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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.

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  • $\begingroup$ Hint: use Cauchy-Schwarz. $\endgroup$ – msteve Jan 12 '16 at 23:31
  • $\begingroup$ 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) $\endgroup$ – dietervdf Jan 12 '16 at 23:37
  • $\begingroup$ 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 $L^2$. $\endgroup$ – msteve Jan 12 '16 at 23:45
  • $\begingroup$ @msteve I think what you have said would be appropriate as an answer. $\endgroup$ – Jason Jan 12 '16 at 23:48
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Using Cauchy-Schwarz and Fubini, we find that \begin{align*} \| \mathcal{K} f \|_{L^2(\Omega)}^2 &\leq \int_{\Omega} \left( \int_{\Omega} |K(x,y)| |f(y)| dy \right)^2 dx\\ &\leq \int_{\Omega} \left( \left( \int_{\Omega} |K(x,y)|^2 dy \right)^{1/2} \| f \|_{L^2(\Omega)} \right)^2 dx \\ &= \| f \|_{L^2(\Omega)}^2 \int_{\Omega} \int_{\Omega} |K(x,y)|^2 dy dx\\ &= \| f \|_{L^2(\Omega)}^2 \| K \|_{L^2(\Omega \times \Omega)}^2 < \infty. \end{align*}

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