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.
