2
$\begingroup$

I want to show the following statement:

Let $(X,\mathcal{A}, \mu)$ be a measurable space, $K: X \times X \rightarrow \mathbb{R}$ measurable. For real-valued functions $f$, we define $Tf(x)=\int_X K(x,y)f(y) \mu(dy)$.

If there exists a constant $C \in \mathbb{R}$ such that

$\int_X |K(x,y)| \mu(dy) \leq C$ for every $x$ and

$\int_X |K(x,y)| \mu(dx) \leq C$ for every $y$,

Show that for $1\leq p <\infty$, $\lVert Tf \rVert_p \leq C \lVert f \rVert_p$.

My attempt:

\begin{align*} \|Tf\|_p &= \left(\int_{X} |Tf(x)|^p \mu(dx) \right)^{\frac{1}{p}} \\ &= \left(\int_{X} \left|\int_X K(x,y) f(y) \mu(dy) \right|^p \mu(dx) \right)^{\frac{1}{p}} \\ &\leq \left(\int_{X} \int_X |K(x,y)|^p |f(y)|^p \mu(dy) \mu(dx) \right)^{\frac{1}{p}} \\ &\leq \left(\int_{X} \int_X C^p |f(y)|^p \mu(dy) \mu(dx) \right)^{\frac{1}{p}} \\ &\leq C \left(\int \int_X |f(y)|^p \mu(dy) \mu(dx) \right)^{\frac{1}{p}}. \end{align*}

My problem is, that I don't know how to get rid of the inner integral $\int_X \mu(dy)$. Further, I didn't use both conditions that, i.e. $\int |K(x,y)| \leq C$ is bounded for all $x$ if $y$ is fixed, and for all $y$ if $x$ is fixed. So to me, it seems that this approach is the wrong one. I have a feeling that I somehow need to use the topic of "bounded linear functionals", to find a solution. But I couldn't think of another one.

$\endgroup$
1

1 Answer 1

1
$\begingroup$

Let $q$ be the conjugate of $p$. Notice \begin{aligned} \int_X\left|K(x,y)f(y)\right|\mu(dy) &=\int_X\left|K(x,y)\right|^{1/q}\left|K(x,y)\right|^{1/p}\left|f(y)\right|\mu(dy)\\ &\le \left(\int_X\left|K(x,y)\right|\mu(dy)\right)^{1/q}\left(\int_X\left|K(x,y)\right|\left|f(y)\right|^p\mu(dy)\right)^{1/p}. \end{aligned}

From here, you may write \begin{aligned} \int_X\left|Tf(x)\right|^p\mu(dx) &\le C^{p/q}\int_X\left(\int_X\left|K(x,y)\right|\left|f(y)\right|^p\mu(dy)\right)\mu(dx)\\ &\le C^{p/q}\int_X\left(\int_X\left|K(x,y)\right|\mu(dx)\right)\left|f(y)\right|^p\mu(dy)\\ &\le C^{p/q}\int_XC\left|f(y)\right|^p\mu(dy)\\ &=C^{p}\int_X\left|f(y)\right|^p\mu(dy), \end{aligned} which yields \begin{aligned} \left\|Tf(x)\right\|_p \le C\left\|f\right\|_p. \end{aligned}

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .