Each worksheet prints cleanly on A4 or Letter. Answer keys are on a separate page.

The Lumenary · First-year graduate

Functional analysis

Name: ______________________

Date: ______________________

You will learn to

Work with Banach and Hilbert spaces and bounded operators, and recognise that linear maps between normed spaces need not be continuous
Prove the projection theorem and the Riesz representation theorem for Hilbert spaces, and use orthonormal bases
State and apply the Hahn–Banach, uniform boundedness, open mapping and closed graph theorems, and prove uniform boundedness from the Baire category theorem
Prove that the closed unit ball of an infinite-dimensional Hilbert space is not norm-compact, and describe the weak topologies that replace compactness
Prove the spectral theorem for compact self-adjoint operators

1.
Which of these normed spaces are complete (Banach spaces)? Select all that apply.
- A. $C[0, 1]$ with $\|f\|_\infty = \max |f|$
- B. $C[0, 1]$ with $\|f\|_1 = \int_0^1 |f|$
- C. $\ell^2$ , the square-summable sequences
- D.The polynomials on $[0, 1]$ with the sup norm
2.
The diagonal operator $T$ on $\ell^2$ is defined by $(Tx)_n = \dfrac{3n}{n+1}\, x_n$ for $n \ge 1$ . Find $\|T\|$ .
3.
On $c_{00}$ (sequences with finitely many non-zero terms) with the sup norm, let $T_n(x) = n x_n$ . Each $x$ has $\sup_n |T_n x| < \infty$ , yet $\|T_n\| = n$ . Why does this not contradict the uniform boundedness principle?
- A. $c_{00}$ is not complete, so the Baire category theorem cannot be applied.
- B.The $T_n$ are not bounded operators.
- C.The family $(T_n)$ is not pointwise bounded.
- D.The uniform boundedness principle only holds in Hilbert spaces.
4.
On $L^2(0, 1)$ let $\ell(f) = \int_0^{1/2} f(x)\, dx$ . By the Riesz representation theorem $\ell(f) = \langle f, g \rangle$ for a unique $g$ . What is $\|\ell\|$ ?
- A. $\tfrac12$
- B. $\tfrac{1}{\sqrt 2}$
- C. $1$
- D. $\ell$ is unbounded.
5.
Which operator on $\ell^2$ is compact?
- A.The diagonal operator $(x_n) \mapsto (x_n / n)$
- B.The identity
- C.The right shift $(x_1, x_2, \ldots) \mapsto (0, x_1, x_2, \ldots)$
- D.The diagonal operator $(x_n) \mapsto ((-1)^n x_n)$
6.
In the proof that a compact self-adjoint $T \ne 0$ has an eigenvalue $\lambda = \pm\|T\|$ , we have unit vectors with $\|Tx_n - \lambda x_n\| \to 0$ . Where is compactness used?
- A.To extract a subsequence along which $Tx_{n_k}$ converges; then $x_{n_k} \to y/\lambda$ is an actual eigenvector.
- B.To show that $\|T\| = \sup_{\|x\| = 1} |\langle Tx, x \rangle|$ .
- C.To show that $\langle Tx, x \rangle$ is real.
- D.To show that $\|Tx_n - \lambda x_n\|^2 \le 2\lambda^2 - 2\lambda \langle Tx_n, x_n \rangle$ .
7.
Which Lean statement correctly formalises "if a real inner product space $E$ contains an orthonormal sequence, its closed unit ball is not compact"?
- A.theorem t {e : ℕ → E} (he : Orthonormal ℝ e) : ¬ IsCompact (Metric.closedBall (0 : E) 1)
- B.theorem t : ¬ IsCompact (Metric.closedBall (0 : E) 1)
- C.theorem t {e : ℕ → E} (he : Orthonormal ℝ e) : ¬ IsClosed (Metric.closedBall (0 : E) 1)
- D.theorem t {e : ℕ → E} (he : Orthonormal ℝ e) : ¬ IsCompact (Metric.ball (0 : E) 1)
8.
Apply Gram–Schmidt to $1, x, x^2, \dots$ in $L^2(-1, 1)$ , keeping the polynomials monic (leading coefficient $1$ ) instead of normalising. A polynomial is a list of Fraction coefficients, constant term first. The helper inner(p, q) computes $\int_{-1}^1 p q \, dx$ exactly. Write monic_orthogonal(n) returning the first n monic orthogonal polynomials (the monic Legendre polynomials).
Computational — Practice

Answer key — Functional analysis

1.
(a) and (c) are Banach spaces. (b) fails: $f_n$ rising linearly from $0$ to $1$ on $[\tfrac12, \tfrac12 + \tfrac1n]$ is $L^1$ -Cauchy, with $L^1$ -limit $\mathbf 1_{(1/2, 1]}$ , which agrees a.e. with no continuous function. (d) fails: the partial sums of $\sum x^k/k!$ are uniformly Cauchy on $[0, 1]$ but converge to $e^x$ . Completeness is exactly what the Baire category theorem, and hence uniform boundedness and open mapping, need.
2.
$\|Tx\|^2 = \sum_n d_n^2 |x_n|^2 \le 3^2 \|x\|^2$ since $d_n = \frac{3n}{n+1} < 3$ , so $\|T\| \le 3$ ; and $\|T e_n\| = d_n \to 3$ , so $\|T\| = 3$ . The supremum is not attained: no unit vector has $\|Tx\| = 3$ . In infinite dimensions the operator norm need not be a maximum. ( $T$ is not compact, since $d_n \not\to 0$ .)
3.
The principle requires the domain to be a Banach space. $c_{00}$ is dense in $c_0$ and not closed, so it is not complete. In the proof, $X = \bigcup F_n$ with closed $F_n$ , and Baire gives some $F_N$ with interior; here every $F_n$ is nowhere dense in $c_{00}$ , and indeed $c_{00} = \bigcup_n \operatorname{span}(e_1, \dots, e_n)$ is a countable union of nowhere dense closed subspaces.
4.
$\ell(f) = \langle f, g \rangle$ with $g = \mathbf 1_{[0, 1/2]}$ (real-valued, so no conjugation issue). By the theorem, $\|\ell\| = \|g\|_2 = \left(\int_0^{1/2} 1 \, dx\right)^{1/2} = \frac{1}{\sqrt 2}$ . The supremum is attained at $f = \sqrt 2\, \mathbf 1_{[0, 1/2]}$ , where $\ell(f) = \sqrt 2 / 2$ .
5.
Only (a). A compact operator maps the bounded sequence $(e_n)$ to a sequence with a convergent subsequence; since $e_n \rightharpoonup 0$ , compact operators in fact send it to $Te_n \to 0$ in norm. For a diagonal operator, $Te_n = d_n e_n$ , so compactness forces $d_n \to 0$ , and conversely $d_n \to 0$ makes $T$ a norm limit of finite-rank operators, hence compact. The identity, the shift and the sign-flip all preserve $\|e_n\| = 1$ .
6.
Compactness gives a subsequence with $Tx_{n_k} \to y$ ; combined with $Tx_{n_k} - \lambda x_{n_k} \to 0$ this forces $x_{n_k} \to y/\lambda$ , a unit vector, which is then an eigenvector by continuity. Every other step works for any bounded self-adjoint operator. Multiplication by $x$ on $L^2(0,1)$ shows the step is essential: $f_n = \sqrt n\, \mathbf 1_{[1 - 1/n, 1]}$ satisfies $\|Mf_n - f_n\| \to 0$ , but $1$ is not an eigenvalue.
7.
Option (a). Option (b) is false for $\R^n$ ; (c) is false because closed balls are closed; (d) is true but trivial in a different way: open balls are never compact in a non-trivial normed space. The Lean proof of (a) follows the lesson: a compact set would give a convergent, hence Cauchy, subsequence of $(e_n)$ , but distinct terms are $\sqrt 2$ apart.

from fractions import Fraction

def inner(p, q):
    total = Fraction(0)
    for i, a in enumerate(p):
        for j, b in enumerate(q):
            if (i + j) % 2 == 0:
                total += a * b * Fraction(2, i + j + 1)
    return total

def monic_orthogonal(n):
    result = []
    for k in range(n):
        p = [Fraction(0)] * k + [Fraction(1)]
        for q in result:
            c = inner(p, q) / inner(q, q)
            p = [a - c * (q[i] if i < len(q) else 0) for i, a in enumerate(p)]
        result.append(p)
    return result

The projections of $x^k$ onto the earlier $P_j$ are subtracted, so $P_k$ is orthogonal to $P_0, \dots, P_{k-1}$ and still monic. One gets $1, x, x^2 - \tfrac13, x^3 - \tfrac35 x, \dots$ , proportional to the Legendre polynomials. Normalised, they form an orthonormal basis of $L^2(-1, 1)$ , because polynomials are dense (Weierstrass).