Find $\vert \vert T_{k} \vert \vert$ where $T_{k}f(s)=\int_{0}^{1}k(s,t)f(t)dt$

Question

Let $T_{k}:L^{p}[0,1] \to L^{p}[0,1]$, where $k \in C([0,1]^{2})$ and

$T_{k}f(s)=\int_{0}^{1}k(s,t)f(t)dt$

Show $\vert \vert T_{k} \vert \vert \leq \sup\limits_{s}(\int^{1}_{0} \vert k(s,t) \vert dt)^{\frac{1}{q}} \sup\limits_{t}(\int^{1}_{0} \vert k(s,t) \vert dt)^{\frac{1}{p}}$

Let $f \in L^{p}[0,1]$:

$\vert \vert T_{k}f\vert \vert_{L^{p}[0,1]}^{p}$=$\int_{0}^{1}\vert T_{k}f(s)\vert^{p}ds=\int_{0}^{1}\vert \int_{0}^{1}k(s,t)f(t)dt\vert^{p}ds\leq \int_{0}^{1}\int_{0}^{1}\vert k(s,t)f(t)dt\vert^{p}ds=*$

It is clear that the argument will be a combination of Fubini and Hölder inequality.

$*\leq \int_{0}^1\sup\limits_{t}\vert k(s,t)\vert^{p}(\int_{0}^{1}\vert f(t)\vert^{p}dt)ds=\int_{0}^{1}\sup\limits_{t}\vert k(s,t)\vert^{p}\vert\vert f\vert\vert^{p}_{p}ds$ So I get to $\vert \vert T_{k} \vert \vert \leq \int_{0}^{1}\sup\limits_{t}\vert k(s,t)\vert^{p}ds$ but not what I was looking for

Answer 1

$$|T_kf(s)| \leq \int |k(s,t)|^{1/q}|k(s,t)|^{1/p}|f(t)| dt$$ $$\leq \int (|k(s,t)|dt)^{1/q} (\int |k(s,t)| |f(t)|^{p} dt)^{1/p}\leq A^{1/q} (\int |k(s,t)| |f(t)|^{p} dt)^{1/p}$$ where $A =\sup_s \int |k(s,t)|dt$. Now integrate w.r.t. $s$, use the inequality $\int |h(s)|^{1/p} ds \leq (\int |h(s)| ds)^{1/p}$ where $h(s)=\int |k(s,t)||f(t)|^{p}dt$ and interchange the order of integration. Can you complete it now?