On Hilbert-Schmidt integral operators

Citing Wikipedia:

Given a domain (an open and connected set) $\Omega$ in $n$-dimensional Euclidean space $\mathbb{R}^n$, a Hilbert–Schmidt kernel is a function $k : \Omega \! \times \! \Omega \to \mathbb C$ with finite $L^2(\Omega \!\times\!\Omega; \mathbb{C})$ norm, that is $$\|k\|_{L^2} := \int _{{\Omega }}\int _{{\Omega }}|k(x,y)|^{{2}}\,dx\,dy<\infty$$ The associated Hilbert–Schmidt integral operator is the operator $K : L^2(\Omega; \mathbb{C}) \to L^2(\Omega; \mathbb{C})$ given by $$(Ku)(x)=\int_{\Omega}k(x,y)\,u(y)\,dy$$ Then $K$ is a Hilbert–Schmidt operator with Hilbert–Schmidt norm $$\Vert K\Vert _{{\mathrm {HS}}}=\Vert k\Vert _{{L^{2}}}$$

Now suppose $n{=}1, \ \Omega\equiv(a, b)$, and consider the function $v \in L^2((a, b); \mathbb{C})$.

Question(s). Is the operator $T_v : L^2((a, b);\mathbb C) \to L^2((a, b);\mathbb C)$ defined by $$(T_vu)(x)= (v \star u)(x) = \int_a^b \!\!v(x{-}y)\,u(y)\,dy$$ a Hilbert-Schmidt operator? In other words, is $\displaystyle\int_a^b\!\!\!\int_a^b|v(x{-}y)|^2\,dx\,dy \lt \infty$ ? If no, is it at least compact?

By the way, Wikipedia doesn't provide proof for the previous statement. Are there some references?

You supposed $v$ is $L^{2}$ on $(a,b)$. The argument $x-y$ in your last expression is not always in $(a,b)$, so that integral does not make sense. $T_{v}$ is then not H-S, since it is not even a defined operator.

You might change the domain or replace the function $v$ on $(a,b)$ with a kernel function $v(x,y)$ on $(a,b)\times(a,b)$.

And what do you mean by a proof of this statement since the statement is a definition? If you are looking to learn more about Hilbert-Schmidt operators I would recommend Reed+Simon Vol. 1.