Do two ways to differentiate Lipschitz functions coincide? Let $f\colon \Omega \to \mathbb{R}$ be a Lipschitz function on an open subset $\Omega\subset \mathbb{R}^n$.
By the Rademacher theorem $f$ has first derivative almost everywhere. We denote it $\nabla f$.
On the other hand $f$ has a derivative in the sense of distributions (with values in $\mathbb{R}^n$). We denote it by $\widetilde{\nabla f}$.
Since $\nabla f$ is bounded and defined almost everywhere, it can be considered as a distribution with values in $\mathbb{R}^n$ as well.
Is it true that $\nabla f=\widetilde{\nabla f}$ in the sense of distributions?
 A: $\newcommand{\g}{\nabla f}\newcommand{\tg}{\widetilde{\nabla f}}\newcommand{\R}{\mathbb R}
\newcommand{\vpi}{\varphi}\newcommand{\Om}{\Omega}$The answer is yes.
For $\Om=\R^n$, this follows from the identity
\begin{equation*}
    \int_{\R^n}\frac{f(x+tu)-f(x)}t\,\vpi(x)\,dx=\int_{\R^n}\frac{\vpi(x-tu)-\vpi(x)}t\,f(x)\,dx, \tag{1}\label{1}
\end{equation*}
where $t\in\R\setminus \{0\}$, $u$ is a unit vector in $\R^n$, and $\vpi$ is any smooth function with support contained in some ball $B_\vpi$ of a strictly positive radius centered at the origin.
(The assumption that $\Om=\R^n$ can be made without loss of generality. Indeed, if $\Om\ne\R^n$, just extend the (say) $L$-Lipschitz function $f$ on $\Om$ to the $L$-Lipschitz function $\bar f$ on $\R^n$, quite naturally, by the infimal convolution formula $\bar f(x):=\inf_{y\in\Om}(f(y)+L\|x-y\|)$ for $x\in\R^n$.)
With $u\in\R^n$ fixed, for almost all $x\in\R^n$ we have
\begin{equation*}
    \frac{f(x+tu)-f(x)}t=\g(x)\cdot u+r_1(x,t),
\end{equation*}
where $\cdot$ is the dot product, $|r_1(x,t)|\le2L$, $L$ is the Lipschitz constant of $f$, and $r_1(x,t)\to0$ as $t\to0$. So, by dominated convergence,
\begin{equation*}
\begin{aligned}
    \int_{\R^n}\frac{f(x+tu)-f(x)}t\,\vpi(x)\,dx
    &=\int_{B_\vpi}\frac{f(x+tu)-f(x)}t\,\vpi(x)\,dx \\ 
    &\to \int_{B_\vpi}\g(x)\cdot u\,\vpi(x)\,dx \\ 
    &=\int_{\R^n}\g(x)\cdot u\,\vpi(x)\,dx \\ 
    &=(\g)(\vpi)\cdot u. 
\end{aligned}
\tag{2}\label{2}
\end{equation*}
On the other hand, for all $x\in\R^n$
\begin{equation*}
    \frac{\vpi(x-tu)-\vpi(x)}t=-\nabla\vpi(x)\cdot u+r_2(x,t),
\end{equation*}
where $r_2(x,t)\to0$ uniformly in $x\in2B_\vpi$ as $t\to0$. So, for $t$ close enough to $0$,
\begin{equation*}
\begin{aligned}
    \int_{\R^n}\frac{\vpi(x-tu)-\vpi(x)}t\,f(x)\,dx
    &=\int_{2B_\vpi}\frac{\vpi(x-tu)-\vpi(x)}t\,f(x)\,dx \\ 
&   \to-\int_{2B_\vpi}\nabla\vpi(x)\cdot u\,f(x)\,dx \\ 
&   =-\int_{\R^n}\nabla\vpi(x)\cdot u\,f(x)\,dx \\ 
&   =(\tg)(\vpi)\cdot u. 
\end{aligned}
\tag{3}\label{3}
\end{equation*}
Thus, by \eqref{1}--\eqref{3}, $\tg=\g$ as distributions, as claimed.

Similarly, $\tg=\g$ even as tempered distributions.
A: Yes, this is true. For example it follows from the proof of Theorem 2.2.1 in the book of Ziemer "Weakly differentiable functions" where it is shown that the distributional gradient gives the classical one, a.e. It is stated for functions defined in the whole space, but this is not a restriction since every Lipschitz function can be extended.
However, I think that the easiest proof (which works for functions in $W^{1,p}_{loc}, p>N$ consists in showing that at every p-Lebesgue point $x_0$ of the distributional gradient $\tilde \nabla f$ the pointwise definition of differentiability (with $\tilde \nabla f(x_0)$) holds.
