# Prove $\int_\Omega K(x,y)f(y) \operatorname \in L^2 (\Omega)$

Let $\Omega \subseteq \mathbb{R}^n$ (open) and $K(x,y)\in L^2(\Omega\times\Omega)$ then define for all $f\in L^2(\Omega)$:

$$\mathcal{K}f(x) = \int_\Omega K(x,y) f(y)\operatorname d y$$

Prove $\mathcal Kf \in L^2(\Omega)$.

We need to prove $\|\mathcal{K}f\|_{L^2(\Omega)}< \infty$, consider:

\begin{aligned} \|\mathcal{K}f\|^2_{L^2(\Omega)} &= \int_\Omega \left|\mathcal K f(x) \right|^2 \operatorname d x\\ &=\int_\Omega \left | \int_\Omega K(x,y) f(y) \operatorname d y\right|^2 \operatorname d x\\ & \leqslant \int_\Omega \left(\int_\Omega | K(x,y) | |f(y)| \operatorname d y\right)^2 \operatorname d x\\ & ?? \end{aligned}

I guess I could write this as $$\int_\Omega \int_\Omega | K(x,y) | |f(y)| \operatorname d y \int_\Omega | K(x,z) | |f(z)| \operatorname d z \operatorname d x$$

This looks like Fubini-Tonelli but not quite, is this a way to go?

Any hints? No full solutions please.

• Hint: use Cauchy-Schwarz. – msteve Jan 12 '16 at 23:31
• You mean like: $$\left | \int_\Omega K(x,y) f(y) \operatorname d y\right | \leqslant \|K\|_{L^2} \|f\|_{L^2}$$ but then I would have $\leqslant \|K\|_{L^2}^2 \|f\|_{L^2}^2 \cdot \mu(\Omega)$ where $\mu(\Omega)$ is the measure of $\Omega$ (possibly infinite) – dietervdf Jan 12 '16 at 23:37
• No, $K$ is a function of both $x$ and $y$, so you get an integral over $\Omega \times \Omega$, which is the space over which $K$ is $L^2$. – msteve Jan 12 '16 at 23:45
• @msteve I think what you have said would be appropriate as an answer. – Jason Jan 12 '16 at 23:48