Norm of an integral operator $L^1 \rightarrow L^\infty$

Question

Let $T:L^1(\mathbb{R}^n)\rightarrow L^\infty(\mathbb{R}^n)$ be an integral operator, i.e. there exist $K:\mathbb{R}^n\times\mathbb{R}^n\rightarrow\mathbb{R}^n$ such that for all $f\in L^1(\mathbb{R}^n)$ it holds $$Tf(x):=\int_{\mathbb{R}^n}K(x,y)f(y)\:dy\quad\text{for all }x\in\mathbb{R}^n.$$

Does anyone knows how to prove that $\|T\|_{L^1\rightarrow L^\infty}=\sup_{x,y}|K(x,y)|$?

Answer 1

Recall the definition of norm of $T$ : $$ \parallel T\parallel = \sup_{f,\ \parallel f\parallel_{L^1} = 1} \parallel Tf\parallel_{L^\infty} $$ $$= \sup_{f,\ \parallel f\parallel_{L^1} = 1} \sup_{x\in \mathbb{R}^n} |Tf(x)| = \sup_{f,\ \parallel f\parallel_{L^1} = 1} \sup_{x\in \mathbb{R}^n} \bigg| \int K(x,y)f(y) dy \bigg| $$

If $k:=\sup_{x,\ y} |K(x,y)|$ (cf. essential sup), then $$ \parallel T\parallel \leq \sup_{f,\ \parallel f\parallel_{L^1} = 1} \sup_{x\in \mathbb{R}^n} k \int |f(y)| dy = k $$

And we have a sequence $(x_n,y_n)$ s.t. $|K(x_n,y_n)| \rightarrow k$ Then let $$f_n= \frac{1}{{\rm vol}\ B(y_n,\epsilon ) } \chi_{B(y_n,\epsilon )} \Rightarrow \parallel f_n\parallel_{L^1}=1 $$ so that $$ \parallel T\parallel \geq \sup_{x\in \mathbb{R}^n} \bigg| \frac{1}{{\rm vol}\ B(y_n,\epsilon ) }\int_{B(y_n,\epsilon )} K(x,y) dy \bigg| \geq \bigg| \frac{1}{{\rm vol}\ B(y_n,\epsilon ) }\int_{B(y_n,\epsilon )} K(x_n,y) dy \bigg| $$

Here last term can be arbitrarily close to $k$ by $n,\ \epsilon$.