Let $f\colon \Omega \to \mathbb{R}$ be a Lipschitz function on an open subset $\Omega\subset \mathbb{R}^n$.


By [the Rademacher theorem](https://en.wikipedia.org/wiki/Rademacher%27s_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?**



Do two ways to differentiate Lipschitz functions coincide?
$\newcommand{\g}{\nabla f}\newcommand{\tg}{\widetilde{\nabla f}}\newcommand{\R}{\mathbb R}
\newcommand{\vpi}{\varphi}\newcommand{\Om}{\Omega}$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.


