What is the Solution to POTW #230 - Oct 25, 2016?

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Here is this week's POTW:

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Show that if $(X,\mathcal{M},\mu)$, $(Y,\mathcal{N},\nu)$ are finite measure spaces, $1 < p < \infty$, and $K$ is a measurable function on $X\times Y$, there is a bounded integral operator $I(K) : \mathscr{L}^p(\nu) \to \mathscr{L}^p(\mu)$ given by

$$I(K)(f) :x \mapsto \int_Y K(x,y)\,f(y)\, d\nu(y)\quad (f\in \mathscr{L}^p(\mu)),$$

provided that the kernel $K$ satisfies the conditions $\sup_x \int_Y \lvert K(x,y)\rvert\, d\nu(y) < \infty$ and $\sup_y \int_X \lvert K(x,y)\rvert \, d\mu(x) < \infty$.
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No one answered this week's problem. That's ok though, because the problem statement originally had some typos -- they have been corrected. :D You can read my solution below.

Spoiler

Assume the conditions on $K$ hold. Let $\alpha$ be the supremum of all the integrals $\int_Y \lvert K(x,y)\vert\, d\nu(y)$ as $x$ ranges over $X$; similarly define $\beta$ as the supremum of all integrals $\int_X \lvert K(x,y)\vert\, d\mu(x)$ as $y$ ranges over $Y$. Let $q$ be the conjugate exponent of $p$. Given $f\in \mathscr{L}^p$ with $\|f\|_p = 1$,

$$\lvert I(K)f\rvert \le \int_Y \lvert K(x,y)\rvert^{1/q} \lvert K(x,y)\rvert^{1/p}\lvert f(y)\rvert\, d\nu(y) \le \alpha^{1/q}\left(\int_Y \lvert K(x,y)\rvert \lvert f(y)\rvert^p\, d\nu(y)\right)^{1/p},$$

using Hölder's inequality, and by Fubini,

$$\|I(K)f\|_p \le \alpha^{1/q}\left(\int_X\int_Y \lvert K(x,y)\rvert \lvert f(y)\rvert^p\, d\nu(y)\, d\mu(x)\right)^{1/p}
= \alpha^{1/q}\left(\int_Y \lvert f(y)\rvert^p \left[\int_X \lvert K(x,y)\rvert\, d\mu(x)\right] d\nu(y)\right)^{1/p}
\le \alpha^{1/q}\beta^{1/p}\|f\|_p = \alpha^{1/q}\beta^{1/p}$$

As $f$ was arbitrary, $T$ is bounded (by $\alpha^{1/q}\beta^{1/p}$). In fact, as $0 \le \alpha, \beta \le \max\{\alpha,\beta\}$, we may deduce $\|I(K)\| \le \max\{\alpha,\beta\}$.