I'm studying some Hilbert-Schmidt integral operator theory. I read that a Hilbert-Schmidt integral operator is uniquely determined by its kernel, and I wonder how exactly that is. I guess what I'm looking for are hints/suggestions on how to show that Hilbert-Schmidt integral operators are uniquely determined by their kernels.
Here are the definitions I'm going by:
$\textit{Let k be a function of two variables }(x,y)\in I \times I=I^2$ $\textit{ where }I\textit{ is a finite or infinite real interval. We define a linear integral operator K with kernel }$ $k(x,y)\textit{ as }$
$$Ku(x)=\int_{I}k(x,y)u(y)\ dy,\ \ x\in I$$
$\textit{whenever this integral makes sense. The domain D(K) will have to be specified in order to accomplish this.}$
$\textit{An integral operator on }L^2(I)\textit{ is called a Hilbert-Schmidt operator if the kernel }k\textit{ is in }L^2(I\times I),\textit{that is if}$
$$||k||_{L^{2}}^2=\int_{I}\int_{I}|k(x,y)|^2\ dxdy<\infty$$
The book I'm using now says it's possible to show that a Hilbert-Schmidt operator is uniquely determined by its kernel, that is, if
$$\int_{I}k(x,y)u(y)\ dy=\int_{I}h(x,y)u(y)\ dy$$
for all $u\in L^2(I)$, then $k=h$ in $L^2(I\times I)$.
This is where I'm unsure of how to show this. Thanks in advance!
Let $T_k$ be the operator $T_ku(x)=\int_{I}k(x,y)u(y)\,dy$. If $T_k=T_h$, then $T_{k-h}=T_k-T_h=0$. So, it suffices to show that if $T_k=0$, then $k=0$.
To prove this, suppose $T_k=0$ and let $k_x(y)=\overline{k(x,y)}$. Note that $k_x\in L^2(I)$ for almost every $x\in I$, and that $T_ku(x)=\langle u, k_x\rangle$. Now pick an orthonormal basis $\{e_n\}$ for $L^2(I)$ (or really just any countable subset whose span is dense). For each $n$, $T_ke_n=0$, meaning that for almost all $x$, $k_x$ is orthogonal to $e_n$. Since a countable union of null sets is null, this means that for almost all $x$, $k_x$ is orthogonal to $e_n$ for all $n$. But since the span of $\{e_n\}$ is dense in $L^2(I)$, this means $k_x=0$ almost everywhere. So $k_x=0$ almost everywhere for almost all $x$, which means $k=0$ almost everywhere.