Integral Operator Satisfying Holmgren Condition is Bounded Consider the integral operator
$$u(x) = kf(x) = \int_{-\infty}^\infty k(x,s)f(s)ds.$$
Assuming the kernel $k(x,s)$ satisfies the Holmgren condition:
$$ \sup_{y \in \mathbb{R}} \int_{-\infty}^\infty \int_{-\infty}^\infty |k(x,s)||k(x,y)|dxds< \infty.$$
Show that $k$ is a bounded linear operator on $L^2(\mathbb{R})$. 
The linear portion is trivial. For the bounded portion, I received a hint to start with $|k(x,s)f(s)| = \sqrt{|k(x,s)|}\sqrt{|k(x,s)|}|f(s)|$. This makes me think that I need to somehow apply Cauchy Schwarz using the Holmgren condition, but I wasn't able to get anything productive. 
 A: By request of the question asker I am posting my comment as a solution.
We have:
\begin{align*}\int_{\mathbb{R}}\left(\left(K(f)\right)(x)\right)^{2}dx&=\int_{\mathbb{R}}\left(\int_{\mathbb{R}}k(x,s)f(s)ds\right)^{2}dx=\int_{\mathbb{R}}\left\lvert\int_{\mathbb{R}}k(x,s)f(s)ds\right\rvert^{2}dx\\
&\le\int_{\mathbb{R}}\left(\int_{\mathbb{R}}\left\lvert k(x,s)\right\rvert\left\lvert f(s)\right\rvert ds\right)^{2}dx=\int_{\mathbb{R}}\left(\int_{\mathbb{R}}\left\{\sqrt{\left\lvert k(x,s)\right\rvert}\right\}\left\{\sqrt{\left\lvert k(x,s)\right\rvert}\left\lvert f(s)\right\rvert\right\}ds\right)^{2}dx\\
&\le\int_{\mathbb{R}}\left(\sqrt{\int_{\mathbb{R}}\left(\sqrt{\left\lvert k(x,s)\right\rvert}\right)^{2}ds}\sqrt{\int_{\mathbb{R}}\left\{\sqrt{\left\lvert k(x,s)\right\rvert}\left\lvert f(s)\right\rvert\right\}^{2}ds}\right)^{2}dx\\
&=\int_{\mathbb{R}}\left(\int_{\mathbb{R}}\left\lvert k(x,s)\right\rvert ds\right)\left(\int_{\mathbb{R}}\left\lvert k(x,s)\right\rvert\left(f(s)\right)^{2}ds\right)dx\\
&=\int_{\mathbb{R}}\left(\int_{\mathbb{R}}\left\lvert k(x,t)\right\rvert dt\right)\left(\int_{\mathbb{R}}\left\lvert k(x,s)\right\rvert\left(f(s)\right)^{2}ds\right)dx\\
&=\int_{\mathbb{R}}\int_{\mathbb{R}}\int_{\mathbb{R}}\left\lvert k(x,t)\right\rvert\left\lvert k(x,s)\right\rvert\left(f(s)\right)^{2}dt\,ds\,dx\\
&=\int_{\mathbb{R}}\int_{\mathbb{R}}\int_{\mathbb{R}}\left\lvert k(x,t)\right\rvert\left\lvert k(x,s)\right\rvert\left(f(s)\right)^{2}dx\,dt\,ds\\
&=\int_{\mathbb{R}}\left(\int_{\mathbb{R}}\int_{\mathbb{R}}\left\lvert k(x,t)\right\rvert\left\lvert k(x,s)\right\rvert dx\,dt\right)\left(f(s)\right)^{2}ds\\
&\le\left(\sup_{s\in\mathbb{R}}\left(\int_{\mathbb{R}}\int_{\mathbb{R}}\left\lvert k(x,t)\right\rvert\left\lvert k(x,s)\right\rvert dx\,dt\right)\right)\left(\int_{\mathbb{R}}\left(f(s)\right)^{2}ds\right)
\end{align*}
Taking square roots shows that:
$$\left\lvert\left\lvert K(f)\right\rvert\right\rvert_{L^{2}\left(\mathbb{R}\right)}\le\sqrt{\sup_{s\in\mathbb{R}}\left(\int_{\mathbb{R}}\int_{\mathbb{R}}\left\lvert k(x,t)\right\rvert\left\lvert k(x,s)\right\rvert dx\,dt\right)}\left\lvert\left\lvert f\right\rvert\right\rvert_{L^{2}\left(\mathbb{R}\right)}$$
