Integral operators

1 Product measures

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

Suppose that $F\in {ℒ}^1(X\times X)$, $F:X\to ℂ$. Fubini’s theorem tells us the following. There are sets $N_1,N_2\in 𝒜$ with $\mu (N_1)=0$ and $\mu (N_2)=0$ such that if $x\in N_1^c$ then $F_x\in {ℒ}^1(X)$ and if $y\in N_2^c$ then $F^y\in {ℒ}^1(X)$. Define

and

$I_1\in {ℒ}^1(X)$ and $I_2\in {ℒ}^1(X)$, 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 $k\in {ℒ}^2(X\times X)$ and let $g\in {ℒ}^2(X)$. By Fubini’s theorem, there is a set $Z\in 𝒜$ with $\mu (Z)=0$ such that if $x\in Z^c$ then $k_x\in {ℒ}^2(X)$. For $x\in Z_n^c$, by the Cauchy-Schwarz inequality,

$${\int }_X|k_{x}g|𝑑\mu \le {\left({\int }_X{|k_x|}^2𝑑\mu \right)}^{1/2}{\left({\int }_X{|g|}^2𝑑\mu \right)}^{1/2}={\parallel k_x\parallel }_{L^2}{\parallel g\parallel }_{L^2},$$

so $k_{x}g\in {ℒ}^1(X)$.

Since $\mu$ is $\sigma$-finite, there are $A_n\in 𝒜$, , with $A_n\uparrow X$. For each $n$, the function $(x,y)\mapsto 1_{A_n}(x)g(y)$ belongs to ${ℒ}^2(X\times X)$ and hence, by the Cauchy-Schwarz inequality, $(x,y)\mapsto k(x,y)1_{A_n}(x)g(y)$ belongs to ${ℒ}^1(X\times X)$. Applying Fubini’s theorem, there is a set $N_n\in 𝒜$ 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 ${ℒ}^1(X)$, and the function $I_n:X\to ℂ$ defined by

belongs to ${ℒ}^1(X)$.

Let $M={\bigcup }_n(Z\cup N_n)$, for which

$$\mu (M)\le \sum _n\mu (Z\cup N_n)\le \sum _n(\mu (Z)+\mu (N_n))=0.$$

We note

$$M^c=\bigcap _n(Z^c\cap N_n^c).$$

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

(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 $X\to ℂ$. 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 $K_{M}g:X\to ℂ$ is measurable.

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

${\displaystyle {\int }_X}{|K_{M}g(x)|}^2𝑑\mu (x)$ $={\displaystyle {\int }_{M^c}}{\left|{\displaystyle {\int }_X}{k}_x(y)g(y)𝑑\mu (y)\right|}^2𝑑\mu (x)$

$\le {\parallel g\parallel }_{L^2}^2\cdot {\displaystyle {\int }_{M^c}}\left({\displaystyle {\int }_X}{|k_x(y)|}^2𝑑\mu (y)\right)𝑑\mu (x)$

$={\parallel g\parallel }_{L^2}^2\cdot {\parallel k\parallel }_{L^2}^2.$

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

$${\parallel K_{M}g\parallel }_{L^2}\le {\parallel k\parallel }_{L^2}\cdot {\parallel g\parallel }_{L^2}.$$

Recapitulating, for $g\in {ℒ}^2(X)$ there is some $M\in 𝒜$ with $\mu (M)=0$ such that for $x\in M^c$, $k_x\in {ℒ}^2(X)$, and such that $K_{M}g:X\to ℂ$ defined by (1) belongs to ${ℒ}^2(X)$. 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 $g\in {ℒ}^2(X)$ it makes sense to define $Kg\in L^2(X)$ by $Kg=K_{M}g$.

If $f,g\in {ℒ}^2(X)$ 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

$$Kg(x)={\int }_X{k}_x(y)g(y)𝑑\mu (y)=⟨g,\overline{k_x}⟩,$$

where

$$⟨f,g⟩={\int }_{X}f\cdot \overline{g}𝑑\mu .$$

Theorem 1.

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

$$Kg(x)={\int }_X{k}_x(y)g(y)𝑑\mu (y)=⟨g,\overline{k_x}⟩.$$

$K:L^2(X)\to L^2(X)$ is a bounded linear operator with $\mathrm{\parallel }K\mathrm{\parallel }\mathrm{\le }{\mathrm{\parallel }k\mathrm{\parallel }}_{L^{\mathrm{2}}}$.

3 Integrals of functions

Suppose that $f:X\to ℂ$ is a function, which we do not ask to be measurable, and that $Z_1,Z_2\in 𝒜$, $\mu (Z_1)=0$, $\mu (Z_2)=0$, satisfy $1_{Z_1^c}\cdot f,1_{Z_2^c}\cdot f\in {ℒ}^1(X)$. We have

${\displaystyle {\int }_X}{1}_{Z_1^c}\cdot f𝑑\mu$ $={\displaystyle {\int }_X}{1}_{Z_1^c}\cdot (1_{Z_2}+1_{Z_2^c})\cdot f𝑑\mu$

$={\displaystyle {\int }_X}{1}_{Z_1^c\cap Z_2}\cdot f𝑑\mu +{\displaystyle {\int }_X}{1}_{Z_1^c\cap Z_2^c}\cdot f𝑑\mu$

$={\displaystyle {\int }_X}{1}_{Z_1^c\cap Z_2^c}\cdot f𝑑\mu$

$={\displaystyle {\int }_X}{1}_{Z_2^c\cap Z_1^c}\cdot f𝑑\mu$

$={\displaystyle {\int }_X}{1}_{Z_2^c}\cdot f𝑑\mu .$

Therefore if there is some $Z\in 𝒜$ with $\mu (Z)=0$ and $1_Z\cdot f\in {ℒ}^1(X)$, 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 $f\in {ℒ}^1(X)$.

4 Self-adjoint operators

Theorem 2.

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

Proof.

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

$⟨Kf,g⟩$ $={\displaystyle {\int }_X}Kf(x)\cdot \overline{g(x)}𝑑\mu (x)$

$={\displaystyle {\int }_X}\left({\displaystyle {\int }_X}{k}_x(y)f(y)𝑑\mu (y)\right)\overline{g(x)}𝑑\mu (x)$

$={\displaystyle {\int }_X}\left({\displaystyle {\int }_X}{k}^y(x)\cdot \overline{g(x)}𝑑\mu (x)\right)f(y)𝑑\mu (y)$

$={\displaystyle {\int }_X}\left({\displaystyle {\int }_X}\overline{k_y(x)}\overline{g(x)}𝑑\mu (x)\right)f(y)𝑑\mu (y)$

$={\displaystyle {\int }_X}\overline{Kg(y)}\cdot f(y)𝑑\mu (y)$

$=⟨f,Kg⟩.$

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

5 Hilbert-Schmidt operators

Let $(X,𝒜,\mu )$ be a measure space and let . It is a fact that if $\mu$ is $\sigma$-finite and $𝒜$ is countably generated, then the Banach space $L^p(X)$ is separable.

Theorem 3.

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

$${\parallel K\parallel }_{\mathrm{HS}}={\parallel k\parallel }_{L^2}.$$

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,

${\displaystyle \sum _n}⟨K{e}_n,K{e}_n⟩$ $={\displaystyle \sum _n}{\displaystyle {\int }_X}{|K{e}_n(x)|}^2𝑑\mu (x)$

$={\displaystyle \sum _n}{\displaystyle {\int }_X}{|⟨e_n,\overline{k_x}⟩|}^2𝑑\mu (x)$

$={\displaystyle {\int }_X}\left({\displaystyle \sum _n}{|⟨e_n,\overline{k_x}⟩|}^2\right)𝑑\mu (x)$

$={\displaystyle {\int }_X}⟨\overline{k_x},\overline{k_x}⟩𝑑\mu (x)$

$={\displaystyle {\int }_X}\left({\displaystyle {\int }_X}{|k_x|}^2𝑑\mu (y)\right)𝑑\mu (x)$

$={\displaystyle {\int }_{X\times X}}{|k|}^{2}d(\mu \otimes \mu )$

$={\parallel k\parallel }_{L^2}^2.$

This shows that

$${\parallel K\parallel }_{\mathrm{HS}}={\left(\sum _n⟨K{e}_n,K{e}_n⟩\right)}^{1/2}={\parallel k\parallel }_{L^2}.$$

∎

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. Let $s_j$ be the nonzero eigenvalues of $|T|$ repeated according to geometric multiplicity, with $s_{j+1}\le s_j$, $j\ge 1$, called the singular values of $T$. By the spectral theorem, there is an orthonormal basis for $\{e_j:j\ge 1\}$ for $L^2(X)$ such that $|T|e_j=s_j{e}_j$ for each $j\ge 1$. Then

${\parallel T\parallel }_{\mathrm{HS}}^2$ $={\displaystyle \sum _{j\ge 1}}⟨T{e}_j,T{e}_j⟩$

$={\displaystyle \sum _{j\ge 1}}⟨T^{*}T{e}_j,e_j⟩$

$={\displaystyle \sum _{j\ge 1}}⟨{|T|}^2{e}_j,e_j⟩$

$={\displaystyle \sum _{j\ge 1}}⟨|T|e_j,|T|e_j⟩$

$={\displaystyle \sum _{j\ge 1}}⟨s_j,s_j⟩$

$={\displaystyle \sum _{j\ge 1}}{|s_j|}^2.$

Summarizing,

$${\parallel k\parallel }_{L^2}^2={\parallel K\parallel }_{\mathrm{HS}}^2=\sum _{j\ge 1}{|s_j(T)|}^2.$$

6 Trace class operators

A compact operator $T$ on $L^2(X)$ is called trace class if , where

$${\parallel T\parallel }_{\mathrm{tr}}=\sum _{j\ge 1}{s}_j(T).$$

For a trace class operator it makes sense to define

$$\mathrm{tr}(T)=\sum _n⟨T{e}_n,e_n⟩,$$

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

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

$$\mu (A)>\mu (U_{ϵ})-ϵ$$

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

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 .

Lemma 4.

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

Proof.

If $L$ is a compact set, $B\in ℬ$, 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 . 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

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

Now let $U$ be an open set with $A\subset U$ and , say . Let $L$ be a compact set with $L\subset U$ and

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

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

and

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

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

∎

Let $X$ be a locally compact Hausdorff space and let $\mu$ be a Radon measure on $X$. An admissible kernel is a function $k\in C(X\times X)\cap {ℒ}^2(X\times X)$ for which there is some $g\in C(X)\cap {ℒ}^2(X)$ such that $|k(x,y)|\le g(x)g(y)$ 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.

Theorem 5.

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

$$Kg(x)={\int }_X{k}_x(y)g(y)𝑑\mu (y).$$

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

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

The following is Mercer’s theorem.

Theorem 6 (Mercer’s theorem).

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

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