Operatornorm inequality

Question

Let $$E:=(C^0([0,1]),\|\cdot\|_1) , \quad F:=(C^0([0,1]),\|\cdot\|_\infty),$$ with $$\|f\|_1:=\int_0^1\vert f(t)\vert \, dt,\quad \|f\|_\infty:=\sup_{t\in [0,1]}\vert f(t)\vert .$$ For $k \in C^0([0,1]\times[0,1])$ and $f\in E$ define $$K(f):=s\mapsto \int_0^1k(s,t)f(t)\,dt, s \in[0,1]$$ Show that $K$ is a continuous linear transformation $E\to F$ and find the operatornorm $C\ge0$, so $x\in E$, so that $\|K(f)\|_\infty\le C\|f\|_1$

I already showed the linearity. So I can use a theorem: Let $X$ and $Y$ be normed spaces and $f:X\to Y$ is linear, then the following are equivalent:

i) $f$ is continuous

ii) $\|f\|_{L(X,Y)}:=\sup_{\|x\|_X{\le1}}\|T_X\|_Y<\infty$

In this context is $\|f\|_{L(X,Y)}$ the Operatornorm $C$. I tried to do the following:

I suppose $\|f\|_1=1$ should be one condition for $C$ in order to be the supremum in $$\|K(f)\|_{L(X,Y)}:=s\rightarrow \left|\int_0^1k(s,t)f(t)\,dt\right|, \quad s\in [0,1]$$ $$\le s\rightarrow\int_0^1|k(s,t)f(t)|\,dt,\quad s\in[0,1]$$ $$\le s\rightarrow\int_0^1|k(s,t)||f(t)|\,dt,\quad s\in [0,1]$$ My question is: Is there a way to get rid of the $f(t)$ in the equation? I guess the final C should be $$C=\sup_{s\in[0,1]}\int_0^1|k(s,t)|\,dt$$ But I don't know how to do the final step(s). Can someone help me?

Answer 1

Let $C = \sup \limits_{s,t \in [0,1]} \lvert k(s,t)\rvert$. Note that for any $f \in E$ we have $$ \|K(f) \|_\infty = \sup_{s \in [0,1]} \big\lvert \int_0^1 k(s,t) f(t) dt \big\rvert. $$ For any $s \in [0,1]$ we have $$ \big\lvert \int_0^1 k(s,t) f(t) dt \big\rvert \leq \int_0^1 \lvert k(s,t) f(t) \rvert dt \leq \int_0^1 C \lvert f(t)\rvert dt= C \|f\|_1. $$ So we have $\|K\|_{L(E,F)} \leq C$.