Is $f\mapsto \int\limits_{\mathbb{R}^n} K(x,y)f(x)dy$ with kernel $K\in C(\mathbb{R}^n\times \mathbb{R}^n)$ compact?

Question

I know if $K\in C([0,1]\times [0,1])$ then $(Tf)(y):=\int\limits_{[0,1]} K(x,y)f(x)dy$ defines a compact operator on $L^2([0,1])$ i.e. $T$ is Hilbert Schmidt.

Is this also true if we replace $[0,1]$ by $\mathbb{R}^n$?

Perhabs it is possible to obtain $T$ as a limit of compact operators.

Answer 1

First, $T$ is a Hilbert-Schmidt operator iff $K\in L^2$. Indeed, continuity is sufficient to ensure that on a bounded set, but not in $\Bbb R^d$. This implies that $T$ is a compact operator.

There is no clear necessary and sufficient criterium in terms of the kernel of $K$ so that $T$ is compact. If $K$ is just continuous (and even bounded) then this is not sufficient. For example if $K=1$, then $Tf$ is a constant function, which is not even in $L^2$, so it does not give you a bounded operator. If $K$ is continuous and compactly supported, or more generally with fast decay at infinity, then it is in $L^2$ and so $T$ is compact, as any other criterium that imply that $K$ is in $L^2$.

Answer 2

For simplicity assume that $K(x,y)$ is continuous. For the boundedness of the integral operator with a kernel $K(x,y)$ a sufficient condition is given by the Schur test. A particular case is the following: if $$\sup_{x\in\mathbb{R}^d}\int\limits_{\mathbb{R}^d}|K(x,y)|\,dy\le A,\quad \sup_{y\in\mathbb{R}^d}\int\limits_{\mathbb{R}^d}|K(x,y)|\,dx\le B\quad (1)$$ then $\|T\|\le \sqrt{AB}.$ Concerning compactness, assume that both integrals are uniformly convergent That means for every $\varepsilon>$ there exists $N>0$ such that $$\sup_{x\in\mathbb{R}^d}\ \int\limits_{\|y\|\ge N}|K(x,y)|\,dy<\varepsilon, \quad \sup_{y\in\mathbb{R}^d}\ \int\limits_{\|x\|\ge N}|K(x,y)|\,dx<\varepsilon\quad (2)$$

A convenient criterion for the uniform convergence of the above integrals is the following $$|K(x,y)|\le f(y),\quad |K(x,y)|\le g(x)$$ where $f,g\in L^1(\mathbb{R}^d).$

The condition $(2)$ implies that the operator $T$ is compact. Indeed, let $$K_N(x,y)=\begin{cases} K(x,y)& \|x\|,\ \|y\|\le N \\ 0 & {\rm otherwise}\end{cases}$$ $$K_{1,N}(x,y)=\begin{cases} K(x,y)& \|x\|\le N \\ 0 & {\rm otherwise}\end{cases}$$ and $T_N,$ $T_{1,N}$ denote the corresponding integral operators. By the Schur test we get $$\|T-T_{1,N}\|\le \sqrt{A\varepsilon},\quad \|T_{1,N}-T_N\|\le \sqrt{B\varepsilon}$$ Hence $$\|T-T_N\|\le (\sqrt{A}+\sqrt{B})\sqrt{\varepsilon}\qquad (3)$$ The operator $T_N$ is compact as $K_N(x,y)$ is square summable. Thus $T$ is compact as by $(3)$ it can be approximated in operator norm by compact operators.

Example Let $d=1$ and $$K(x,y)=(x^2+y^2+1)^{-\alpha},\quad \alpha> {1\over 2}$$ Then $K$ satisfies $(2)$ as $K(x,y)=K(y,x)\le (x^2+1)^{-\alpha}.$ For $\alpha\le 1$ the kernel $K(x,y)$ is not square summable over $\mathbb{R}^2,$ hence the corresponding operator $T$ does not belong to the Hilbert-Schmidt class, nevertheless the operator $T$ is compact.