Let $I \subset \mathbb{R}^{N}$ be a convex, bounded open set with Lipschitz boundary $\partial I$. Let $\lbrace u_{n} \rbrace_{n}$ and $u$ be such that $$ u_{n} \rightharpoonup^{*} u~~ \text{ in }~ W^{1,\infty}(I), $$ $$ \Vert \nabla u_{n} \Vert_{L^{\infty}(I)} \leq \sigma, $$ where $\sigma > 0$ is fixed (we denote the weak * convergence by the symbol '$\rightharpoonup^{*}$').

Let $$ \delta_{n} := \frac{1}{\sigma}\Vert u_{n}-u \Vert_{L^{\infty}} $$ and $$ I_{n} := \lbrace x \in I: \text{dist}(x,\partial I) > \delta_{n} \rbrace. $$ Now define $$ {\bar{v}}_{n}(x) := \begin{cases} & u_{n}(x) ~~~~~~~~~~~~\text{ if } x \in {\bar{I}}_{n},\\ &u(x) ~~~~~~~~~~~~~~~\text{if } x \in \partial I. \end{cases} $$ It can be shown that $\bar{v}_{m}$ is locally Lipschitz with constant $2\sigma$ on $\bar{I}_{m} \cup \partial I$. Using Mac Shanes's Lemma we can find an extension of $\bar{v}_{m}$ to $\bar{I}$, which we call $v_{m}$. So by Mac Shane's Lemma, $v_{m}$ is Lipshitz continuous on $\bar{I}$, with constant $2\sigma$. i.e. $|v_{m}(x)-v_{m}(y)| \leq 2\sigma|x-y|$.

Question: How does it follow that $\| \nabla v_{m}\|_{L^{\infty}(I)} \leq 2\sigma$?

Thanks a lot for any assistance.

Mac Shane's Lemma is stated as follows: Consider metric space $(V,d_{V})$. Let $f:A \subset (V,d_{V}) \rightarrow \mathbb{R}$ be a Lipschitz continuous function i.e. $\exists K > 0$ such that $$ |f(x)-f(y)| \leq Kd_{V}(x,y) ~~~~\forall x,y \in A, $$ then $\exists \bar{f}: V \rightarrow \mathbb{R}$ such that $\text{lip}(\bar{f}) \leq K$, where $\text{lip}(\bar{f})$ denotes the Lipschitz constant of $\bar{f}$.


Note that since $v_{m}$ is Lipshitz continuous, it is also almost everywhere differentiable (by Rademachar's Theorem). It follows then that $$\lvert \partial_k v_m(x)\rvert = \lim_{h\to 0} \left\lvert\frac{v_m(x+h\cdot e_k) - v_m(x)}{h}\right\rvert \leqslant 2\sigma$$ for $1 \leqslant k \leqslant n$, it follows then that $\lVert \nabla v_m\rVert_{L^\infty(I)} \leqslant 2\sigma$.


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