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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.

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  • $\begingroup$ You're correct about the first one--just fixed it. I believe the second is correct--it's from a practice exam and I just double checked the problem. It makes sense to me to be $dxds$ though, as the sup is over $y$... $\endgroup$ – Chriz26 Jan 2 '17 at 20:16
  • $\begingroup$ oh yep you're right - just a slightly odd condition. will give it some thought $\endgroup$ – πr8 Jan 2 '17 at 20:17
  • $\begingroup$ $\int_{\mathbb{R}}((K(f))(x))^{2}dx=\int_{\mathbb{R}}\left(\int_{\mathbb{R}}k(x,s)f(s)ds\right)^{2}dx\le\int_{\mathbb{R}}\left(\int_{\mathbb{R}}\left\lvert k(x,s)f(s)\right\rvert ds\right)^{2}dx$ $=\int_{\mathbb{R}}\left(\int_{\mathbb{R}}\sqrt{\left\lvert k(x,s)\right\rvert}\left(\sqrt{\left\lvert k(x,s)\right\rvert}\left\lvert f(s)\right\rvert\right)\right)^{2}\le\int_{\mathbb{R}}\left(\int_{\mathbb{R}}\left\lvert k(x,s)\right\rvert ds\right)\left(\int_{\mathbb{R}}\left\lvert k(x,s)\right\rvert (f(s))^{2}ds\right)dx$ $\endgroup$ – user71352 Jan 2 '17 at 20:27
  • $\begingroup$ $=\int_{\mathbb{R}}\left(\int_{\mathbb{R}}\left\lvert k(x,t)\right\rvert dt)\right)\left(\int_{\mathbb{R}}\left\lvert k(x,s)\right\rvert(f(s))^{2}ds\right)dx=\int_{\mathbb{R}}\int_{\mathbb{R}}\int_{\mathbb{R}}\left\lvert k(x,t)\right\rvert\left\lvert k(x,s)\right\rvert(f(s))^{2}dt\,ds\,dx$ $\endgroup$ – user71352 Jan 2 '17 at 20:27
  • $\begingroup$ $\int_\mathbb{R}\left(\int_{\mathbb{R}}\int_{\mathbb{R}}\left\lvert k(x,t)\right\rvert \left\lvert k(x,s)\right\rvert dt\,dx\right)(f(s)^{2})ds\le\sup_{s\in\mathbb{R}}\left(\int_{\mathbb{R}}\int_{\mathbb{R}}\left\lvert k(x,t)\right\rvert \left\lvert k(x,s)\right\rvert dt\,dx\right)\left(\int_{\mathbb{R}}(f(s))^{2}ds\right)$ $\endgroup$ – user71352 Jan 2 '17 at 20:27
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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:

$$\left\lvert\left\lvert K(f)\right\rvert\right\rvert_{L^{2}\left(\mathbb{R}\right)}\le\sqrt{\sup_{s\in\mathbb{R}}\left(\int_{\mathbb{R}}\int_{\mathbb{R}}\left\lvert k(x,t)\right\rvert\left\lvert k(x,s)\right\rvert dx\,dt\right)}\left\lvert\left\lvert f\right\rvert\right\rvert_{L^{2}\left(\mathbb{R}\right)}$$

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