A step in the proof of Characterization of $W^{1,\infty}$ In the book of PDE, evans, the proof of the characterization of $W^{1,\infty}$ uses the following argument:
Suppose $u$ has compact support and Lipchitz continuous. Then 
$$
\Vert D_i^{-h}u\Vert_{L^\infty(\mathbb{R}^n)}\leq Lip(u),
$$
where $D_i^{-h}u$ the difference quotient defined by 
$$
D_i^hu=\frac{u(x+he_i)-u(x)}{h}.
$$
Then there exists a function $v_i\in L^\infty(\mathbb{R}^n)$ and a subsequence $h_k\rightarrow0$ such that 
$$
D_i^{-h_k}u\rightharpoonup v_i \quad\text{weakly in }L^2_{loc}(\mathbb{R}^n).
$$
Question: I don't know why such a subsequence exists. Since $u$ has some compact support $\Omega$, then $u\in L^2(\Omega)$. Then there exists a function $v_i\in L^2(\Omega)$ such that 
$$
D_i^{-h_k}u\rightharpoonup v_i \quad\text{weakly in }L^2(\Omega).
$$
But I don't know why this $v_i\in L^\infty$.
 A: Note that the $D_i^{-h_k}$ are all bounded in $L^\infty(\Omega)$ by the Lipschitz constant of $u$; call this constant $M$.
Claim. Suppose $f_k \in L^2(\Omega)$, $f_k \rightharpoonup f$ in $L^2(\Omega)$,  and $\|f_k\|_{L^\infty} \le M$ for all $k$.  Then $\|f\|_{L^\infty} \le M$ as well.
Proof.  Fix $\epsilon > 0$ and let $A = \{ f \ge M+\epsilon \}$.  Since $1_A$ is bounded, it is therefore in $L^2$, so by weak convergence we have
$$\int_A f_k = \int_\Omega 1_A f_k \to \int_\Omega 1_A f = \int_A f.$$
However, since $\|f_k\|_{L^\infty} \le M$, we have $\int_A f_k \le M \mu(A)$.  And since $f \ge M+\epsilon$ on $A$, we have $\int_A f \ge (M+\epsilon) \mu(A)$.  The only way to reconcile these statements is to have $\mu(A) = 0$, which is to say that $f < M+\epsilon$ almost everywhere.  Letting $\epsilon \downarrow 0$ along a sequence, we have $f \le M$ almost everywhere.  A similar argument will show that $f \ge -M$ almost everywhere.
Alternate proof.  Since $L^1 \cap L^2$ is dense in $L^1$, by $L^p$ duality we have $$\|f\|_{L^\infty} = \sup\left\{\int_\Omega fg : g \in L^1 \cap L^2, \|g\|_{L^1} \le 1\right\}.$$ 
So suppose $g \in L^1 \cap L^2$ with $\|g\|_{L^1} \le 1$.  We have $\int_\Omega f_k g \le M$ for every $k$, so by weak convergence, $\int_\Omega fg \le M$ also.  Taking the supremum over $g$, we have $\|f\|_{L^\infty} \le M$.
Yet another proof.  Let $B = \{h \in L^2 : \|h\|_{L^\infty} \le M\}$.  I claim first $B$ is strongly closed in $L^2$.  For suppose $\|h_k\|_{L^\infty} \le M$ and $h_k \to h$ in $L^2$.  There is a subsequence $h_{k_j}$ which converges to $h$ almost everywhere.  Since $|h_{k_j}| \le M$ almost everywhere, the same is therefore true of $h$.  Now since $M$ is strongly closed and convex, by Mazur's lemma it is also weakly closed, which is the desired statement.
