Let $\Omega\subset\mathbb{R}^n$ be a bounded domain and suppose $u\in W^{2,1}_{\operatorname{loc}}(\Omega)\cap C(\Omega)$. The book I am reading claims that the normal mapping of $u$ satisfies
\begin{equation} \chi_u(y) := \{p\in\mathbb{R}^n:u(x)\leq u(y) + p\cdot(x-y)~\forall x\in\Omega\} = \{Du(y)\} \quad \text{for a.e. }y\in\Gamma_u := \{z\in\Omega:\chi_u(z)\not=\emptyset\}. \end{equation}
The given argument is as follows. We know that $w(x):= u(y)+p\cdot(x-y) - u(x)$ takes its minimum value (of zero) on $\Gamma_u\subset\Omega$. It is a fact that for all $\xi\in\mathbb{R}^n$ with $|\xi|=1$, we have \begin{equation} \frac{w(y+h\xi)-w(y)}{h} \rightarrow \frac{\partial w}{\partial \xi} \quad\text{in }L_{\operatorname{loc}}^1(\Omega), \end{equation} where $\frac{\partial w}{\partial \xi}$ denotes the weak derivative of $w$ in the direction $\xi$. We therefore know that, up to a subsequence, this convergence holds a.e. in $\Omega$. The authors then say -- and this is the step I do not understand -- that by taking $h\rightarrow 0^+$ and $h\rightarrow 0^-$ along a suitable subsequence, we conclude that \begin{equation} \frac{\partial w}{\partial \xi}=0 \quad\text{a.e. in }\Gamma_u. \end{equation} (Of course, the definition of $w$ and taking $\xi$ to be the coordinate directions then gives the desired result).
How do the authors conclude that the weak derivative of $w$ in the direction $\xi$ is zero a.e. in $\Gamma_u$? Is it a general fact that if the weak derivative of a continuous function exists, then the weak derivative is zero almost everywhere on the set where the function attains its minimum? Thanks in advance!
$\newcommand{\R}{\mathbb{R}}$ Fix $\xi \in \R^n$ of length one and denote $$ w_h(y) := \frac{w(y+h\xi)-w(y)}{h} \quad \text{for } h \in \R. $$ If the rest of the argument is fine, we have $w_h \to \frac{\partial w}{\partial \xi}$ in $L_{\operatorname{loc}}^1(\Omega)$ as $h \to 0$.
Focus first on $h > 0$. Since $w = 0$ on $\Gamma_u$ and $w \ge 0$ in general, we have $w_h \ge 0$ on $\Gamma_u$. This last property is preserved by $L^1$-convergence (one can choose an a.e. convergent subsequence), so also $\frac{\partial w}{\partial \xi} \ge 0$ a.e. on $\Gamma_u$. Considering $h < 0$, we obtain the opposite inequality.