A question in the proof that Lipchitz continuous functions implies $W^{1,\infty}$.

Question

In $\S 5.8.2$ of Evan's PDE book, there is a theorem about characterization of $W^{1,\infty}$.

Here it says

On the other hand assume now $u$ is Lipschitz continuous; we must prove that $u$ has essentially bounded weak first derivatives. Since $u$ is Lipschitz, we see \begin{equation*} \Vert D_i^{-h}u \Vert_{L^{\infty}(\mathbb{R}^n)} \le \mbox{Lip}(u), \end{equation*} and thus there exists a function $v_i \in L^{\infty}(\mathbb{R}^n)$ and a subsequence $h_k \to 0$ such that \begin{equation*} D_i^{-h_k}u \rightharpoonup v_i \quad {\text{weakly in}}\quad L^2_{\text{loc}}(\mathbb{R}^n) \end{equation*}

Here, $u$ has compact support and $D_i^h u = \frac{u(x+ he_i) - u(x)}{h}$.

Since $D_i^{-h}$ has compact support and are bounded in $L^\infty$, $D_i^{-h} \in L^2(\mathbb{R}^n)$. Also, since $L^2(\mathbb{R}^n)$ is reflexive, we have a subsequence $D_i^{-h_k}u \rightharpoonup v_i$ in $L^2_{\text{loc}}(\mathbb{R}^n)$ as the book says. However, I don't understand why $v_i \in L^\infty(\mathbb{R}^n)$, which is the key point of proving $u \in W^{1,\infty}$. Could you help me understand this? Any help would be appreciated!

Answer 1

Your sequence that weakly converges in $L^2_{\text{loc}}$ is still bounded in $L^\infty:=L^\infty(\mathbb R^n) $, $$ \| D^{-h_k}_i u \|_{L^\infty}\le \text{Lip}(u).$$ Since $L^\infty= (L^1)^*$, the Banach-Alaoglu theorem tells us that there is a further subsequence (which I will also call $h_k$) and some $w_i \in L^\infty$ such that

$$ D^{-h_k}_i u \overset{*}\rightharpoonup w_i \text{ in } L^\infty $$ Combined with $D^{-h_k}_i u \rightharpoonup v_i $ in $L^2$, we need a weaker notion of convergence, implied by both weak* convergence in $L^\infty$ and weak convergence in $L^2_{\text{loc}}$, that still has uniqueness of limits: this will imply $v_i = w_i$ and in particular $v_i \in L^\infty$. One example is convergence in the space of distributions.