University Mathematics — Year 3 · Bachelor Year 3

15Compact 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 in general — multiplication by $x$ on $L^2(\intcc01)$ has no eigenvalues at all (Exercise 15.6) — but it survives, in perfect form, for the operators that are almost finite dimensional: the compact ones. The spectral theorem for compact self-adjoint operators is the single most used theorem of applied functional analysis: it diagonalizes integral equations, drives the Fredholm alternative, 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 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 $\mathcal K(E, F)$ form a closed subspace of $\mathcal L(E, F)$ , and a two-sided ideal: $S$ compact $\Rightarrow$ $AS$ and $SB$ compact for bounded $A, B$ . Moreover, in a Hilbert space, every compact operator is a norm-limit of finite-rank operators.

Proof. Subspace: clear from the sequence characterization. Ideal: bounded maps send convergent sequences to convergent ones and bounded to bounded. Closedness: let $T_n \to T$ with $T_n$ 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: let $T$ be compact, $K = \overline{T(B)}$ 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 iff $\lambda_n \to 0$ (Exercise 15.2). (b) Kernel operators on $\mathcal C(\intcc01)$ : compact by Ascoli (Exercise 7.7). (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 on $L^2(\intcc01)$ with $\vertiii{T_k} \leq \norm k_{L^2}$ (Exercise 15.4: 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), 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 $T$ :

\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) 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 and self-adjoint. Then $\vertiii T$ or $-\vertiii T$ is an eigenvalue of $T$ .

Proof. By Proposition 15.5, 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 gives $Tx = \mu x$ . ∎

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

Let $T$ be a compact self-adjoint operator on a Hilbert space $H$ .

$H$ 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 $H = \ker T \,\oplus^\perp\, \overline{\operatorname{Vect}}(e_n : n \in N)$ .
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.
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) 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 self-adjoint on the Hilbert space $H_0^\perp$ ; if $T' \neq 0$ , Lemma 15.6 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(1) on $H_0$ ), 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); the union is an orthonormal basis of $H$ by the decomposition in (1). ∎

Theorem 15.8 (Fredholm alternative)

Let $T$ be compact self-adjoint and $\lambda \in \R\setminus \{0\}$ .

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 on $f$ .
If $\lambda$ is an eigenvalue, $Tx - \lambda x = f$ is solvable 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 ( $z, w \in \ker T$ ). 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 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 with real symmetric kernel: compact and self-adjoint. 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 $f$ , and $Tf$ is continuous 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 (Hilbert–Schmidt kernel, or Ascoli) and, if possible, self-adjoint (symmetric real kernel); (3) diagonalize with the spectral theorem or invoke the Fredholm alternative for solvability; (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: always invert first.

15.4 Exercises

Exercise 15.1 ★

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

Solution

Solution of Exercise 15.1.

(a) $T(B)$ is a bounded subset of the finite-dimensional $\operatorname{im}T$ : relatively compact by Heine–Borel (Corollary 6.17, transported by a linear homeomorphism with $\R^n$ ). (b) $I$ compact means the closed unit ball is compact, which by Riesz’s theorem (Year 2) happens exactly in finite dimension. If a compact $T$ had bounded inverse $T^{-1}$ , then $I = T^{-1}T$ would be compact (Proposition 15.2): 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 iff $\lambda_n \to 0$ . (For $\Leftarrow$ , truncate; for $\Rightarrow$ , test on $(e_n)$ .) (c) When is $T$ self-adjoint? Verify the spectral theorem by inspection in that case.

Solution

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 by Proposition 15.2. ( $\Rightarrow$ ) 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 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 property (Proposition 15.2): if $S$ is compact and $A, B$ bounded, then $ASB$ is compact. Deduce that if $ST = TS = I$ for some bounded $S$ , and $\dim H = \infty$ , then $T$ is not compact — and reconcile with Exercise 15.1(b).

Solution

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 of $A$ gives $ASBx_{n_k} \to Ay$ : $ASB$ is compact. If $ST = TS = I$ with $T$ compact and $\dim H = \infty$ : $I = ST$ would be compact, contradicting Exercise 15.1(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 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 there), and show that truncating the sum gives finite-rank operators converging to $T_k$ in operator norm: $T_k$ is compact.

Solution

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 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; 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 (Proposition 15.2).

Exercise 15.5 ★★

Let $T$ be self-adjoint 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

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 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, with $\vertiii M = 1$ , but has no eigenvalues. (b) Show $M$ is not 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’s proof break for $M$ ?

Solution

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 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, 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 (Hilbert–Schmidt with kernel $\mathbf 1_{y < x}$ ) but not self-adjoint; compute $V^*$ . (b) Show that $V$ has no nonzero eigenvalue. (From $Vf = \lambda f$ : $f$ has a continuous representative, then is $\mathcal C^1$ , and solves $\lambda f' = f$ , $f(0) = 0$ .) (c) Conclude that compactness alone yields no eigenvectors: self-adjointness in Theorem 15.7 is essential.

Solution

Solution of Exercise 15.7.

(a) $V = T_k$ with $k(x, y) = \mathbf 1_{y < x} \in L^2(\intcc01^2)$ : compact (Exercise 15.4). 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 on $\intcc01$ (dominated convergence in $\int_0^x f$ ), so $f = \frac1\lambda Vf$ has a continuous representative; then $Vf$ is $\mathcal C^1$ (fundamental theorem of calculus for continuous 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 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, not just compactness.

Exercise 15.8 ★★★

(Courant–Fischer) Let $T$ be compact, self-adjoint, 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

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 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 for $Tf(x) = \int_0^1\min(x,y)f(y)\dd y$ (Example 15.9): 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

Solution of Exercise 15.9.

Rewrite $f - \lambda Tf = g$ . For $\lambda = 0$ : $f = g$ , always uniquely solvable. For $\lambda \neq 0$ : this is $(T - \frac1\lambda)f = -\frac g\lambda$ , and by the Fredholm alternative (Theorem 15.8) with the eigenvalues $\lambda_n = \bigl((n + \frac12)\pi\bigr)^{-2}$ of $T$ (Example 15.9): unique solvability 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). (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: verify via Exercise 15.2-style testing on $(e_n)$ . (c) Comment: for non-self-adjoint, non-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

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. (c) For compact self-adjoint operators the eigenvalues capture everything (Theorem 15.7); 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, self-adjoint, positive ( $\langle x, Tx\rangle \geq 0$ ) on a Hilbert space $H$ , 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, self-adjoint, positive, with $S^2 = T$ . (b) Prove uniqueness: any compact positive self-adjoint $R$ 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: 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

Solution of Exercise 15.11.

(a) $S$ is the diagonal operator with coefficients $\sqrt{\mu_n} \to 0$ : compact (Exercise 15.2(b), transported to the basis $(e_n)$ completed by $\ker T$ , where $S = 0$ ), self-adjoint (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): $R = S$ .

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, not necessarily self-adjoint. (a) Show that $T^*T$ is compact, self-adjoint, 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’s converse for Hilbert spaces); and for the Volterra operator $V$ of Exercise 15.7, 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’s territory).

Solution

Solution of Exercise 15.12.

(a) $T^*T$ is compact (product of a bounded and a compact operator, Exercise 15.3), self-adjoint ( $(T^*T)^* = T^*T$ ), 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 in the closed span plus kernel); applying the continuous $T$ :

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), 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’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 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.

Show that $g$ is continuous, symmetric, with $0 \leq g \leq \frac14$ , and that $G$ is compact and self-adjoint (Example 15.3(c)).
For $f$ continuous, 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.
Show $\ker G = \{0\}$ (if $Gf = 0$ with $f \in L^2$ : test against continuous $\varphi$ , transfer $G$ by symmetry/Fubini onto $\varphi$ , and use the fundamental lemma Corollary 12.11 — or regularize), and that $G$ is a positive operator: $\langle f, Gf\rangle \geq 0$ . (For continuous $f$ : $\langle f, Gf\rangle = \int_0^1 (u')^2$ with $u = Gf$ , by parts; conclude by density.)

Part II — Diagonalization: the sine basis.

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 representative — $Gf$ is continuous for $f \in L^2$ , why? — hence is $\mathcal C^2$ by bootstrapping question 2).
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.
Conclude from Theorem 15.7 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 for it.

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

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 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.
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.
Re-derive $\zeta(2)$ a third way: apply 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 applied to the whole kernel at once”?

Part IV — The string vibrates.

(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 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 in one paragraph.

Part V — Variational dividends: the power method, Weyl stability, and a rigorous bound on $\pi$ . Let $A$ be a compact self-adjoint 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; here we spend them.

(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.
(Weyl stability) For compact self-adjoint positive $A, B$ , deduce from Exercise 15.8 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 — eigenvalues of big symmetric systems can be computed from approximations with guaranteed error.
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. $\pi \leq \sqrt{168/17} < 3.1437$ .
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 $\zeta(4)$ .

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, cf. Exercise 15.5 — and apply Parseval on the square; question 7 identifies the coefficients).
Compute the double integral:

$\iint g^2 = 2\int_0^1(1-x)^2\Bigl(\int_0^x y^2\,\dd y\Bigr)\dd x = \frac23\int_0^1x^3(1-x)^2\,\dd x = \frac1{90} .$
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.
( $\pi$ 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 $c_n = \langle e_n, f\rangle$ .

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 (Theorem 15.8), with uniqueness.
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.
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, self-adjoint, and positive: the whole spectral analysis restarts, shifted by $\nu$ .
(Synthesis) Compile the dictionary of this problem: eigenvalue $\leftrightarrow$ squared frequency (harmonics); trace $\leftrightarrow$ $\zeta(2)$ ; Hilbert–Schmidt norm $\leftrightarrow$ $\zeta(4)$ ; Fredholm alternative $\leftrightarrow$ 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.

( $\zeta(6)$ , for free) Question 6’s 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 on one well-chosen function is the cheaper route here.
(The ground state is positive) Let $A$ be a compact self-adjoint positive operator on $L^2(\intcc01)$ given by a continuous 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 — 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 zeros).
(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, 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

Solution of Problem 15.1.

1. Continuity: $\min$ and $\max$ are continuous; 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 (Exercise 15.4); the kernel is real symmetric: $G$ self-adjoint.

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 $f$ , 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): by the fundamental lemma (Corollary 12.11), $f = 0$ a.e. Positivity: for continuous $f$ , 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 $f_n$ : both sides pass to the limit ( $G$ bounded).

4. If $Gu = \lambda u$ , $\lambda \neq 0$ : $Gu$ is continuous ( $\abs{Gu(x) - Gu(x')} \leq \norm{g(x,\cdot) - g(x',\cdot)}_2\norm u_2$ , and the kernel is uniformly continuous), so $u$ has a continuous 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(1) gives $H = \overline{\operatorname{Vect}}(e_n)$ : the sines are a Hilbert basis 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: $\int_0^1x^2(1-x)^2\dd x = \frac1{30} = \sum_{n \text{ odd}}\frac{32}{n^6\pi^6}$ , 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, and for each $x$ it has the same $L^2(\dd y)$ -coefficients as $g(x, \cdot)$ : the two continuous 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: $1 = \sum_{n\text{ odd}}\frac{8}{n^2\pi^2}$ , 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 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 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(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 in $t$ (finitely many cosines): $t \mapsto u(t,\cdot)$ is continuous 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: $\mu_n(A) \leq \mu_n(B) + \vertiii{A - B}$ , 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 of $L^2(\intcc01^2)$ (orthonormality by Tonelli; totality as in Exercise 15.5). 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 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 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 $\frac1{\pi^2 - \nu}$ . 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 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 $\leftrightarrow$ 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 self-adjoint operator, 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 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, then on $P\times N$ the integrand equals $2k\abs{u(x)}\abs{u(y)} > 0$ on a set of positive 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 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 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.

15.1 Compact operators

15.2 Self-adjoint operators

15.3 The spectral theorem

15.4 Exercises

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

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