Hilbert Schmidt Operator

Hilbert-Schmidt Operator

On $L^2({\mathbb{R}}^d)$, we can define an operator $T_K:L^2({\mathbb{R}}^d)\to L^2({\mathbb{R}}^d)$ by the formula $T_K(f):={\int }_{{\mathbb{R}}^d}K(x,y)f(y)\text{dd}y,$ we call it an integral operator with associated kernel $K:{\mathbb{R}}^d\times {\mathbb{R}}^d\to \mathbb{R}$, where $K$ is a measurable function such that ${\int }_{{\mathbb{R}}^d}|K(x,y){|}^2\text{dd}y<\infty$ for almost every $x\in {\mathbb{R}}^d$.

An integral operator $T_K$ is called a Hilbert-Schmidt operator if the associated kernel $K$ is an element of $L^2({\mathbb{R}}^d\times {\mathbb{R}}^d)$.

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Proposition

Let $T$ be a Hilbert-Schmidt operator on $L^2({\mathbb{R}}^d)$ with kernel $K$. Then

• for every $f\in L^2({\mathbb{R}}^d)$, and almost every $x\in {\mathbb{R}}^d$, the function $y\mapsto K(x,y)f(y)$ is integrable.
• $T$ is bounded, and $\Vert T\Vert \le \Vert K{\Vert }_{L^2({\mathbb{R}}^d\times {\mathbb{R}}^d)}$.
• the adjoint of $T$ is also a Hilbert-Schmidt operator with kernel $\overline{K(y,x)}$.

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Corollary

Every Hilbert-Schmidt operator is compact.