Conditions on integral operator to be in $L^{2}$

Question

Suppose we have an open set $\Omega$ in $\mathbb{R}^{n}$ and for every $x\in\Omega$ a function $T\left(x,\cdot\right)\in L^{2}\left(\Omega\right)$. If for $f\in L^{2}\left(\Omega \right)$ we consider $M\left[f\right]$ defined as $$M\left[f\right]\left(x\right)=\int_{\Omega}f\left(y\right)T\left(x,y\right)dy,$$ is $M\left[f\right]\in L^{2}\left(\Omega\right)$? Are there any equivalent conditions that need to be satisfied?

Answer 1

Perhaps you mean $\int_{\Omega} f(y) T(x,y)\; dy$? The way you have it, $M$ is multiplication by the function $g(x) = \int_{\Omega} T(x,y)\; dy$ (which may not even be finite unless $\Omega$ has finite measure), and that takes $L^2$ into itself iff $g \in L^\infty$.

EDIT: For the changed question, it is true if $T \in L^2(\Omega \times \Omega)$, as then $|M[f](x)| \le \|f\|_2 \|T(x,\cdot)\|_2$, and $\int_{\Omega} \|T(x,.)\|_2^2\; dx = \|T\|_2^2$.