12345

UNIVERSITY OF YORK

BA and BSc Examinations 1996 MATHEMATICS Functional Analysis Sample Paper

Time Allowed: 1\frac{1}{2} hours.

Answer not more than \underline{3} questions. Calculators may be used but should be unnecessary.

1. Let H be a complex vector space. What is meant by an inner product on H? Define the norm ∥x∥ of a vector in H. What does it mean to say that H is a Hilbert space?

Consider the vector space ℓ^{2}(ℂ) of square-summable complex sequences, with

⟨x, y⟩ = ∑_{j=1}^{∞} x_{j}\overline{y_{j}}

Show that this sum converges absolutely for all x, y ∈ ℓ^{2}(ℂ) and defines an inner product which makes ℓ^{2}(ℂ) into a Hilbert space. (You may assume the finite-dimensional Cauchy-Schwarz inequality.)

2. Let U be a closed subspace of a Hilbert space H. Define the orthogonal complement U^{⊥} of U and show that U^{⊥} is also a closed subspace of H, with U ∩ U^{⊥} = {0}.

Given that for any x ∈ H there exist vectors u ∈ U and v ∈ U^{⊥} with x = u + v, show that these are the unique vectors in U and U^{⊥} whose sum is x, and show that (U^{⊥})^{⊥} = U.

Define the orthogonal projections P and Q of H onto U and U^{⊥}. Show that P + Q = I, that P and Q are linear, that P^{2} = P, Q^{2} = Q and P Q = QP = 0.

12345

3. Let H be a Hilbert space and let (e_{j}) be a sequence in H. What does it mean to say that (e_{j}) is a complete orthonormal sequence? State the Riesz-Fischer Theorem.

In the Hilbert space L^{2}([−π, π], ℂ), what is the exponential basis? Show that it is orthonormal.

Compute the Fourier coefficients of the function f given by f(t) = t with respect to the exponential basis and use the Riesz-Fischer Theorem to show that

∑_{j=1}^{∞} \frac{1}{j^{2}} = \frac{π^{2}}{6}

4. Let T be a linear operator on a separable Hilbert space H. What does it mean to say that T is a Hilbert-Schmidt operator? Define the Hilbert-Schmidt norm ∥T ∥_{H-S} of T.

Let M be the multiplication operator on ℓ^{2}(ℂ) defined by

(Mx)_{j} = m_{j}x_{j}

where (m_{j}) is a bounded sequence of complex numbers. Show that M is a Hilbert-Schmidt operator if and only if m ∈ ℓ^{2}(ℂ), in which case

∥M∥_{H-S}^{2} = ∥m∥_{2}^{2} = ∑_{j=1}^{∞} |m_{j}|^{2}

Now suppose T is a compact self-adjoint operator on H. Define the terms eigenvalue and eigenvector of T and state without proof the Spectral Theorem for T. Use the Spectral Theorem to show that if (λ_{j}) is the sequence of eigenvalues of T, repeated according to multiplicity, then T is a Hilbert-Schmidt operator if and only if λ ∈ ℓ^{2}(ℝ), in which case

∥T ∥_{H-S}^{2} = ∑_{j=1}^{∞} λ_{j}^{2} = ∥λ∥_{2}^{2}

SOLUTIONS 12345

1. An inner product on H is a map ⟨·, ·⟩ : H × H → ℂ satisfying

(i) For x, y ∈ H, ⟨x, y⟩ = \overline{⟨y, x⟩};

(ii) for x, y, z ∈ H and λ, μ ∈ ℂ,

⟨λx + μy, z⟩ = λ ⟨x, z⟩ + μ ⟨y, z⟩

(iii) for x ∈ H, ⟨x, x⟩ ≥ 0 and ⟨x, x⟩ = 0 if and only if x = 0

The norm is defined by

∥x∥ = \sqrt{⟨x, x⟩}

taking the non-negative square root. H is a Hilbert space if it is complete in this norm.

Write ∥x∥^{2} = ∑_{j=1}^{∞} |x_{j}|^{2}, and similarly ∥y∥^{2}. For any N ∈ ℕ, the finite-dimensional Cauchy-Schwarz inequality implies that

∑_{j=1}^{N} |x_{j}\overline{y_{j}}| ≤ (∑_{j=1}^{N} |x_{j}|^{2})^{1/2} (∑_{j=1}^{N} |y_{j}|^{2})^{1/2} ≤ ∥x∥∥y∥

so the partial sums of ∑ |x_{j}\overline{y_{j}}| are bounded above. Since it is a sum of non-negative terms, it is convergent and hence the original sum is absolutely convergent for any x, y ∈ ℓ^{2}(ℂ).

To show it is an inner product,

(i) ⟨x, y⟩ = ∑_{j=1}^{∞} x_{j}\overline{y_{j}} = ∑_{j=1}^{∞} \overline{y_{j}\overline{x_{j}}} = \overline{∑_{j=1}^{∞} y_{j}\overline{x_{j}}} = \overline{⟨y, x⟩} since conjugation is continuous.

(ii) ⟨λx + μy, z⟩ = ∑_{j=1}^{∞}(λx_{j}+μy_{j})\overline{z_{j}} = λ ∑_{j=1}^{∞} x_{j}\overline{z_{j}}+μ ∑_{j=1}^{∞} y_{j}\overline{z_{j}} = λ ⟨x, z⟩ + μ ⟨y, z⟩

(iii) ⟨x, x⟩ = ∑_{j=1}^{∞} |x_{j}|^{2}; this is a sum of non-negative terms, so ⟨x, x⟩ ≥ 0. If ⟨x, x⟩ = 0 then each term must be zero, so x_{j} = 0 for all j showing that x = 0.

Completeness: let (x^{(n)}) be a Cauchy sequence in ℓ^{2}(ℂ), so given ε > 0 there exists N ∈ ℕ such that if n, m > N then

∑_{j=1}^{∞} |x_{j}^{(n)} − x_{j}^{(m)}|^{2} < ε^{2}  (*)

so in particular for each j,

|x_{j}^{(n)} − x_{j}^{(m)}|^{2} < ε^{2}

This shows that if j is fixed, then (x_{j}^{(n)}) is a Cauchy sequence in ℂ, hence convergent to x_{j} say. From (*), for any M ∈ ℕ,

∑_{j=1}^{M} |x_{j}^{(n)} − x_{j}^{(m)}|^{2} < ε^{2}

Since this is a finite sum, we can let m → ∞, to deduce that

∑_{j=1}^{M} |x_{j}^{(n)} − x_{j}|^{2} ≤ ε^{2}

and now it follows that

∑_{j=1}^{∞} |x_{j}^{(n)} − x_{j}|^{2} ≤ ε^{2}  (**)

so for n > N, x^{(n)} − x ∈ ℓ^{2}(𝕂) and ∥x^{(n)} − x∥ ≤ ε. Thus, x^{(n)} → x in the ℓ^{2}(ℂ) metric.

Finally, x ∈ ℓ^{2}(ℂ) since x = (x−x^{(n)})+x^{(n)}, x^{(n)} ∈ ℓ^{2}(ℂ) by choice, x − x^{(n)} ∈ ℓ^{2}(ℂ) by (**) and ℓ^{2}(ℂ) is a vector space, hence closed under addition.

2. The orthogonal complement of U is defined by

U^{⊥} = {x ∈ U : ⟨x, u⟩ = 0 for all u ∈ U}

Clearly, 0 ∈ U^{⊥} so U^{⊥} ≠ ∅. Suppose x, y ∈ U^{⊥} and λ, μ are scalars. Then for u ∈ U,

⟨λx + μy, u⟩ = λ ⟨x, u⟩ + μ ⟨y, u⟩ = 0

so U^{⊥} is a subspace.

Now let (x_{n}) be a sequence in U^{⊥} converging to x ∈ H. For any u ∈ U, ⟨x_{n}, u⟩ = 0. But x_{n} → x and for fixed u the function ϕ defined by ϕ(v) = ⟨v, u⟩ is continuous (by the Cauchy-Schwarz inequality), so ⟨x_{n}, u⟩ → ⟨x, u⟩, showing that ⟨x, u⟩ = 0. Since u was an arbitrary element of U, we have x ∈ U^{⊥} showing that U^{⊥} is closed.

Clearly 0 ∈ U ∩ U^{⊥}. Suppose x ∈ U ∩ U^{⊥}. Since x ∈ U^{⊥}, it is orthogonal to every element of U. But x ∈ U, so ⟨x, x⟩ = 0 and hence x = 0, showing that U ∩ U^{⊥} = {0}.

Suppose x = u + v and x = u′ + v′ where u, u′ ∈ U and v, v′ ∈ U^{⊥}. Since u + v = u′ + v′, we have u − u′ = v′ − v = y say. But U and U^{⊥} are both subspaces, so u − u′ ∈ U and v′ − v ∈ U^{⊥}. Thus y ∈ U ∩U^{⊥} so y = 0 by the previous part. Thus, u−u′ = v′ −v = 0 so u′ = u and v′ = v showing that u and v are the only elements of U and U^{⊥} which sum to x.

Suppose u ∈ U. Then for all x ∈ U^{⊥}, ⟨x, u⟩ = 0 so x ∈ (U^{⊥})^{⊥} and hence U ⊆ (U^{⊥})^{⊥}.

Now suppose x ∈ (U^{⊥})^{⊥}. Since x ∈ H, we have x = u + v where u ∈ U and v ∈ U^{⊥}. Since x ∈ (U^{⊥})^{⊥}, we must have ⟨u + v, v⟩ = 0, so ⟨u, v⟩ + ∥v∥^{2} = 0. But ⟨u, v⟩ = 0, so v = 0 and x = u. This shows that x ∈ U, so (U^{⊥})^{⊥} ⊆ U.

The orthogonal projection of H onto U is defined by P x = u where u is the unique element of U for which x = u + v for some v ∈ U^{⊥}. Similarly, the orthogonal projection onto U^{⊥} is defined by Qx = w where w is the unique element of U^{⊥} for which x = w + z for some z ∈ (U^{⊥})^{⊥}.

If x ∈ H then x = u + v where u ∈ U and v ∈ U^{⊥}. By definition, P x = u. But since (U^{⊥})^{⊥} = U, we also have x = v + u with v ∈ U^{⊥} and u ∈ (U^{⊥})^{⊥} = U. Again by definition, Qx = v. Thus P x + Qx = u + v = x, or equivalently P + Q = I.

If x, y ∈ H and λ, μ are scalars, then x = P x+Qx and y = P y+Qy, so

λx+μy = λ(P x+Qx)+μ(P y +Qy) = (λP x+μP y)+(λQx+μQy)

Since U and U^{⊥} are both subspaces, λP x + μP y ∈ U and λQx + μQy ∈ U^{⊥}. But the only representation of λx + μy as a sum of elements from U and U^{⊥} is

λx + μy = P (λx + μy) + Q(λx + μy)

so we must have

P (λx + μy) = λP x + μP y; Q(λx + μy) = λQx + μQy

showing that P and Q are linear.

For any x, P x = P x+0. Since P x ∈ U and 0 ∈ U^{⊥}, the uniqueness of the decomposition implies that P (P (x)) = P (x) and Q(P (x)) = 0 so P^{2} = P and QP = 0. Similarly, Qx = 0 + Qx so Q(Q(x)) = Q(x) and P (Q(x)) = 0.

3. (e_{j}) is an orthonormal sequence in U if ⟨e_{j}, e_{k}⟩ = 0 if j ≠ k and ⟨e_{j}, e_{j}⟩ = 1 for all j. It is complete if in addition the only vector x ∈ U for which ⟨x, e_{j}⟩ = 0 for all j is x = 0.

The Riesz-Fischer Theorem states that if (e_{j}) is an orthonormal sequence then the following are equivalent:

(i) (e_{j}) is complete;

(ii) for all x ∈ H, x = ∑_{j=1}^{∞} ⟨x, e_{j}⟩ e_{j};

(iii) for all x ∈ H, ∥x∥^{2} = ∑_{j=1}^{∞} | ⟨x, e_{j}⟩ |^{2};

(iv) the linear span of {e_{j} : j ∈ ℕ} is dense in H.

The exponential basis in L^{2}([−π, π], ℂ) is given by the functions

\frac{1}{\sqrt{2π}} e^{int}

for n ∈ ℤ.

For any n, m ∈ ℤ,

⟨e_{n}, e_{m}⟩ = \frac{1}{2π} ∫_{−π}^{π} e^{int}\overline{e^{imt}}dt = \frac{1}{2π} ∫_{−π}^{π} e^{i(n−m)t}dt

If n ≠ m this gives

⟨e_{n}, e_{m}⟩ = \frac{1}{2πi(n − m)} [(−1)^{n−m} − (−1)^{m−n}] = 0

and if n = m it gives

⟨e_{n}, e_{n}⟩ = \frac{1}{2π} ∫_{−π}^{π} dt = 1

so (e_{n}) is orthonormal.

SOLUTIONS 12345

The nth Fourier coefficient c_{n} of f with respect to this basis is

\sqrt{2π}c_{n} = ∫_{−π}^{π} te^{−int}dt

= [\frac{te^{−int}}{−in} ]_{−π}^{π} − ∫_{−π}^{π} \frac{e^{−int}}{−in} dt

= \frac{2π(−1)^{n}}{−in} − [\frac{e^{−int}}{n^{2}} ]_{−π}^{π}

= \frac{2πi(−1)^{n}}{n}

provided n ≠ 0; c_{0} satisfies

\sqrt{2π}c_{0} = ∫_{−π}^{π} tdt = 0

so c_{0} = 0. By the Riesz-Fischer Theorem,

∫_{−π}^{π} |f(t)|^{2}dt = ∑_{n=−∞}^{∞} |c_{n}|^{2}

But c_{−n} = c_{n} and c_{0} = 0, so

∫_{−π}^{π} t^{2}dt = 2 ∑_{n=1}^{∞} \frac{2π}{n^{2}}

\frac{2π^{3}}{3} = 4π ∑_{n=1}^{∞} \frac{1}{n^{2}}

∑_{n=1}^{∞} \frac{1}{n^{2}} = \frac{π^{2}}{6}

4. T is a Hilbert-Schmidt operator if there exists a complete orthonor-

mal sequence (e_{j}) such that

∑_{j=1}^{∞} ∥T e_{j}∥^{2} < ∞

The Hilbert-Schmidt norm is defined to be

∥T ∥_{H-S}^{2} = ∑_{j=1}^{∞} ∥T e_{j}∥^{2}

which can be shown to be independent of the choice of (e_{j}).

Let e^{(n)} be the nth standard basis sequence in ℓ^{2}(ℂ), i.e. e^{(n)} is 1 in the nth term and zero elsewhere. This is a complete orthonormal sequence in ℓ^{2}(ℂ), so

∥M∥_{H-S}^{2} = ∑_{n=1}^{∞} ∥Me^{(n)}∥^{2}

but ∥Me^{(n)}∥ = ∥m_{n}e^{(n)}∥ = |m_{n}|, so

∥M∥_{H-S}^{2} = ∑_{n=1}^{∞} |m_{n}|^{2} = ∥m∥_{2}^{2}

λ ∈ ℂ is an eigenvalue of T if T u = λu for some non-zero u ∈ H; such a vector u is an eigenvector of T.

The Spectral Theorem for compact self-adjoint operators states that there is an complete orthonormal sequence (e_{j}) in H consisting of eigenvectors of T and a sequence (λ_{j}) of corresponding eigenvalues such that for all x,

T x = ∑_{j=1}^{∞} λ_{j} ⟨x, e_{j}⟩ e_{j}.

Each eigenvalue of T occurs exactly m times, where m is its multiplicity.

Using (e_{j}) as the complete orthonormal sequence in the definition of the Hilbert-Schmidt norm,

∥T ∥_{H-S}^{2} = ∑_{j=1}^{∞} ∥T e_{j}∥^{2} = ∑_{j=1}^{∞} ∥λ_{j}e_{j}∥^{2} = ∑_{j=1}^{∞} |λ_{j}|^{2} = ∑_{j=1}^{∞} λ_{j}^{2} = ∥λ∥_{2}^{2}

using the fact that λ_{j} is real, so T is Hilbert-Schmidt if and only if λ ∈ ℓ^{2}(ℝ) in which case ∥T ∥_{H-S} = ∥λ∥_{2}.