# Integral operators

Jordan Bell
April 26, 2016

## 1 Product measures

Let $(X,\mathscr{A},\mu)$ be a $\sigma$-finite measure space. Then with $\mathscr{A}\otimes\mathscr{A}$ the product $\sigma$-algebra and $\mu\otimes\mu$ the product measure on $\mathscr{A}\otimes\mathscr{A}$, $(X\times X,\mathscr{A}\otimes\mathscr{A},\mu\otimes\mu)$ is itself a $\sigma$-finite measure space.

Write $F_{x}(y)=F(x,y)$ and $F^{y}(x)=F(x,y)$. For any measurable space $(X^{\prime},\mathscr{A}^{\prime})$, it is a fact that if $F:X\times X\to X^{\prime}$ is measurable then $F_{x}$ is measurable for each $x\in X$ and $F^{y}$ is measurable for each $y\in X$.11 1 Heinz Bauer, Measure and Integration Theory, p. 138, Lemma 23.5.

Suppose that $F\in\mathscr{L}^{1}(X\times X)$, $F:X\to\mathbb{C}$. Fubini’s theorem tells us the following.22 2 Heinz Bauer, Measure and Integration Theory, p. 139, Corollary 23.7. 33 3 Suppose that $F:X\times X\to[0,\infty]$ is measurable. Tonelli’s theorem, Heinz Bauer, Measure and Integration Theory, p. 138, Theorem 23.6, tells us that the functions $x\mapsto\int_{X}F_{x}d\mu,\qquad y\mapsto\int_{X}F^{y}d\mu$ are measurable $X\to[0,\infty]$, and that $\int_{X\times X}Fd(\mu\otimes\mu)=\int_{X}\left(\int_{X}F^{y}d\mu\right)d\mu(y% )=\int_{X}\left(\int_{X}F_{x}d\mu\right)d\mu(x).$ There are sets $N_{1},N_{2}\in\mathscr{A}$ with $\mu(N_{1})=0$ and $\mu(N_{2})=0$ such that if $x\in N_{1}^{c}$ then $F_{x}\in\mathscr{L}^{1}(X)$ and if $y\in N_{2}^{c}$ then $F^{y}\in\mathscr{L}^{1}(X)$. Define

 $I_{1}(x)=\begin{cases}\int_{X}F_{x}(y)d\mu(y)&x\in N_{1}^{c}\\ 0&x\in N_{1}\end{cases}$

and

 $I_{2}(y)=\begin{cases}\int_{X}F^{y}(x)d\mu(x)&y\in N_{2}^{c}\\ 0&y\in N_{2}.\end{cases}$

$I_{1}\in\mathscr{L}^{1}(X)$ and $I_{2}\in\mathscr{L}^{1}(X)$, and

 $\int_{X\times X}Fd(\mu\otimes\mu)=\int_{X}I_{2}(y)d\mu(y)=\int_{X}I_{1}(x)d\mu% (x).$

## 2 Integral operators in L2

Let $k\in\mathscr{L}^{2}(X\times X)$ and let $g\in\mathscr{L}^{2}(X)$. By Fubini’s theorem, there is a set $Z\in\mathscr{A}$ with $\mu(Z)=0$ such that if $x\in Z^{c}$ then $k_{x}\in\mathscr{L}^{2}(X)$. For $x\in Z_{n}^{c}$, by the Cauchy-Schwarz inequality,

 $\int_{X}|k_{x}g|d\mu\leq\left(\int_{X}|k_{x}|^{2}d\mu\right)^{1/2}\left(\int_{% X}|g|^{2}d\mu\right)^{1/2}=\left\|k_{x}\right\|_{L^{2}}\left\|g\right\|_{L^{2}},$

so $k_{x}g\in\mathscr{L}^{1}(X)$.

Since $\mu$ is $\sigma$-finite, there are $A_{n}\in\mathscr{A}$, $\mu(A_{n})<\infty$, with $A_{n}\uparrow X$. For each $n$, the function $(x,y)\mapsto 1_{A_{n}}(x)g(y)$ belongs to $\mathscr{L}^{2}(X\times X)$ and hence, by the Cauchy-Schwarz inequality, $(x,y)\mapsto k(x,y)1_{A_{n}}(x)g(y)$ belongs to $\mathscr{L}^{1}(X\times X)$. Applying Fubini’s theorem, there is a set $N_{n}\in\mathscr{A}$ with $\mu(N_{n})=0$ such that if $x\in N_{n}^{c}$ then $y\mapsto k(x,y)1_{A_{n}}(x)g(y)$ belongs to $\mathscr{L}^{1}(X)$, and the function $I_{n}:X\to\mathbb{C}$ defined by

 $I_{n}(x)=\begin{cases}\int_{X}k_{x}(y)1_{A_{n}}(x)g(y)d\mu(y)&x\in N_{n}^{c}\\ 0&x\in N_{n}\end{cases}$

belongs to $\mathscr{L}^{1}(X)$.

Let $M=\bigcup_{n}(Z\cup N_{n})$, for which

 $\mu(M)\leq\sum_{n}\mu(Z\cup N_{n})\leq\sum_{n}(\mu(Z)+\mu(N_{n}))=0.$

We note

 $M^{c}=\bigcap_{n}(Z^{c}\cap N_{n}^{c}).$

For $g\in\mathscr{L}^{2}(X)$, define $K_{M}g:X\to\mathbb{C}$ by

 $K_{M}g(x)=\begin{cases}\int_{X}k_{x}(y)g(y)d\mu(y)&x\in M^{c}\\ 0&x\in M.\end{cases}$ (1)

For $x\in M^{c}$,

 $I_{n}(x)=\int_{X}k_{x}(y)1_{A_{n}}(x)g(y)d\mu(y)=1_{A_{n}}(x)\int_{X}k_{x}(y)g% (y)d\mu(y)=1_{A_{n}}(x)\cdot K_{M}g(x).$

Then

 $1_{A_{n}}\cdot K_{M}g=1_{M^{c}}\cdot 1_{A_{n}}\cdot Kg=1_{M^{c}}\cdot I_{n},$

which shows that $f_{n}=1_{A_{n}}\cdot K_{M}g$ is measurable $X\to\mathbb{C}$. For any $x\in X$, for sufficiently large $n$ we have $f_{n}(x)=K_{M}g(x)$, thus $f_{n}\to K_{M}g$ pointwise, which implies that $K_{M}g:X\to\mathbb{C}$ is measurable.44 4 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 142, Lemma 4.29.

Using the Cauchy-Schwarz inequality and then Fubini’s theorem,

 $\displaystyle\int_{X}|K_{M}g(x)|^{2}d\mu(x)$ $\displaystyle=\int_{M^{c}}\left|\int_{X}k_{x}(y)g(y)d\mu(y)\right|^{2}d\mu(x)$ $\displaystyle\leq\left\|g\right\|_{L^{2}}^{2}\cdot\int_{M^{c}}\left(\int_{X}|k% _{x}(y)|^{2}d\mu(y)\right)d\mu(x)$ $\displaystyle=\left\|g\right\|_{L^{2}}^{2}\cdot\left\|k\right\|_{L^{2}}^{2}.$

This shows that $K_{M}g\in\mathscr{L}^{2}(X)$, with

 $\left\|K_{M}g\right\|_{L^{2}}\leq\left\|k\right\|_{L^{2}}\cdot\left\|g\right\|% _{L^{2}}.$

Recapitulating, for $g\in\mathscr{L}^{2}(X)$ there is some $M\in\mathscr{A}$ with $\mu(M)=0$ such that for $x\in M^{c}$, $k_{x}\in\mathscr{L}^{2}(X)$, and such that $K_{M}g:X\to\mathbb{C}$ defined by (1) belongs to $\mathscr{L}^{2}(X)$. If $N$ is any set satisfying these conditions, then for $x\in M^{c}\cap N^{c}$,

 $K_{M}g(x)=\int_{X}k_{x}(y)g(y)d\mu(y)=K_{N}g(x),$

and $\mu((M^{c}\cap N^{c})^{c})=\mu(M\cup N)=0$. Therefore, for $g\in\mathscr{L}^{2}(X)$ it makes sense to define $Kg\in L^{2}(X)$ by $Kg=K_{M}g$.

If $f,g\in\mathscr{L}^{2}(X)$ and $f=g$ in $L^{2}(X)$, check that $Kf=Kg$ in $L^{2}(X)$. We thus define $K:L^{2}(X)\to L^{2}(X)$ for $g\in L^{2}(X)$ as

 $Kg(x)=\int_{X}k_{x}(y)g(y)d\mu(y)=\left\langle g,\overline{k_{x}}\right\rangle,$

where

 $\left\langle f,g\right\rangle=\int_{X}f\cdot\overline{g}d\mu.$
###### Theorem 1.

Let $(X,\mathscr{A},\mu)$ be a $\sigma$-finite measure space. For $k\in L^{2}(X\times X)$, it makes sense to define $Kg\in L^{2}(X)$ by

 $Kg(x)=\int_{X}k_{x}(y)g(y)d\mu(y)=\left\langle g,\overline{k_{x}}\right\rangle.$

$K:L^{2}(X)\to L^{2}(X)$ is a bounded linear operator with $\left\|K\right\|\leq\left\|k\right\|_{L^{2}}$.

## 3 Integrals of functions

Suppose that $f:X\to\mathbb{C}$ is a function, which we do not ask to be measurable, and that $Z_{1},Z_{2}\in\mathscr{A}$, $\mu(Z_{1})=0$, $\mu(Z_{2})=0$, satisfy $1_{Z_{1}^{c}}\cdot f,1_{Z_{2}^{c}}\cdot f\in\mathscr{L}^{1}(X)$. We have

 $\displaystyle\int_{X}1_{Z_{1}^{c}}\cdot fd\mu$ $\displaystyle=\int_{X}1_{Z_{1}^{c}}\cdot(1_{Z_{2}}+1_{Z_{2}^{c}})\cdot fd\mu$ $\displaystyle=\int_{X}1_{Z_{1}^{c}\cap Z_{2}}\cdot fd\mu+\int_{X}1_{Z_{1}^{c}% \cap Z_{2}^{c}}\cdot fd\mu$ $\displaystyle=\int_{X}1_{Z_{1}^{c}\cap Z_{2}^{c}}\cdot fd\mu$ $\displaystyle=\int_{X}1_{Z_{2}^{c}\cap Z_{1}^{c}}\cdot fd\mu$ $\displaystyle=\int_{X}1_{Z_{2}^{c}}\cdot fd\mu.$

Therefore if there is some $Z\in\mathscr{A}$ with $\mu(Z)=0$ and $1_{Z}\cdot f\in\mathscr{L}^{1}(X)$, it makes sense to define

 $\int_{X}fd\mu=\int_{X}1_{Z}\cdot fd\mu.$

However, only if $f$ is itself measurable do we write $f\in\mathscr{L}^{1}(X)$.

###### Theorem 2.

Let $(X,\mathscr{A},\mu)$ be a $\sigma$-finite measure space. For $k\in L^{2}(X\times X)$ satisfying $k_{x}=\overline{k^{x}}$, $K:L^{2}(X)\to L^{2}(X)$ is self-adjoint.

###### Proof.

For $f,g\in L^{2}(X)$,

 $\displaystyle\left\langle Kf,g\right\rangle$ $\displaystyle=\int_{X}Kf(x)\cdot\overline{g(x)}d\mu(x)$ $\displaystyle=\int_{X}\left(\int_{X}k_{x}(y)f(y)d\mu(y)\right)\overline{g(x)}d% \mu(x)$ $\displaystyle=\int_{X}\left(\int_{X}k^{y}(x)\cdot\overline{g(x)}d\mu(x)\right)% f(y)d\mu(y)$ $\displaystyle=\int_{X}\left(\int_{X}\overline{k_{y}(x)}\overline{g(x)}d\mu(x)% \right)f(y)d\mu(y)$ $\displaystyle=\int_{X}\overline{Kg(y)}\cdot f(y)d\mu(y)$ $\displaystyle=\left\langle f,Kg\right\rangle.$

It follows that $K:L^{2}(X)\to L^{2}(X)$ is self-adjoint. ∎

## 5 Hilbert-Schmidt operators

Let $(X,\mathscr{A},\mu)$ be a measure space and let $1\leq p<\infty$. It is a fact that if $\mu$ is $\sigma$-finite and $\mathscr{A}$ is countably generated, then the Banach space $L^{p}(X)$ is separable.55 5 Donald L. Cohn, Measure Theory, second ed., p. 102, Proposition 3.4.5.

###### Theorem 3.

Let $(X,\mathscr{A},\mu)$ be a $\sigma$-finite countably generated measure space. For $k\in L^{2}(X\times X)$, $K:L^{2}(X)\to L^{2}(X)$ is a Hilbert-Schmidt operator with

 $\left\|K\right\|_{\mathrm{HS}}=\left\|k\right\|_{L^{2}}.$
###### Proof.

$L^{2}(X)$ is separable, so there is an orthonormal basis $\{e_{n}\}$ for $L^{2}(X)$. Using Parseval’s formula and then Fubini’s theorem,

 $\displaystyle\sum_{n}\left\langle Ke_{n},Ke_{n}\right\rangle$ $\displaystyle=\sum_{n}\int_{X}|Ke_{n}(x)|^{2}d\mu(x)$ $\displaystyle=\sum_{n}\int_{X}|\left\langle e_{n},\overline{k_{x}}\right% \rangle|^{2}d\mu(x)$ $\displaystyle=\int_{X}\left(\sum_{n}|\left\langle e_{n},\overline{k_{x}}\right% \rangle|^{2}\right)d\mu(x)$ $\displaystyle=\int_{X}\left\langle\overline{k_{x}},\overline{k_{x}}\right% \rangle d\mu(x)$ $\displaystyle=\int_{X}\left(\int_{X}|k_{x}|^{2}d\mu(y)\right)d\mu(x)$ $\displaystyle=\int_{X\times X}|k|^{2}d(\mu\otimes\mu)$ $\displaystyle=\left\|k\right\|_{L^{2}}^{2}.$

This shows that

 $\left\|K\right\|_{\mathrm{HS}}=\left(\sum_{n}\left\langle Ke_{n},Ke_{n}\right% \rangle\right)^{1/2}=\left\|k\right\|_{L^{2}}.$

If $T$ is a compact linear operator on $L^{2}(X)$, then $T^{*}T$ is a positive compact operator on $L^{2}(X)$. Then $|T|=\sqrt{T^{*}T}$ is a positive compact operator.66 6 See Anton Deitmar and Siegfried Echterhoff, Principles of Harmonic Analysis, second ed., p. 109, Theorem 5.1.3 Let $s_{j}$ be the nonzero eigenvalues of $|T|$ repeated according to geometric multiplicity, with $s_{j+1}\leq s_{j}$, $j\geq 1$, called the singular values of $T$. By the spectral theorem, there is an orthonormal basis for $\{e_{j}:j\geq 1\}$ for $L^{2}(X)$ such that $|T|e_{j}=s_{j}e_{j}$ for each $j\geq 1$. Then

 $\displaystyle\left\|T\right\|_{\mathrm{HS}}^{2}$ $\displaystyle=\sum_{j\geq 1}\left\langle Te_{j},Te_{j}\right\rangle$ $\displaystyle=\sum_{j\geq 1}\left\langle T^{*}Te_{j},e_{j}\right\rangle$ $\displaystyle=\sum_{j\geq 1}\left\langle|T|^{2}e_{j},e_{j}\right\rangle$ $\displaystyle=\sum_{j\geq 1}\left\langle|T|e_{j},|T|e_{j}\right\rangle$ $\displaystyle=\sum_{j\geq 1}\left\langle s_{j},s_{j}\right\rangle$ $\displaystyle=\sum_{j\geq 1}|s_{j}|^{2}.$

Summarizing,

 $\left\|k\right\|_{L^{2}}^{2}=\left\|K\right\|_{\mathrm{HS}}^{2}=\sum_{j\geq 1}% |s_{j}(T)|^{2}.$

## 6 Trace class operators

A compact operator $T$ on $L^{2}(X)$ is called trace class if $\left\|T\right\|_{\mathrm{tr}\,}<\infty$, where

 $\left\|T\right\|_{\mathrm{tr}\,}=\sum_{j\geq 1}s_{j}(T).$

For a trace class operator it makes sense to define

 $\mathrm{tr}\,(T)=\sum_{n}\left\langle Te_{n},e_{n}\right\rangle,$

which does not depend on the orthonormal basis $\{e_{n}\}$ of $L^{2}(X)$.

Let $X$ be a locally compact Hausdorff space and let $\mathscr{B}$ be the Borel $\sigma$-algebra of $X$. A Borel measure on $X$ is a measure on $\mathscr{B}$. We say that a Borel measure $\mu$ on $X$ is locally finite if for each $x\in X$ there is an open set $U_{x}$ with $x\in U_{x}$ and $\mu(U_{x})<\infty$. A Radon measure on $X$ is a locally finite Borel measure $\mu$ on $X$ such that for each $A\in\mathscr{B}$ and for any $\epsilon>0$ there is an open set $U_{\epsilon}$ with $A\subset U_{\epsilon}$ and

 $\mu(A)>\mu(U_{\epsilon})-\epsilon$

and for each open set $U$ and for any $\epsilon>0$ there is a compact set $K_{\epsilon}$ with $K_{\epsilon}\subset U$ and

 $\mu(U)<\mu(K_{\epsilon})+\epsilon.$

By definition, if $\mu$ is a Radon measure then $\mu(U)$ can be approximated by $\mu(K)$ for compact sets $K$ contained in $U$. We prove that this holds for $\mu(A)$ if $\mu(A)<\infty$.77 7 Anton Deitmar and Siegfried Echterhoff, Principles of Harmonic Analysis, second ed., p. 291, Lemma B.2.1.

###### Lemma 4.

Let $X$ be a locally compact Hausdorff space and let $\mu$ be a Radon measure on $X$. If $A\in\mathscr{B}$ with $\mu(A)<\infty$, there for any $\epsilon>0$ there is a compact set $K_{\epsilon}$, $K_{\epsilon}\subset A$, such that

 $\mu(A)<\mu(K_{\epsilon})+\epsilon.$
###### Proof.

If $L$ is a compact set, $B\in\mathscr{B}$, and $B\subset L$, let $T=L\setminus B$. For $\delta>0$ there is an open set $W_{\delta}$, $T\subset W_{\delta}$, such that $\mu(W_{\delta})<\mu(T)+\delta$. Let $K_{\delta}=L\setminus W_{\delta}$, and because $X$ is Hausdorff, $L$ is closed and hence $K_{\delta}$ is closed and therefore compact. Now, as $B\subset L$,

 $L\setminus W_{\delta}\subset L\setminus T=L\setminus(L\setminus B)=B$

and

 $\mu(B\setminus K_{\delta})=\mu(B\setminus(L\setminus W_{\delta}))\leq\mu(W_{% \delta}\setminus(L\setminus B))=\mu(W_{\delta}\setminus T)<\delta.$

We have proved that if $L$ is a compact set and $B$ is a Borel set contained in $L$, then for any $\delta>0$ then there is a compact set $K_{\delta}$ with $K_{\delta}\subset B$ and

 $\mu(B\setminus K_{\delta})<\delta.$

Now let $U$ be an open set with $A\subset U$ and $\mu(U)<\infty$, say $\mu(U)<\mu(A)+1$. Let $L$ be a compact set with $L\subset U$ and

 $\mu(U)<\mu(L)+\epsilon.$

$A=(A\cap L)\cup(A\setminus L)$, so

 $\mu(A)=\mu(A\cap L)+\mu(A\setminus L),$

and

 $\mu(A\setminus L)\leq\mu(U\setminus L)<\epsilon.$

Let $B=A\cap L$. Because $B$ is a Borel set contained in a compact set $L$, there is a compact set $K$ contained in $B$ such that

 $\mu(B\setminus K)<\epsilon.$

As $A=B\cup(A\setminus L)$ and $K\subset B$,

 $\mu(A\setminus K)=\mu((B\setminus K)\cup(A\setminus L))=\mu(B\setminus K)+\mu(% A\setminus L)<2\epsilon.$

Let $X$ be a locally compact Hausdorff space and let $\mu$ be a Radon measure on $X$. An admissible kernel is a function $k\in C(X\times X)\cap\mathscr{L}^{2}(X\times X)$ for which there is some $g\in C(X)\cap\mathscr{L}^{2}(X)$ such that $|k(x,y)|\leq g(x)g(y)$ for all $(x,y)\in X\times X$. We call $S:L^{2}(X)\to L^{2}(X)$ an admissible integral operator if there is an admissible kernel $k$ such that

 $Sg(x)=\int_{X}k_{x}(y)g(y)d\mu(y).$

The following gives conditions under which we can calculate the trace of an integral operator.88 8 Anton Deitmar and Siegfried Echterhoff, Principles of Harmonic Analysis, second ed., p. 172, Proposition 9.3.1.

###### Theorem 5.

Let $X$ be a first-countable locally compact Hausdorff space and let $\mu$ be a Radon measure on $X$. Let $k\in C(X\times X)\cap\mathscr{L}^{2}(X)$ and let

 $Kg(x)=\int_{X}k_{x}(y)g(y)d\mu(y).$

If there are admissible integral operators $S_{1}$ and $S_{2}$ such that $K=S_{1}S_{2}$, then $K$ is of trace class and

 $\mathrm{tr}\,(K)=\int_{X}k(x,x)d\mu(x).$

The following is Mercer’s theorem.99 9 E. Brian Davies, Linear Operators and their Spectra, p. 156, Proposition 5.6.9.

###### Theorem 6 (Mercer’s theorem).

If $k\in C(X\times X)\cap\mathscr{L}^{2}(X\times X)$ and $K:L^{2}(X)\to L^{2}(X)$ is a positive operator, then

 $\mathrm{tr}\,(K)=\int_{X}k(x,x)d\mu(x).$