# $\lVert Tf\rVert_{L^{p}}\le \gamma\lVert f\rVert_{L^{p}}$

For $s,t\geq0,$ let $K(s,t)$ satisfy $K\geq0$ and $K(\lambda s,\lambda t)=\lambda^{-1}K(s,t)$ for every $\lambda>0$ and suppose that $\int_{0}^{\infty}t^{\frac{-1}{p}}K(1,t)dt=\gamma<\infty$ for some $p\in[1,\infty]$.

Show that if $$(Tf)(s)=\int_{0}^{\infty}f(t)K(s,t)dt~~~,(f\ge 0)$$ then $$\lVert Tf\rVert_{L^{p}}\le \gamma\lVert f\rVert_{L^{p}}$$

Here's my working :

Now we may assume $\lVert f\rVert_{L^{p}}<\infty$ for $p\in[1,\infty],$otherwise our conclusion holds trivially.

Case i) : $1\le p<\infty$

Note that for $s>0$ , we have

$$Tf(s)=\int_{0}^{\infty}f(t)K(s,t)~dt=\int_{0}^{\infty}f(t)s^{-1}K(1,\frac{t}{s})~dt=\int_{0}^{\infty}f(ts)K(1,t)~dt$$ Then, one has

\begin{align} \lVert Tf\rVert_{L^{p}}&=\bigg\lVert \int_{0}^{\infty}f(ts)K(1,t)~dt\bigg\lVert_{L^{p}(\mathbb{R_{>0}}~~,~ds)}\\ &\le\int_{0}^{\infty}\lVert f(ts)K(1,t)\rVert_{L^{p}(\mathbb{R_{>0}}~~,~ds)}~dt\\ &=\int_{0}^{\infty} K(1,t)\lVert f(ts) \rVert_{L^{p}(\mathbb{R_{>0}}~~,~ds)}~dt\\ &=\int_{0}^{\infty} K(1,t) t^{\frac{-1}{p}}\lVert f \rVert_{L^{p}}~dt\\ &=\lVert f \rVert_{L^{p}}\int_{0}^{\infty}t^{\frac{-1}{p}}K(1,t)dt\\ &=\gamma\lVert f \rVert_{L^{p}}\ \end{align}

,where the first inequality holds by Minkowski's integral inequality and $\mathbb{R_{>0}}=\{x>0 :x\in\mathbb{R}\}$.

Case ii) $p=\infty,$ for almost everywhere $s\in (0,\infty)$ one has,

$$|Tf(s)|=\bigg|\int_{0}^{\infty} f(ts)K(1,t)dt\bigg|\le \lVert f \rVert_{L^{\infty}} \bigg|\int_{0}^{\infty}K(1,t)\bigg|=\gamma\lVert f \rVert_{L^{\infty}}~.$$

Henceforth, $$\lVert Tf\rVert_{L^{\infty}}\le \gamma\lVert f\rVert_{L^{\infty}}~.$$

Is there anyone checking my attempt for validity ? any advice or comment will be appreciated. Thanks for considering my request.

• The words "hence" and "henceforth" mean different things. You meant "hence".... – David C. Ullrich Dec 28 '17 at 19:01

For the last one, you should write \begin{align*} \left|\int_{0}^{\infty}f(ts)K(1,t)dt\right|\leq\int_{0}^{\infty}|f(ts)|\cdot|K(1,t)|dt\leq\|f\|_{L^{\infty}}\int_{0}^{\infty}K(1,t)dt=\gamma\|f\|_{L^{\infty}} \end{align*} since $K\geq 0$.