---
title: "Compact Operators and the Spectral Theorem"
book: "University Mathematics — Year 3"
subject: math
language: en
chapter: 15
exercises: 12
source: https://one-course.com/books/math/5/en/chapter/15-compact-operators-and-the-spectral-theorem
---

# Chapter 15 — Compact Operators and the Spectral Theorem

Diagonalization is finite-dimensional linear algebra’s crown jewel: a symmetric matrix has an orthonormal basis of eigenvectors. In infinite dimension this fails for bounded [self-adjoint operators](#def-b3-spectral-selfadjoint) in general — multiplication by $x$ on $L^2(\intcc01)$ has *no* eigenvalues at all ([Exercise 15.6](#exo-b3-spectral-6)) — but it survives, in [perfect](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#prop-b3-galois-perfect) form, for the operators that are *almost finite dimensional*: the [compact](#def-b3-spectral-compact) ones. The spectral theorem for [compact](#def-b3-spectral-compact) [self-adjoint operators](#def-b3-spectral-selfadjoint) is the single most used theorem of applied functional analysis: it diagonalizes integral equations, drives the [Fredholm alternative](#thm-b3-spectral-fredholm), and (weekend problem) solves the vibrating string, producing the sine basis of Fourier analysis from pure operator theory — with Euler’s $\zeta(2) = \frac{\pi^2}6$ falling out of a trace formula as a parting gift. Throughout, $H$ is a [Hilbert space](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#def-b3-hilbert-inner) over $\C$ (or $\R$; statements adapt), and operators are bounded.

## 15.1 Compact operators

**Definition 15.1.**

$T \in \mathcal L(E, F)$ ($E, F$ Banach) is *compact* if the image $T(B)$ of the unit ball is relatively compact in $F$ — equivalently, every bounded sequence $(x_n)$ has a subsequence with $(Tx_{n_k})$ convergent. Finite-rank operators are compact (bounded sets in finite dimension); the identity of an infinite-dimensional space never is (Riesz’s theorem, Year 2).

**Proposition 15.2.**

The [compact operators](#def-b3-spectral-compact) $\mathcal K(E, F)$ form a closed subspace of $\mathcal L(E, F)$, and a two-sided [ideal](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-ideal): $S$ [compact](#def-b3-spectral-compact) $\Rightarrow$ $AS$ and $SB$ [compact](#def-b3-spectral-compact) for bounded $A, B$. Moreover, in a [Hilbert space](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#def-b3-hilbert-inner), every [compact operator](#def-b3-spectral-compact) is a norm-limit of finite-rank operators.

**Proof.** Subspace: clear from the sequence characterization. [Ideal](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-ideal): bounded maps send convergent sequences to convergent ones and bounded to bounded. Closedness: let $T_n \to T$ with $T_n$ [compact](#def-b3-spectral-compact), and $(x_k)$ bounded by $1$; a diagonal extraction makes $(T_nx_{k_j})_j$ convergent for every $n$; then $(Tx_{k_j})$ is Cauchy, since

$$
\norm{Tx_{k_j} - Tx_{k_l}} \leq 2\vertiii{T - T_n} +
\norm{T_nx_{k_j} - T_nx_{k_l}} ,
$$

choosing first $n$ then the indices. Approximation in [Hilbert spaces](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#def-b3-hilbert-inner): let $T$ be [compact](#def-b3-spectral-compact), $K = \overline{T(B)}$ [compact](#def-b3-spectral-compact); given $\varepsilon$, cover $K$ by finitely many balls $B(y_i, \varepsilon)$ and let $P$ be the orthogonal projection onto $V = \operatorname{Vect}(y_1, \dots, y_m)$ (closed: finite-dimensional). Then $PT$ has finite rank, and for $\norm x \leq 1$: picking $y_i$ with $\norm{Tx - y_i} <
\varepsilon$,

$$
\norm{Tx - PTx} \leq \norm{Tx - y_i} + \norm{P(y_i - Tx)}
\leq 2\varepsilon
$$

($y_i = Py_i$; $\vertiii P \leq 1$): $\vertiii{T - PT} \leq
2\varepsilon$. ∎

**Example 15.3.**

(a) Diagonal operators on $\ell^2$: $T(x_n) = (\lambda_nx_n)$ is [compact](#def-b3-spectral-compact) iff $\lambda_n \to 0$ ([Exercise 15.2](#exo-b3-spectral-2)). (b) Kernel operators on $\mathcal C(\intcc01)$: [compact](#def-b3-spectral-compact) by Ascoli ([Exercise 7.7](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#exo-b3-complete-7)). (c) *Hilbert–Schmidt operators*: for $k \in L^2(\intcc01^2)$,

$$
(T_kf)(x) = \int_0^1k(x, y)\,f(y)\,\dd y
$$

defines a [compact operator](#def-b3-spectral-compact) on $L^2(\intcc01)$ with $\vertiii{T_k} \leq \norm k_{L^2}$ ([Exercise 15.4](#exo-b3-spectral-4): truncating the basis expansion of $k$ exhibits $T_k$ as a limit of finite-rank operators).

## 15.2 Self-adjoint operators

**Definition 15.4.**

$T \in \mathcal L(H)$ is *self-adjoint* if $T = T^*$ ([Exercise 13.8](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#exo-b3-hilbert-8)), i.e. $\langle Tx, y\rangle = \langle x, Ty\rangle$ for all $x, y$. Then $\langle x, Tx\rangle \in \R$ for every $x$ (equal to its conjugate).

**Proposition 15.5.**

For [self-adjoint](#def-b3-spectral-selfadjoint) $T$:

$$
\vertiii T = \sup_{\norm x \leq 1}\ \abs{\langle x,
Tx\rangle} .
$$

Eigenvalues of $T$ are real, and eigenvectors for distinct eigenvalues are orthogonal.

**Proof.** Let $M$ be the supremum; $M \leq \vertiii T$ by Cauchy–Schwarz. Conversely, the polarization-type identity

$$
\langle x{+}y, T(x{+}y)\rangle - \langle x{-}y,
T(x{-}y)\rangle = 4\operatorname{Re}\langle y, Tx\rangle
$$

(expand; the cross terms $\langle y, Tx\rangle + \langle x,
Ty\rangle = 2\operatorname{Re}\langle y, Tx\rangle$ by [self-adjointness](#def-b3-spectral-selfadjoint)) gives, with the parallelogram law,

$$
4\operatorname{Re}\langle y, Tx\rangle \leq
M\bigl(\norm{x{+}y}^2 + \norm{x{-}y}^2\bigr) = 2M\bigl(\norm
x^2 + \norm y^2\bigr).
$$

For $\norm x = 1$ with $Tx \neq 0$, take $y = Tx/\norm{Tx}$: $4\norm{Tx} \leq 4M$. So $\vertiii T \leq M$. Eigenvalues: $Tx = \lambda x$, $x \ne 0$ gives $\lambda\norm x^2 = \langle
x, Tx\rangle \in \R$. Orthogonality: $\lambda\langle x,
y\rangle = \langle Tx, y\rangle = \langle x, Ty\rangle =
\mu\langle x, y\rangle$ with $\lambda \neq \mu$ real. ∎

## 15.3 The spectral theorem

**Lemma 15.6 (Existence of an extreme eigenvalue).**

Let $T \neq 0$ be [compact](#def-b3-spectral-compact) and [self-adjoint](#def-b3-spectral-selfadjoint). Then $\vertiii T$ or $-\vertiii T$ is an eigenvalue of $T$.

**Proof.** By [Proposition 15.5](#prop-b3-spectral-sanorm), pick unit vectors $x_n$ with $\langle x_n, Tx_n\rangle \to \mu$, where $\abs\mu =
\vertiii T > 0$ (pass to a subsequence to fix the sign). Then

$$
\norm{Tx_n - \mu x_n}^2
= \norm{Tx_n}^2 - 2\mu\langle x_n, Tx_n\rangle + \mu^2
\leq \vertiii T^2 - 2\mu\langle x_n, Tx_n\rangle + \mu^2
\longrightarrow 2\mu^2 - 2\mu\cdot\mu = 0 .
$$

By compactness, a subsequence $Tx_{n_k} \to y$; then $\mu
x_{n_k} = Tx_{n_k} - (Tx_{n_k} - \mu x_{n_k}) \to y$, so $x_{n_k} \to x = y/\mu$, a unit vector, and [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) gives $Tx = \mu x$. ∎

**Theorem 15.7 (Spectral theorem for compact self-adjoint operators).**

Let $T$ be a [compact](#def-b3-spectral-compact) [self-adjoint operator](#def-b3-spectral-selfadjoint) on a [Hilbert space](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#def-b3-hilbert-inner) $H$.

1. $H$ admits an orthonormal system $(e_n)_{n \in N}$ ($N$ finite or countable) of eigenvectors of $T$, with real nonzero eigenvalues $(\lambda_n)$, such that $$Tx = \sum_{n\in N}\lambda_n\,\langle e_n, x\rangle\,  e_n \qquad (x \in H),$$ and $H = \ker T \,\oplus^\perp\,  \overline{\operatorname{Vect}}(e_n : n \in N)$.
2. If $N$ is infinite, $\lambda_n \to 0$ ; for each $\delta > 0$ only finitely many $n$ have $\abs{\lambda_n} \geq \delta$ , and each eigenspace $\ker(T - \lambda)$ , $\lambda \neq 0$ , is finite-dimensional.
3. Completing $(e_n)$ by an orthonormal basis of $\ker T$ yields, when $H$ is separable, an orthonormal basis of $H$ made of eigenvectors: $T$ is diagonalized.

**Proof.** (2) first. If infinitely many orthonormal eigenvectors $x_k$ had $\abs{\lambda_{(k)}} \geq \delta$: $\norm{Tx_k - Tx_l}^2 =
\lambda_{(k)}^2 + \lambda_{(l)}^2 \geq 2\delta^2$ (orthogonality, Pythagoras): no convergent subsequence of $(Tx_k)$, contradicting compactness of $T$ on the bounded $(x_k)$. This bounds by a finite number, for each $\delta$, the total multiplicity of eigenvalues outside $\intoo{-\delta}\delta$; countability and $\lambda_n \to 0$ follow.

(1) Let $H_0$ be the closed span of *all* eigenvectors with nonzero eigenvalues, organized (by (2) and Gram–Schmidt within each finite-dimensional eigenspace, orthogonality across eigenspaces from [Proposition 15.5](#prop-b3-spectral-sanorm)) into an orthonormal system $(e_n)$ with eigenvalues $\lambda_n \neq 0$. $T$ maps $H_0$ into $H_0$, and also $H_0^\perp$ into $H_0^\perp$: for $y
\perp H_0$ and $e$ an eigenvector, $\langle e, Ty\rangle =
\langle Te, y\rangle = \lambda\langle e, y\rangle = 0$. The restriction $T' = T\restriction_{H_0^\perp}$ is [compact](#def-b3-spectral-compact) [self-adjoint](#def-b3-spectral-selfadjoint) on the [Hilbert space](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#def-b3-hilbert-inner) $H_0^\perp$; if $T' \neq
0$, [Lemma 15.6](#lem-b3-spectral-existence) produces an eigenvector of $T$ with nonzero eigenvalue inside $H_0^\perp$ — impossible, such vectors live in $H_0$. So $T' = 0$: $H_0^\perp \subseteq \ker T$. Conversely $\ker T \perp$ every $e_n$ ($\langle e_n, z\rangle = \frac1{\lambda_n}\langle
Te_n, z\rangle = \frac1{\lambda_n}\langle e_n, Tz\rangle =
0$): $\ker T \subseteq H_0^\perp$, whence $\ker T =
H_0^\perp$ and the orthogonal decomposition. The expansion: for $x = z + \sum_nc_ne_n$ ($z \in \ker T$, $c_n = \langle
e_n, x\rangle$; [Theorem 13.7](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#thm-b3-hilbert-parseval)(1) on $H_0$), [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) of $T$ gives $Tx = \sum_nc_n\lambda_ne_n$.

(3) $\ker T$, a closed subspace of a separable space, is separable: it has an orthonormal basis ([Proposition 13.8](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#prop-b3-hilbert-gramschmidt)); the union is an orthonormal basis of $H$ by the decomposition in (1). ∎

**Theorem 15.8 (Fredholm alternative).**

Let $T$ be [compact](#def-b3-spectral-compact) [self-adjoint](#def-b3-spectral-selfadjoint) and $\lambda \in \R\setminus
\{0\}$.

1. If $\lambda$ is not an eigenvalue, then $T - \lambda I$ is bijective with bounded inverse: for every $f$ , the equation $Tx - \lambda x = f$ has exactly one solution, depending [continuously](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) on $f$ .
2. If $\lambda$ is an eigenvalue, $Tx - \lambda x = f$ is [solvable](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived) iff $f \perp \ker(T - \lambda I)$ , and the solution is unique up to that (finite-dimensional) kernel.

**Proof.** Decompose $x = z + \sum c_ne_n$ and $f = w + \sum d_ne_n$ along [Theorem 15.7](#thm-b3-spectral-spectral) ($z, w \in \ker T$). The equation reads

$$
-\lambda z = w, \qquad (\lambda_n - \lambda)\,c_n = d_n\
(n \in N).
$$

(1) $\lambda \notin \{\lambda_n\}\cup\{0\}$: by (2) of the spectral theorem, $\inf_n\abs{\lambda_n - \lambda} = \delta >
0$ (eigenvalues accumulate only at $0 \neq \lambda$). Solve: $z = -w/\lambda$, $c_n = d_n/(\lambda_n - \lambda)$, with $\sum\abs{c_n}^2 \leq \delta^{-2}\sum\abs{d_n}^2$: a unique solution with $\norm x \leq C\norm f$. (2) $\lambda = \lambda_{n}$ for $n$ in a finite set $F$: [solvability](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived) of $(\lambda_n - \lambda)c_n = d_n$ for $n \in F$ requires $d_n = 0$, i.e. $f \perp e_n$ ($n \in F$), i.e. $f
\perp \ker(T - \lambda I)$; the $c_n$, $n \in F$, are then free. ∎

**Example 15.9.**

On $L^2(\intcc01)$, let $Tf(x) = \int_0^1\min(x, y)f(y)\dd y$: a [Hilbert–Schmidt operator](#ex-b3-spectral-examples) with real symmetric kernel: [compact](#def-b3-spectral-compact) and [self-adjoint](#def-b3-spectral-selfadjoint). Solving $Tf = \lambda f$: the relation $\bigl(Tf\bigr)(x) = \int_0^xyf(y)\dd y +
x\int_x^1f(y)\dd y$ shows $u = Tf$ satisfies $u'' = -f$ (two differentiations, legitimate for [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $f$, and $Tf$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) for $f \in L^2$: dominated convergence), with $u(0)
= 0$ and $u'(1) = 0$. So eigenfunctions solve $\lambda u'' =
-u$, $u(0) = 0$, $u'(1) = 0$:

$$
u_n(x) = \sin\Bigl(\bigl(n + \tfrac12\bigr)\pi x\Bigr),
\qquad
\lambda_n = \frac{1}{\bigl(n + \frac12\bigr)^2\pi^2}
\quad (n \geq 0),
$$

and the spectral theorem asserts — with no Fourier theory — that these sines form an orthonormal basis of $L^2(\intcc01)$ after normalization (the kernel of $T$ is $0$: $Tf = 0$ forces, by the two differentiations, $f = 0$ a.e.). The weekend problem runs the same circle of ideas for the vibrating string and extracts $\zeta(2)$ from the trace.

**Method 15.10.**

Given an integral or differential equation: (1) recast it as $(I - \lambda K)u = f$ or $Ku = \lambda u$ with $K$ an integral operator; (2) verify $K$ [compact](#def-b3-spectral-compact) (Hilbert–Schmidt kernel, or Ascoli) and, if possible, [self-adjoint](#def-b3-spectral-selfadjoint) (symmetric real kernel); (3) diagonalize with the spectral theorem or invoke the [Fredholm alternative](#thm-b3-spectral-fredholm) for [solvability](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived); (4) read off existence, uniqueness, stability, and series formulas for solutions in the eigenbasis. Differential operators are unbounded, but their *inverses* (Green operators) are [compact](#def-b3-spectral-compact): always invert first.

## 15.4 Exercises

**Exercise 15.1 ★.**

(a) Show that a bounded operator with finite-dimensional range is [compact](#def-b3-spectral-compact). (b) Show that the identity of a normed space is [compact](#def-b3-spectral-compact) iff the dimension is finite (Riesz, Year 2). Deduce that a [compact operator](#def-b3-spectral-compact) on an infinite-dimensional space is never invertible with bounded inverse.

**Solution of Exercise 15.1.**

(a) $T(B)$ is a bounded subset of the finite-dimensional $\operatorname{im}T$: relatively [compact](#def-b3-spectral-compact) by Heine–Borel ([Corollary 6.17](https://one-course.com/books/math/5/en/chapter/6-general-topology#cor-b3-topology-heineborel), transported by a linear [homeomorphism](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) with $\R^n$). (b) $I$ [compact](#def-b3-spectral-compact) means the closed unit ball is [compact](#def-b3-spectral-compact), which by Riesz’s theorem (Year 2) happens exactly in finite dimension. If a [compact](#def-b3-spectral-compact) $T$ had bounded inverse $T^{-1}$, then $I = T^{-1}T$ would be [compact](#def-b3-spectral-compact) ([Proposition 15.2](#prop-b3-spectral-ideal)): impossible in infinite dimension.

**Exercise 15.2 ★.**

Let $T(x_1, x_2, \dots) = (\lambda_1x_1, \lambda_2x_2, \dots)$ on $\ell^2$, with $(\lambda_n)$ bounded. (a) Show $\vertiii T = \sup\abs{\lambda_n}$. (b) Show that $T$ is [compact](#def-b3-spectral-compact) iff $\lambda_n \to 0$. *(For $\Leftarrow$, truncate; for $\Rightarrow$, test on $(e_n)$.)* (c) When is $T$ [self-adjoint](#def-b3-spectral-selfadjoint)? Verify the spectral theorem by inspection in that case.

**Solution of Exercise 15.2.**

(a) $\norm{Tx}^2 = \sum\abs{\lambda_n}^2\abs{x_n}^2 \leq
\sup\abs{\lambda_n}^2\norm x^2$, with near-equality on the $e_n$ realizing the sup. (b) ($\Leftarrow$) The truncations $T_N$ (keep $n \leq N$, zero beyond) have finite rank and $\vertiii{T - T_N} =
\sup_{n>N}\abs{\lambda_n} \to 0$: [compact](#def-b3-spectral-compact) by [Proposition 15.2](#prop-b3-spectral-ideal). ($\Rightarrow$) If $\abs{\lambda_{n_k}} \geq \delta > 0$ along a subsequence: $\norm{Te_{n_k} - Te_{n_l}}^2 = \abs{\lambda_{n_k}}^2 +
\abs{\lambda_{n_l}}^2 \geq 2\delta^2$: no convergent subsequence of $(Te_{n_k})$. (c) $T^* =$ diagonal with $(\bar\lambda_n)$: [self-adjoint](#def-b3-spectral-selfadjoint) iff all $\lambda_n \in \R$. Then the standard basis $(e_n)$ is an orthonormal basis of eigenvectors, eigenvalues $\lambda_n \to
0$: the spectral theorem verbatim.

**Exercise 15.3 ★★.**

Give the details of the [ideal](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-ideal) property ([Proposition 15.2](#prop-b3-spectral-ideal)): if $S$ is [compact](#def-b3-spectral-compact) and $A, B$ bounded, then $ASB$ is [compact](#def-b3-spectral-compact). Deduce that if $ST = TS = I$ for some bounded $S$, and $\dim H = \infty$, then $T$ is not [compact](#def-b3-spectral-compact) — and reconcile with [Exercise 15.1](#exo-b3-spectral-1)(b).

**Solution of Exercise 15.3.**

Let $(x_n)$ be bounded. Then $(Bx_n)$ is bounded ($\vertiii B < \infty$); compactness of $S$ extracts $SBx_{n_k} \to y$; [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) of $A$ gives $ASBx_{n_k} \to
Ay$: $ASB$ is [compact](#def-b3-spectral-compact). If $ST = TS = I$ with $T$ [compact](#def-b3-spectral-compact) and $\dim H = \infty$: $I = ST$ would be [compact](#def-b3-spectral-compact), contradicting [Exercise 15.1](#exo-b3-spectral-1)(b) — which is the same statement seen from the other side.

**Exercise 15.4 ★★.**

(Hilbert–Schmidt) Let $k \in L^2(\intcc01^2)$ and $(e_n)$ a [Hilbert basis](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#def-b3-hilbert-onb) of $L^2(\intcc01)$. (a) Show that $\vertiii{T_k} \leq \norm k_{L^2}$ *(Cauchy–Schwarz in the $y$-variable, then Tonelli)*. (b) Expand $k(x,y) = \sum_{m,n}c_{mn}e_m(x)\overline{e_n(y)}$ in $L^2$ of the square (justify that the products form a [Hilbert basis](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#def-b3-hilbert-onb) there), and show that truncating the sum gives finite-rank operators converging to $T_k$ in [operator norm](https://one-course.com/books/math/5/en/chapter/8-banach-spaces-and-the-fundamental-theorems#def-b3-banach-operator): $T_k$ is [compact](#def-b3-spectral-compact).

**Solution of Exercise 15.4.**

(a) By Cauchy–Schwarz in $y$: $\abs{T_kf(x)}^2 \leq
\bigl(\int\abs{k(x,y)}^2\dd y\bigr)\norm f_2^2$; integrate in $x$ (Tonelli): $\norm{T_kf}_2 \leq \norm k_{L^2(\square)}
\norm f_2$.

(b) The family $e_{mn}(x,y) = e_m(x)\overline{e_n(y)}$ is orthonormal in $L^2(\intcc01^2)$ (Tonelli separates the double integral). Total: if $h \perp$ all $e_{mn}$, then for each $m$, the function $y \mapsto \int h(x,y)\overline{e_m(x)}\dd
x$ (in $L^2$ by Cauchy–Schwarz and Tonelli) is orthogonal to every $\overline{e_n}$ — and the conjugates $(\overline{e_n})$ form a [Hilbert basis](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#def-b3-hilbert-onb) whenever $(e_n)$ does (conjugation is an isometric bijection of $L^2$ preserving orthogonality and totality) — so it is $0$ a.e.; then for a.e. $y$, $h(\cdot, y) \perp$ every $e_m$: $h(\cdot, y) = 0$ a.e.: $h = 0$ (Tonelli). So $(e_{mn})$ is a [Hilbert basis](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#def-b3-hilbert-onb); expand $k = \sum c_{mn}e_{mn}$. The truncation $k_N$ (indices $\leq N$) gives $T_{k_N}$ of finite rank (range in $\operatorname{Vect}(e_1, \dots,
e_N)$), and by (a),

$$
\vertiii{T_k - T_{k_N}} \leq \norm{k - k_N}_{L^2} \to 0 :
$$

$T_k$ is a norm-limit of finite-rank operators: [compact](#def-b3-spectral-compact) ([Proposition 15.2](#prop-b3-spectral-ideal)).

**Exercise 15.5 ★★.**

Let $T$ be [self-adjoint](#def-b3-spectral-selfadjoint) with $\langle x, Tx\rangle \geq 0$ for all $x$ (*positive* operator). (a) Show that eigenvalues are $\geq 0$ and that $\vertiii T =
\sup_{\norm x\leq1}\langle x, Tx\rangle$. (b) Prove the generalized Cauchy–Schwarz inequality $\abs{\langle x, Ty\rangle}^2 \leq \langle x, Tx\rangle\langle
y, Ty\rangle$.

**Solution of Exercise 15.5.**

(a) $\lambda\norm x^2 = \langle x, Tx\rangle \geq 0$ on an eigenvector. The formula is [Proposition 15.5](#prop-b3-spectral-sanorm) with all values $\langle x,
Tx\rangle \geq 0$: the absolute value is redundant. (b) $(x, y) \mapsto \langle x, Ty\rangle$ is a Hermitian positive (possibly degenerate) sesquilinear form; the usual Cauchy–Schwarz proof (expand $\langle x + t\eu^{\iu\theta}y,
T(x + t\eu^{\iu\theta}y)\rangle \geq 0$ and take the discriminant) never uses definiteness.

**Exercise 15.6 ★★.**

On $L^2(\intcc01)$, let $(Mf)(x) = x\,f(x)$. (a) Show that $M$ is bounded, [self-adjoint](#def-b3-spectral-selfadjoint), with $\vertiii M =
1$, but has *no* eigenvalues. (b) Show $M$ is not [compact](#def-b3-spectral-compact) *(exhibit a bounded sequence whose image has no convergent subsequence, e.g. normalized indicators of shrinking intervals near $1$ — or invoke the spectral theorem)*. (c) Where does [Lemma 15.6](#lem-b3-spectral-existence)’s proof break for $M$?

**Solution of Exercise 15.6.**

(a) $\norm{Mf}_2 \leq \norm f_2$, and on $f_n = \sqrt n\,
\mathbf 1_{\intcc{1 - 1/n}1}$ (unit vectors), $\norm{Mf_n}_2
\geq 1 - \frac1n$: $\vertiii M = 1$; [self-adjoint](#def-b3-spectral-selfadjoint) since the multiplier is real. Eigenvalues: $xf(x) = \lambda f(x)$ a.e. forces $f = 0$ a.e. off the null set $\{x = \lambda\}$: $f =
0$ in $L^2$. (b) With the same $f_n$: $\norm{Mf_n - f_n}_2 \leq \frac1n \to
0$, while $f_n \rightharpoonup 0$ (for fixed $g \in L^2$, $\abs{\langle g, f_n\rangle} \leq \norm{g\,\mathbf
1_{\intcc{1-1/n}1}}_2 \to 0$ by DCT). If $Mf_{n_k} \to h$ in norm, then $f_{n_k} \to h$, forcing $h = 0$ (weak limit) yet $\norm h = 1$: no convergent subsequence. (c) In [Lemma 15.6](#lem-b3-spectral-existence), precisely the extraction “$Tx_{n_k} \to y$” uses compactness; for $M$ the maximizing sequences concentrate near $x = 1$ and their images converge weakly to $0$, never in norm: the eigenvector at the top of the numerical range simply fails to exist.

**Exercise 15.7 ★★.**

(Volterra) On $L^2(\intcc01)$, let $Vf(x) = \int_0^xf(y)\dd
y$. (a) Show $V$ is [compact](#def-b3-spectral-compact) (Hilbert–Schmidt with kernel $\mathbf
1_{y < x}$) but not [self-adjoint](#def-b3-spectral-selfadjoint); compute $V^*$. (b) Show that $V$ has no nonzero eigenvalue. *(From $Vf =
\lambda f$: $f$ has a [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) representative, then is $\mathcal C^1$, and solves $\lambda f' = f$, $f(0) = 0$.)* (c) Conclude that compactness alone yields no eigenvectors: [self-adjointness](#def-b3-spectral-selfadjoint) in [Theorem 15.7](#thm-b3-spectral-spectral) is essential.

**Solution of Exercise 15.7.**

(a) $V = T_k$ with $k(x, y) = \mathbf 1_{y < x} \in
L^2(\intcc01^2)$: [compact](#def-b3-spectral-compact) ([Exercise 15.4](#exo-b3-spectral-4)). Its adjoint is the kernel operator with kernel $\overline{k(y,
x)} = \mathbf 1_{y > x}$: $V^*f(x) = \int_x^1f$; $V \neq V^*$ (test on $f = \mathbf 1$). (b) If $Vf = \lambda f$, $\lambda \ne 0$: $Vf$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) on $\intcc01$ (dominated convergence in $\int_0^x f$), so $f
= \frac1\lambda Vf$ has a [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) representative; then $Vf$ is $\mathcal C^1$ (fundamental theorem of calculus for [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) integrands), so $f$ is $\mathcal C^1$, and $\lambda f' = f$ with $f(0) = \frac1\lambda Vf(0) = 0$: $f =
C\eu^{x/\lambda}$ with $C = f(0) = 0$. (c) $V$ is [compact](#def-b3-spectral-compact) with no eigenvalue at all except possibly $0$ ($Vf = 0$ forces $f = 0$ a.e. by differentiating the integral — so not even $0$): the spectral machinery genuinely requires [self-adjointness](#def-b3-spectral-selfadjoint), not just compactness.

**Exercise 15.8 ★★★.**

(Courant–Fischer) Let $T$ be [compact](#def-b3-spectral-compact), [self-adjoint](#def-b3-spectral-selfadjoint), *positive*, with eigenvalues $\mu_1 \geq \mu_2 \geq \dots
> 0$ (repeated by multiplicity, eigenvectors $e_1, e_2,
\dots$). Show:

$$
\mu_{k} = \max_{\substack{V \subseteq H \\ \dim V = k}}\
\min_{\substack{x \in V\\ \norm x = 1}}\ \langle x, Tx\rangle
= \min_{\substack{W \subseteq H\\ \operatorname{codim}W =
k-1}}\ \max_{\substack{x\in W\\ \norm x = 1}}\ \langle x,
Tx\rangle .
$$

*(Test $V = \operatorname{Vect}(e_1,\dots,e_k)$; for the upper bound intersect any $V$ with $\operatorname{Vect}(e_k, e_{k+1}, \dots)$-type spaces: dimension counting forces a nonzero intersection.)* Deduce that eigenvalues depend monotonically on $T$ ($T \leq S
\Rightarrow \mu_k(T) \leq \mu_k(S)$).

**Solution of Exercise 15.8.**

Write $x = \sum_ic_ie_i + z$, $z \in \ker T$, so $\langle x,
Tx\rangle = \sum_i\mu_i\abs{c_i}^2$. *Lower bound*: on the unit sphere of $V_k = \operatorname{Vect}(e_1, \dots,
e_k)$, $\langle x, Tx\rangle = \sum_{i\leq
k}\mu_i\abs{c_i}^2 \geq \mu_k$: the max over $V$ of the min is $\geq \mu_k$. *Upper bound*: let $\dim V = k$ and $W_k
= \overline{\operatorname{Vect}}(e_k, e_{k+1}, \dots) +
\ker T$, of codimension $k - 1$ (its [orthogonal complement](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#thm-b3-hilbert-decomposition) is $V_{k-1}$); $V \cap W_k \neq \{0\}$ (a linear map $V \to
H/W_k \cong V_{k-1}$ of rank $\leq k - 1$ has nontrivial kernel), and a unit $x \in V\cap W_k$ has $\langle x,
Tx\rangle = \sum_{i \geq k}\mu_i\abs{c_i}^2 \leq \mu_k$: the min over $V$ is $\leq \mu_k$. Together: the first formula; the second is proved symmetrically (test $W = W_k$; for arbitrary $W$ of codimension $k-1$, $W \cap V_k \neq 0$ gives a unit vector with $\langle x, Tx\rangle \geq \mu_k$). Monotonicity: $\langle x, Tx\rangle \leq \langle x,
Sx\rangle$ pointwise transfers through $\max\min$.

**Exercise 15.9 ★★.**

Using [Theorem 15.8](#thm-b3-spectral-fredholm) for $Tf(x) =
\int_0^1\min(x,y)f(y)\dd y$ ([Example 15.9](#ex-b3-spectral-minxy)): for which $\lambda \in \R$ does the integral equation

$$
f(x) - \lambda\int_0^1\min(x,y)\,f(y)\,\dd y = g(x)
$$

have a unique solution $f \in L^2$ for every $g \in L^2$? What happens at the exceptional values?

**Solution of Exercise 15.9.**

Rewrite $f - \lambda Tf = g$. For $\lambda = 0$: $f = g$, always uniquely [solvable](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived). For $\lambda \neq 0$: this is $(T - \frac1\lambda)f = -\frac g\lambda$, and by the [Fredholm alternative](#thm-b3-spectral-fredholm) ([Theorem 15.8](#thm-b3-spectral-fredholm)) with the eigenvalues $\lambda_n = \bigl((n + \frac12)\pi\bigr)^{-2}$ of $T$ ([Example 15.9](#ex-b3-spectral-minxy)): unique [solvability](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived) for all $g$ iff $\frac1\lambda \neq \lambda_n$ for every $n$, i.e.

$$
\lambda \neq \Bigl(n + \tfrac12\Bigr)^2\pi^2
\qquad (n = 0, 1, 2, \dots).
$$

At an exceptional $\lambda = (n+\frac12)^2\pi^2$: solutions exist iff $g \perp \sin\bigl((n{+}\frac12)\pi x\bigr)$, and are then unique up to adding multiples of that sine.

**Exercise 15.10 ★★★.**

Let $S$ be the shift on $\ell^2$ ([Exercise 8.1](https://one-course.com/books/math/5/en/chapter/8-banach-spaces-and-the-fundamental-theorems#exo-b3-banach-1)). (a) Show that $S$ has no eigenvalues, while every $\lambda$ with $\abs\lambda < 1$ is an eigenvalue of $S^*$ (find the eigenvectors explicitly, geometric sequences). (b) Neither $S$ nor $S^*$ is [compact](#def-b3-spectral-compact): verify via [Exercise 15.2](#exo-b3-spectral-2)-style testing on $(e_n)$. (c) Comment: for [non-self-adjoint](#def-b3-spectral-selfadjoint), [non-compact](#def-b3-spectral-compact) operators, the eigenvalue landscape can be anything from empty to a full disc — the notion that survives is the *spectrum*, studied in a later course.

**Solution of Exercise 15.10.**

(a) $Sx = \lambda x$: comparing coordinates, $0 = \lambda
x_1$ and $x_n = \lambda x_{n+1}$; if $\lambda \ne 0$ then $x_1 = 0$ and inductively $x = 0$; if $\lambda = 0$, $Sx = 0$ forces $x = 0$ ($S$ isometric). No eigenvalues. $S^*x =
\lambda x$ reads $x_{n+1} = \lambda x_n$: $x =
x_1(1, \lambda, \lambda^2, \dots)$, in $\ell^2$ exactly when $\abs\lambda < 1$: a full open disc of eigenvalues. (b) $\norm{Se_n - Se_m} = \norm{e_{n+1} - e_{m+1}} = \sqrt2$: the image of the bounded $(e_n)$ has no Cauchy subsequence; likewise $S^*e_{n+1} = e_n$. Neither is [compact](#def-b3-spectral-compact). (c) For [compact](#def-b3-spectral-compact) [self-adjoint operators](#def-b3-spectral-selfadjoint) the eigenvalues capture everything ([Theorem 15.7](#thm-b3-spectral-spectral)); dropping either hypothesis, eigenvalues may vanish entirely ($S$, Volterra) or fill a disc ($S^*$): the robust object is the spectrum $\{\lambda : T - \lambda I \text{ not invertible}\}$, whose theory belongs to a later course.

**Exercise 15.11 ★★.**

(Square roots) Let $T$ be [compact](#def-b3-spectral-compact), [self-adjoint](#def-b3-spectral-selfadjoint), positive ($\langle x, Tx\rangle \geq 0$) on a [Hilbert space](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#def-b3-hilbert-inner) $H$, with spectral decomposition $Tx = \sum_n\mu_n\langle e_n,
x\rangle e_n$ ($\mu_n > 0$). (a) Define $Sx = \sum_n\sqrt{\mu_n}\,\langle e_n, x\rangle
e_n$; show $S$ is [compact](#def-b3-spectral-compact), [self-adjoint](#def-b3-spectral-selfadjoint), positive, with $S^2
= T$. (b) Prove uniqueness: any [compact](#def-b3-spectral-compact) positive [self-adjoint](#def-b3-spectral-selfadjoint) $R$ with $R^2 = T$ preserves the eigenspaces of $T$ *($RT = R^3 = TR$: $R$ commutes with $T$, so $R(\ker(T - \mu)) \subseteq \ker(T - \mu)$)*, and on $\ker(T - \mu)$, $R$ is a positive operator squaring to $\mu\,\mathrm{id}$ on a finite-dimensional space: diagonalize it there and conclude $R =
\sqrt\mu\,\mathrm{id}$ on each eigenspace, hence $R = S$. (c) Compute $\sqrt G$ for the string operator $G$ of [Problem 15.1](#pb-b3-spectral-1): which kernel has eigenvalues $\frac1{n\pi}$ on the sine basis? (Express $\sqrt G$ as the $L^2$-limit of kernels; no closed form is required.)

**Solution of Exercise 15.11.**

(a) $S$ is the diagonal operator with coefficients $\sqrt{\mu_n} \to 0$: [compact](#def-b3-spectral-compact) ([Exercise 15.2](#exo-b3-spectral-2)(b), transported to the basis $(e_n)$ completed by $\ker T$, where $S = 0$), [self-adjoint](#def-b3-spectral-selfadjoint) (real diagonal), positive ($\langle x, Sx\rangle =
\sum\sqrt{\mu_n}\abs{\langle e_n, x\rangle}^2$), and $S^2 =
T$ termwise.

(b) $R$ commutes with $T = R^2$; for an eigenvector $x$ of $T$ with eigenvalue $\mu$: $T(Rx) = RTx = \mu Rx$, so the finite-dimensional eigenspace $E_\mu = \ker(T - \mu)$ is $R$-stable. On $E_\mu$, $R$ is symmetric positive with $R^2
= \mu\,\mathrm{id}$: its eigenvalues $\rho$ satisfy $\rho^2
= \mu$, $\rho \geq 0$: all equal $\sqrt\mu$, and a diagonalizable operator with a single eigenvalue is scalar: $R = \sqrt\mu\,\mathrm{id}$ on $E_\mu$. On $\ker T$: $\norm
{Rx}^2 = \langle x, R^2x\rangle = \langle x, Tx\rangle = 0$. So $R$ agrees with $S$ on $\ker T$ and on every eigenspace, whose closed span is $H$ (spectral theorem): $R = S$.

(c) $\sqrt G$ acts as $\frac1{n\pi}$ on $e_n =
\sqrt2\sin(n\pi x)$: it is the kernel operator with

$$
k(x, y) = \sum_{n\geq1}\frac{2\sin(n\pi x)\sin(n\pi
y)}{n\pi},
$$

the series converging in $L^2(\intcc01^2)$ (coefficients $\frac1{n\pi} \in \ell^2$; the partial-sum kernels give the finite-rank approximations). No elementary closed form is needed: the spectral side *is* the operator.

**Exercise 15.12 ★★★.**

(Singular value decomposition) Let $T \in \mathcal L(H)$ be *[compact](#def-b3-spectral-compact)*, not necessarily [self-adjoint](#def-b3-spectral-selfadjoint). (a) Show that $T^*T$ is [compact](#def-b3-spectral-compact), [self-adjoint](#def-b3-spectral-selfadjoint), positive; let $(e_n)$ be an orthonormal family of eigenvectors with $T^*Te_n = s_n^2e_n$, $s_n > 0$ (the *singular values*), completed by $\ker(T^*T) = \ker T$ (prove this equality). (b) Set $f_n = \frac{Te_n}{s_n}$; show $(f_n)$ is orthonormal, and establish the *SVD*:

$$
Tx = \sum_n s_n\,\langle e_n, x\rangle\,f_n
\qquad (x \in H),
$$

with convergence in $H$. (c) Deduce: $\vertiii T = \max_ns_n$; $T$ is a norm-limit of finite-rank operators (re-proving [Proposition 15.2](#prop-b3-spectral-ideal)’s converse for [Hilbert spaces](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#def-b3-hilbert-inner)); and for the Volterra operator $V$ of [Exercise 15.7](#exo-b3-spectral-7), which has no eigenvalues, explain why the SVD nevertheless exists and what its ingredients are (identify $V^*V$ as a string-type kernel operator — computing its eigenvalues explicitly is [Exercise 15.9](#exo-b3-spectral-9)’s territory).

**Solution of Exercise 15.12.**

(a) $T^*T$ is [compact](#def-b3-spectral-compact) (product of a bounded and a [compact operator](#def-b3-spectral-compact), [Exercise 15.3](#exo-b3-spectral-3)), [self-adjoint](#def-b3-spectral-selfadjoint) ($(T^*T)^* = T^*T$), positive ($\langle x, T^*Tx\rangle =
\norm{Tx}^2$). Kernel: $T^*Tx = 0 \Rightarrow \norm{Tx}^2 =
\langle x, T^*Tx\rangle = 0 \Rightarrow Tx = 0$, and conversely: $\ker T^*T = \ker T$. The spectral theorem supplies the orthonormal $(e_n)$ with $T^*Te_n = s_n^2e_n$, $s_n > 0$, spanning $(\ker T)^\perp$.

(b) $\langle f_m, f_n\rangle = \frac{\langle Te_m,
Te_n\rangle}{s_ms_n} = \frac{\langle e_m,
T^*Te_n\rangle}{s_ms_n} = \frac{s_n^2}{s_ms_n}\delta_{mn} =
\delta_{mn}$. Expand $x = x_0 + \sum_n\langle e_n, x\rangle
e_n$ with $x_0 \in \ker T$ ([Parseval](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#thm-b3-hilbert-parseval) in the closed span plus kernel); applying the [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $T$:

$$
Tx = \sum_n\langle e_n, x\rangle\,Te_n =
\sum_ns_n\langle e_n, x\rangle\,f_n,
$$

the series converging because its partial sums are Cauchy ($\norm{\sum_{N<n\leq M}s_n\langle e_n, x\rangle f_n}^2 =
\sum s_n^2\abs{\langle e_n, x\rangle}^2$, dominated by $\sup_{n>N}s_n^2\cdot\norm x^2$, and $s_n \to 0$).

(c) $\norm{Tx}^2 = \sum_ns_n^2\abs{\langle e_n, x\rangle}^2
\leq (\max s_n)^2\norm x^2$, attained at the maximizing $e_n$: $\vertiii T = \max s_n$. Truncating the SVD at rank $N$ leaves an operator of norm $\sup_{n>N}s_n \to 0$: finite-rank approximation. The Volterra operator has no eigenvalues ([Exercise 15.7](#exo-b3-spectral-7)), but $V^*V$ does: $V^*Vf(x) = \int_x^1\int_0^tf(s)\,\dd s\,\dd t$ is a symmetric positive kernel operator (kernel $1 - \max(x,y)$, a string-type Green kernel), whose eigenpairs — computed via the boundary value problem $-u'' = \lambda^{-1}u$, $u'(0)
= u(1) = 0$, i.e. [Exercise 15.9](#exo-b3-spectral-9)’s family — give singular values $s_n = \bigl((n + \frac12)\pi\bigr)^{-1}$. The SVD lives on *two* orthonormal families precisely because $V$ rotates its eigengeometry away: no eigenvectors, yet [perfect](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#prop-b3-galois-perfect) diagonal structure between two different bases.

## 15.5 Problem: the vibrating string and $\zeta(2)$

**Problem 15.1.**

Weekend problem — Green’s operator, the sine basis, and a trace formula

We solve the eigenvalue problem of the vibrating string with fixed ends — $-u'' = \nu u$, $u(0) = u(1) = 0$ — by operator theory, obtain the sine orthonormal basis without any Fourier computation, and evaluate $\zeta(2)$ by comparing two expressions for the *trace* of the Green operator. Define, on $L^2(\intcc01)$,

$$
(Gf)(x) = \int_0^1 g(x,y)\,f(y)\,\dd y,
\qquad
g(x, y) = \min(x,y)\,\bigl(1 - \max(x,y)\bigr).
$$

**Part I — The Green operator.**

1. Show that $g$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) , symmetric, with $0 \leq g  \leq \frac14$ , and that $G$ is [compact](#def-b3-spectral-compact) and [self-adjoint](#def-b3-spectral-selfadjoint) ( [Example 15.3](#ex-b3-spectral-examples) (c)).
2. For $f$ [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), show that $u = Gf$ is $\mathcal  C^2$ with $$-u'' = f, \qquad u(0) = u(1) = 0$$ *(write $u(x) = (1-x)\int_0^xyf(y)\dd y +  x\int_x^1(1-y)f(y)\dd y$ and differentiate twice)*. Conversely, if $u \in \mathcal C^2$ with $u(0) = u(1)  = 0$, then $G(-u'') = u$: $G$ inverts the string operator.
3. Show $\ker G = \{0\}$ *(if $Gf = 0$ with $f \in  L^2$: test against [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $\varphi$, transfer $G$ by symmetry/Fubini onto $\varphi$, and use the fundamental lemma [Corollary 12.11](https://one-course.com/books/math/5/en/chapter/12-the-lp-spaces#cor-b3-lp-fundlemma) — or regularize)* , and that $G$ is a positive operator: $\langle f, Gf\rangle \geq 0$ . *(For [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $f$: $\langle f, Gf\rangle = \int_0^1 (u')^2$ with $u  = Gf$, by parts; conclude by [density](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#ex-b3-lebesgue-gamma).)*

**Part II — Diagonalization: the sine basis.**

4. Show that the eigenfunctions of $G$ with eigenvalue $\lambda \ne 0$ are, up to scalars, the solutions of $-\lambda u'' = u$ , $u(0) = u(1) = 0$ *(an eigenfunction has a [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) representative — $Gf$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) for $f \in L^2$, why? — hence is $\mathcal C^2$ by bootstrapping question 2)* .
5. Solve the boundary value problem: the eigenvalues of $G$ are $\lambda_n = \frac1{n^2\pi^2}$ ( $n \geq 1$ ), with normalized eigenfunctions $e_n(x) =  \sqrt2\,\sin(n\pi x)$ ; check orthonormality by direct integration as a sanity test.
6. Conclude from [Theorem 15.7](#thm-b3-spectral-spectral) and question 3 that $\bigl(\sqrt2\sin(n\pi x)\bigr)_{n\geq1}$ is an orthonormal *basis* of $L^2(\intcc01)$ — no Stone–Weierstrass, no Fourier series needed. Expand $f(x) = x(1-x)$ in this basis and write [Parseval](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#thm-b3-hilbert-parseval) for it.

**Part III — The trace formula and $\zeta(2)$.**

7. Prove the two identities $$\langle e_n, Ge_n\rangle = \lambda_n  \quad\text{and}\quad  \sum_{n\geq1}\lambda_n = \int_0^1 g(x,x)\,\dd x .$$ For the second (the *trace formula*): expand $g(x, \cdot)$, for fixed $x$, in the basis $(e_n)$ — show that the coefficients are $\lambda_ne_n(x)$, so that $g(x, \cdot) = \sum_n\lambda_ne_n(x)\,e_n$ in $L^2$. Here the sines are explicit: verify directly that $\sum_n\lambda_ne_n(x)e_n(y)$ converges *uniformly* on the square (compare with $\sum  \frac2{n^2\pi^2}$), so its sum is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) and, having the same $L^2$-expansions in $y$ for each $x$, equals $g(x,y)$ everywhere. Set $y = x$ and integrate term by term.
8. Compute $\int_0^1g(x,x)\dd x = \int_0^1x(1-x)\dd x =  \frac16$, and conclude $$\sum_{n\geq1}\frac{1}{n^2\pi^2} = \frac16,  \qquad\text{i.e.}\qquad  \boxed{\ \zeta(2) = \frac{\pi^2}{6}\ } :$$ Euler’s sum from an operator trace.
9. Re-derive $\zeta(2)$ a third way: apply [Parseval](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#thm-b3-hilbert-parseval) in the sine basis to the constant function $\mathbf 1$ , compute $\sum_{n \text{ odd}}\frac1{n^2}$ , and conclude. Then compare mechanisms: in what sense is the trace argument of questions 7–8 “ [Parseval](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#thm-b3-hilbert-parseval) applied to the whole kernel at once”?

**Part IV — The string vibrates.**

10. (Separation of variables, synthesized) For $f \in  L^2$, define $$u(t, x) = \sum_{n\geq1}\;c_n\,  \cos(n\pi t)\,\sqrt2\sin(n\pi x),  \qquad c_n = \langle e_n, f\rangle .$$ Show the series converges in $L^2(\intcc01)$ for each $t$, that $t\mapsto u(t, \cdot)$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) into $L^2$, and that for $f$ in the span of finitely many $e_n$ it solves the wave equation $\partial_t^2u =  \partial_x^2u$ with $u(0) = f$, $\partial_tu(0) = 0$, fixed ends. The eigenvalues $n^2\pi^2$ are the squared frequencies: the string’s harmonics — explain the musical interpretation of [Theorem 15.7](#thm-b3-spectral-spectral) in one paragraph.

**Part V — Variational dividends: the power method, Weyl stability, and a rigorous bound on $\pi$.** Let $A$ be a [compact](#def-b3-spectral-compact) [self-adjoint](#def-b3-spectral-selfadjoint) *positive* operator with eigenvalues $\mu_1 \geq \mu_2 \geq \cdots > 0$ and orthonormal eigenvectors $(u_n)$; $R(x) = \frac{\langle x,
Ax\rangle}{\norm x^2}$. The min–max formulas are [Exercise 15.8](#exo-b3-spectral-8); here we spend them.

11. (Power method) For $x \neq 0$ write $m_p =  \sum_n\mu_n^p\abs{\langle u_n, x\rangle}^2$. Show $m_pm_{p+2} \geq m_{p+1}^2$ (Cauchy–Schwarz), deduce the chain $$R(x) \;\leq\; \frac{\langle Ax, Ax\rangle}{\langle x,  Ax\rangle} \;\leq\; R(Ax) \;\leq\; \mu_1,$$ and prove that if $\langle u_1, x\rangle \neq 0$, then $R(A^kx) \to \mu_1$: iterating the operator on any generic vector computes the top eigenvalue — the power method of numerical analysis, certified.
12. (Weyl stability) For [compact](#def-b3-spectral-compact) [self-adjoint](#def-b3-spectral-selfadjoint) positive $A, B$, deduce from [Exercise 15.8](#exo-b3-spectral-8) that $$\abs{\mu_n(A) - \mu_n(B)} \;\leq\; \vertiii{A - B}  \qquad\text{for every } n :$$ the full spectrum is $1$-Lipschitz in the [operator norm](https://one-course.com/books/math/5/en/chapter/8-banach-spaces-and-the-fundamental-theorems#def-b3-banach-operator) — eigenvalues of big symmetric systems can be computed from approximations with guaranteed error.
13. Apply the Rayleigh bound to $G$ with the test function $u(x) = x(1-x)$: solve $-w'' = u$, $w(0) =  w(1) = 0$ to get $Gu = w = \frac{x(1-x)(1 + x -  x^2)}{12}$, compute $$\norm u_2^2 = \frac1{30}, \qquad  \langle u, Gu\rangle = \frac1{12}\Bigl(\frac1{30} +  \frac1{140}\Bigr) = \frac{17}{5040},  \qquad R(u) = \frac{17}{168},$$ and conclude the *rigorous* bound $\frac1{\pi^2}  = \lambda_1 \geq \frac{17}{168}$, i.e. $\pi \leq  \sqrt{168/17} < 3.1437$.
14. One step of question 11’s chain, on the same test function: using $\int_0^1(x - x^2)^4\dd x =  \frac1{630}$, compute $$\norm{Gu}_2^2 = \frac1{144}\Bigl(\frac1{30} +  \frac2{140} + \frac1{630}\Bigr) = \frac{31}{90720},  \qquad  \frac{\langle Gu, Gu\rangle}{\langle u, Gu\rangle} =  \frac{31}{306},$$ and conclude $\pi \leq \sqrt{306/31} < 3.1419$: two integrals, four correct digits. (Each further iteration roughly squares the accuracy: the eigenvector gap $\lambda_1/\lambda_2 = 4$ drives geometric convergence.)

**Part VI — The trace of $G^2$, and $\zeta(4)$.**

15. Show that $\sum_n\lambda_n^2 =  \iint_{\intcc01^2}g(x,y)^2\,\dd x\,\dd y$ *(expand $g$ on the product basis $(e_m(x)e_n(y))_{m,n}$ of $L^2(\intcc01^2)$ — a [Hilbert basis](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#def-b3-hilbert-onb), cf. [Exercise 15.5](#exo-b3-spectral-5) — and apply [Parseval](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#thm-b3-hilbert-parseval) on the square; question 7 identifies the coefficients)* .
16. Compute the double integral: $$\iint g^2 = 2\int_0^1(1-x)^2\Bigl(\int_0^x  y^2\,\dd y\Bigr)\dd x  = \frac23\int_0^1x^3(1-x)^2\,\dd x = \frac1{90} .$$
17. Conclude $\zeta(4) = \dfrac{\pi^4}{90}$ ; explain, without computation, how traces of higher powers $G^k$ produce $\zeta(2k) \in \pi^{2k}\,\Q$ for all $k \geq 1$ , and why the odd values $\zeta(3),  \zeta(5), \dots$ are structurally out of this machine’s reach.
18. ($\pi$ from below) From $\lambda_1^2 \leq  \sum_n\lambda_n^2 = \frac1{90}$ deduce $\pi \geq  90^{1/4} > 3.080$, and assemble with question 14 the two-sided verdict $$3.080 \;<\; \pi \;<\; 3.1419,$$ obtained entirely from the vibrating string’s arithmetic. Which side converges faster if one uses higher traces $(\operatorname{tr}G^{2k})^{-1/4k}$, and why?

**Part VII — Forcing and resonance.** Fix $\nu \in \R$ and consider the forced string $-u'' - \nu u =
f$, $u(0) = u(1) = 0$, with $f \in L^2$ and $c_n = \langle
e_n, f\rangle$.

19. Suppose $\nu \notin \{n^2\pi^2 : n \geq 1\}$. Show that $$u = \sum_{n\geq1}\frac{c_n}{n^2\pi^2 - \nu}\,e_n$$ converges in $L^2$ and uniformly on $\intcc01$ *(Cauchy–Schwarz between $(c_n)$ and the tails of $\sum n^{-4}$, with $\norm{e_n}_\infty =  \sqrt2$)*, and that it satisfies $u = Gf + \nu Gu$ — the explicit-coordinates form of the [Fredholm alternative](#thm-b3-spectral-fredholm) ([Theorem 15.8](#thm-b3-spectral-fredholm)), with uniqueness.
20. Suppose $\nu = m^2\pi^2$ . Show that $u = Gf + \nu  Gu$ has a solution iff $c_m = 0$ , unique up to adding multiples of $e_m$ . Physical reading: pushing a swing exactly at its own frequency.
21. For $\nu < \pi^2$ , show that the solution operator $R_\nu\colon f \mapsto u$ is bounded on $L^2$ with norm $\frac1{\pi^2 - \nu}$ , [compact](#def-b3-spectral-compact) , [self-adjoint](#def-b3-spectral-selfadjoint) , and positive: the whole spectral analysis restarts, shifted by $\nu$ .
22. (Synthesis) Compile the dictionary of this problem: eigenvalue $\leftrightarrow$ squared frequency (harmonics); trace $\leftrightarrow$ $\zeta(2)$ ; Hilbert–Schmidt norm $\leftrightarrow$ $\zeta(4)$ ; [Fredholm alternative](#thm-b3-spectral-fredholm) $\leftrightarrow$ resonance; min–max $\leftrightarrow$ variational bounds ( $\pi <  3.1437$ from one polynomial). One integral operator, five chapters of analysis cashed in.

**Part VIII — Three last echoes.**

23. ($\zeta(6)$, for free) Question 6’s [Parseval](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#thm-b3-hilbert-parseval) identity gave $\sum_{n\text{ odd}}n^{-6} =  \frac{\pi^6}{960}$. Split $\zeta(6)$ into odd and even $n$ and conclude $$\zeta(6) = \frac{\pi^6}{945},$$ with no new integral: the machine of question 17 (traces of $G^3$) would have produced the same value at the cost of an iterated kernel — [Parseval](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#thm-b3-hilbert-parseval) on one well-chosen function is the cheaper route here.
24. (The ground state is positive) Let $A$ be a [compact](#def-b3-spectral-compact) [self-adjoint](#def-b3-spectral-selfadjoint) positive operator on $L^2(\intcc01)$ given by a [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) symmetric kernel $k > 0$ on $\intoo01^2$ , with largest eigenvalue $\mu_1$ . Show: (a) any maximizer of the Rayleigh quotient is a $\mu_1$ -eigenfunction; (b) if $u$ is one, then $\langle\abs u, A\abs u\rangle \geq \langle u,  Au\rangle$ , with *strict* inequality if $u$ takes both signs on sets of positive [measure](https://one-course.com/books/math/5/en/chapter/9-measure-theory#def-b3-measure-measure) — hence $u$ has a.e. constant sign, and $u = \mu_1^{-1}Au$ never vanishes on $\intoo01$ ; (c) $\mu_1$ is a *simple* eigenvalue. Verify every claim on $G$ : $e_1 = \sqrt2\sin(\pi x) > 0$ , and each $e_n$ , $n  \geq 2$ , being orthogonal to $e_1$ , must change sign (it does: $n - 1$ [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) zeros).
25. (Distance to the spectrum, and the price of resonance) For $\nu \notin \{n^2\pi^2\}$, show that the solution operator $R_\nu$ of question 19 is bounded, [self-adjoint](#def-b3-spectral-selfadjoint), [compact](#def-b3-spectral-compact), with $$\vertiii{R_\nu} =  \frac1{\min_{n\geq1}\,\abs{n^2\pi^2 - \nu}}  = \frac1{\operatorname{dist}\bigl(\nu,  \{n^2\pi^2\}\bigr)},$$ the norm being attained on the nearest mode. Then quantify question 20’s swing: forcing with $f = e_1$ at $\nu = (1 - \varepsilon)\pi^2$ produces $u =  \frac{1}{\varepsilon\pi^2}\,e_1$, an amplification by $\frac1\varepsilon$ of the static response $Ge_1 =  \frac1{\pi^2}e_1$ — at one percent below the fundamental ($\varepsilon = 10^{-2}$), the string answers a hundred times louder.

**Solution of Problem 15.1.**

**1.** [Continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity): $\min$ and $\max$ are [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity); symmetry: swapping $x, y$ swaps neither $\min(x,y)$ nor $1 -
\max(x,y)$. Bounds: $0 \leq g$, and $g(x,y) \leq
\max\cdot(1-\max)$-type bounds give $g \leq \frac14$ (for $u
= \max$: $\min \leq u$ so $g \leq u(1-u) \leq \frac14$). $g
\in L^2(\square)$: Hilbert–Schmidt, hence $G$ [compact](#def-b3-spectral-compact) ([Exercise 15.4](#exo-b3-spectral-4)); the kernel is real symmetric: $G$ [self-adjoint](#def-b3-spectral-selfadjoint).

**2.** Splitting at $y = x$:

$$
u(x) = (1 - x)\int_0^x y\,f(y)\,\dd y +
x\int_x^1(1 - y)\,f(y)\,\dd y .
$$

For [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $f$, differentiate (product and fundamental theorem):

$$
u'(x) = -\int_0^xyf + \int_x^1(1-y)f
\qquad\text{(the boundary terms cancel)},
$$

and $u''(x) = -xf(x) - (1 - x)f(x) = -f(x)$; clearly $u(0) =
u(1) = 0$. Conversely if $u \in \mathcal C^2$ vanishes at both ends, $w = u - G(-u'')$ satisfies $w'' = 0$, $w(0) = w(1) =
0$: $w$ is affine and vanishes twice, so $w = 0$.

**3.** Let $Gf = 0$, $f \in L^2$. For $\psi \in \mathcal
C_c^\infty(\intoo01)$: $\psi = G(-\psi'')$ by question 2, so

$$
\langle f, \psi\rangle = \langle f, G(-\psi'')\rangle
= \langle Gf, -\psi''\rangle = 0
$$

($G$ [self-adjoint](#def-b3-spectral-selfadjoint)): by the fundamental lemma ([Corollary 12.11](https://one-course.com/books/math/5/en/chapter/12-the-lp-spaces#cor-b3-lp-fundlemma)), $f = 0$ a.e. Positivity: for [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $f$, with $u = Gf$,

$$
\langle f, Gf\rangle = \int_0^1 fu = \int_0^1(-u'')u
= \bigl[-u'u\bigr]_0^1 + \int_0^1(u')^2 = \int_0^1(u')^2
\geq 0 ;
$$

for $f \in L^2$, approximate in $L^2$ by [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $f_n$: both sides pass to the limit ($G$ bounded).

**4.** If $Gu = \lambda u$, $\lambda \neq 0$: $Gu$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) ($\abs{Gu(x) - Gu(x')} \leq \norm{g(x,\cdot) -
g(x',\cdot)}_2\norm u_2$, and the kernel is uniformly [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity)), so $u$ has a [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) representative; then question 2’s formulas show $Gu \in \mathcal C^2$, so $u =
\frac1\lambda Gu \in \mathcal C^2$ with $-\lambda u'' =
-(Gu)'' = u$ and $u(0) = u(1) = 0$.

**5.** $-\lambda u'' = u$, $u(0) = 0$: $u = A\sin(x/\sqrt
\lambda)$ (positive $\lambda$: by question 3, $\lambda =
\langle u, Gu\rangle/\norm u^2 > 0$ on eigenvectors). $u(1) =
0$ forces $\frac1{\sqrt\lambda} = n\pi$: $\lambda_n =
\frac1{n^2\pi^2}$, eigenfunctions $\sin(n\pi x)$, normalized $e_n = \sqrt2\sin(n\pi x)$ ($\int_0^12\sin^2(n\pi x)\dd x =
1$). Orthogonality check: $2\sin(m\pi x)\sin(n\pi x) =
\cos((m-n)\pi x) - \cos((m+n)\pi x)$ integrates to $0$ for $m
\neq n$.

**6.** $\ker G = \{0\}$ (question 3), so [Theorem 15.7](#thm-b3-spectral-spectral)(1) gives $H =
\overline{\operatorname{Vect}}(e_n)$: the sines are a [Hilbert basis](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#def-b3-hilbert-onb) of $L^2(\intcc01)$. For $f(x) = x(1 - x)$:

$$
c_n = \sqrt2\int_0^1x(1-x)\sin(n\pi x)\,\dd x
= \sqrt2\;\frac{2\bigl(1 - (-1)^n\bigr)}{n^3\pi^3}
= \begin{cases}\dfrac{4\sqrt2}{n^3\pi^3} & n \text{ odd},\\
0 & n \text{ even},\end{cases}
$$

(two integrations by parts). [Parseval](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#thm-b3-hilbert-parseval): $\int_0^1x^2(1-x)^2\dd
x = \frac1{30} = \sum_{n \text{ odd}}\frac{32}{n^6\pi^6}$, i.e. $\sum_{n\text{ odd}}n^{-6} = \frac{\pi^6}{960}$.

**7.** $\langle e_n, Ge_n\rangle = \lambda_n\norm{e_n}^2
= \lambda_n$. For fixed $x$, the coefficients of $g(x,
\cdot)$: $\langle e_n, g(x,\cdot)\rangle = (Ge_n)(x) =
\lambda_ne_n(x)$, so $g(x,\cdot) =
\sum_n\lambda_ne_n(x)\,e_n$ in $L^2$. The explicit series $\sum_n\lambda_ne_n(x)e_n(y) = \sum_n\frac{2\sin(n\pi
x)\sin(n\pi y)}{n^2\pi^2}$ converges normally on the square ($\abs{\text{term}} \leq \frac2{n^2\pi^2}$): its sum is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), and for each $x$ it has the same $L^2(\dd y)$-coefficients as $g(x, \cdot)$: the two [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) functions agree for every $(x, y)$. Setting $y = x$ and integrating (normal convergence allows term-by-term integration):

$$
\int_0^1g(x,x)\,\dd x =
\sum_n\lambda_n\int_0^1e_n(x)^2\dd x = \sum_n\lambda_n .
$$

**8.** $\int_0^1g(x,x)\dd x = \int_0^1x(1 - x)\dd x =
\frac16$, so $\sum_{n\geq1}\frac1{n^2\pi^2} = \frac16$:

$$
\zeta(2) = \sum_{n\geq1}\frac1{n^2} = \frac{\pi^2}6 .
$$

**9.** For $f = \mathbf 1$: $c_n =
\sqrt2\int_0^1\sin(n\pi x)\dd x = \sqrt2\,\frac{1 -
(-1)^n}{n\pi}$: $c_n = \frac{2\sqrt2}{n\pi}$ for odd $n$, $0$ for even. [Parseval](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#thm-b3-hilbert-parseval): $1 = \sum_{n\text{
odd}}\frac{8}{n^2\pi^2}$, so $\sum_{n\text{ odd}}n^{-2} =
\frac{\pi^2}8$, and $\zeta(2) = \frac{\pi^2}8\cdot\frac{1}{1
- \frac14} = \frac{\pi^2}6$ (even terms are $\frac14\zeta(2)$). Comparison: [Parseval](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#thm-b3-hilbert-parseval) for one $f$ sums $\abs{\langle e_n, f\rangle}^2$; the trace formula integrates the diagonal of the *kernel*, which amounts to summing [Parseval](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#thm-b3-hilbert-parseval) over an entire orthonormal family at once — $\sum_n\langle e_n, Ge_n\rangle$ — and is therefore blind to any particular choice of test function.

**10.** $\abs{c_n\cos(n\pi t)} \leq \abs{c_n}$ with $\sum\abs{c_n}^2 < \infty$: for each $t$ the series converges in $L^2$ (orthonormal expansion, [Theorem 13.7](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#thm-b3-hilbert-parseval)(3)); the tail bound $\norm{u(t) - u_N(t)}_2^2 \leq \sum_{n>N}\abs{c_n}^2$ is uniform in $t$, and each partial sum is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) in $t$ (finitely many cosines): $t \mapsto u(t,\cdot)$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) into $L^2$. For $f = \sum_{n\leq N}c_ne_n$: each mode $\cos(n\pi t)\sin(n\pi x)$ satisfies $\partial_t^2 =
-n^2\pi^2 = \partial_x^2$ applied to it, vanishes at $x = 0,
1$, has value $\sin(n\pi x)$ and time-derivative $0$ at $t =
0$: the finite sum solves everything. Musically: the string’s motion is a superposition of *standing waves* $e_n$, whose frequencies $n\pi$ are the fundamental and its overtones; the spectral theorem says every initial shape decomposes uniquely into these pure tones, the coefficients $c_n$ being the timbre. Hearing a string is computing an orthonormal expansion.

**11.** With $a_n = \abs{\langle u_n, x\rangle}^2$: $m_{p+1} = \sum_n\mu_n^{p+1}a_n = \sum_n\bigl(\mu_n^{p/2}
\sqrt{a_n}\bigr)\bigl(\mu_n^{p/2+1}\sqrt{a_n}\bigr) \leq
\sqrt{m_p\,m_{p+2}}$ (Cauchy–Schwarz in $\ell^2$). Hence the ratios $m_{p+1}/m_p$ are nondecreasing in $p$; since $R(x) = \frac{m_1}{m_0}$, $\frac{\langle Ax, Ax\rangle}
{\langle x, Ax\rangle} = \frac{m_2}{m_1}$ and $R(Ax) =
\frac{m_3}{m_2}$, the chain follows, each term $\leq \mu_1$ because $m_{p+1} \leq \mu_1m_p$ termwise. Convergence: if $a_1 > 0$ (writing the top eigenvalue’s total weight as $a_1$), then

$$
\mu_1 \geq R(A^kx) = \frac{m_{2k+1}}{m_{2k}} =
\mu_1\,\frac{a_1 + \sum_{\mu_n<\mu_1}(\mu_n/\mu_1)^{2k+1}
a_n}{a_1 + \sum_{\mu_n<\mu_1}(\mu_n/\mu_1)^{2k}a_n}
\longrightarrow \mu_1,
$$

by dominated convergence of the sums (ratios $< 1$): the power method converges for every starting vector not orthogonal to the top eigenspace.

**12.** Pointwise, $\abs{\langle x, (A - B)x\rangle}
\leq \vertiii{A - B}\,\norm x^2$, so $R_A(x) \leq R_B(x) +
\vertiii{A - B}$ for every $x$. Feeding this into the max–min formula of [Exercise 15.8](#exo-b3-spectral-8): $\mu_n(A)
\leq \mu_n(B) + \vertiii{A - B}$, and symmetrically in $A,
B$: $\abs{\mu_n(A) - \mu_n(B)} \leq \vertiii{A - B}$ for all $n$ at once.

**13.** $-w'' = x - x^2$ integrates to $w = -\frac{x^3}6
+ \frac{x^4}{12} + cx$ (with $w(0) = 0$), and $w(1) = 0$ gives $c = \frac1{12}$:

$$
w = \frac{x^4 - 2x^3 + x}{12} = \frac{x(1-x)(1 + x -
x^2)}{12} = Gu .
$$

Then $\norm u_2^2 = \int_0^1x^2(1-x)^2 = \frac1{30}$, and with $\int_0^1x^3(1-x)^3 = B(4,4) = \frac1{140}$:

$$
\langle u, Gu\rangle = \frac1{12}\Bigl(\frac1{30} +
\frac1{140}\Bigr) = \frac{17}{5040},
\qquad
R(u) = \frac{17/5040}{1/30} = \frac{17}{168} .
$$

So $\frac1{\pi^2} = \lambda_1 \geq \frac{17}{168}$, i.e. $\pi^2 \leq \frac{168}{17} = 9.8824$: $\pi \leq 3.14364 <
3.1437$ (true value $\pi^2 = 9.8696$). A polynomial, an integral, a digit.

**14.** Write $u = x - x^2$, so $Gu =
\frac{u(1 + u)}{12}$ and, using $\int u^2 = \frac1{30}$, $\int u^3 = \frac1{140}$, $\int u^4 = B(5,5) =
\frac{4!\,4!}{9!} = \frac1{630}$:

$$
\norm{Gu}_2^2 = \frac1{144}\int u^2(1+u)^2 =
\frac1{144}\Bigl(\frac1{30} + \frac2{140} +
\frac1{630}\Bigr) = \frac1{144}\cdot\frac{62}{1260} =
\frac{31}{90720} .
$$

Hence $\frac{\langle Gu, Gu\rangle}{\langle u, Gu\rangle} =
\frac{31/90720}{17/5040} = \frac{31}{306}$, and by question 11 this is still $\leq \mu_1 = \frac1{\pi^2}$: $\pi^2 \leq
\frac{306}{31} = 9.87097$, i.e. $\pi \leq 3.14181 <
3.1419$ — four digits (and the next iterate would give about eight, the error contracting by $(\lambda_2/\lambda_1)^2 = \frac1{16}$ per step).

**15.** The family $(e_m \otimes e_n)(x,y) =
e_m(x)e_n(y)$ is a [Hilbert basis](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#def-b3-hilbert-onb) of $L^2(\intcc01^2)$ (orthonormality by Tonelli; totality as in [Exercise 15.5](#exo-b3-spectral-5)). By question 7, for fixed $x$: $g(x, \cdot) = \sum_n\lambda_ne_n(x)e_n$, so the coefficient of $g$ on $e_m\otimes e_n$ is

$$
\langle e_m\otimes e_n,\ g\rangle
= \int_0^1 e_m(x)\,\lambda_n e_n(x)\,\dd x
= \lambda_n\,\delta_{mn} .
$$

[Parseval](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#thm-b3-hilbert-parseval) in the square:

$$
\iint g^2 = \sum_{m,n}\abs{\langle e_m\otimes e_n,
g\rangle}^2 = \sum_n\lambda_n^2 .
$$

**16.** By symmetry of $g$,

$$
\iint g^2 = 2\iint_{y<x}y^2(1-x)^2
= 2\int_0^1(1-x)^2\,\frac{x^3}3\,\dd x
= \frac23\,B(4, 3) = \frac23\cdot\frac{3!\,2!}{6!}
= \frac23\cdot\frac1{60} = \frac1{90} .
$$

**17.** Combining: $\sum_n\frac1{n^4\pi^4} =
\frac1{90}$, i.e. $\zeta(4) = \frac{\pi^4}{90}$. In general, $\operatorname{tr}(G^k) = \sum\lambda_n^k =
\frac{\zeta(2k)}{\pi^{2k}}$ equals an iterated integral of products of the rational-polynomial kernel $g$ over the $k$-cube: a rational number. Hence $\zeta(2k) \in
\pi^{2k}\Q$ for every $k$. The machine only reaches even arguments because eigenvalues enter through their *powers* — $\sum\lambda_n^k$ — and $\lambda_n =
\frac1{n^2\pi^2}$: no combination of traces produces $\sum
n^{-3}$; the arithmetic nature of $\zeta(3)$ (irrational by Apéry, transcendence open) lies beyond spectral bookkeeping.

**18.** The top term of a sum of positive terms is at most the sum: $\lambda_1^2 \leq \sum\lambda_n^2 =
\frac1{90}$, so $\frac1{\pi^2} \leq \frac1{\sqrt{90}}$ and $\pi \geq 90^{1/4} = 3.0801\ldots$ With question 14: $3.080 < \pi < 3.1419$, by string arithmetic alone. Higher traces sharpen the lower bound geometrically: $\lambda_1 \leq (\operatorname{tr}G^{2k})^{1/2k} =
\lambda_1\bigl(1 + \sum_{n\geq2}(\lambda_n/\lambda_1)^{2k}
\bigr)^{1/2k}$, and the parasitic factor dies like $\bigl(\tfrac14\bigr)^{2k}\cdot\frac1{2k}$-fast — the same spectral-gap mechanism as the power method’s convergence (question 11), seen from the trace side.

**19.** $\abs{n^2\pi^2 - \nu} \geq \delta > 0$ for all $n$ (the sequence $n^2\pi^2 \to \infty$ avoids $\nu$ by a margin), and $\abs{n^2\pi^2 - \nu} \geq \frac{n^2\pi^2}2$ for $n$ large. $L^2$ convergence: the coefficients $\frac{c_n}{n^2\pi^2 - \nu}$ are square-summable (dominated by $\frac{\abs{c_n}}\delta$). Uniform convergence: the tail sup-norms are bounded by $\sqrt2\sum_{n>N}
\frac{\abs{c_n}}{\abs{n^2\pi^2 - \nu}} \leq
\frac{2\sqrt2}{\pi^2}\bigl(\sum\abs{c_n}^2\bigr)^{1/2}
\bigl(\sum_{n>N}n^{-4}\bigr)^{1/2} \to 0$ (Cauchy–Schwarz). Verification: $Gf + \nu Gu$ has $e_n$-coefficient

$$
\lambda_nc_n + \frac{\nu\lambda_nc_n}{n^2\pi^2 - \nu}
= \frac{c_n}{n^2\pi^2}\Bigl(1 + \frac{\nu}{n^2\pi^2 -
\nu}\Bigr) = \frac{c_n}{n^2\pi^2 - \nu} :
$$

exactly the coefficients of $u$, so $u = Gf + \nu Gu$; uniqueness because a difference $v$ of solutions satisfies $v = \nu
Gv$, i.e. $\langle e_n, v\rangle(n^2\pi^2 - \nu) = 0$ for all $n$: $v = 0$.

**20.** As in question 19, the equation $u = Gf + \nu
Gu$ is equivalent to the family of coefficient equations $(n^2\pi^2 - \nu)\,\langle e_n, u\rangle = c_n$, $n \geq
1$. For $n = m$ the left side is $0$: [solvability](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived) forces $c_m = 0$, and then $\langle e_m,
u\rangle$ is free while all other coefficients are determined: solutions form the line $u_0 + \R e_m$. Resonance: a forcing with a component on the eigenmode pumps energy into it without bound — the swing pushed at its own frequency.

**21.** From question 19’s formula, $\norm{R_\nu f}_2^2
= \sum\frac{\abs{c_n}^2}{(n^2\pi^2 - \nu)^2} \leq
\frac{\norm f^2}{(\pi^2 - \nu)^2}$ (for $\nu < \pi^2$ the closest eigenvalue is $\pi^2$), with equality approached on $f = e_1$: [operator norm](https://one-course.com/books/math/5/en/chapter/8-banach-spaces-and-the-fundamental-theorems#def-b3-banach-operator) $\frac1{\pi^2 - \nu}$. Compactness: $R_\nu$ is the norm-limit of its finite-rank truncations (the tail coefficients $\frac1{n^2\pi^2 - \nu} \to 0$); [self-adjointness](#def-b3-spectral-selfadjoint) and positivity are read off the diagonal form (all coefficients $\frac1{n^2\pi^2 - \nu} > 0$). $R_\nu$ has eigenvalues $\frac1{n^2\pi^2 - \nu}$: the analysis of Parts I–VI restarts verbatim.

**22.** The dictionary: *eigenvalue* $\lambda_n =
\frac1{n^2\pi^2}$ $\leftrightarrow$ squared frequency $n^2\pi^2$ of the $n$-th harmonic; *trace* $\sum
\lambda_n = \frac16$ $\leftrightarrow$ $\zeta(2) =
\frac{\pi^2}6$; *Hilbert–Schmidt norm* $\iint g^2 =
\frac1{90}$ $\leftrightarrow$ $\zeta(4) = \frac{\pi^4}{90}$; *[Fredholm alternative](#thm-b3-spectral-fredholm)* $\leftrightarrow$ resonance of the forced string; *min–max* $\leftrightarrow$ variational estimates, down to $\pi < 3.1437$ from one polynomial. Behind each pairing, the same object: one [compact](#def-b3-spectral-compact) [self-adjoint operator](#def-b3-spectral-selfadjoint), diagonalized once, exploited five ways.

**23.** Splitting over parity and substituting $n = 2m$ in the even part,

$$
\zeta(6) = \sum_{n\text{ odd}}\frac1{n^6} +
\sum_{m\geq1}\frac1{(2m)^6}
= \frac{\pi^6}{960} + \frac{\zeta(6)}{64},
$$

so $\frac{63}{64}\zeta(6) = \frac{\pi^6}{960}$ and $\zeta(6)
= \frac{64\,\pi^6}{63\cdot960} = \frac{\pi^6}{945}$. The trace route would compute $\operatorname{tr}G^3 =
\sum\lambda_n^3 = \zeta(6)/\pi^6$ as $\iint g\,g_2$ with the iterated kernel $g_2(x,y) = \int_0^1g(x,z)g(z,y)\dd z$ — three integrations of piecewise polynomials; [Parseval](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#thm-b3-hilbert-parseval) on $x(1-x)$ needed only one.

**24.** (a) Diagonalize: $v = \sum_na_nu_n$ (plus a possible kernel component, on which $\langle v, Av\rangle$ gains nothing and $\norm v^2$ grows, so a maximizer has none). Then $\langle v, Av\rangle = \sum\mu_na_n^2 \leq
\mu_1\sum a_n^2$, with equality iff $a_n = 0$ whenever $\mu_n < \mu_1$: a maximizer lies in the $\mu_1$-eigenspace. (b) For any $u$,

$$
\langle\abs u, A\abs u\rangle - \langle u, Au\rangle
= \iint k(x,y)\,\bigl(\abs{u(x)}\abs{u(y)} -
u(x)u(y)\bigr)\dd x\,\dd y \;\geq\; 0,
$$

the integrand being pointwise nonnegative. If $P = \{u >
0\}$ and $N = \{u < 0\}$ both have positive [measure](https://one-course.com/books/math/5/en/chapter/9-measure-theory#def-b3-measure-measure), then on $P\times N$ the integrand equals $2k\abs{u(x)}\abs{u(y)} >
0$ on a set of positive [measure](https://one-course.com/books/math/5/en/chapter/9-measure-theory#def-b3-measure-measure): strict inequality. A $\mu_1$-eigenfunction $u$ maximizes the Rayleigh quotient, and $\abs u$ has the same norm, so strictness would exhibit $R(\abs u) > \mu_1$ — impossible; hence $u$ has constant sign a.e., say $u \geq 0$. Then $u(x) = \mu_1^{-1}(Au)(x) =
\mu_1^{-1}\int k(x,y)u(y)\dd y > 0$ for every $x \in
\intoo01$ ($k(x,\cdot) > 0$ and $u \neq 0$). (c) If the eigenspace had dimension $\geq 2$, it would contain two *orthogonal* eigenfunctions $u, v$, each of constant sign and interior-nonvanishing by (b); but then $\abs{\langle u, v\rangle} = \int\abs u\,\abs v > 0$ — contradiction. On the string: $k = g > 0$ on the open square, $\mu_1 = \lambda_1 = \frac1{\pi^2}$ is indeed simple, $e_1 = \sqrt2\sin(\pi x)$ is positive on $\intoo01$; and each $e_n = \sqrt2\sin(n\pi x)$, $n \geq 2$, orthogonal to the positive $e_1$, must integrate to zero against it, hence changes sign — as its $n - 1$ [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) zeros $\frac kn$ confirm.

**25.** Since $n^2\pi^2 \to \infty$, the minimum $d =
\min_n\abs{n^2\pi^2 - \nu}$ is attained, at some mode $m$, and $d > 0$ because $\nu$ avoids the spectrum. Question 19’s diagonal formula gives

$$
\norm{R_\nu f}_2^2 =
\sum_n\frac{\abs{c_n}^2}{(n^2\pi^2 - \nu)^2}
\leq \frac1{d^2}\,\norm f_2^2,
$$

with equality for $f = e_m$: $\vertiii{R_\nu} = \frac1d$, the reciprocal of the distance from $\nu$ to the spectrum — the general resolvent principle, here in explicit coordinates. [Self-adjointness](#def-b3-spectral-selfadjoint) is read off the real diagonal coefficients; compactness follows as in question 21 (the coefficients tend to $0$, so finite-rank truncations converge in norm). Resonance price: for $f = e_1$ and $\nu =
(1-\varepsilon)\pi^2$, the formula gives $u =
\frac{c_1}{\pi^2 - \nu}e_1 = \frac1{\varepsilon\pi^2}e_1$, against the static response $Ge_1 = \frac1{\pi^2}e_1$: amplification $\frac1\varepsilon$. At $\varepsilon =
10^{-2}$ the response is $100$ times the static one — and it diverges as $\varepsilon \to 0$, which is question 20’s alternative seen from the bounded side: the closer the forcing frequency to a natural one, the less bounded the inverse.