Existence of integral kernel

Question

I know the following statement ture.

Let $T \in B(L^1(\mathbb{R}^d), L^\infty(\mathbb{R}^d))$, where $B(X, Y)$ denotes all bounded linear operoters from $X$ to $Y$. Then, $T$ has the integral kernel $k(x,y) \in L^\infty(\mathbb{R}^d \times \mathbb{R}^d)$, that is, for all $f \in L^1(\mathbb{R}^d)$ and for almost all $x \in \mathbb{R}^d$, \begin{equation} Tf(x) = \int_{\mathbb{R}^d} k(x, y)f(y) dy. \end{equation} Furthermore, $\|T\|_{B(L^1, L^\infty)} = \|k\|_{L^\infty(\mathbb{R}^{2d})}$.

This is a corollary of Theorem 2.2.5 of the following article:

N. Dunford and B. J. Pettis "Linear operations on summable functions", Trans. Amer. Math. Soc. 47, (1940), 323–392. https://www.jstor.org/stable/1989960?seq=1#metadata_info_tab_contents

However, this article is written in a very abstract style. I would like to know more elementary proof.

Answer 1

If $f(x,y)=\sum_i c_i \chi_{E_i \times F_i}(x,y)=\sum_i c_i \chi_{E_i}(x) \chi_{F_i}(y)$ with $(E_i\times F_i)\cap (E_j \times F_j) =\emptyset$ for $i\neq j$, define $$\phi (f)=\sum_i c_i \int_{R^d} (T\chi_{E_i})(y) \chi_{F_i}(y)\, dy. $$ Then one checks that $\phi$ is well defined and continuous with respect the $L^1(R^d\times R^d)$ norm. Hence there is $K \in L^\infty (R^d\times R^d)$ such that $\phi(f)=\langle K,f\rangle$. Writing this equality when $f(x,y)=\chi_E(x)\chi_F(y)$ you get $T=T_K$, where $T_K$ is the integral operator with kernel $K$. This is basically the proof, some details as density arguments are left, and probably the only boring fact is that $\phi$ is well-defined, that is its definition does not depend on the representation of $f$.