$K(x,y)$ is a continuous no negative function on $[0,1]$x$[0,1]$
$T:\mathcal{L^1}[0,1]\to \mathcal{C}[0,1]$
$T(f)(x)=\int_0^{1}K(x,y)f(y)dy$
I know T is linear and bounded. I want to know the norm.
I could get $\|T\|\leq max_{x,y\in[0,1]\text{x}[0,1]} K(x,y)$, but I don't know how to get the equality working with $\| \cdot \|_{\infty}$
Sorry if it's a duplicate, I have seen this operator (or a similar one) in many places.
Thanks in advance.
Let $||\cdot||_1$ denote the usual $L^1$ and $||\cdot||$ denote the max-norm. We will show that for any $\varepsilon>0$ there exists a function $f\in L^1[0,1]$ with $||f||_1=1$ s.t. $$||Kf|| > \max_{x,y\in[0,1]} K(x,y)-\varepsilon.$$ This will prove that $||K||\geq \max_{x,y\in[0,1]} K(x,y)$. Let $(x_0,y_0):=\mathrm{argmax}_{x,y\in[0,1]}\ K(x,y)$ and let $\varepsilon>0$ be fixed. Choose $f_\delta(y):= \frac{1}{2\delta}\chi_{[y_0-\delta,y_0+\delta]}(y).$ It is clear that for any $f\geq 0$ $$ |Kf(x)| = \int_0^1 K(x,y)f(y)dy \leq \int_0^1K(x_0,y)f(y)dy \quad \forall x\in [0,1] $$
hence $$ \max_x |Kf(x)| = \int_0^1 K(x_0,y)f(y)dy. $$ Now let us return to $f_\delta$ which is positive. $$ Kf(x_0) = \int_0^1 K(x_0,y)f(y)dy = \frac1{2\delta}\int_{y_0-\delta}^{y_0+\delta} K(x_0,y)dy = K(x_0,\eta_\delta) $$ for some $\eta_\delta \in (y_0-\delta,y_0+\delta) $. Since $K$ is continuous we can choose $\delta$ s.t. $K(x_0,y_0)-K(x_0,\eta_\delta)<\varepsilon$ hence $$ \max_x Kf_\delta(x) = Kf_\delta(x_0)=K(x_0,\eta_\delta)> K(x_0,y_0)-\varepsilon. $$ This all means that $||K|| \geq \max_{x,y\in[0,1]}K(x,y), $ where $$ ||K|| = \sup\{||Kf||: f\in L^1[0,1], ||f||_1 =1\} $$