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.
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}