Lecture Notes in Functional Analysis

by William L. Paschke

edition 0.9




I. Hilbert Space

II. Bounded Operators

III. Compact Operators

IV. The Spectral Theorem

Index and References

B. Additional Exercises

Exercise IV.44

Let XBanach space=equals{setfcontinuous functionelement ofCspace of continuous functions([interval0zero, 1one]interval)|such that fcontinuous function(1one)=equals0zero }set \( X= \mathopen{}\left\{\, f\in \mathrm{C}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}\,\middle\vert\, , f\mathopen{}\left( 1\right)\mathclose{}= 0, \,\right\}\mathclose{} \), made into a Banach space by endowing it with the uniform norm: fcontinuous function=equalsmaxmaximum {set|modulusfcontinuous function(treal number)|modulus|such that 0zeroless than or equal totreal numberless than or equal to1one }set \( \mathopen{}\left\lVert{}f\right\rVert_\infty\mathclose{}= \max{} \mathopen{}\left\{\, \mathopen{}\left\lvert{}f\mathopen{}\left( t\right)\mathclose{}\right\rvert\mathclose{}\,\middle\vert\, , 0\leq t\leq 1, \,\right\}\mathclose{} \). Consider Eclosed subspace=equals{setgcontinuous functionelement ofCspace of continuous functions([interval0zero, 1one]interval)|such that gcontinuous function(1one)=equals0zero integral0zero1onetreal numbertimesgcontinuous function(treal number)dtreal number=equals1one }set. \[ E= \mathopen{}\left\{\, g\in \mathrm{C}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}\,\middle\vert\, , g\mathopen{}\left( 1\right)\mathclose{}= 0, , \int _{0}^{1}{}tg\mathopen{}\left( t\right)\mathclose{}\,\mathrm{d}t= 1, \,\right\}\mathclose{}\text{.} \]

(a)  Show that Eclosed subspace\( E \) is a closed convex subset of XBanach space\( X \).

(b)  Find the distance from 0zero\( 0 \) to Eclosed subspace\( E \); that is, find infinfimumgcontinuous functionelement ofEclosed subspacegcontinuous functioninfinity \( \inf_{g\in E}{}\mathopen{}\left\lVert{}g\right\rVert\mathclose{}_{\infty} \).

(c)  Show that there is no nearest point in Eclosed subspace\( E \) to 0zero\( 0 \); that is, show that there is no gcontinuous function\( g \) in Eclosed subspace\( E \) for which gcontinuous functioninfinity=equalsinfinfimumgcontinuous functionelement ofEclosed subspacegcontinuous functioninfinity \( \mathopen{}\left\lVert{}g\right\rVert\mathclose{}_{\infty}= \inf_{g\in E}{}\mathopen{}\left\lVert{}g\right\rVert\mathclose{}_{\infty} \).

Exercise IV.45

Let Xseparable normed linear space\( X \) be a separable normed linear space and let (sequencexvectoriinteger)sequenceiinteger=1oneinfinity \( \mathopen{}\left({x}_{i}\right)\mathclose{}_{i=1}^{\infty} \) be a sequence whose closure is the unit ball of Xseparable normed linear space\( X \). For ffunctional\( f \) and gfunctional\( g \) in Xseparable normed linear space* \( X^{*} \), define d(ffunctionalgfunctional)=equalssummation2twonintegertimes|modulusffunctional(xvectorninteger)-minusgfunctional(xvectorninteger)|modulus \( \operatorname{d}\mathopen{}\left( f, g\right)\mathclose{}= \sum{}{2}^{{-}n}\mathopen{}\left\lvert{}f\mathopen{}\left( {x}_{n}\right)\mathclose{}-g\mathopen{}\left( {x}_{n}\right)\mathclose{}\right\rvert\mathclose{} \). Show that d\( \operatorname{d} \) is a metric on Xseparable normed linear space* \( X^{*} \) and that d\( \operatorname{d} \) gives the w*\( \mathop{\mathrm{w}^*} \) topology on the unit ball of Xseparable normed linear space* \( X^{*} \).

Exercise IV.46

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.

Exercise IV.47

Let TVolterra operator\( T \) be the Volterra operator on HHilbert space=equalsL2Lebesgue space([interval0zero, 1one]interval) \( H= \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{} \), so (TVolterra operator(fcontinuous function))(xvector)=equalsintegral0zeroxvectorfcontinuous function(treal number)dtreal number \( \mathopen{}\left(T\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}= \int _{0}^{x}{}f\mathopen{}\left( t\right)\mathclose{}\,\mathrm{d}t \).

(a)  Show that TVolterra operator\( T \) has no eigenvalues.

(b)  Show that σ(TVolterra operator)=equals{set0zero}set \( \mathop{\sigma}\mathopen{}\left( T\right)\mathclose{}= \mathopen{}\left\{\, 0\,\right\}\mathclose{} \).

(c)  Prove the integration by parts formula TVolterra operator(ψfunctiontimesTVolterra operator(hfunction))=equalsψfunctiontimesTVolterra operator(hfunction)-minusTVolterra operator(ψfunctiontimeshfunction) \( T\mathopen{}\left( ψ'T\mathopen{}\left( h\right)\mathclose{}\right)\mathclose{}= ψT\mathopen{}\left( h\right)\mathclose{}-T\mathopen{}\left( ψh\right)\mathclose{} \) for ψfunctionelement ofC1space of continuously differentiable functions([interval0zero, 1one]interval) \( ψ\in \mathrm{C}^{1}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{} \) and hfunctionelement ofL2Lebesgue space([interval0zero, 1one]interval) \( h\in \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{} \). Advice: C1space of continuously differentiable functions([interval0zero, 1one]interval) \( \mathrm{C}^{1}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{} \) is dense in L2Lebesgue space([interval0zero, 1one]interval) \( \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{} \).

(d)  Define φfunctionλcomplex number \( {φ}_{λ} \) for non-zero λcomplex number\( λ \) by φfunctionλcomplex number(xvector)=equals eEuler's constant xvector λcomplex number \( {φ}_{λ}\mathopen{}\left( x\right)\mathclose{}= {\mathrm{e}}^{\frac{{-}x}{λ}} \). Show that (λcomplex number-minusTVolterra operator) 1inverse \( { \mathopen{}\left(λ-T\right)\mathclose{} }^{-1} \) is given on L2Lebesgue space([interval0zero, 1one]interval) \( \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{} \) by the formula (λcomplex number-minusTVolterra operator) 1inverse(fcontinuous function)=equals1oneλcomplex numbertimesfcontinuous function+plus1oneλcomplex number2twotimes1oneφfunctionλcomplex numbertimesTVolterra operator(φfunctionλcomplex numbertimesfcontinuous function). \[ { \mathopen{}\left(λ-T\right)\mathclose{} }^{-1}\mathopen{}\left( f\right)\mathclose{}= \frac{1}{λ}f+\frac{1}{{λ}^{2}}\frac{1}{{φ}_{λ}}T\mathopen{}\left( {φ}_{λ}f\right)\mathclose{}\text{.} \]

Exercise IV.48

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

(a)  Find an orthonormal basis for HHilbert space\( H \) consisting of eigenvectors for TVolterra operator*timesTVolterra operator \( T^{*}T \).

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

(c)  Calculate Trtrace(TVolterra operator*timesTVolterra operator) \( \operatorname{Tr}\mathopen{}\left( T^{*}T\right)\mathclose{} \) by summing a series. Use summationkinteger=1oneinfinitykinteger2two=equals πpi2two 6six \( \sum_{k=1}^{\infty}{}{k}^{{-}2}= \frac{{\mathrm{\pi}}^{2}}{6} \).

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

(e)  Write a series for |modulusTVolterra operator|modulus \( \mathopen{}\left\lvert{}T\right\rvert\mathclose{} \) that converges in the norm of HHilbert space\( H \) at each fcontinuous functionelement ofHHilbert space \( f\in H \). Express |modulusTVolterra operator|modulus \( \mathopen{}\left\lvert{}T\right\rvert\mathclose{} \) of the constant function 1identity function \( \mathbf{1} \), i.e. |modulusTVolterra operator|modulustimes1identity function \( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathbf{1} \), as the sum of a uniformly convergent trigonometric series. (As a partial check, you should get (|modulusTVolterra operator|modulustimes1identity function)(0zero)=equals8eightπpi2twotimessummationninteger=0zeroinfinity 1oneninteger (2twotimesninteger+plus1one) 2two =equals8eightπpi2twotimesGCatalan's constant \( \mathopen{}\left(\mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathbf{1}\right)\mathclose{}\mathopen{}\left( 0\right)\mathclose{}= \frac{8}{{\mathrm{\pi}}^{2}}\sum_{n=0}^{\infty}{} \frac{{{-}1}^{n}}{{\mathopen{}\left(2n+1\right)\mathclose{}}^{2}} = \frac{8}{{\mathrm{\pi}}^{2}}G \), where GCatalan's constant\( G \) is Catalan's constant.)

(f)  For fcontinuous functionelement ofHHilbert space \( f\in H \) and αcomplex numberelement ofCcomplex numbersset differenceσ(TVolterra operator*timesTVolterra operator) \( α\in \mathbb{C}\setminus \mathop{\sigma}\mathopen{}\left( T^{*}T\right)\mathclose{} \), let gcontinuous function=equals (1one-minusαcomplex numbertimesTVolterra operator*timesTVolterra operator) 1inversetimesfcontinuous function \( g= { \mathopen{}\left(1-α T^{*}T\right)\mathclose{} }^{-1}f \). (i) Write gcontinuous function\( g \) as the sum of a trigonometric series whose coefficients depend on αcomplex number\( α \) and fcontinuous function\( f \). (ii) Verify that gcontinuous function\( g \) is given by gcontinuous function(treal number)=equalsfcontinuous function(treal number)+plusαcomplex numbertimes( coscosine(αcomplex numbertimestreal number) coscosine(αcomplex number) timesintegral0zero1one sinsine(αcomplex numbertimes(1one-minusθreal number))timesfcontinuous function(θreal number) dθreal number-minusintegral0zerotreal number sinsine(αcomplex numbertimes(treal number-minusθreal number))timesfcontinuous function(θreal number) dθreal number). \[ g\mathopen{}\left( t\right)\mathclose{}= f\mathopen{}\left( t\right)\mathclose{}+\sqrt{α}\mathopen{}\left(\frac{\cos\mathopen{}\left( \sqrt{α}t\right)\mathclose{}}{\cos\mathopen{}\left( \sqrt{α}\right)\mathclose{}}\int _{0}^{1}{} \sin\mathopen{}\left( \sqrt{α}\mathopen{}\left(1-θ\right)\mathclose{}\right)\mathclose{}f\mathopen{}\left( θ\right)\mathclose{} \,\mathrm{d}θ-\int _{0}^{t}{} \sin\mathopen{}\left( \sqrt{α}\mathopen{}\left(t-θ\right)\mathclose{}\right)\mathclose{}f\mathopen{}\left( θ\right)\mathclose{} \,\mathrm{d}θ\right)\mathclose{}\text{.} \]

Advice for (a), (b), (d): This is like the example in Section G, with TVolterra operator*timesTVolterra operator \( T^{*}T \) in place of Koperator\( K \) and the differential operator fcontinuous functionis mapped tofcontinuous functionsecond derivative \( f\mapsto {-}f'' \) with boundary conditions fcontinuous functionderivative(0zero)=equals0zero=equalsfcontinuous function(1one) \( f' \mathopen{}\left( 0\right)\mathclose{}= 0= f\mathopen{}\left( 1\right)\mathclose{} \) in place of Loperator\( L \).

Exercise IV.49

(a)  Let Rring with identity\( R \) be a ring with identity, and let Jproper ideal\( J \) be a proper ideal of Rring with identity\( R \). Show that if xelement of idealelement ofJproper ideal \( x\in J \) and 1one-minusxelement of ideal \( 1-x \) is invertible, then 1one-minus(1one-minusxelement of ideal)1inverse \( 1-{\mathopen{}\left(1-x\right)\mathclose{}}^{-1} \) is invertible. Deduce that if the Hilbert space operator Kcompact operator\( K \) is compact (resp. Hilbert-Schmidt, trace class, finite-dimensional) and 1onenot an element ofσ(Kcompact operator) \( 1\notin \mathop{\sigma}\mathopen{}\left( K\right)\mathclose{} \), then (1one-minusKcompact operator)1inverse-minus1one \( {\mathopen{}\left(1-K\right)\mathclose{}}^{-1}-1 \) is compact (resp. Hilbert-Schmidt, trace class, finite-dimensional).

(b)   In the situation of part (f) of Exercise IV.48, calculate the trace of (1one-minusαcomplex numbertimesToperator*timesToperator) 1inverse-minus1one \( { \mathopen{}\left(1-α T^{*}T\right)\mathclose{} }^{-1}-1 \) 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.)

Exercise IV.50

Let Soperatorelement ofbounded linear operators(HHilbert space) \( S\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{} \).

(a)  Show that if ηvectorelement ofHHilbert space \( η\in H \) satisfies |modulusξvector, ηvector|modulusless than or equal toMpositive real numbertimesSoperatortimesξvector \( \mathopen{}\left\lvert{}\mathopen{}\left\langle{}ξ, η\right\rangle\mathclose{}\right\rvert\mathclose{}\leq M\mathopen{}\left\lVert{}Sξ\right\rVert\mathclose{} \) for some positive constant Mpositive real number\( M \) and all ξvectorelement ofHHilbert space \( ξ\in H \), then ηvectorelement ofSoperator*timesHHilbert space \( η\in S^{*}H \). (Advice: That ηvectorelement of Soperator*timesHHilbert space ¯ \( η\in \overline{ S^{*}H } \) is easy. If you can get ηvector=equalslimlimitninteger Soperator*timesρvectorninteger \( η= \lim_{n}{} S^{*}{ρ}_{n} \) for a bounded sequence (sequence ρvectorninteger )sequence \( \mathopen{}\left( {ρ}_{n} \right)\mathclose{} \), Alaoglu's Theorem will then close the sale.)

(b)  Suppose that Kerkernel(Soperator)=equals{set0zero}set=equalsKerkernel(Soperator*) \( \operatorname{Ker}\mathopen{}\left( S\right)\mathclose{}= \mathopen{}\left\{\, 0\,\right\}\mathclose{}= \operatorname{Ker}\mathopen{}\left( S^{*}\right)\mathclose{} \). Define Loperator\( L \) on the dense subspace SoperatortimesHHilbert space \( SH \) by LoperatortimesSoperatortimesξvector=equalsξvector \( LSξ= ξ \). Show that Loperator* \( L^{*} \) has domain Soperator*timesHHilbert space \( S^{*}H \), and is given there by Loperator*timesSoperator*timesξvector=equalsξvector \( L^{*} S^{*}ξ= ξ \). (In particular, the inverse of a bounded self-adjoint operator with kernel {set0zero}set \( \mathopen{}\left\{\, 0\,\right\}\mathclose{} \) is self-adjoint on its natural domain.)

Exercise IV.51

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

Theorem S: Given a densely defined self-adjoint operator hdensely defined self-adjoint operator\( h \) in the Hilbert space HHilbert space\( H \), there exist a measure space (tupleXmeasure space, Mσ-algebra, μreal number)tuple \( \mathopen{}\left(X, M, μ\right)\mathclose{} \), a measurable function αmeasureable function:mapsXmeasure spaceto (intervalinfinity, infinity)interval \( α : X \to \mathopen{}\left({-}\infty, \infty\right)\mathclose{} \), and a unitary map Uunitary map:mapsHHilbert spaceto L2Lebesgue space(Xmeasure spaceMσ-algebraμreal number) \( U : H \to \mathrm{L}^{\mathrm{2}}\mathopen{}\left( XMμ\right)\mathclose{} \) such that Uunitary map(Ddomain(hdensely defined self-adjoint operator))=equals{setξfunctionelement ofL2Lebesgue space(Xmeasure space)|such that αmeasureable function(ξfunction)element ofL2Lebesgue space(Xmeasure space) }set \( U\mathopen{}\left( \mathop{\mathcal{D}}\mathopen{}\left( h\right)\mathclose{}\right)\mathclose{}= \mathopen{}\left\{\, ξ\in \mathrm{L}^{\mathrm{2}}\mathopen{}\left( X\right)\mathclose{}\,\middle\vert\, , α\mathopen{}\left( ξ\right)\mathclose{}\in \mathrm{L}^{\mathrm{2}}\mathopen{}\left( X\right)\mathclose{}, \,\right\}\mathclose{} \) and Uunitary map(hdensely defined self-adjoint operator(Uunitary map*(ξfunction)))=equalsαmeasureable functiontimesξfunction \( U\mathopen{}\left( h\mathopen{}\left( U^{*}\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}\right)\mathclose{}= αξ \) for all ξfunctionelement ofUunitary map(Ddomain(hdensely defined self-adjoint operator)) \( ξ\in U\mathopen{}\left( \mathop{\mathcal{D}}\mathopen{}\left( h\right)\mathclose{}\right)\mathclose{} \).

In other words, HHilbert space\( H \) is unitarily equivalent to multiplication by a measurable real function in an L2Lebesgue space\( \mathrm{L}^{\mathrm{2}} \)-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 Ωcompact Hausdorff space\( Ω \) is a compact Hausdorff space and Φpositive linear functional:mapsCspace of continuous functions(Ωcompact Hausdorff space)toCcomplex numbers \( Φ : \mathrm{C}\mathopen{}\left( Ω\right)\mathclose{} \to \mathbb{C} \) is a positive linear functional (i.e. fcontinuous functiongreater than or equal to0zero \( f\geq 0 \) implies Φpositive linear functional(fcontinuous function)greater than or equal to0zero \( Φ\mathopen{}\left( f\right)\mathclose{}\geq 0 \), then there is a σ\( σ \)-algebra Mσ-algebra\( M \) of subsets of Ωcompact Hausdorff space\( Ω \) and a measure μreal number\( μ \) defined on Ωcompact Hausdorff space\( Ω \) such that (i) Mσ-algebra\( M \) contains every open subset of Ωcompact Hausdorff space\( Ω \) (so every function in Cspace of continuous functions(Ωcompact Hausdorff space) \( \mathrm{C}\mathopen{}\left( Ω\right)\mathclose{} \) is measurable); (ii) Cspace of continuous functions(Ωcompact Hausdorff space) \( \mathrm{C}\mathopen{}\left( Ω\right)\mathclose{} \) is dense in L2Lebesgue space( Ωcompact Hausdorff space Mσ-algebra μreal number ) \( \mathrm{L}^{\mathrm{2}}\mathopen{}\left( Ω M μ \right)\mathclose{} \); and (iii) Φpositive linear functional(fcontinuous function)=equalsintegralΩcompact Hausdorff spacefcontinuous functiondμreal number \( Φ\mathopen{}\left( f\right)\mathclose{}= \int _{Ω}{}f\,\mathrm{d}μ \) for all fcontinuous functionelement ofCspace of continuous functions(Ωcompact Hausdorff space) \( f\in \mathrm{C}\mathopen{}\left( Ω\right)\mathclose{} \).

Start with the case of a bounded self-adjoint operator Aself-adjoint operator\( A \) on HHilbert space\( H \). For a vector ξfunctionelement ofHHilbert space \( ξ\in H \), call the closure of the linear span of {setAself-adjoint operatorninteger(ξfunction)|such that nintegergreater than or equal to0zero nintegerelement ofNnatural numbers (including zero) }set \( \mathopen{}\left\{\, {A}^{n}\mathopen{}\left( ξ\right)\mathclose{}\,\middle\vert\, , n\geq 0, , n\in \mathbb{N}, \,\right\}\mathclose{} \) the cyclic subspace generated by Aself-adjoint operator\( A \). If this subspace is all of HHilbert space\( H \), i.e. {setfcontinuous function(Aself-adjoint operator)timesξfunction|such that fcontinuous functionelement ofCspace of continuous functions(σ(Aself-adjoint operator)) }set ¯=equalsHHilbert space \( \overline{ \mathopen{}\left\{\, f\mathopen{}\left( A\right)\mathclose{}ξ\,\middle\vert\, , f\in \mathrm{C}\mathopen{}\left( \mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}\right)\mathclose{}, \,\right\}\mathclose{} }= H \), we say that ξfunction\( ξ \) is a cyclic vector for Aself-adjoint operator\( A \).

(a)  Prove Theorem S in the case of a bounded self-adjoint operator Aself-adjoint operator\( A \) on HHilbert space\( H \) with a cyclic vector. (Advice: Ωcompact Hausdorff space=equalsσ(Aself-adjoint operator) \( Ω= σ\mathopen{}\left( A\right)\mathclose{} \), Φpositive linear functional(fcontinuous function)=equalsfcontinuous function(Aself-adjoint operator)timesξfunction, ξfunction \( Φ\mathopen{}\left( f\right)\mathclose{}= \mathopen{}\left\langle{}f\mathopen{}\left( A\right)\mathclose{}ξ, ξ\right\rangle\mathclose{} \), and fcontinuous function(Aself-adjoint operator)timesξfunction 2two =equalsΦpositive linear functional(fcontinuous function2two)=equals (fcontinuous function L2Lebesgue space(Ωcompact Hausdorff spaceμreal number) ) 2two \( {\mathopen{}\left\lVert{}f\mathopen{}\left( A\right)\mathclose{}ξ\right\rVert\mathclose{}}^{2}= Φ\mathopen{}\left( {\mathopen{}\left\lVert{}f\right\rVert\mathclose{}}^{2}\right)\mathclose{}= {\mathopen{}\left(\mathopen{}\left\lVert{}f\right\rVert\mathclose{}_{ \mathrm{L}^{\mathrm{2}}\mathopen{}\left( Ωμ\right)\mathclose{} }\right)\mathclose{}}^{2} \).)

(b)  Show that in general, HHilbert space\( H \) is the direct sum of mutually orthogonal cyclic subspaces for Aself-adjoint operator\( A \).

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

Now let hdensely defined self-adjoint operator\( h \) be a self-adjoint operator in HHilbert space\( H \). Form the bounded operator Aself-adjoint operator\( A \) on HHilbert space\( H \) as in Section A, that is, Aself-adjoint operator=equalshdensely defined self-adjoint operatortimes (1one+plushdensely defined self-adjoint operator2two) 1one2two \( A= h{\mathopen{}\left(1+{h}^{2}\right)\mathclose{}}^{{-}\frac{1}{2}} \), which as we have seen entails Aself-adjoint operator=equalsAself-adjoint operator* \( A= A^{*} \), 1oneless than or equal toAself-adjoint operatorless than or equal to1one \( {-}1\leq A\leq 1 \), hdensely defined self-adjoint operator=equalsAself-adjoint operatortimes (1one-minusAself-adjoint operator2two) 1one2two \( h= A{\mathopen{}\left(1-{A}^{2}\right)\mathclose{}}^{\frac{1}{2}} \), and Kerkernel(1one-minusAself-adjoint operator2two)=equals{set0zero}set \( \operatorname{Ker}\mathopen{}\left( 1-{A}^{2}\right)\mathclose{}= \mathopen{}\left\{\, 0\,\right\}\mathclose{} \). Invoke (c) for Aself-adjoint operator\( A \) to get (tupleXmeasure space, Mσ-algebra, μreal number)tuple \( \mathopen{}\left(X, M, μ\right)\mathclose{} \), αmeasureable function\( α \), and Uunitary map\( U \) doing what they're supposed to do. Move everything over to L2Lebesgue space(Xmeasure spaceμreal number) \( \mathrm{L}^{\mathrm{2}}\mathopen{}\left( X, μ\right)\mathclose{} \). Relabel Uunitary maptimeshdensely defined self-adjoint operatortimesUunitary map \( UhU \) as hdensely defined self-adjoint operator\( h \), now acting on HHilbert space=equalsL2Lebesgue space(Xmeasure spaceμreal number) \( H= \mathrm{L}^{\mathrm{2}}\mathopen{}\left( X, μ\right)\mathclose{} \), and let Aself-adjoint operator\( A \) be multiplication by αmeasureable function\( α \) on the new HHilbert space\( H \). Notice that 1one<less thanαmeasureable function<less than1one \( {-}1\lt α\lt 1 \) (after redefining on a set of measure zero, if one is fussy). Consider now the measurable function βmeasurable function\( β \) defined by βmeasurable function=equalsαmeasureable functiontimes (1one-minusαmeasureable function2two) 1one2two :mapsXmeasure spaceto (intervalinfinity, infinity)interval \( β= α{\mathopen{}\left(1-{α}^{2}\right)\mathclose{}}^{{-}\frac{1}{2}} : X \to \mathopen{}\left({-}\infty, \infty\right)\mathclose{} \). Let Mβmeasurable functionmultiplication operatorβmeasurable function\( \mathrm{M}_{β} \) be multiplication by βmeasurable function\( β \), with domain Ddomain(Mβmeasurable functionmultiplication operatorβmeasurable function)=equals{setξfunctionelement ofL2Lebesgue space(Xmeasure spaceμreal number)|such that βmeasurable functiontimesξfunctionelement ofL2Lebesgue space(Xmeasure spaceμreal number) }set \( \mathop{\mathcal{D}}\mathopen{}\left( \mathrm{M}_{β}\right)\mathclose{}= \mathopen{}\left\{\, ξ\in \mathrm{L}^{\mathrm{2}}\mathopen{}\left( Xμ\right)\mathclose{}\,\middle\vert\, , βξ\in \mathrm{L}^{\mathrm{2}}\mathopen{}\left( Xμ\right)\mathclose{}, \,\right\}\mathclose{} \). It is useful at this point to introduce the subspaces HHilbert spaceτpositive real number=equalsL2Lebesgue space(βmeasurable function1inverse([intervalτpositive real number, τpositive real number]interval)) \( {H}_{τ}= \mathrm{L}^{\mathrm{2}}\mathopen{}\left( {β}^{-1}\mathopen{}\left( \mathopen{}\left[{-}τ, τ\right]\mathclose{}\right)\mathclose{}\right)\mathclose{} \) for τpositive real number>greater than0zero \( τ\gt 0 \).

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

(e)   Show that G(Mβmeasurable functionmultiplication operatorβmeasurable function)=equalsG(hdensely defined self-adjoint operator) \( \mathop{\mathcal{G}}\mathopen{}\left( \mathrm{M}_{β}\right)\mathclose{}= \mathop{\mathcal{G}}\mathopen{}\left( h\right)\mathclose{} \), thus finishing the proof of Theorem S. (Same hint.)

Exercise IV.52

For compactly supported continuous functions fcontinuous function:mapsRreal numbers2twotoCcomplex numbers \( f : {\mathbb{R}}^{2} \to \mathbb{C} \), define Soperatorfcontinuous function \( {S}_{f} \) on L2Lebesgue space(Rreal numbers) \( \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathbb{R}\right)\mathclose{} \) by (Soperatorfcontinuous function(ξfunction))(treal number)=equalsintegralRreal numbers fcontinuous function(treal numbertreal number-minussreal number)timesξfunction(sreal number) dsreal number \( \mathopen{}\left({S}_{f}\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( t\right)\mathclose{}= \int _{\mathbb{R}}{} f\mathopen{}\left( t, t-s\right)\mathclose{}ξ\mathopen{}\left( s\right)\mathclose{} \,\mathrm{d}s \).

(a)  Show that Soperatorfcontinuous functionelement ofbounded linear operators(L2Lebesgue space(Rreal numbers)) \( {S}_{f}\in \mathcal{L}\mathopen{}\left( \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathbb{R}\right)\mathclose{}\right)\mathclose{} \), with Soperatorfcontinuous functionless than or equal tofcontinuous functionL2Lebesgue space(Rreal numbers2two) \( \mathopen{}\left\lVert{}{S}_{f}\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{}f\right\rVert\mathclose{}_{\mathrm{L}^{\mathrm{2}}\mathopen{}\left( {\mathbb{R}}^{2}\right)\mathclose{}} \).

(b)  Let Kset of operators=equals{setSoperatorfcontinuous function|such that fcontinuous functionelement ofCspace of continuous functions0zero(Rreal numbers2two) }set \( K= \mathopen{}\left\{\, {S}_{f}\,\middle\vert\, , f\in {\mathrm{C}}_{0}\mathopen{}\left( {\mathbb{R}}^{2}\right)\mathclose{}, \,\right\}\mathclose{} \). Show that the closure of Kset of operators\( K \) (in the operator norm) is 𝒦compact linear operators(L2Lebesgue space(Rreal numbers)) \( \mathcal{K}\mathopen{}\left( \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathbb{R}\right)\mathclose{}\right)\mathclose{} \).

Exercise IV.53

Let Sunilateral shift\( S \) be the unilateral shift on l2\( \mathrm{l}^{0} \). That is, Sunilateral shift(avector)=equalsSunilateral shift(areal number1oneareal number2twoareal number3three)=equals(tuple0zero, areal number1one, areal number2two, )tuple \( S\mathopen{}\left( \mathbf{a}\right)\mathclose{}= S\mathopen{}\left( {a}_{1}, {a}_{2}, {a}_{3}, \dotsc\right)\mathclose{}= \mathopen{}\left(0, {a}_{1}, {a}_{2}, \dotsc\right)\mathclose{} \). Show that σ(Sunilateral shift+plusSunilateral shift*)=equals[interval2two, 2two]interval \( \mathop{\sigma}\mathopen{}\left( S+ S^{*}\right)\mathclose{}= \mathopen{}\left[{-}2, 2\right]\mathclose{} \). (Advice: The inclusion σ(Sunilateral shift+plusSunilateral shift*)subset[interval2two, 2two]interval \( \mathop{\sigma}\mathopen{}\left( S+ S^{*}\right)\mathclose{}\subseteq \mathopen{}\left[{-}2, 2\right]\mathclose{} \) is clear. To show [interval2two, 2two]intervalsubsetσ(Sunilateral shift+plusSunilateral shift*) \( \mathopen{}\left[{-}2, 2\right]\mathclose{}\subseteq \mathop{\sigma}\mathopen{}\left( S+ S^{*}\right)\mathclose{} \), try solving (Sunilateral shift+plusSunilateral shift*-minusλcomplex number)(avector)=equals(tuple1one, 0zero, 0zero, )tuple \( \mathopen{}\left(S+ S^{*}-λ\right)\mathclose{}\mathopen{}\left( \mathbf{a}\right)\mathclose{}= \mathopen{}\left(1, 0, 0, \dotsc\right)\mathclose{} \), say, for avectorelement ofl2 \( \mathbf{a}\in \mathrm{l}^{0} \). 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 (tupleareal numberninteger+plus2two, areal numberninteger+plus1one)tuple \( \mathopen{}\left({a}_{n+2}, {a}_{n+1}\right)\mathclose{} \) to (tupleareal numberninteger+plus1one, areal numberninteger)tuple \( \mathopen{}\left({a}_{n+1}, {a}_{n}\right)\mathclose{} \).)

Exercise IV.54

This exercise has to do with the order 0zero\( 0 \) Bessel function of the first kind, that is, the function JBessel function of the first kind0zero \( {J}_{0} \) given as a power series by JBessel function of the first kind0zero(xreal number)=equalssummationninteger=0zeroinfinity (1one) ninteger (ninteger!) 2two times (xreal number2two) 2twotimesninteger \( {J}_{0}\mathopen{}\left( x\right)\mathclose{}= \sum_{n=0}^{\infty}{} \frac{{\mathopen{}\left({-}1\right)\mathclose{}}^{n}}{{\mathopen{}\left({n!}\right)\mathclose{}}^{2}}{\mathopen{}\left(\frac{x}{2}\right)\mathclose{}}^{2n} \). It is up to scalar multiple the only solution without singularity at xreal number=equals0zero \( x= 0 \) of Bessel's equation of order 0zero\( 0 \): xreal numbertimesyfunctionsecond derivative(xreal number)+plusyfunctionderivative(xreal number)+plusxreal numbertimesyfunction(xreal number)=equals0zero \( xy'' \mathopen{}\left( x\right)\mathclose{}+y' \mathopen{}\left( x\right)\mathclose{}+xy\mathopen{}\left( x\right)\mathclose{}= 0 \). The main thing here is to show that the sum of the squared reciprocals of the positive zeros of JBessel function of the first kind0zero \( {J}_{0} \) is 1one4four \( \frac{1}{4} \). Along the way, an orthonormal basis for L2Lebesgue space([interval0zero, 1one]interval) \( \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{} \) 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 Dsubset of functions0zero=equals{setfcontinuous functionelement ofC1space of continuously differentiable functions([interval0zero, 1one]interval)|such that xreal numberis mapped toxreal numbertimesfcontinuous functionderivative(xreal number)element ofC1space of continuously differentiable functions([interval0zero, 1one]interval) fcontinuous function(1one)=equals0zero }set \( {D}_{0}= \mathopen{}\left\{\, f\in \mathrm{C}^{1}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}\,\middle\vert\, , x\mapsto xf' \mathopen{}\left( x\right)\mathclose{}\in \mathrm{C}^{1}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}, , f\mathopen{}\left( 1\right)\mathclose{}= 0, \,\right\}\mathclose{} \). Define Loperator:maps Dsubset of functions0zero to Cspace of continuous functions([interval0zero, 1one]interval) \( L : {D}_{0} \to \mathrm{C}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{} \) by (Loperator(fcontinuous function))(xreal number)=equalsxreal numbertimesfcontinuous functionsecond derivative(xreal number)-minusfcontinuous functionderivative(xreal number)=equals ddxreal numberderivative with respect to x (xreal numbertimesfcontinuous functionderivative(xreal number)) . \[ \mathopen{}\left(L\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}= {-}xf'' \mathopen{}\left( x\right)\mathclose{}-f' \mathopen{}\left( x\right)\mathclose{}= {-} \frac{\mathrm{d}}{\mathrm{d}x}\mathopen{}\left( xf' \mathopen{}\left( x\right)\mathclose{}\right)\mathclose{} \text{.} \] Show that the eigenvalues of Loperator\( L \) are {set βzero of Bessel function2two 4four |such that JBessel function of the first kind0zero(βzero of Bessel function)=equals0zero }set \( \mathopen{}\left\{\, \frac{{β}^{2}}{4}\,\middle\vert\, , {J}_{0}\mathopen{}\left( β\right)\mathclose{}= 0, \,\right\}\mathclose{} \), with corresponding eigenfunctions xreal numberis mapped toJBessel function of the first kind0zero(βzero of Bessel functiontimesxreal number) \( x\mapsto {J}_{0}\mathopen{}\left( β\sqrt{x}\right)\mathclose{} \). (Advice: Loperator(fcontinuous function)=equalsλpositive real numbertimesfcontinuous function \( L\mathopen{}\left( f\right)\mathclose{}= λf \) makes λpositive real numbertimesfcontinuous function2two=equalsLoperator(fcontinuous function), fcontinuous function \( λ{\mathopen{}\left\lVert{}f\right\rVert\mathclose{}}^{2}= \mathopen{}\left\langle{}L\mathopen{}\left( f\right)\mathclose{}, f\right\rangle\mathclose{} \). Integrate by parts to show the eigenvalues are positive. For λpositive real number>greater than0zero \( λ\gt 0 \), show that xreal numbertimesfcontinuous functionsecond derivative(xreal number)+plusfcontinuous functionderivative(xreal number)+plusλpositive real numbertimesfcontinuous function(xreal number)=equals0zero \( xf'' \mathopen{}\left( x\right)\mathclose{}+f' \mathopen{}\left( x\right)\mathclose{}+λf\mathopen{}\left( x\right)\mathclose{}= 0 \) if and only if yfunction=equalsfcontinuous function(2twotimesλpositive real numbertimesxreal number) \( y= f\mathopen{}\left( 2\sqrt{λx}\right)\mathclose{} \) satisfies Bessel's equation of order 0zero\( 0 \).)

(b)  For ξfunctionelement ofHHilbert space=equalsL2Lebesgue space([interval0zero, 1one]interval) \( ξ\in H= \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{} \), define KHilbert-Schmidt operator(ξfunction) \( K\mathopen{}\left( ξ\right)\mathclose{} \) on [interval0zero, 1one]interval \( \mathopen{}\left[0, 1\right]\mathclose{} \) by (KHilbert-Schmidt operator(ξfunction))(xreal number)=equals loglogarithm(xreal number) timesintegral0zeroxreal numberξfunction(treal number)dtreal number-minusintegralxreal number1one loglogarithm(treal number)timesξfunction(treal number) dtreal number. \[ \mathopen{}\left(K\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}= {-} \log\mathopen{}\left( x\right)\mathclose{} \int _{0}^{x}{}ξ\mathopen{}\left( t\right)\mathclose{}\,\mathrm{d}t-\int _{x}^{1}{} \log\mathopen{}\left( t\right)\mathclose{}ξ\mathopen{}\left( t\right)\mathclose{} \,\mathrm{d}t\text{.} \] Show that KHilbert-Schmidt operator\( K \) is a self-adjoint Hilbert-Schmidt operator on HHilbert space\( H \) mapping Cspace of continuous functions([interval0zero, 1one]interval) \( \mathrm{C}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{} \) to Dsubset of functions0zero \( {D}_{0} \), that Loperator(KHilbert-Schmidt operator(ξfunction))=equalsξfunction \( L\mathopen{}\left( K\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}= ξ \) for ξfunctionelement ofCspace of continuous functions([interval0zero, 1one]interval) \( ξ\in \mathrm{C}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{} \), and that KHilbert-Schmidt operator(Loperator(fcontinuous function))=equalsfcontinuous function \( K\mathopen{}\left( L\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}= f \) for fcontinuous functionelement ofDsubset of functions0zero \( f\in {D}_{0} \). (Advice: Exhibit KHilbert-Schmidt operator\( K \) as an integral operator with kernel kkernelelement ofL2Lebesgue space( [interval0zero, 1one]interval 2two ) \( k\in \mathrm{L}^{\mathrm{2}}\mathopen{}\left( {\mathopen{}\left[0, 1\right]\mathclose{}}^{2}\right)\mathclose{} \) satisfying kkernel(xreal numbertreal number)=equalskkernel(treal numberxreal number) \( k\mathopen{}\left( x, t\right)\mathclose{}= k\mathopen{}\left( t, x\right)\mathclose{} \). Notice loglogarithmelement ofHHilbert space \( \log\in H \) and use L'Hôpital's Rule at xreal number=equals0zero \( x= 0 \) when necessary.)

(c)  Let βzero of Bessel functioniinteger \( {β}_{i} \), iintegerelement of{set1one2two}set \( i\in \mathopen{}\left\{\, 1, 2, \dotsc\,\right\}\mathclose{} \), denote the positive zeros of JBessel function of the first kind0zero \( {J}_{0} \). Define φfunctionjinteger \( {φ}_{j} \) on [interval0zero, 1one]interval \( \mathopen{}\left[0, 1\right]\mathclose{} \) by φfunctionjinteger(xreal number)=equals JBessel function of the first kind0zero(βzero of Bessel functionjintegertimesxreal number) (integral0zero1oneJBessel function of the first kind0zerotimes (βzero of Bessel functionjintegertimesxreal number) 2two dxreal number) 1one2two . \[ {φ}_{j}\mathopen{}\left( x\right)\mathclose{}= \frac{{J}_{0}\mathopen{}\left( {β}_{j}\sqrt{x}\right)\mathclose{}}{{\mathopen{}\left(\int _{0}^{1}{}{J}_{0}{\mathopen{}\left({β}_{j}\sqrt{x}\right)\mathclose{}}^{2}\,\mathrm{d}x\right)\mathclose{}}^{\frac{1}{2}}}\text{.} \] Show that {setφfunction1oneφfunction2two}set \( \mathopen{}\left\{\, {φ}_{1}, {φ}_{2}, \dotsc\,\right\}\mathclose{} \) is an orthonormal basis for HHilbert space\( H \) with KHilbert-Schmidt operator(φfunctionjinteger)=equals 4four βzero of Bessel functionjinteger2two timesφfunctionjinteger \( K\mathopen{}\left( {φ}_{j}\right)\mathclose{}= \frac{4}{{{β}_{j}}^{2}}{φ}_{j} \) for every jinteger\( j \). (This includes showing that Kerkernel(KHilbert-Schmidt operator)=equals{set0zero}set \( \operatorname{Ker}\mathopen{}\left( K\right)\mathclose{}= \mathopen{}\left\{\, 0\,\right\}\mathclose{} \), and accounting for all of the nonzero eigenvalues of KHilbert-Schmidt operator\( K \).)

It is experimentally obvious, and not very difficult to show, that βzero of Bessel functionjinteger>greater thanjinteger \( {β}_{j}\gt j \) for every jinteger\( j \). It follows that KHilbert-Schmidt operator\( K \) is a trace class operator, with trace timessummationjinteger βzero of Bessel functionjinteger2two \( \sum_{j}{} {{β}_{j}}^{{-}2} \). 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 (interval0zero, 0zero)interval \( \mathopen{}\left(0, 0\right)\mathclose{} \), and proof of the relevant theorem strongly uses uniform continuity of the kernel. To save the calculation, we have to approximate KHilbert-Schmidt operator\( K \) by operators with kernels that are continuous everywhere on the square.

(d)  For nintegergreater than or equal to2two \( n\geq 2 \), nintegerelement ofNnatural numbers (including zero) \( n\in \mathbb{N} \), let Λfunctionninteger \( {Λ}_{n} \) be the function on [interval0zero, 1one]interval \( \mathopen{}\left[0, 1\right]\mathclose{} \) that is loglogarithm(ninteger) \( \log\mathopen{}\left( n\right)\mathclose{} \) on [interval0zero, 1oneninteger]interval \( \mathopen{}\left[0, \frac{1}{n}\right]\mathclose{} \) and loglogarithm(ninteger) \( {-}\log\mathopen{}\left( n\right)\mathclose{} \) on (interval1oneninteger, 1one]interval \( \mathopen{}\left(\frac{1}{n}, 1\right]\mathclose{} \). Define the operator KHilbert-Schmidt operatorninteger \( {K}_{n} \) by (KHilbert-Schmidt operatorninteger, ξfunction)(xreal number)=equalsΛfunctionninteger(xreal number)timesintegral0zeroxreal numberξfunction(treal number)dtreal number+plusintegralxreal number1one Λfunctionninteger(treal number)timesξfunction(treal number) dtreal number. \[ \mathopen{}\left({K}_{n}, ξ\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}= {Λ}_{n}\mathopen{}\left( x\right)\mathclose{}\int _{0}^{x}{}ξ\mathopen{}\left( t\right)\mathclose{}\,\mathrm{d}t+\int _{x}^{1}{} {Λ}_{n}\mathopen{}\left( t\right)\mathclose{}ξ\mathopen{}\left( t\right)\mathclose{} \,\mathrm{d}t\text{.} \] You can easily write down the kernel for KHilbert-Schmidt operatorninteger \( {K}_{n} \) and see thereby that KHilbert-Schmidt operatorninteger \( {K}_{n} \) is a self-adjoint Hilbert-Schmidt operator on HHilbert space\( H \). Consider also Δoperatorninteger \( {Δ}_{n} \) defined by (Δoperatorninteger(ξfunction))(xreal number)=equals{cases0zero, xreal number>greater than1oneninteger ; loglogarithm(xreal number)timesintegral0zeroxreal numberξfunction(treal number)dtreal number-minusintegralxreal number1oneninteger loglogarithm(treal number)timesξfunction(treal number) dtreal number, xreal numberless than or equal to1oneninteger }. \[ \mathopen{}\left({Δ}_{n}\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}= \begin{cases}0, & x\gt \frac{1}{n} ; \\ {-}\log\mathopen{}\left( x\right)\mathclose{}\int _{0}^{x}{}ξ\mathopen{}\left( t\right)\mathclose{}\,\mathrm{d}t-\int _{x}^{\frac{1}{n}}{} \log\mathopen{}\left( t\right)\mathclose{}ξ\mathopen{}\left( t\right)\mathclose{} \,\mathrm{d}t, & x\leq \frac{1}{n} \end{cases}\text{.} \] Show that KHilbert-Schmidt operator=equalsKHilbert-Schmidt operatorninteger+plusΔoperatorninteger-minusloglogarithm(ninteger)timesχ [interval0zero, 1oneninteger]interval characteristic function of [0, 1ninteger] χ [interval0zero, 1oneninteger]interval characteristic function of [0, 1ninteger] \( K= {K}_{n}+{Δ}_{n}-\log\mathopen{}\left( n\right)\mathclose{}\chi_{ \mathopen{}\left[0, \frac{1}{n}\right]\mathclose{} }\otimes \chi_{ \mathopen{}\left[0, \frac{1}{n}\right]\mathclose{} } \).

(e)  Show that Δoperatorninteger \( {Δ}_{n} \) is trace class and calculate its trace in terms of the Bessel zeros βzero of Bessel functionjinteger \( {β}_{j} \). (Advice: Notice that Δoperatorninteger \( {Δ}_{n} \) is the operator obtained by rescaling KHilbert-Schmidt operator\( K \) from [interval0zero, 1one]interval \( \mathopen{}\left[0, 1\right]\mathclose{} \) to [interval0zero, 1oneninteger]interval \( \mathopen{}\left[0, \frac{1}{n}\right]\mathclose{} \).)

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

(g)  Show that summationjinteger βzero of Bessel functionjinteger2two =equals1one4four \( \sum_{j}{} {{β}_{j}}^{{-}2} = \frac{1}{4} \).

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