Let $\( X= \mathopen{}\left\{\, f\in \mathrm{C}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}\,\middle\vert\, , f\mathopen{}\left( 1\right)\mathclose{}= 0, \,\right\}\mathclose{} \)$, made into a Banach space by endowing it with the uniform norm: $\( \mathopen{}\left\lVert{}f\right\rVert_\infty\mathclose{}= \max{} \mathopen{}\left\{\, \mathopen{}\left\lvert{}f\mathopen{}\left( t\right)\mathclose{}\right\rvert\mathclose{}\,\middle\vert\, , 0\leq t\leq 1, \,\right\}\mathclose{} \)$. Consider $$\[ E= \mathopen{}\left\{\, g\in \mathrm{C}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}\,\middle\vert\, , g\mathopen{}\left( 1\right)\mathclose{}= 0, , \int _{0}^{1}{}tg\mathopen{}\left( t\right)\mathclose{}\,\mathrm{d}t= 1, \,\right\}\mathclose{}\text{.} \]$$

(a)  Show that $\( E \)$ is a closed convex subset of $\( X \)$.

(b)  Find the distance from $\( 0 \)$ to $\( E \)$; that is, find $\( \inf_{g\in E}{}\mathopen{}\left\lVert{}g\right\rVert\mathclose{}_{\infty} \)$.

(c)  Show that there is no nearest point in $\( E \)$ to $\( 0 \)$; that is, show that there is no $\( g \)$ in $\( E \)$ for which $\( \mathopen{}\left\lVert{}g\right\rVert\mathclose{}_{\infty}= \inf_{g\in E}{}\mathopen{}\left\lVert{}g\right\rVert\mathclose{}_{\infty} \)$.

Let $\( X \)$ be a separable normed linear space and let $\( \mathopen{}\left({x}_{i}\right)\mathclose{}_{i=1}^{\infty} \)$ be a sequence whose closure is the unit ball of $\( X \)$. For $\( f \)$ and $\( g \)$ in $\( X^{*} \)$, define $\( \operatorname{d}\mathopen{}\left( f, g\right)\mathclose{}= \sum{}{2}^{{-}n}\mathopen{}\left\lvert{}f\mathopen{}\left( {x}_{n}\right)\mathclose{}-g\mathopen{}\left( {x}_{n}\right)\mathclose{}\right\rvert\mathclose{} \)$. Show that $\( \operatorname{d} \)$ is a metric on $\( X^{*} \)$ and that $\( \operatorname{d} \)$ gives the $\( \mathop{\mathrm{w}^*} \)$ topology on the unit ball of $\( X^{*} \)$.

Show that a bounded linear operator on a Hilbert space is compact if and only if it takes bounded weakly convergent nets to norm-convergent nets.

Let $\( T \)$ be the Volterra operator on $\( H= \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{} \)$, so $\( \mathopen{}\left(T\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}= \int _{0}^{x}{}f\mathopen{}\left( t\right)\mathclose{}\,\mathrm{d}t \)$.

(a)  Show that $\( T \)$ has no eigenvalues.

(b)  Show that $\( \mathop{\sigma}\mathopen{}\left( T\right)\mathclose{}= \mathopen{}\left\{\, 0\,\right\}\mathclose{} \)$.

(c)  Prove the integration by parts formula $\( T\mathopen{}\left( ψ'T\mathopen{}\left( h\right)\mathclose{}\right)\mathclose{}= ψT\mathopen{}\left( h\right)\mathclose{}-T\mathopen{}\left( ψh\right)\mathclose{} \)$ for $\( ψ\in \mathrm{C}^{1}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{} \)$ and $\( h\in \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{} \)$. Advice: $\( \mathrm{C}^{1}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{} \)$ is dense in $\( \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{} \)$.

(d)  Define $\( {φ}_{λ} \)$ for non-zero $\( λ \)$ by $\( {φ}_{λ}\mathopen{}\left( x\right)\mathclose{}= {\mathrm{e}}^{\frac{{-}x}{λ}} \)$. Show that $\( { \mathopen{}\left(λ-T\right)\mathclose{} }^{-1} \)$ is given on $\( \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{} \)$ by the formula $$\[ { \mathopen{}\left(λ-T\right)\mathclose{} }^{-1}\mathopen{}\left( f\right)\mathclose{}= \frac{1}{λ}f+\frac{1}{{λ}^{2}}\frac{1}{{φ}_{λ}}T\mathopen{}\left( {φ}_{λ}f\right)\mathclose{}\text{.} \]$$

Let $\( T \)$ be the Volterra operator.

(a)  Find an orthonormal basis for $\( H \)$ consisting of eigenvectors for $\( T^{*}T \)$.

(b)  Find $\( \mathopen{}\left\lVert{}T\right\rVert\mathclose{} \)$.

(c)  Calculate $\( \operatorname{Tr}\mathopen{}\left( T^{*}T\right)\mathclose{} \)$ by summing a series. Use $\( \sum_{k=1}^{\infty}{}{k}^{{-}2}= \frac{{\mathrm{\pi}}^{2}}{6} \)$.

(d)  Write $\( T^{*}T \)$ as an integral operator on $\( H \)$ with a continuous kernel, and then use Theorem III.42 to calculate $\( \operatorname{Tr}\mathopen{}\left( T^{*}T\right)\mathclose{} \)$.

(e)  Write a series for $\( \mathopen{}\left\lvert{}T\right\rvert\mathclose{} \)$ that converges in the norm of $\( H \)$ at each $\( f\in H \)$. Express $\( \mathopen{}\left\lvert{}T\right\rvert\mathclose{} \)$ of the constant function $\( \mathbf{1} \)$, i.e. $\( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathbf{1} \)$, as the sum of a uniformly convergent trigonometric series. (As a partial check, you should get $\( \mathopen{}\left(\mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathbf{1}\right)\mathclose{}\mathopen{}\left( 0\right)\mathclose{}= \frac{8}{{\mathrm{\pi}}^{2}}\sum_{n=0}^{\infty}{} \frac{{{-}1}^{n}}{{\mathopen{}\left(2n+1\right)\mathclose{}}^{2}} = \frac{8}{{\mathrm{\pi}}^{2}}G \)$, where $\( G \)$ is Catalan's constant.)

(f)  For $\( f\in H \)$ and $\( α\in \mathbb{C}\setminus \mathop{\sigma}\mathopen{}\left( T^{*}T\right)\mathclose{} \)$, let $\( g= { \mathopen{}\left(1-α T^{*}T\right)\mathclose{} }^{-1}f \)$. (i) Write $\( g \)$ as the sum of a trigonometric series whose coefficients depend on $\( α \)$ and $\( f \)$. (ii) Verify that $\( g \)$ is given by $$\[ g\mathopen{}\left( t\right)\mathclose{}= f\mathopen{}\left( t\right)\mathclose{}+\sqrt{α}\mathopen{}\left(\frac{\cos\mathopen{}\left( \sqrt{α}t\right)\mathclose{}}{\cos\mathopen{}\left( \sqrt{α}\right)\mathclose{}}\int _{0}^{1}{} \sin\mathopen{}\left( \sqrt{α}\mathopen{}\left(1-θ\right)\mathclose{}\right)\mathclose{}f\mathopen{}\left( θ\right)\mathclose{} \,\mathrm{d}θ-\int _{0}^{t}{} \sin\mathopen{}\left( \sqrt{α}\mathopen{}\left(t-θ\right)\mathclose{}\right)\mathclose{}f\mathopen{}\left( θ\right)\mathclose{} \,\mathrm{d}θ\right)\mathclose{}\text{.} \]$$

Advice for (a), (b), (d): This is like the example in Section G, with $\( T^{*}T \)$ in place of $\( K \)$ and the differential operator $\( f\mapsto {-}f'' \)$ with boundary conditions $\( f' \mathopen{}\left( 0\right)\mathclose{}= 0= f\mathopen{}\left( 1\right)\mathclose{} \)$ in place of $\( L \)$.

(a)  Let $\( R \)$ be a ring with identity, and let $\( J \)$ be a proper ideal of $\( R \)$. Show that if $\( x\in J \)$ and $\( 1-x \)$ is invertible, then $\( 1-{\mathopen{}\left(1-x\right)\mathclose{}}^{-1} \)$ is invertible. Deduce that if the Hilbert space operator $\( K \)$ is compact (resp. Hilbert-Schmidt, trace class, finite-dimensional) and $\( 1\notin \mathop{\sigma}\mathopen{}\left( K\right)\mathclose{} \)$, then $\( {\mathopen{}\left(1-K\right)\mathclose{}}^{-1}-1 \)$ is compact (resp. Hilbert-Schmidt, trace class, finite-dimensional).

(b)   In the situation of part (f) of Exercise IV.48, calculate the trace of $\( { \mathopen{}\left(1-α T^{*}T\right)\mathclose{} }^{-1}-1 \)$ as the sum of a series using (i), and in closed form using (ii). (As a check, have Mathematica or the like sum the series exactly.)

Let $\( S\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{} \)$.

(a)  Show that if $\( η\in H \)$ satisfies $\( \mathopen{}\left\lvert{}\mathopen{}\left\langle{}ξ, η\right\rangle\mathclose{}\right\rvert\mathclose{}\leq M\mathopen{}\left\lVert{}Sξ\right\rVert\mathclose{} \)$ for some positive constant $\( M \)$ and all $\( ξ\in H \)$, then $\( η\in S^{*}H \)$. (Advice: That $\( η\in \overline{ S^{*}H } \)$ is easy. If you can get $\( η= \lim_{n}{} S^{*}{ρ}_{n} \)$ for a bounded sequence $\( \mathopen{}\left( {ρ}_{n} \right)\mathclose{} \)$, Alaoglu's Theorem will then close the sale.)

(b)  Suppose that $\( \operatorname{Ker}\mathopen{}\left( S\right)\mathclose{}= \mathopen{}\left\{\, 0\,\right\}\mathclose{}= \operatorname{Ker}\mathopen{}\left( S^{*}\right)\mathclose{} \)$. Define $\( L \)$ on the dense subspace $\( SH \)$ by $\( LSξ= ξ \)$. Show that $\( L^{*} \)$ has domain $\( S^{*}H \)$, and is given there by $\( L^{*} S^{*}ξ= ξ \)$. (In particular, the inverse of a bounded self-adjoint operator with kernel $\( \mathopen{}\left\{\, 0\,\right\}\mathclose{} \)$ is self-adjoint on its natural domain.)

Your mission is to prove the following version of the spectral theorem.

Theorem S: Given a densely defined self-adjoint operator $\( h \)$ in the Hilbert space $\( H \)$, there exist a measure space $\( \mathopen{}\left(X, M, μ\right)\mathclose{} \)$, a measurable function $\( α : X \to \mathopen{}\left({-}\infty, \infty\right)\mathclose{} \)$, and a unitary map $\( U : H \to \mathrm{L}^{\mathrm{2}}\mathopen{}\left( XMμ\right)\mathclose{} \)$ such that $\( U\mathopen{}\left( \mathop{\mathcal{D}}\mathopen{}\left( h\right)\mathclose{}\right)\mathclose{}= \mathopen{}\left\{\, ξ\in \mathrm{L}^{\mathrm{2}}\mathopen{}\left( X\right)\mathclose{}\,\middle\vert\, , α\mathopen{}\left( ξ\right)\mathclose{}\in \mathrm{L}^{\mathrm{2}}\mathopen{}\left( X\right)\mathclose{}, \,\right\}\mathclose{} \)$ and $\( U\mathopen{}\left( h\mathopen{}\left( U^{*}\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}\right)\mathclose{}= αξ \)$ for all $\( ξ\in U\mathopen{}\left( \mathop{\mathcal{D}}\mathopen{}\left( h\right)\mathclose{}\right)\mathclose{} \)$.

In other words, $\( H \)$ is unitarily equivalent to multiplication by a measurable real function in an $\( \mathrm{L}^{\mathrm{2}} \)$-space.

The proof envisioned here requires an important result from measure theory at a level somewhat beyond the assumed prerequisite knowledge for this course. One of several theorems called the Riesz representation theorem says that if $\( Ω \)$ is a compact Hausdorff space and $\( Φ : \mathrm{C}\mathopen{}\left( Ω\right)\mathclose{} \to \mathbb{C} \)$ is a positive linear functional (i.e. $\( f\geq 0 \)$ implies $\( Φ\mathopen{}\left( f\right)\mathclose{}\geq 0 \)$, then there is a $\( σ \)$-algebra $\( M \)$ of subsets of $\( Ω \)$ and a measure $\( μ \)$ defined on $\( Ω \)$ such that (i) $\( M \)$ contains every open subset of $\( Ω \)$ (so every function in $\( \mathrm{C}\mathopen{}\left( Ω\right)\mathclose{} \)$ is measurable); (ii) $\( \mathrm{C}\mathopen{}\left( Ω\right)\mathclose{} \)$ is dense in $\( \mathrm{L}^{\mathrm{2}}\mathopen{}\left( Ω M μ \right)\mathclose{} \)$; and (iii) $\( Φ\mathopen{}\left( f\right)\mathclose{}= \int _{Ω}{}f\,\mathrm{d}μ \)$ for all $\( f\in \mathrm{C}\mathopen{}\left( Ω\right)\mathclose{} \)$.

Start with the case of a bounded self-adjoint operator $\( A \)$ on $\( H \)$. For a vector $\( ξ\in H \)$, call the closure of the linear span of $\( \mathopen{}\left\{\, {A}^{n}\mathopen{}\left( ξ\right)\mathclose{}\,\middle\vert\, , n\geq 0, , n\in \mathbb{N}, \,\right\}\mathclose{} \)$ the cyclic subspace generated by $\( A \)$. If this subspace is all of $\( H \)$, i.e. $\( \overline{ \mathopen{}\left\{\, f\mathopen{}\left( A\right)\mathclose{}ξ\,\middle\vert\, , f\in \mathrm{C}\mathopen{}\left( \mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}\right)\mathclose{}, \,\right\}\mathclose{} }= H \)$, we say that $\( ξ \)$ is a cyclic vector for $\( A \)$.

(a)  Prove Theorem S in the case of a bounded self-adjoint operator $\( A \)$ on $\( H \)$ with a cyclic vector. (Advice: $\( Ω= σ\mathopen{}\left( A\right)\mathclose{} \)$, $\( Φ\mathopen{}\left( f\right)\mathclose{}= \mathopen{}\left\langle{}f\mathopen{}\left( A\right)\mathclose{}ξ, ξ\right\rangle\mathclose{} \)$, and $\( {\mathopen{}\left\lVert{}f\mathopen{}\left( A\right)\mathclose{}ξ\right\rVert\mathclose{}}^{2}= Φ\mathopen{}\left( {\mathopen{}\left\lVert{}f\right\rVert\mathclose{}}^{2}\right)\mathclose{}= {\mathopen{}\left(\mathopen{}\left\lVert{}f\right\rVert\mathclose{}_{ \mathrm{L}^{\mathrm{2}}\mathopen{}\left( Ωμ\right)\mathclose{} }\right)\mathclose{}}^{2} \)$.)

(b)  Show that in general, $\( H \)$ is the direct sum of mutually orthogonal cyclic subspaces for $\( A \)$.

(c)  Prove Theorem S for any bounded self-adjoint $\( A \)$. (Hint: The direct sum of measure spaces $\( \mathopen{}\left({X}_{i}, {M}_{i}, {μ}_{i}\right)\mathclose{} \)$ is the disjoint union of the $\( {X}_{i} \)$, with the obvious measure concocted from the $\( {μ}_{i} \)$ on the obvious $\( σ \)$-algebra determined by the $\( {M}_{i} \)$.)

Now let $\( h \)$ be a self-adjoint operator in $\( H \)$. Form the bounded operator $\( A \)$ on $\( H \)$ as in Section A, that is, $\( A= h{\mathopen{}\left(1+{h}^{2}\right)\mathclose{}}^{{-}\frac{1}{2}} \)$, which as we have seen entails $\( A= A^{*} \)$, $\( {-}1\leq A\leq 1 \)$, $\( h= A{\mathopen{}\left(1-{A}^{2}\right)\mathclose{}}^{\frac{1}{2}} \)$, and $\( \operatorname{Ker}\mathopen{}\left( 1-{A}^{2}\right)\mathclose{}= \mathopen{}\left\{\, 0\,\right\}\mathclose{} \)$. Invoke (c) for $\( A \)$ to get $\( \mathopen{}\left(X, M, μ\right)\mathclose{} \)$, $\( α \)$, and $\( U \)$ doing what they're supposed to do. Move everything over to $\( \mathrm{L}^{\mathrm{2}}\mathopen{}\left( X, μ\right)\mathclose{} \)$. Relabel $\( UhU \)$ as $\( h \)$, now acting on $\( H= \mathrm{L}^{\mathrm{2}}\mathopen{}\left( X, μ\right)\mathclose{} \)$, and let $\( A \)$ be multiplication by $\( α \)$ on the new $\( H \)$. Notice that $\( {-}1\lt α\lt 1 \)$ (after redefining on a set of measure zero, if one is fussy). Consider now the measurable function $\( β \)$ defined by $\( β= α{\mathopen{}\left(1-{α}^{2}\right)\mathclose{}}^{{-}\frac{1}{2}} : X \to \mathopen{}\left({-}\infty, \infty\right)\mathclose{} \)$. Let $\( \mathrm{M}_{β} \)$ be multiplication by $\( β \)$, with domain $\( \mathop{\mathcal{D}}\mathopen{}\left( \mathrm{M}_{β}\right)\mathclose{}= \mathopen{}\left\{\, ξ\in \mathrm{L}^{\mathrm{2}}\mathopen{}\left( Xμ\right)\mathclose{}\,\middle\vert\, , βξ\in \mathrm{L}^{\mathrm{2}}\mathopen{}\left( Xμ\right)\mathclose{}, \,\right\}\mathclose{} \)$. It is useful at this point to introduce the subspaces $\( {H}_{τ}= \mathrm{L}^{\mathrm{2}}\mathopen{}\left( {β}^{-1}\mathopen{}\left( \mathopen{}\left[{-}τ, τ\right]\mathclose{}\right)\mathclose{}\right)\mathclose{} \)$ for $\( τ\gt 0 \)$.

(d)   Show that $\( \mathrm{M}_{β} \)$ is self-adjoint. (Hint: Modify certain of the arguments in Section A.)

(e)   Show that $\( \mathop{\mathcal{G}}\mathopen{}\left( \mathrm{M}_{β}\right)\mathclose{}= \mathop{\mathcal{G}}\mathopen{}\left( h\right)\mathclose{} \)$, thus finishing the proof of Theorem S. (Same hint.)

For compactly supported continuous functions $\( f : {\mathbb{R}}^{2} \to \mathbb{C} \)$, define $\( {S}_{f} \)$ on $\( \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathbb{R}\right)\mathclose{} \)$ by $\( \mathopen{}\left({S}_{f}\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( t\right)\mathclose{}= \int _{\mathbb{R}}{} f\mathopen{}\left( t, t-s\right)\mathclose{}ξ\mathopen{}\left( s\right)\mathclose{} \,\mathrm{d}s \)$.

(a)  Show that $\( {S}_{f}\in \mathcal{L}\mathopen{}\left( \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathbb{R}\right)\mathclose{}\right)\mathclose{} \)$, with $\( \mathopen{}\left\lVert{}{S}_{f}\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{}f\right\rVert\mathclose{}_{\mathrm{L}^{\mathrm{2}}\mathopen{}\left( {\mathbb{R}}^{2}\right)\mathclose{}} \)$.

(b)  Let $\( K= \mathopen{}\left\{\, {S}_{f}\,\middle\vert\, , f\in {\mathrm{C}}_{0}\mathopen{}\left( {\mathbb{R}}^{2}\right)\mathclose{}, \,\right\}\mathclose{} \)$. Show that the closure of $\( K \)$ (in the operator norm) is $\( \mathcal{K}\mathopen{}\left( \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathbb{R}\right)\mathclose{}\right)\mathclose{} \)$.

Let $\( S \)$ be the unilateral shift on $\( \mathrm{l}^{0} \)$. That is, $\( S\mathopen{}\left( \mathbf{a}\right)\mathclose{}= S\mathopen{}\left( {a}_{1}, {a}_{2}, {a}_{3}, \dotsc\right)\mathclose{}= \mathopen{}\left(0, {a}_{1}, {a}_{2}, \dotsc\right)\mathclose{} \)$. Show that $\( \mathop{\sigma}\mathopen{}\left( S+ S^{*}\right)\mathclose{}= \mathopen{}\left[{-}2, 2\right]\mathclose{} \)$. (Advice: The inclusion $\( \mathop{\sigma}\mathopen{}\left( S+ S^{*}\right)\mathclose{}\subseteq \mathopen{}\left[{-}2, 2\right]\mathclose{} \)$ is clear. To show $\( \mathopen{}\left[{-}2, 2\right]\mathclose{}\subseteq \mathop{\sigma}\mathopen{}\left( S+ S^{*}\right)\mathclose{} \)$, try solving $\( \mathopen{}\left(S+ S^{*}-λ\right)\mathclose{}\mathopen{}\left( \mathbf{a}\right)\mathclose{}= \mathopen{}\left(1, 0, 0, \dotsc\right)\mathclose{} \)$, say, for $\( \mathbf{a}\in \mathrm{l}^{0} \)$. The algebraic solution of the second-order linear recurrence you encounter is governed by the eigenvalues and eigenvectors of the two-by-two matrix relating $\( \mathopen{}\left({a}_{n+2}, {a}_{n+1}\right)\mathclose{} \)$ to $\( \mathopen{}\left({a}_{n+1}, {a}_{n}\right)\mathclose{} \)$.)

This exercise has to do with the order $\( 0 \)$ Bessel function of the first kind, that is, the function $\( {J}_{0} \)$ given as a power series by $\( {J}_{0}\mathopen{}\left( x\right)\mathclose{}= \sum_{n=0}^{\infty}{} \frac{{\mathopen{}\left({-}1\right)\mathclose{}}^{n}}{{\mathopen{}\left({n!}\right)\mathclose{}}^{2}}{\mathopen{}\left(\frac{x}{2}\right)\mathclose{}}^{2n} \)$. It is up to scalar multiple the only solution without singularity at $\( x= 0 \)$ of Bessel's equation of order $\( 0 \)$: $\( xy'' \mathopen{}\left( x\right)\mathclose{}+y' \mathopen{}\left( x\right)\mathclose{}+xy\mathopen{}\left( x\right)\mathclose{}= 0 \)$. The main thing here is to show that the sum of the squared reciprocals of the positive zeros of $\( {J}_{0} \)$ is $\( \frac{1}{4} \)$. Along the way, an orthonormal basis for $\( \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{} \)$ will emerge that is indispensable for modeling the small oscillations of a flexible chain hung from a hook. (See Chapter 9 of Young's book [12].)

(a)  Let $\( {D}_{0}= \mathopen{}\left\{\, f\in \mathrm{C}^{1}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}\,\middle\vert\, , x\mapsto xf' \mathopen{}\left( x\right)\mathclose{}\in \mathrm{C}^{1}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}, , f\mathopen{}\left( 1\right)\mathclose{}= 0, \,\right\}\mathclose{} \)$. Define $\( L : {D}_{0} \to \mathrm{C}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{} \)$ by $$\[ \mathopen{}\left(L\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}= {-}xf'' \mathopen{}\left( x\right)\mathclose{}-f' \mathopen{}\left( x\right)\mathclose{}= {-} \frac{\mathrm{d}}{\mathrm{d}x}\mathopen{}\left( xf' \mathopen{}\left( x\right)\mathclose{}\right)\mathclose{} \text{.} \]$$ Show that the eigenvalues of $\( L \)$ are $\( \mathopen{}\left\{\, \frac{{β}^{2}}{4}\,\middle\vert\, , {J}_{0}\mathopen{}\left( β\right)\mathclose{}= 0, \,\right\}\mathclose{} \)$, with corresponding eigenfunctions $\( x\mapsto {J}_{0}\mathopen{}\left( β\sqrt{x}\right)\mathclose{} \)$. (Advice: $\( L\mathopen{}\left( f\right)\mathclose{}= λf \)$ makes $\( λ{\mathopen{}\left\lVert{}f\right\rVert\mathclose{}}^{2}= \mathopen{}\left\langle{}L\mathopen{}\left( f\right)\mathclose{}, f\right\rangle\mathclose{} \)$. Integrate by parts to show the eigenvalues are positive. For $\( λ\gt 0 \)$, show that $\( xf'' \mathopen{}\left( x\right)\mathclose{}+f' \mathopen{}\left( x\right)\mathclose{}+λf\mathopen{}\left( x\right)\mathclose{}= 0 \)$ if and only if $\( y= f\mathopen{}\left( 2\sqrt{λx}\right)\mathclose{} \)$ satisfies Bessel's equation of order $\( 0 \)$.)

(b)  For $\( ξ\in H= \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{} \)$, define $\( K\mathopen{}\left( ξ\right)\mathclose{} \)$ on $\( \mathopen{}\left[0, 1\right]\mathclose{} \)$ by $$\[ \mathopen{}\left(K\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}= {-} \log\mathopen{}\left( x\right)\mathclose{} \int _{0}^{x}{}ξ\mathopen{}\left( t\right)\mathclose{}\,\mathrm{d}t-\int _{x}^{1}{} \log\mathopen{}\left( t\right)\mathclose{}ξ\mathopen{}\left( t\right)\mathclose{} \,\mathrm{d}t\text{.} \]$$ Show that $\( K \)$ is a self-adjoint Hilbert-Schmidt operator on $\( H \)$ mapping $\( \mathrm{C}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{} \)$ to $\( {D}_{0} \)$, that $\( L\mathopen{}\left( K\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}= ξ \)$ for $\( ξ\in \mathrm{C}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{} \)$, and that $\( K\mathopen{}\left( L\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}= f \)$ for $\( f\in {D}_{0} \)$. (Advice: Exhibit $\( K \)$ as an integral operator with kernel $\( k\in \mathrm{L}^{\mathrm{2}}\mathopen{}\left( {\mathopen{}\left[0, 1\right]\mathclose{}}^{2}\right)\mathclose{} \)$ satisfying $\( k\mathopen{}\left( x, t\right)\mathclose{}= k\mathopen{}\left( t, x\right)\mathclose{} \)$. Notice $\( \log\in H \)$ and use L'Hôpital's Rule at $\( x= 0 \)$ when necessary.)

(c)  Let $\( {β}_{i} \)$, $\( i\in \mathopen{}\left\{\, 1, 2, \dotsc\,\right\}\mathclose{} \)$, denote the positive zeros of $\( {J}_{0} \)$. Define $\( {φ}_{j} \)$ on $\( \mathopen{}\left[0, 1\right]\mathclose{} \)$ by $$\[ {φ}_{j}\mathopen{}\left( x\right)\mathclose{}= \frac{{J}_{0}\mathopen{}\left( {β}_{j}\sqrt{x}\right)\mathclose{}}{{\mathopen{}\left(\int _{0}^{1}{}{J}_{0}{\mathopen{}\left({β}_{j}\sqrt{x}\right)\mathclose{}}^{2}\,\mathrm{d}x\right)\mathclose{}}^{\frac{1}{2}}}\text{.} \]$$ Show that $\( \mathopen{}\left\{\, {φ}_{1}, {φ}_{2}, \dotsc\,\right\}\mathclose{} \)$ is an orthonormal basis for $\( H \)$ with $\( K\mathopen{}\left( {φ}_{j}\right)\mathclose{}= \frac{4}{{{β}_{j}}^{2}}{φ}_{j} \)$ for every $\( j \)$. (This includes showing that $\( \operatorname{Ker}\mathopen{}\left( K\right)\mathclose{}= \mathopen{}\left\{\, 0\,\right\}\mathclose{} \)$, and accounting for all of the nonzero eigenvalues of $\( K \)$.)

It is experimentally obvious, and not very difficult to show, that $\( {β}_{j}\gt j \)$ for every $\( j \)$. It follows that $\( K \)$ is a trace class operator, with trace $\( \sum_{j}{} {{β}_{j}}^{{-}2} \)$. We would like to identify this number in a more vivid way by integrating the kernel you found in part (b) along the diagonal. Unfortunately, though, the kernel has a singularity at $\( \mathopen{}\left(0, 0\right)\mathclose{} \)$, and proof of the relevant theorem strongly uses uniform continuity of the kernel. To save the calculation, we have to approximate $\( K \)$ by operators with kernels that are continuous everywhere on the square.

(d)  For $\( n\geq 2 \)$, $\( n\in \mathbb{N} \)$, let $\( {Λ}_{n} \)$ be the function on $\( \mathopen{}\left[0, 1\right]\mathclose{} \)$ that is $\( \log\mathopen{}\left( n\right)\mathclose{} \)$ on $\( \mathopen{}\left[0, \frac{1}{n}\right]\mathclose{} \)$ and $\( {-}\log\mathopen{}\left( n\right)\mathclose{} \)$ on $\( \mathopen{}\left(\frac{1}{n}, 1\right]\mathclose{} \)$. Define the operator $\( {K}_{n} \)$ by $$\[ \mathopen{}\left({K}_{n}, ξ\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}= {Λ}_{n}\mathopen{}\left( x\right)\mathclose{}\int _{0}^{x}{}ξ\mathopen{}\left( t\right)\mathclose{}\,\mathrm{d}t+\int _{x}^{1}{} {Λ}_{n}\mathopen{}\left( t\right)\mathclose{}ξ\mathopen{}\left( t\right)\mathclose{} \,\mathrm{d}t\text{.} \]$$ You can easily write down the kernel for $\( {K}_{n} \)$ and see thereby that $\( {K}_{n} \)$ is a self-adjoint Hilbert-Schmidt operator on $\( H \)$. Consider also $\( {Δ}_{n} \)$ defined by $$\[ \mathopen{}\left({Δ}_{n}\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}= \begin{cases}0, & x\gt \frac{1}{n} ; \\ {-}\log\mathopen{}\left( x\right)\mathclose{}\int _{0}^{x}{}ξ\mathopen{}\left( t\right)\mathclose{}\,\mathrm{d}t-\int _{x}^{\frac{1}{n}}{} \log\mathopen{}\left( t\right)\mathclose{}ξ\mathopen{}\left( t\right)\mathclose{} \,\mathrm{d}t, & x\leq \frac{1}{n} \end{cases}\text{.} \]$$ Show that $\( K= {K}_{n}+{Δ}_{n}-\log\mathopen{}\left( n\right)\mathclose{}\chi_{ \mathopen{}\left[0, \frac{1}{n}\right]\mathclose{} }\otimes \chi_{ \mathopen{}\left[0, \frac{1}{n}\right]\mathclose{} } \)$.

(e)  Show that $\( {Δ}_{n} \)$ is trace class and calculate its trace in terms of the Bessel zeros $\( {β}_{j} \)$. (Advice: Notice that $\( {Δ}_{n} \)$ is the operator obtained by rescaling $\( K \)$ from $\( \mathopen{}\left[0, 1\right]\mathclose{} \)$ to $\( \mathopen{}\left[0, \frac{1}{n}\right]\mathclose{} \)$.)

(f)  It now follows that $\( {K}_{n} \)$ is trace class. Find its trace by integrating its kernel function along the diagonal.

(g)  Show that $\( \sum_{j}{} {{β}_{j}}^{{-}2} = \frac{1}{4} \)$.