Does an integral operator with a symmetric integrable kernel have to be bounded on $L^2$?

Question

Suppose $K(x,y)$ is a symmetric kernel. Let $\phi\in L^2(\Omega)$, where $\Omega$ everywhere is a domain in $R^n$. Can $\int_{\Omega}K(x,y)\,\phi(y)\,dy$ belong to $L^2$? In other words can an inequality of the type $$ \left\|\int_{\Omega}K(x,y)\,\phi(y)\,dy\right\|_{2} \leq \|K\|_{L^1(\Omega\times\Omega)} \|\varphi\|_{2} $$ exist?

Answer 1

Another counterexample even if $K \in L^1(\Omega \times \Omega)$.

Consider $K(x,y) = \frac{1}{\sqrt{x}} \frac{1}{\sqrt{y}}$. And $\Omega \times \Omega = (0,1) \times (0,1)$. Put $\varphi \equiv 1$ in $(0,1)$. It is clear that $\varphi \in L^2(0,1)$. Then $$ \int_0^1 \left( \int_0^1 \frac{1}{\sqrt{x}} \frac{1}{\sqrt{y}} dy\right)^2 dx = \int_0^1 \frac{1}{x} \left( \int_0^1 \frac{1}{\sqrt{y}} dy\right)^2 dx = 4 \int_0^1 \frac{1}{x} \left( \sqrt{y}|_0^1 \right)^2 dx = \\ 4 \int_0^1 \frac{1}{x} dx = +\infty. $$

(it is also clear, that $K \in L^1(\Omega \times \Omega)$, since $$ \int_0^1 \int_0^1 \frac{1}{\sqrt{x}} \frac{1}{\sqrt{y}} dx dy = \left(\int_0^1 \frac{1}{\sqrt{x}} dx \right)^2 = constant) $$