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.



Existence of integral kernel
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 $f$.


