TITLE: Computing the $L^p$ norm of an integral operator
QUESTION [5 upvotes]: Let $T : L^{p}(\mathbb{R}_{>0}) \to L^{p}(\mathbb{R}_{>0})$ be given by:
$$
(Tf)(x) = \int_0^\infty \frac{f(y)}{x+y} \mathrm{d}y
$$
I would like to show that this is a bounded operator for $ p \in (1,\infty)$. I was given a hint for the question, namely to make a linear change of variables $y = xu$.  Thus, to compute $\|T_f\|_p$, we calculate:
$$
\|Tf\|
_p = \left(\int_0^\infty\left(\int_0^\infty \frac{f(y)}{x + y} \mathrm{d}y\right)^p\mathrm{d}x\right)^{1/p} = \left(\int_0^\infty\left(\int_0^\infty \frac{f(xu)}{1 + u} \mathrm{d}u\right)^p\mathrm{d}x\right)^{1/p}
$$
From here, I am stuck. The natural thing to me seems to apply Holder's inequality to the inner integral, and compute:
$$
\left(\int_0^\infty\left(\int_0^\infty \frac{f(xu)}{1 + u} \mathrm{d}u\right)^p\mathrm{d}x\right)^{1/p} \leq \left(\int_0^\infty \|f(x\cdot)\|_{L^p}^p \left\|\frac{1}{1+ \cdot }\right\|_{L^q}\mathrm{d}x\right)^{1/p} \leq \left\|\frac{1}{1+ . }\right\|_{L^q}^{1/p} \left(\int_0^{\infty}\|f(x \cdot)\|_{L^p}\mathrm{d}x\right)^{1/p}
$$
But the integral on the right hand integral does not seem to converge. Does anyone have any ideas? 
Another question I have is, how is the operator above related to the Laplace transform:
$$
(\mathcal{L}f)(s) := \int_0^\infty e^{-sx}f(x) dx
$$
?
Many thanks.

REPLY [6 votes]: You could use Minkowski's Integral Inequality. Then
\begin{align}
\left(\int_0^\infty\left(\int_0^\infty \frac{f(xu)}{1 + u} \mathrm{d}u\right)^p\mathrm{d}x\right)^{1/p}
&\leq \int_0^\infty\left(\int_0^\infty \frac{|f(xu)|^p}{(1 + u)^p} \mathrm{d}x\right)^{1/p}\mathrm{d}u\\[0.3cm]
&= \|f\|_p\,\int_0^\infty\frac{1}{(1 + u)^p}\left(\int_0^\infty  |f(xu)|^p\,\mathrm{d}x\right)^{1/p}\mathrm{d}u\\[0.3cm]
&= \int_0^\infty\frac{1}{(1 + u)^p\,u^{1/p}}\,\mathrm{d}u<\infty
\end{align}
As for the Laplace transform, this is one possibility. You have 
$$
\frac1{x+y}=\int_0^\infty e^{-(x+y)t}\,dt.
$$
Then (since Fubini applies for any $x>0$ here), 
\begin{align}
Tf(x)
&=\int_0^\infty f(y)\,\int_0^\infty e^{-(x+y)t}\,dt\,dy\\[0.3cm]
&=\int_0^\infty \int_0^\infty f(y)\,e^{-(x+y)t}\,dy\,dt\\[0.3cm]
&=\int_0^\infty e^{-xt}\int_0^\infty f(y)\,e^{-yt}\,dy\,dt\\[0.3cm]
&=\int_0^\infty e^{-xt}\,\mathscr L[f](t)\,dt\\[0.3cm]
&=\mathscr L[\mathscr L[f]](x).
\end{align}