$L^1$ and $L^{\infty}$ are not reflexive I want some proof for the following statement :
$L^1$ and $L^{\infty}$ are not reflexive.
Can anyone help me, please? or reference me? 
 A: Brezis "Fuctional analysis Sobolev spaces and Partial Differential Equations" pag 99-102
There is everything you need
A: If $V$ is a Banach space we call $V'$ the dual space (see continuous dual space on wikipedia), i.e. the space of linear continuous functionals $\xi \colon V\to \mathbb R$. Then it is well known that there exists a natural injection
$$
 J \colon V \to V''
$$
defined by
$$
  J(v)(\xi) = \xi(v)
$$
for all $\xi \in V'$. We know that $J$ is an isometry, in particular: linear, continuous and injective (see double dual). We say that $V$ is reflexive if $J$ is also surjective i.e. if every $T\in V''$ can be written as $J(v)$ for some $v\in V$ (see reflexivity on wikipedia). It is not difficult to prove that if $V$ is reflexive, so is $V'$ (in fact being $V''$ isomorphic to $V$ one finds that $V'''$ is isomorphic to $V'$ see properties).
We know that $(L^1)'$ can be represented by $L^\infty$. This means that given any $\xi \in (L^1)'$ there exists $f\in L^\infty$ such that
$$
\xi(g) = \int f \cdot g
$$
for all $g\in L^1$ (see dual spaces of $L^p$ on wikipedia). What we want to prove is that there exists a linear continuous functional $T\colon L^\infty\to \mathbb R$ which cannot be written as $T = J(f)$ for any $f\in L^1$.
Let $W$ be the subspace of $L^\infty$ composed by continuous functions. Define the linear operator $T:W\to \mathbb R$ by
$$
T(f) = f(0)
$$
and notice that $T$ is continuous with respect to the norm of $L^\infty$: in fact $|T(f)| = |f(0)| \le \sup |f| = \Vert f\Vert_{L^\infty}$. Hence by Hahn-Banach theorem you can extend it to a continuous linear function on the whole space $T:L^\infty \to \mathbb R$. 
Suppose now that $T=J(f)$ for some $f\in L^1$. This means that $T$ can be represented as:
$$
  T(g) = \int f \cdot g
$$
for some $f\in L^1$. Then we know that $\int f\cdot g = g(0)$ for all continuous and bounded $g$. Now notice that given any $x_0\neq 0$ and $\epsilon <|x_0|$, it is possible to find continuous functions $g_k$ with support in $[x_0-\epsilon,x_0+\epsilon]$ which converge in $L^\infty$ to the characteristic function of the interval $[x_0-\epsilon,x_0+\epsilon]$. We hence find that 
$$
 0 = \frac{1}{2\epsilon}\int f \cdot g_k \to \frac{1}{2\epsilon}\int_{x_0-\epsilon}^{x_0+\epsilon} f(x)\,dx
$$
for all $\epsilon>0$. But if $x_0$ is a Lebesgue point (wikipedia) for $f$ such an integral must converge to $f(x_0)$ when $\epsilon\to 0$. Hence $f(x_0)=0$ for all Lebesgue points $x_0\neq 0$ which means (since almost every point is a Lebesgue point) that $f=0$ almost everywhere. But then we should conclude that $T=0$ which is a contradiction.
Hence we have found $T\in (L^\infty)'$ which is not represented by any $f \in L^1$. This means that $(L^\infty)' \supsetneq L^1$ and hence $L^1$ and $L^\infty$ are not reflexive.
