.]

Solution

Take a Cauchy sequence (yn) in Y which does not converge and consider thesequence of operators (Bn),

Bnλ := λyn, ∀λ ∈ F.

It is easy to see that Bn ∈ B(F, Y ) and Bn = yn, n ∈ N. Since(Bn − Bm)λ = λ(yn − ym), we have Bn − Bm = yn − ym. Therefore(Bn) is a Cauchy sequence in B(F, Y ). Suppose there exists B ∈ B(F, Y )

such that Bn − B → 0 as n → +∞. Let y := B1 ∈ Y . Then yn − y =Bn1−B1 ≤ Bn−B → 0 as n → +∞, i.e. the sequence (yn) converges toy. This contradiction proves that (Bn) cannot be convergent. Hence B(F, Y )is not a Banach space.

2. Give an example of a bounded linear operator A such that Ran(A) isnot closed. [Hint: consider the embedding X → Y , where X is the spaceC ([0, 1]) equipped with the norm · ∞ and Y = L p([0, 1]) is the completion

of the normed space (C ([0, 1]), · p), 1 ≤ p < ∞.]

Solution

Let X be the space C ([0, 1]) equipped with the norm ·∞ and Y = L p([0, 1])be the completion of the normed space (C ([0, 1]), · p), 1 ≤ p < ∞. LetA : X → Y be the embedding: Af = f , ∀f ∈ X . Then Ran(A) = C ([0, 1])is dense in Y = L p([0, 1]) but does not coincide with it. Therefore Ran(A)cannot be closed.

1



3. Give an example of a normed space and an absolutely convergent seriesin it, which is not convergent.

Solution

Let X be an arbitrary normed non-Banach space (e.g., X = (C ([0, 1]), · p),1 ≤ p < ∞). Then there exists a Cauchy sequence (xn) in X which does

not converge. Since (xn) is Cauchy, for any k ∈ N there exists nk ∈ N such

that xn − xm ≤ 2−k, ∀n, m ≥ nk. Consider the series

∞k=1

(xnk+1 − xnk) . (1)

This series is absolutely convergent because

∞k=1

xnk+1 − xnk ≤∞k=1

2−k = 1 < +∞.

On the other hand, (1) is not convergent in X . Indeed, if the sequence of partial sums

j

k=1(xnk+1 − xnk) = xnj+1 − xn1 is convergent as j → +∞, thenso is the sequence (xnj+1) and also (xn), since (xn) is Cauchy. Contradiction!

4. Let X be the Banach space C ([0, 1]) and Y be the space of all continuouslydifferentiable functions on [0, 1] which equal 0 at 0. Both of the spaces areequipped with the norm · ∞. Show that the linear operator B : X → Y ,

(Bf )(t) := t0

f (τ )dτ,

is bounded, one–to–one and onto, but the inverse operator B−1 : Y → X

is not bounded. Compare this with the Banach theorem (bounded inversetheorem).

Solution

2



It is clear that Bf ∞ ≤ f ∞, ∀f ∈ C ([0, 1]). So, B is bounded.

Suppose Bf 1 = Bf 2, i.e. t0

f 1(τ )dτ =

t0

f 2(τ )dτ, ∀t ∈ [0, 1].

Then differentiation gives f 1(t) = f 2(t), ∀t ∈ [0, 1], i.e. f 1 = f 2. So, B isone–to–one.Let ϕ be continuously differentiable and ϕ(0) = 0. Then ϕ ∈ X and

ϕ(t) =

t0

ϕ(τ )dτ, ∀t ∈ [0, 1],

i.e. ϕ = Bϕ. So, B is onto.It follows from the above that B is invertible and

B−1ϕ(t) = ϕ(t), ∀t ∈ [0, 1].

It is easy to see that B−1 : Y → X is not bounded. Indeed,

B−1 sin(nt)∞ = n cos(nt)∞ = n ≥ n sin(nt)∞, ∀n ∈ N.

This does not contradict the Banach theorem (bounded inverse theorem) be-

cause Y equipped with the norm · ∞ is not a Banach space.

5. Denote by C 2([0, 1]) the space of twice continuously differentiable func-tions on the interval [0, 1] equipped with the norm

u := max0≤s≤1

|u(s)| + max0≤s≤1

|u(s)| + max0≤s≤1

|u(s)|.

Let X denote the subspace of C 2([0, 1]) containing functions satisfying bound-

ary conditions u(0) = u(1) = 0. Prove that the operator A = −d2

ds2 is abounded operator acting from X to C ([0, 1]).

Solution

The boundedness of A : X → C ([0, 1]) follows directly from the definitionof the norm in X . (Note that C ([0, 1]) is assumed to be equipped with thenorm · ∞.)

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6. Show that the operator A from the previous question has a bounded in-verse A−1 : C ([0, 1]) → X and construct it effectively.

Solution

Consider the equation Au = f , where f ∈ C ([0, 1]) is given and u ∈ X

is unknown. It is equivalent to −d2u(s)ds2

= f (s), ∀s ∈ [0, 1]. Successiveintegration gives

du(s)ds

= −

s

0

f (τ )dτ + c1,

u(s) = −

s0

t0

f (τ )dτdt + c1s + c0,

where c1 and c0 are some constants. Now the boundary conditions u(0) =u(1) = 0 imply c0 = 0 and

c1 =

10

t0

f (τ )dτdt.

Hence u = Rf , where

(Rf )(s) := −

s0

t0

f (τ )dτdt + s

10

t0

f (τ )dτdt =

(s − 1)

s0

t0

f (τ )dτdt + s

1s

t0

f (τ )dτdt.

It is easy to see that Rf ∈ X , ∀f ∈ C ([0, 1]) and

(ARf )(s) = −d2

ds2−

s

0

t

0

f (τ )dτdt + s 1

0

t

0

f (τ )dτdt =

d

ds

s0

f (τ )dτ −

10

t0

f (τ )dτdt

= f (s), ∀s ∈ [0, 1],

(RAu)(s) = −

s0

t0

(−u(τ ))dτdt + s

10

t0

(−u(τ ))dτdt =

s0

(u(t) − u(0))dt − s

10

(u(t) − u(0))dt = u(s) − u(0) − u(0)s

−su(1) + su(0) + su(0) = u(s), ∀s ∈ [0, 1], ∀u ∈ X,

4



since u(0) = u(1) = 0. Hence, AR = I and RA = I , i.e. A is invertible and

A−1 = R.The boundedness of A−1 = R : C ([0, 1]) → X can be obtained directly fromthe definition of the norm in X . It also follows immediately from the Banachtheorem on the inverse operator. (Note that the completeness of X followsfrom the well known results on uniform convergence.)

5



OPERATOR THEORY

Solution II

1. Let B ∈ B(C ([0, 1])) be defined by the formula

Bf (t) = tf (t), t ∈ [0, 1].

Find σ(B) and the set of all eigenvalues of B.

Solution

σ(B) = [ 0, 1] and B does not have eigenvalues. This is a special case of Question 2.

2. Let g ∈ C ([0, 1]) be a fixed function and let A ∈ B(C ([0, 1])) be definedby the formula

Af (t) = g(t)f (t), t ∈ [0, 1].

Find σ(A) and construct effectively the resolvent R(A; λ). Find the eigen-

values and eigenvectors of A.

Solution

Let λ ∈ C, λ ∈ g([0, 1]) := {g(t)| t ∈ [0, 1]}. Then since g ∈ C ([0, 1]),1/(g − λ) ∈ C ([0, 1]) and A − λI has an inverse R(A; λ) = (A − λI )−1 ∈B(C ([0, 1])) defined by

R(A; λ)f (t) = (g(t) − λ)−1f (t), t ∈ [0, 1].

Hence σ(A) ⊂ g([0, 1]).Suppose now λ ∈ g([0, 1]), i.e. λ = g(t0) for some t0 ∈ [0, 1]. Then (A −λI )f (t0) = (g(t0) − λ)f (t0) = 0, i.e. Ran(A − λI ) consist of functionsvanishing at t0. Consequently Ran(A − λI ) = C ([0, 1]) and A − λI is notinvertible. Therefore g([0, 1]) ⊂ σ(A). Finally, σ(A) = g([0, 1]).Take an arbitrary λ ∈ g([0, 1]). Let g−1(λ) := {τ ∈ [0, 1] : g(τ ) = λ}.The equation Af = λf , i.e. (g(t) − λ)f (t) = 0 is equivalent to f (t) = 0,∀t ∈ [0, 1] \ g−1(λ). If g−1(λ) contains an interval of positive length, then itis easy to see that the set {f ∈ C ([0, 1]) \ {0} : f (t) = 0, ∀t ∈ [0, 1] \ g−1(λ)}is non-empty and coincides with the set of all eigenvectors corresponding to

1



the eigenvalue λ. If g−1(λ) does not contain an interval of positive length,

then [0, 1] \ g−1(λ) is dense in [0, 1] and f (t) = 0, ∀t ∈ [0, 1] \ g−1(λ) impliesby continuity that f ≡ 0. In this case λ is not an eigenvalue.

3. Let K ⊂ C be an arbitrary nonempty compact set. Construct an operatorB ∈ B(l p), 1 ≤ p ≤ ∞, such that σ(B) = K .

Solution

Let {λk}k∈N be a dense subset of K . Consider the operator B : l p → l p

defined by

Bx = (λ1x1, λ2x2, . . . , λkxk, . . . ), ∀x = (x1, x2, . . . ) ∈ l p.

Then B is a bounded linear operator and λk’s are its eigenvalues. (Why?)Consequently {λk}k∈N ⊂ σ(B). Since σ(B) is closed and {λk}k∈N is dense inK ,

K ⊂ σ(B).

On the other hand, let λ ∈ C \ K . Then d := inf k∈N |λk − λ| > 0 and B − λI

has a bounded inverse (B − λI )−1 : l p → l p defined by

(B − λI )−1x =

1

λ1 − λx1,

1

λ2 − λx2, . . . ,

1

λk − λxk, . . .

,

∀x = (x1, x2, . . . ) ∈ l p.

Hence λ ∈ σ(B). Thereforeσ(B) ⊂ K.

Finally,σ(B) = K

4. Let k ∈ C ([0, 1]) be a given function. Consider the operator B ∈B(C ([0, 1])) defined by the formula

(Bu)(s) =

s0

k(t)u(t)dt.

2



Construct effectively (not as a power series!) the resolvent of A. How does

this resolvent R(B; λ) behave when λ → 0?

Solution

It is clear that Bu is continuously differentiable for any u ∈ C ([0, 1]). Hence,Ran(B) = C ([0, 1]) and B is not invertible, i.e. 0 ∈ σ(B).

Suppose now λ = 0. Consider the equation (B − λI )u = f , f ∈ C ([0, 1]), i.e.

s

0

k(t)u(t)dt − λu(s) = f (s), s ∈ [0, 1]. (1)

Suppose this equation has a solution u ∈ C ([0, 1]). Then

s0

k(t)u(t)dt = f (s) + λu(s), s ∈ [0, 1].

Therefore f + λu is continuously differentiable and

k(s)u(s) = (f (s) + λu(s)), s ∈ [0, 1].

If f is continuously differentiable, then

k(s)u(s) = f (s) + λu(s), s ∈ [0, 1],

i.e.

u(s) −1

λk(s)u(s) = −

1

λf (s), s ∈ [0, 1]. (2)

Taking s = 0 in (1) gives

u(0) = −1

λf (0).

Solving (2) with this initial condition we obtain

u(s) = e1λ

s0k(τ )dτ

− 1

λf (0) −

s0

1λ

f (t)e−1λ

t0k(τ )dτ dt

=

e1

λ

s

0k(τ )dτ

−

1

λf (0) −

1

λf (t)e−

1

λ

t

0k(τ )dτ

s0

−

1

λ2

s0

f (t)k(t)e−1

λ

t

0k(τ )dτ dt

= −

1

λf (s) −

1

λ2

s0

f (t)k(t)e1

λ

s

tk(τ )dτ dt.

Let

Aλf (s) := −1

λf (s) −

1

λ2

s0

f (t)k(t)e1

λ

s

tk(τ )dτ dt. (3)

3



It is easy to see that Aλ : C ([0, 1]) → C ([0, 1]) is a bounded linear operator.

The above argument shows that if f is continuously differentiable and (1)has a solution u ∈ C ([0, 1]), then u = Aλf . In particular, (1) with f = 0 hasonly a trivial solution u = 0, i.e. Ker(B − λI ) = {0}.For any f ∈ C ([0, 1]) and u = Aλf the function

f (s) + λu(s) = −1

λ

s0

f (t)k(t)e1

λ

s

tk(τ )dτ dt

is continuously differentiable and

(f (s) + λu(s)) = −1

λ

f (s)k(s) −k(s)

λ2

s

0

f (t)k(t)e1

λ

s

tk(τ )dτ dt =

k(s)Aλf (s) = k(s)u(s)

(see (3)). Hence

f (s) + λu(s) =

s0

k(t)u(t)dt + const.

It follows from (3) that f (0) + λu(0) = f (0) + λAλf (0) = 0. Therefore

f (s) + λu(s) =

s0

k(t)u(t)dt,

i.e. f = (B − λI )Aλf , ∀f ∈ C ([0, 1]), i.e. (B − λI )Aλ = I . Consequently Aλ

is a right inverse of B −λI and Ran(B −λI ) = C ([0, 1]). Since Ker(B −λI ) ={0}, the operator B − λI is invertible for any λ = 0 and (B − λI )−1 = Aλ.Thus

σ(B) = {0} and R(B; λ) = Aλ, ∀λ = 0,

where Aλ is given by (3).It follows from the well known property of the resolvent that R(B; λ) ≥1/|λ|, ∀λ = 0. So R(B; λ) → ∞ as λ → 0. If k ≡ 0, the above inequalitybecomes an equality. Let us show that if k ≡ 0, then R(B; λ) grows much

faster than 1/|λ| as λ → 0 in a certain direction. Indeed,

(R(B; λ)1)(s) = (Aλ1)(s) = −1

λ−

1

λ2

s0

k(t)e1

λ

s

tk(τ )dτ dt =

−1

λ+

1

λe1

λ

s

tk(τ )dτ

s0

= −1

λe1

λ

s

0k(τ )dτ .

Further, k ≡ 0 implies s0

k(τ )dτ ≡ 0. Let

C := max[0,1]

s0

k(τ )dτ

=

s00

k(τ )dτ

> 0.

4



Then for λ such that 1λ

s00

k(τ )dτ > 0 we have

R(B; λ) ≥ R(B; λ)1 ≥1

|λ|eC/|λ| as λ → 0.

5. Let A, B ∈ B(X ). Show that for any λ ∈ ρ(A)

ρ(B),

R(B; λ) − R(A; λ) = R(B; λ)(A − B)R(A; λ).

Solution

R(B; λ)(A − B)R(A; λ) = R(B; λ)((A − λI ) − (B − λI ))R(A; λ) =

R(B; λ) − R(A; λ).

5



OPERATOR THEORY

Solution II I

1. Let X be a Banach space and A, B ∈ B(X ).(a) Show that if I − AB is invertible, then I − BA is also invertible. [Hint:

consider B(I − AB)−1A + I .](b) Prove that if λ ∈ σ(AB) and λ = 0, then λ ∈ σ(BA).(c) Give an example of operators A and B such that 0 ∈ σ(AB) but 0 ∈σ(BA).(d) Show that σ(AB)

{0} = σ(BA)

{0}.

(e) Prove that r(AB) = r(BA).

Solution

(a) Suppose I − AB is invertible. Then since

(I − BA)

B(I − AB)−1A + I

= (I − BA)B(I − AB)−1A + (I − BA) =

B(I − AB)(I − AB)−1A + (I − BA) = BA + (I − BA) = I,

and

B(I − AB)

−1

A + I

(I − BA) = B(I − AB)−1

A(I − BA) + (I − BA) =B(I − AB)−1(I − AB)A + (I − BA) = BA + (I − BA) = I,

the operator I − BA is also invertible and (I − BA)−1 = B(I − AB)−1A + I .

(b) Take an arbitrary λ ∈ σ(AB) \ {0}. Suppose BA − λI = −λ(I − 1λ

BA) isinvertible. Then I − 1

λBA is invertible and (a) implies that I − 1

λAB is also

invertible. Therefore −λ(I − 1λ

AB) = AB − λI is invertible, i.e. λ ∈ σ(AB).This contradiction shows that BA − λI cannot be invertible, i.e. λ ∈ σ(BA).

(c) Consider the right and left shift operators R, L : l p → l p, 1 ≤ p ≤ ∞,

Rx = (0, x1, x2, . . . ), Lx = (x2, x3, . . . ), ∀x = (x1, x2, . . . ) ∈ l p.

It is easy to see that

LRx = L(0, x1, x2, . . . ) = x, RLx = R(x2, x3, . . . ) = (0, x2, x3, . . . ), ∀x ∈ l p.

Hence LR = I , while Ran(RL) = l p. Therefore LR is invertible while RL isnot, i.e. 0 ∈ σ(RL) but 0 ∈ σ(LR).

1



(d) Follows from (b).

(e) Follows from (d) and the definition of the spectral radius.

2. Let X be a Banach space and let operators A, B ∈ B(X ) commute:AB = BA. Prove that r(A + B) ≤ r(A) + r(B).

Solution

Take an arbitrary ε > 0. The spectral radius formula implies that An ≤(r(A) + ε)n, Bn ≤ (r(B) + ε)n for sufficiently large n ∈ N. Therefore thereexists M ≥ 1 such that

An ≤ M (r(A) + ε)n, Bn ≤ M (r(B) + ε)n, ∀n ∈ N.

Since A and B commute, we have

(A + B)n =n

k=0

n!

k!(n − k)!An−kBk.

Hence,

(A + B)n ≤n

k=0

n!

k!(n − k)!

An−k Bk

≤

M 2n

k=0

n!

k!(n − k)!(r(A) + ε)n−k(r(B) + ε)k = M 2(r(A) + r(B) + 2ε)n

for any n ∈ N. Consequently

r(A + B) = limn→+∞

(A + B)n1/n ≤ r(A) + r(B) + 2ε, ∀ε > 0,

i.e. r(A + B) ≤ r(A) + r(B).

3. Let k ∈ C ([0, 1] × [0, 1]) be a given function. Consider the operatorB ∈ B(C ([0, 1])) defined by the formula

(Bu)(s) =

s0

k(s, t)u(t)dt.

2



Find the spectral radius of B. What is the spectrum of B? [Hint: prove by

induction that|(Bnu)(s)| ≤

M n

n!snu∞, ∀n ∈ N,

for some constant M > 0.]

Solution

Let us prove by induction that

|(Bnu)(s)| ≤M n

n!snu∞, ∀s ∈ [0, 1], ∀n ∈ N, (1)

whereM := max

(s,t)∈[0,1]2|k(s, t)|.

For n = 0 inequality (1) is trivial. Suppose (1) holds for n = k. Then forn = k + 1 we have

Bk+1u

(s) =

s0

k(s, t)(Bku)(t)dt

≤

s0

|k(s, t)|(Bku)(t)

dt ≤

M s

0(Bku)(t) dt ≤ M

s

0

M k

k!tku∞dt =

M k+1

k!u∞

s

0

tkdt =

M k+1

(k + 1)!sk+1u∞, ∀s ∈ [0, 1].

Hence, (1) is proved by induction.It follows from (1) that

Bnu∞ ≤M n

n!u∞, ∀u ∈ C ([0, 1]),

i.e.

Bn ≤ M

n

n!, ∀n ∈ N.

Therefore

r(B) = limn→+∞

Bn1/n ≤ limn→+∞

M

(n!)1/n= 0.

Since r(B) = 0, σ(B) cannot contain nonzero elements. Taking into accountthat σ(B) is nonempty we conclude σ(B) = {0}.

3



OPERATOR THEORY

Solution IV

1. Suppose X is a Banach space and let A ∈ B(X ) be nilpotent , i.e. An = 0for some n ∈ N. Find σ(A).

Solution

The Spectral Mapping Theorem implies

{λn : λ ∈ σ(A)} = σ(An) = σ(0) = {0}.

Therefore, λ ∈ σ(A) ⇐⇒ λ

n

= 0 ⇐⇒ λ = 0, i.e. σ(A) = {0}.

2. Show that the range of the operator B : l p → l p, 1 ≤ p < ∞,

(Bx)n :=1

1 + n2xn, n ∈ N, x = (x1, x2, . . . ),

is not closed.

Solution

Take an arbitrary y = (y1, y2, . . . ) ∈ l p. Let y(N ) = (y(N )1 , y

(N )2 , . . . ),

y(N )n :=

yn if n ≤ N,

0 if n > N,N ∈ N.

Then y − y(N ) → 0 as N → +∞ and the vectors x(N ) defined by x(N )n :=

(1 + n2)y(N )n belong to l p. (Why?) It is clear that y(N ) = Bx(N ) ∈ Ran(B).

So, Ran(B) is dense in l p. On the other hand Ran(B) does not coincide withl p. Indeed, consider y ∈ l p, yn = 1

1+n2, n ∈ N. Suppose y ∈ Ran(B). Then

there exists x ∈ l p such that Bx = y, i.e.

11 + n2

xn = 11 + n2

, n ∈ N,

i.e. x = (1, 1, . . . ) ∈ l p. This contradiction shows that y ∈ Ran(B). Hence,Ran(B) = l p. Since Ran(B) is dense in l p, it cannot be closed.

3. Let P ∈ B(X ) be a projection, i.e. P 2 = P . Construct R(P ; λ).

1



Solution

If P = 0, then obviously P − λI = −λI , σ(P ) = {0} and R(P ; λ) = −λ−1I .

If P = I , then P − λI = (1 − λ)I , σ(P ) = {1} and R(P ; λ) = (1 − λ)−1I .

Suppose now P is non-trivial, i.e. P = 0, I . Take any λ = 0, 1. Then usingthe equalities P 2 = P , Q2 = Q and QP = P Q = 0, where Q = I − P , weobtain

(1 − λ)−1P − λ−1Q

(P − λI ) =

(1 − λ)−1P − λ−1Q

((1 − λ)P − λQ) =

P + Q = I

and similarly(P − λI )

(1 − λ)−1P − λ−1Q

= I.

ThusR(P ; λ) = (1 − λ)−1P − λ−1Q.

4. Let A−1l

∈ B(Y, X ) be a left inverse of A ∈ B(X, Y ), i.e. A−1l

A = I X

.Find σ(AA−1l ).Let B−1

r ∈ B(Y, X ) be a right inverse of B ∈ B(X, Y ), i.e. BB−1r = I Y . Find

σ(B−1r B).

Solution

It is easy to see that AA−1l is a projection. Indeed,

(AA−1l )2 = AA−1l AA−1l = AI XA−1l = AA−1l .

It is also clear that AA

−1

l = 0. Indeed, if AA

−1

l = 0 then A

−1

l (AA

−1

l )A =A−1l 0A = 0, i.e. I X = 0. Contradiction! (We assume of course thatX = {0}.) So, AA−1l is a nonzero projection. Hence σ(AA−1l ) = {0, 1}if AA−1l = I , i.e. if A is not invertible. If A is invertible, i.e. AA−1l = I , thenobviously σ(AA−1l ) = {1}.

Similarly

σ(B−1r B) =

{0, 1} if B is not invertible,

{1} if B is invertible.

2



OPERATOR THEORY

Solution V

1. Let B ∈ B(X ) and let T ∈ B(Y, X ) be invertible: T −1 ∈ B(X, Y ). Provethat

σ(B) = σ(T −1BT ).

Solution

Since T is invertible,

λ ∈ σ(B) ⇐⇒ B − λI is invertible ⇐⇒ T −1(B − λI )T is invertible

⇐⇒ T −1BT − λI is invertible ⇐⇒ λ ∈ σ(T −1BT ),

i.e. σ(B) = σ(T −1BT ).

2. Consider the right-shift operator R : l∞ → l∞ defined by

Rx = (0, x1, x2, . . . ), x = (x1, x2, . . . ) ∈ l∞.

Find the eigenvalues and the spectrum of this operator. Is this operatorcompact?

Solution

It is clear that Ker(R) = {0}, i.e. 0 is not an eigenvalue of R. Suppose λ = 0is an eigenvalue of R. Then Rx = λx for some non-zero x. So,

0 = λx1

x1 = λx2

x2 = λx3

· · ·

Solving the last system we obtain x = 0. Contradiction! Thus R does nothave eigenvalues.

1



It is clear that Rx = x for any x ∈ l∞. Therefore R = 1 and σ(R) ⊂

{λ ∈ C : |λ| ≤ 1}. Let us take an arbitrary λ ∈ C such that 0 < |λ| < 1.Suppose λ ∈ σ(R). Then the equation

(R − λI )x = y (1)

has a unique solution x ∈ l∞ for any y ∈ l∞. For y = (1, 0, 0, . . . ), (1) takesthe form

−λx1 = 1

x1 − λx2 = 0

x2 − λx3 = 0

· · ·

Solving the last system we obtain xk = −λ−k. Since |λ| < 1, the elementx = (x1, x2, . . . ) does not belong to l∞. The obtained contradiction showsthat λ ∈ σ(R). Hence, {λ ∈ C : 0 < |λ| < 1} ⊂ σ(R). Taking into accountthat σ(R) is closed we obtain

σ(R) = {λ ∈ C : |λ| ≤ 1}.

The operator R is not compact. This follows from the fact that R is an isom-

etry or from the fact that its spectrum cannot be the spectrum of a compactoperator.

3. Consider the set

M = {x ∈ l∞ : |xn| ≤ n−α, n ∈ N} ⊂ l∞,

where α > 0 is a fixed number. Prove that M is compact.

Solution

It is clear that M is a closed set. Take an arbitrary ε > 0. There exists N ∈ Nsuch that n−α < ε, ∀n > N . Let x(l) = (x

(l)1 , . . . , x

(l)N ) ∈ CN , l = 1, . . . , L be

a finite ε–net of the set

M N := {x = (x1, . . . , xN ) ∈ CN : |xn| ≤ n−α, n = 1, . . . , N }.

Such an ε–net can be easily constructed with the help of ε–nets of the disks{xn ∈ C : |xn| ≤ n−α}. (The existence of such an ε–net also follows

2



from the fact that M N is relatively compact as a bounded subset of the

finite–dimensional space CN .) Now it is easy to see that

x(l) := (x(l)1 , . . . , x

(l)N , 0, 0, . . . ), l = 1, . . . , L

is an ε–net of M . Hence M is a closed relatively compact set, i.e. a compactset.

(An alternative proof: it is clear that M = T (S ∞), where T ∈ B(l∞),

T x := (x1, 2−αx2, 3−αx3, . . . , n−αxn, . . . ), x = (x1, x2, . . . ) ∈ l∞

and S ∞ is the unit ball of l∞:

S ∞ = {x ∈ l∞ : |xn| ≤ 1, n ∈ N}.

It is easy to see that T − T N → 0 as N → +∞, where T N is a finite rankoperator defined by

T N x := (x1, 2−αx2, . . . , N −αxN , 0, 0 . . . ), x = (x1, x2, . . . ) ∈ l∞.

Hence T is a compact operator and M = T (S ∞) is a closed relatively com-pact set, i.e. a compact set.)

4. Let g ∈ C ([0, 1]) be a fixed function. Consider the operator A ∈B(C ([0, 1])) defined by the formula

(Au)(s) := g(s)u(s),

i.e. the operator of multiplication by g. Is this operator compact?

Solution

It is clear that if g ≡ 0 then A is compact. Let us prove that t if g ≡ 0 then

A is not compact. Indeed, since g ≡ 0, there exists a subinterval [a, b] ⊂ [0, 1]such that m := mins∈[a,b] |g(s)| > 0. Consider the sequence un ∈ C ([0, 1]),n ∈ N, un(s) := sin(2n s−a

b−aπ), s ∈ [0, 1]. It is clear that (un) is a bounded

sequence. On the other hand (Aun) does not have Cauchy subsequences.Indeed, take arbitrary k, n ∈ N. Assume for definiteness that k > n, i.e.k ≥ n + 1. Let sn := a + 2−(n+1)(b − a). Then sn ∈ [a, b] and

Auk − Aun = maxs∈[0,1]

|g(s)(uk(s) − un(s))| ≥ m maxs∈[a,b]

|uk(s) − un(s)| ≥

m|uk(sn) − un(sn)| = m| sin(2k−n−1π) − sin(π/2)| = m|0 − 1| = m > 0.

3



Since (Aun) does not have Cauchy subsequences, A is not compact.

(An alternative proof: according to the solution of Exercise 2, Sheet II,σ(A) = g([0, 1]). If g ≡ 0 is a constant, then 0 ∈ g([0, 1]) and σ(A) cannotbe the spectrum of a compact operator. If g is nonconstant, then g([0, 1]) isa connected subset of C consisting of more than one point and σ(A) cannotbe the spectrum of a compact operator.)

5. Let X be an infinite-dimensional Banach space and B, T ∈ B(X ). Which

of the following statements are true?(i) If BT is compact then either B or T is compact.(ii) If T 2 = 0 then T is compact.(iii) If T n = I for some n ∈ N then T is not compact.

Solution

(i) is false. This follows from the fact that (ii) is false.

(ii) is false. Indeed, let X = l p, 1 ≤ p ≤ +∞ and

T x = (0, x1, 0, x3, 0, x5, 0, . . . ), x = (x1, x2, x3, . . . ) ∈ l p.

Then T 2 = 0 and it is easy to see that T is not compact (why?).

(iii) is true. Indeed, suppose T is compact. Then I = T n is also compact,which is impossible, since X is infinite-dimensional. Contradiction!

4



OPERATOR THEORY

Solution VI

1. Prove that the norm · p on l p, p = 2, is not induced by an inner product.(Hint: Prove that for x = (1, 1, 0, . . . ) ∈ l p and y = (1, −1, 0, . . . ) ∈ l p theparallelogram law fails.)

Solution

It is easy to see that

x + y p = 2 = x − y p, x p = 21/p = y p.

Hence the parallelogram law

x + y2 + x − y2 = 2(x2 + y2)

is equivalent to 8 = 4 · 22/p, i.e. to p = 2. So, if p = 2, the parallelogram lawfails, i.e. the norm · p is not induced by an inner product.

2. Prove that the norm · p, p = 2 on C ([0, 1]) is not induced by an innerproduct. (Hint: Prove that for functions f (t) = 1/2 − t and

g(t) =

1/2 − t if 0 ≤ t ≤ 1/2 ,t − 1/2 if 1/2 < t ≤ 1 ,

the parallelogram law fails).

Solution

We have

f (t) + g(t) =

1 − 2t if 0 ≤ t ≤ 1/2 ,0 if 1/2 < t ≤ 1 ,

f (t) − g(t) =

0 if 0 ≤ t ≤ 1/2 ,1 − 2t if 1/2 < t ≤ 1 ,

1



f + g p = 1/2

0(1 − 2t) pdt

1/p

=1

(2( p + 1))1/p ,

f − g p =

11/2

|1 − 2t| pdt

1/p

=

11/2

(2t − 1) pdt

1/p

=

1

(2( p + 1))1/p,

f p =

10

|1/2 − t| pdt

1/p

=

1/20

(1/2 − t) pdt + 11/2

(t − 1/2) pdt1/p

=

21

( p + 1)2 p+11/p

=

1

2( p + 1)1/p= g p .

Hence the parallelogram law is equivalent to 2 · 2−2/p = 1, i.e. to p = 2. So,if p = 2, the parallelogram law fails, i.e. the norm · p is not induced by aninner product.

3. Let {en}n∈N be an orthonormal set in an inner product space H. Provethat

∞n=1

|(x, en)(y, en)| ≤ xy, ∀x, y ∈ H.

Solution

Using the Cauchy–Schwarz inequality for l2 and Bessel’s inequality for H weobtain

∞n=1

|(x, en)(y, en)| ≤

∞n=1

|(x, en)|2

1/2 ∞n=1

|(y, en)|2

1/2

≤ xy.

2



4. Prove that in a complex inner product space H the following equalities

hold:

(x, y) =1

N

N k=1

x + e2πik/N y2e2πik/N for N ≥ 3,

(x, y) =1

2π

2π0

x + eiθy2eiθdθ, ∀x, y ∈ H.

Solution

1

N

N k=1

x + e2πik/N y2e2πik/N =1

N

N k=1

(x + e2πik/N y, x + e2πik/N y)e2πik/N =

1

N

N k=1

x2e2πik/N + (y, x)e4πik/N + (x, y) + y2e2πik/N

=

1

N x2

e2πi(N +1)/N − e2πi/N

e2πi/N − 1+

1

N (y, x)

e4πi(N +1)/N − e4πi/N

e4πi/N − 1+ (x, y) +

1

N

y2e2πi(N +1)/N − e2πi/N

e2πi/N

− 1

= (x, y),

since e2πi/N = 1 and e4πi/N = 1 for N ≥ 3.Similarly

1

2π

2π0

x + eiθy2eiθdθ =1

2π

2π0

(x + eiθy, x + eiθy)eiθdθ =

1

2πx2

2π0

eiθdθ +1

2π(y, x)

2π0

e2iθdθ +1

2π(x, y)

2π0

1dθ +

1

2πy2

2π

0

eiθdθ = (x, y).

5. Show that A⊥⊥ = spanA for any subset of a Hilbert space.

Solution

3



Claim I: For any subset M of a Hilbert space H the orthogonal complement

M ⊥ is a closed linear subspace of H (see Proposition 3.15(i)). Indeed, if z1, z2 ∈ M ⊥, then

(αz1 + βz2, y) = α(z1, y) + β (z2, y) = α0 + β 0 = 0, ∀y ∈ M, ∀α, β ∈ F

and hence, αz1 + βz2 ∈ M ⊥. So, M ⊥ is a linear subspace of H.Suppose z ∈ Cl(M ⊥). Then there exist zn ∈ M ⊥, n ∈ N such that zn → zas n → +∞. Therefore

(z, y) = limn→+∞

(zn, y) = limn→+∞

0 = 0, ∀y ∈ M,

i.e. z ∈ M ⊥. Thus M ⊥ is a closed linear subspace of H.Claim II: A ⊂ A⊥⊥ (see Proposition 3.17(ii)). Indeed, for any x ∈ A we have

(x, y) = (y, x) = 0, ∀y ∈ A⊥,

i.e. x ∈ A⊥⊥.

Since A⊥⊥ = (A⊥)⊥ is a closed linear subspace of H, we obtain

spanA ⊂ A⊥⊥.

Now, take any x ∈ A⊥⊥. Since H = spanA ⊕ (spanA)⊥, we have

x = z + y, z ∈ spanA, y ∈ (spanA)⊥,

and therefore, (x, y) = (z, y) + y2. Since y ∈ A⊥, we obtain 0 = y2, i.e.y = 0, i.e. x = z, i.e. x ∈ spanA. Consequently A⊥⊥ ⊂ spanA. Finally,

A⊥⊥ = spanA.

6. Let M and N be closed subspaces of a Hilbert space. Show that(M + N )⊥ = M ⊥ ∩ N ⊥, (M ∩ N )⊥ = Cl(M ⊥ + N ⊥).

Solution

4



Let z ∈ (M + N )⊥, i.e.

(z, x + y) = 0, ∀x ∈ M, ∀y ∈ N. (1)

Taking x = 0 or y = 0 we obtain

(z, x) = 0, ∀x ∈ M, (z, y) = 0, ∀y ∈ N, (2)

i.e. z ∈ M ⊥ ∩ N ⊥. Hence (M + N )⊥ ⊂ M ⊥ ∩ N ⊥.Suppose now z ∈ M ⊥ ∩ N ⊥, i.e. (2) holds. Then obviously, (1) holds, i.ez ∈ (M + N )⊥. Therefore M ⊥ ∩ N ⊥ ⊂ (M + N )⊥. Thus

(M + N )⊥ = M ⊥ ∩ N ⊥. (3)

Writing (3) for M ⊥ and N ⊥ instead of M and N and using Exercise 5, weobtain

(M ⊥ + N ⊥)⊥ = M ⊥⊥ ∩ N ⊥⊥ = M ∩ N,

since M and N are closed linear subspaces. Taking the orthogonal com-plements of the LHS and the RHS and using Exercise 5 again, we arriveat

Cl(M ⊥ + N ⊥) = (M ∩ N )⊥,

since M ⊥ + N ⊥ is a linear subspace.

7. Show that M := {x = (xn) ∈ l2 : x2n = 0, ∀n ∈ N} is a closed subspaceof l2. Find M ⊥.

Solution

Take any x, y ∈ M . It is clear that for any α, β ∈F

(αx + βy)2n = αx2n + βy2n = 0,

i.e. αx + βy ∈ M . Hence M is a linear subspace of l2. Let us prove that itis closed.Take any x ∈ Cl(M ). There exist x(k) ∈ M such that x(k) → x as k → +∞.

Since x(k)2n = 0, we obtain

x2n = limk→+∞

x(k)2n = 0,

5



i.e. x ∈ M . Hence M is closed.

Further,

z ∈ M ⊥ ⇐⇒ (z, x) = 0, ∀x ∈ M ⇐⇒∞n=0

z2n+1x2n+1 = 0 for all x2n+1 ∈ F, n ∈ N

such that∞n=0

|x2n+1|2 < +∞ ⇐⇒

z2n+1 = 0, ∀n = 0, 1, . . .

Therefore

M ⊥ = {z = (zn) ∈ l2 : z2n+1 = 0, ∀n = 0, 1, . . . }.

8. Show that vectors x1, . . . , xN in an inner product space H are linearlyindependent iff their Gram matrix (a jk)N

j,k=1 = ((xk, x j))N j,k=1 is nonsingular,

i.e. iff the corresponding Gram determinant det((xk, x j)) does not equal zero.

Take an arbitrary x ∈ H and set b j = (x, x j). Show that, whether or not x jare linearly independent, the system of equations

N k=1

a jkck = b j, j = 1, . . . , N ,

is solvable and that for any solution (c1, . . . , cN ) the vectorN

j=1 c jx j is thenearest to x point of lin{x1, . . . , xN }.

Solution

Let c1, . . . , cN ∈ F. It is clear that

N k=1

ckxk = 0 ⇐⇒

N k=1

ckxk, y

= 0, ∀y ∈ lin{x1, . . . , xN } ⇐⇒

N k=1

ckxk, x j

= 0, ∀ j = 1, . . . , N ⇐⇒

N k=1

a jkck = 0, ∀ j = 1, . . . , N .

6



Therefore

the vectors x1, . . . , xN are linearly independent ⇐⇒N k=1

ckxk = 0 iff c1 = · · · = cN = 0 ⇐⇒

the systemN k=1

a jkck = 0, j = 1, . . . , N has only

the trivial solution c1 = · · · = cN = 0 ⇐⇒

det((xk, x j)) = det(a jk) = 0.

For any x ∈ H and b j = (x, x j), j = 1, . . . , N we have

the vectorN k=1

ckxk is the nearest to x point of

lin{x1, . . . , xN } ⇐⇒

x −N

k=1

ckxk ∈ lin{x1, . . . , xN }⊥ ⇐⇒

x −

N k=1

ckxk, x j

= 0, j = 1, . . . , N ⇐⇒

N k=1

a jkck = b j , j = 1, . . . , N .

7



CM414Z/CMMS08 OPERATOR THEORY

Solutions to 2007 Exam

Each question will be marked out of 25. The tentative marks for partsare indicated in square brackets.

1. Let X be a Banach space and B ∈ B(X ), i.e. let B be a boundedlinear operator acting on X . Define the kernel Ker(B) and the rangeRan(B) of B.

Suppose there exists a constant c > 0 such that

Bx ≥ cx, ∀x ∈ X. (1)

Show that Ker(B) = {0} and Ran(B) is closed.

Suppose A ∈ B(X ) is left invertible and C ∈ B(X ) is right invertible,i.e. there exist A−1l , C −1r ∈ B(X ) such that A−1l A = I and CC −1r = I ,where I is the identity operator. Prove that Ran(A) and Ran(C ) areclosed. Show that Ker(A) = {0}. Give an example of a right invertibleC ∈ B(X ) such that Ker(C ) = {0}.

Explain what is meant by saying that P ∈ B(X ) is a projection .

Let A and C be as above. Prove that AA−1l and C −1r C are projections.

Solution

Ker(B) := {x ∈ X | Bx = 0}, Ran(B) := {Bx| x ∈ X }.

Bookwork. [2]

Suppose (1) holds with a positive constant c. If x ∈ Ker(B), then itfollows from (1) that x = 0, i.e. x = 0. Hence Ker(B) = {0}.

Let us now prove that Ran(B) is closed. For any y ∈ Cl(Ran(B)) thereexist xn ∈ X such that Bxn → y as n → ∞. Then

xn − xm ≤ c−1Bxn − Bxm → 0 as n, m → ∞.

Hence (xn) is a Cauchy sequence in the Banach space X . Let us denoteits limit by x. We have

Bx = B

limn→∞

xn

= lim

n→∞Bxn = y.

1



Thus y ∈ Ran(B), i.e. Ran(B) is closed.Bookwork. [7]

Sincex = A−1l Ax ≤ A−1l Ax, ∀x ∈ X,

A satisfies (1) with c = 1/A−1l . Hence Ran(A) is closed.Unseen. [3]

Sincex = CC −1r x ∈ Ran(C ), ∀x ∈ X,

Ran(C ) = X . Hence Ran(C ) is clearly closed.Unseen. [3]

Suppose x ∈ Ker(A). Then

x = A−1

l Ax = A−1

l (Ax) = A−1

l 0 = 0.

So, Ker(A) = {0}.Unseen. [3]

Let X = l2 and C be the left shift operator:

Cx = (x2, x3, . . . ), ∀x = (x1, x2, . . . ) ∈ l2.

It is easy to see that C is right invertible with a right inverse

C −1r x = (0, x1, x2, x3, . . . ),

∀x = (x1, x2, . . . )

∈l2,

andKer(C ) = {(λ, 0, 0, 0, . . . )| λ ∈ C} = {0}.

Unseen in this form. [3]

P is called a projection if P 2 = P .Bookwork. [1]

(AA

−1

l )

2

= AA

−1

l AA

−1

l = AIA

−1

l = AA

−1

l

(C −1r C )2 = C −1r CC −1r C = C −1r IC = C −1r C.

Homework. [3]

2



2. Let B be a bounded linear operator acting in a Banach space X . Forthe operator B define the following notions: norm, eigenvalue, eigenvec-tor, resolvent set, spectrum, resolvent.

Let C ([−π, π]) be the Banach space of all continuous complex valuedfunctions on [

−π, π] equipped with the standard norm

f = maxt∈[−π,π]

|f (t)|.

Consider the operator A : C ([−π, π]) → C ([−π, π]) defined by

(Af )(t) = (sin t)f (t), −π ≤ t ≤ π, ∀f ∈ C ([−π, π]).

Find A.

Find the spectrum σ(A) and eigenvalues of A.

Consider the bounded linear operator V : C ([−π, π]) → C ([−π, π]) de-fined by

(V f )(t) = f (−t), t ∈ [−π, π], f ∈ C ([−π, π]).

Find the spectrum of AV + V A.

Solution

If B is bounded we define its norm by the equality

B

:= inf

{C :

Bx

≤C

x

,

∀x

∈X

}.

(Any of the following equivalent definitions will be accepted as a correct one:

B = inf {C : Bx ≤ C, all x s.t. x ≤ 1}= inf {C : Bx ≤ C, all x s.t. x = 1} = sup

Bxx : x = 0

= sup{Bx : x ≤ 1} = sup{Bx : x = 1}. )

A number λ ∈ C is called an eigenvalue of B if there exists an x ∈ X ,x = 0, called an eigenvector of B, such that Bx = λx. The resolvent set ρ(B) is the set of all λ

∈C for which there exists a bounded lin-

ear operator R(B; λ), called the resolvent , such that R(B; λ)(B − λI ) =(B − λI )R(B; λ) = I , i.e. R(B; λ) = (B − λI )−1. The spectrum σ(B) isdefined as σ(B) = C \ ρ(B), i.e. σ(B) is the set of all λ ∈ C such thatB − λI is not invertible on X .Bookwork. [4]

It is clear that

Af = maxt∈[−π,π]

|(sin t)f (t)| ≤ maxt∈[−π,π]

|f (t)| = f , ∀f ∈ C ([−π, π]).

3



So, A ≤ 1. On the other hand, we have for f ≡ 1

Af = maxt∈[−π,π]

|sin t| = 1 = f .

Hence A ≥ 1. Finally, A = 1.

Unseen. [4]

Let λ ∈ C, λ ∈ [−1, 1] = {sin t| t ∈ [−π, π]}. Then 1/(sin t − λ) ∈C ([−π, π]) and A−λI has an inverse R(A; λ) = (A−λI )−1 ∈ B(C ([−π, π]))defined by

R(A; λ)g(t) = (sin t − λ)−1g(t), t ∈ [−π, π].

Hence σ(A) ⊂ [−1, 1].Unseen but similar to problems from homework and from 2006 paper. [4]

Suppose now λ ∈ [−1, 1], i.e. λ = sin tλ for tλ = arcsin λ ∈ [−π/2, π/2].

Then (A − λI )f (tλ) = (sin tλ − λ)f (tλ) = 0, i.e. Ran(A − λI ) consistsof functions vanishing at tλ. Consequently Ran(A − λI ) = C ([−π, π])and A − λI is not invertible. Therefore [−1, 1] ⊂ σ(A). Finally, σ(A) =[−1, 1].Unseen but similar to problems from homework and from 2006 paper. [4]

Let λ ∈ [−1, 1]. The equation Af = λf , i.e. (sin t − λ)f (t) ≡ 0 is equiv-alent to f (t) = 0, ∀t ∈ [−π, π] \ {tλ, t̃λ}, where t̃λ is the second solutionof sin t = λ, t ∈ [−π, π]. So, f ≡ 0 by continuity. Hence λ cannot be aneigenvalue of A, i.e. A has no eigenvalues.Unseen but similar to problems from homework and from 2006 paper. [4]

Since

(AV + V A)f (t) = (sin t)f (−t) + (sin(−t))f (−t)

= (sin t)f (−t) − (sin t)f (−t) = 0,∀t ∈ [−π, π], ∀f ∈ C ([−π, π]),

AV + V A = 0 and σ(AV + V A) = {0}.Unseen. [5]

4



3. Let X and Y be normed spaces. Explain what is meant by sayingthat T : X → Y is a compact operator.

Let X be a Banach space and K : X → X be a compact operator.Suppose that λ = 0 belongs to the spectrum of K and µ, ζ ∈ C belongto the resolvent set of K .

Show that the operator

A := µ(K − µI )−1 − ζ (K − ζI )−1

is compact. (Here and below, you may use without justification anytextbook material on compact operators.)

Show that (λ − µ)−1 is an eigenvalue of (K − µI )−1.

Show that

α :=λ(µ − ζ )

(λ − µ)(λ − ζ )

is an eigenvalue of finite multiplicity of the operator A if µ = ζ .

Solution

Let X and Y be normed spaces. A linear operator T : X → Y is calledcompact if it maps bounded sets of X into relatively compact sets of Y ,i.e. if for any bounded sequence zn ∈ X , n ∈ N, the sequence (T zn)n∈Nhas a Cauchy subsequence.Bookwork. [2]

Let X be a Banach space and K : X → X be compact.

Suppose that µ, ζ ∈ C belong to the resolvent set of K . Then

µ(K − µI )−1 − ζ (K − ζI )−1

= (K − µI )−1(µ(K − ζI ) − ζ (K − µI ))(K − ζI )−1

= (µ − ζ )(K − µI )−1K (K − ζI )−1

and this operator is compact as a product of the compact operator T andbounded operators (K − µI )−1 and (K − ζI )−1.Unseen. [7]

Since λ = 0 belongs to the spectrum of K and K is compact, λ is an eigen-value K . Let x ∈ X \ {0} be a corresponding eigenvector: Kx = λx.

Then (K − µI )x = (λ − µ)x. Hence (λ − µ)−1

x = (K − µI )−1

x, i.e.(λ − µ)−1 is an eigenvalue of (K − µI )−1.Unseen. [7]

Similarly (λ − ζ )−1x = (K − ζI )−1x. Therefore

Ax = µ(K − µI )−1x − ζ (K − ζI )−1x = µ(λ − µ)−1x − ζ (λ − ζ )−1x

=

µ

λ − µ− ζ

λ − ζ

x =

λ(µ − ζ )

(λ − µ)(λ − ζ )x = αx.

5



Hence α is an eigenvalue of A.Unseen. [7]Since λ = 0, this is a nonzero eigenvalue if µ = ζ . Since A is a compactoperator, α has a finite multiplicity.Unseen. [2]

6



4. Let H be a Hilbert space.

Show that vectors x1, . . . , xN ∈ H are linearly independent if and only if their Gram matrix (a jk)N

j,k=1 = ((xk, x j))N j,k=1 is nonsingular, i.e. if and

only if the corresponding Gram determinant det((xk, x j)) does not equalzero.

Define the orthogonal complement of a set M ⊂ H.

Let H = l2. Show that L := {x = (xn)n∈N ∈ l2| x2k−1 = x2k, k ∈ N} is aclosed subspace of l2. Find the orthogonal complement of L.

Solution

Let c1, . . . , cN ∈ F. It is clear that

N

k=1

ckxk = 0

⇐⇒ N

k=1

ckxk, y = 0,

∀y

∈lin

{x1, . . . , xN

} ⇐⇒N k=1

ckxk, x j

= 0, ∀ j = 1, . . . , N ⇐⇒

N k=1

a jkck = 0, ∀ j = 1, . . . , N .

Therefore

the vectors x1, . . . , xN are linearly independent ⇐⇒N k=1

ckxk = 0 iff c1 = · · · = cN = 0 ⇐⇒

the systemN k=1

a jkck = 0, j = 1, . . . , N has only

the trivial solution c1 = · · · = cN = 0 ⇐⇒det((xk, x j)) = det(a jk) = 0.

Homework. [10]

The orthogonal complement M ⊥ of a set M ⊂ H is the set

M ⊥ :={

z∈ H

: (z, x) = 0,∀

x∈

M }

.

Bookwork. [1]

Take any x, y ∈ L ⊂ l2. It is clear that for any scalars α and β ,

(αx + βy)2k−1 = αx2k−1 + βy2k−1 = αx2k + βy2k = (αx + βy)2k, k ∈ N,

i.e. αx + βy ∈ L. Hence L is a linear subspace of l2. Let us prove thatit is closed.

7



Take any x ∈ Cl(L). There exist x( j) ∈ L such that x( j) → x as j → +∞.

Since x( j)2k−1 = x

( j)2k , we obtain

x2k−1 = lim j→+∞

x( j)2k−1 = lim

j→+∞x( j)2k = x2k, k ∈ N,

i.e. x ∈ L. Hence L is closed.Unseen. [7]

Further,

z ∈ L⊥ ⇐⇒ (z, x) = 0, ∀x ∈ L

⇐⇒∞k=1

(z2k−1 + z2k)x2k = 0 for all x2k ∈ F, k ∈ N

such that∞

k=1

|x2k|2 < +∞

⇐⇒ z2k−1 = −z2k, k ∈ N.

Therefore

M ⊥ = {z = (zn)n∈N ∈ l2| z2k−1 = −z2k, k ∈ N}.

Unseen. [7]

8



5. Let H be a complex Hilbert space.

Let B : H → H be a bounded linear operator. Explain what is meantby saying thati) B is self-adjoint , ii) B is unitary , iii) B is normal .

Show that for any bounded linear operator A :

H → Hthere exist unique

self-adjoint operators B and C such that A = B + iC . Prove that A isnormal if and only if B and C commute: BC = CB.

Define the numerical range Num(A) of A.

Consider the operator A : l2 → l2 defined by

Ax = (−x1,x2

2, −x3

3, . . . , (−1)k

xk

k, . . . ), ∀x = (x1, x2, . . . ) ∈ l2.

Find Num(A).

Solution

The operator B is called(i) self-adjoint if B∗ = B, i.e. if

(Bx,y) = (x,By), ∀x, y ∈ H;

(ii) unitary if B∗B = BB∗ = I , i.e. if B−1 = B∗;(iii) normal if BB∗ = B∗B.Bookwork. [2]

Suppose there exist self-adjoint operators B and C such that A = B +iC .

Then A∗ = B − iC and we have

B =1

2(A + A∗), C =

1

2i(A − A∗). (2)

It is clear that the above operators are self-adjoint and A = B + iC .Homework and 2005 paper. [4]Suppose B and C commute. Then

AA∗ = (B + iC )(B − iC ) = B2 + iCB − iBC + C 2

= B2 + iBC − iCB + C 2 = (B − iC )(B + iC ) = A∗A,

i.e. A is normal.Unseen. [4]Suppose now A is normal. Then (B + iC )(B − iC ) = (B − iC )(B + iC ),i.e.

B2 + iCB − iBC + C 2 = B2 + iBC − iCB + C 2,

i.e. CB − BC = BC − CB , i.e. CB = BC .(An alternative way of proving this is to use (2).)Unseen. [4]

9



Num(A) := {(Ax,x) : x = 1, x ∈ H}.

Bookwork. [1]

Take an arbitrary x

∈l2 such that

x

2 = 1. Then

(Ax,x) =∞k=1

(−1)k

k|xk|2 ≤

∞ j=1

1

2 j|x2 j |2 ≤ 1

2

∞ j=1

|x2 j|2 ≤ 1

2

∞k=1

|xk|2 =1

2,

(Ax,x) =∞k=1

(−1)k

k|xk|2 ≥ −

∞ j=1

1

2 j − 1|x2 j−1|2 ≥ −

∞ j=1

|x2 j−1|2

≥ −∞k=1

|xk|2 = −1.

Hence Num(A)⊂

[−

1, 1

2].

Unseen. [5]

Take an arbitrary µ ∈ [−1, 12

]. Let t = 13

(1 − 2µ). Then 0 ≤ t ≤ 1

and −t + 12

(1 − t) = µ. Let x = (√

t,√

1 − t, 0, 0, . . . ) ∈ l2. Thenx2 = t + 1 − t = 1 and

(Ax,x) = −t +1

2(1 − t) = µ.

Therefore µ ∈ Num(A), ∀µ ∈ [−1, 12

], i.e. [−1, 12

] ⊂ Num(A). Finally,Num(A) = [−1, 1

2].

Unseen. [5]

10



6. Let H be a complex Hilbert space. State the Hilbert-Schmidt theoremfor a compact self-adjoint linear operator on H.

Let T : H → H be a compact self-adjoint linear operator.

Use the Hilbert-Schmidt theorem to prove that there exists a compactlinear operator S :

H → Hsuch that S 2 = T .

Suppose additionally that T is not a finite rank operator. Which of thefollowing statements is correct?(i) There is a unique S : H → H such that S 2 = T .(ii) There are exactly two operators S : H → H such that S 2 = T .(iii) There are infinitely many operators S : H → H such that S 2 = T .Justify your answer.

Suppose H is separable. Explain what it means to say that K : H → His a Hilbert-Schmidt operator .

Give an example of a non-Hilbert-Schmidt operator S : H → H suchthat T = S 2 is Hilbert-Schmidt.

Solution

The Hilbert-Schmidt theorem: Let T : H → H be a compact self-adjointoperator. Then there exists a finite or a countable orthonormal set{en}N

n=1, N ∈ N ∪ {∞}, of eigenvectors of T such that any x ∈ Hhas a unique representation of the form

x =N n=1

cnen + y, y ∈ Ker(T ), cn ∈ C. (3)

One then has

T x =N n=1

λncnen, (4)

where λn = 0 is the eigenvalue of T corresponding to the eigenvector en.Moreover,

σ(T ) \ {0} = {λn}N n=1 ⊂ R, |λ1| ≥ |λ2| ≥ · · ·

andlimn→∞

λn = 0 if N = ∞.

Bookwork. [4]

Consider the operator

Sx :=N n=1

λn cnen

11



(see (3)). It is clear that S 2 = T (see (4)). If N < +∞ then S is obviouslya bounded finite rank operator. Hence S is compact. If N = +∞,consider the bounded finite rank operators

S jx :=

j

n=1 λn cnen, j

∈N.

We have

(S − S j)x2 =

∞

n= j+1

λn cnen

2

=∞

n= j+1

|λn||cn|2

≤ supn≥ j+1

|λn|∞

n= j+1

|cn|2 = |λ j+1|∞

n= j+1

|cn|2 ≤ |λ j+1| x2.

Hence

S − S j ≤ |λ j+1| → 0 as j → +∞,and S is compact as the norm limit of a sequence of compact (finite rank)operators.Unseen. [7]

If T is not a finite rank operator, then N = +∞ in (3), (4). It is clearthat for any choice of “ + ” or “ − ” for each n, the operator

S̃x :=∞n=1

±

λn cnen

satisfies the equality˜S

2

= T . Since the set of such operators is infinite(in fact, has cardinality of the continuum), there are infinitely many op-erators S : H → H such that S 2 = T .Unseen. [6]

Let H be separable and let {en}n∈N be an arbitrary complete orthonormalset in H. K is called a Hilbert-Schmidt operator if

∞n=1

Ken2 < ∞.

Bookwork. [2]

Let

Sx :=N n=1

1√n

(x, en) en.

Then

T = S 2 =N n=1

1

n(x, en) en,

12



and∞n=1

Sen2 =

∞n=1

1

n= ∞,

∞n=1

T en2 =

∞n=1

1

n2< ∞.

So, T is Hilbert-Schmidt and S is not.Unseen. [6]

13

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