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?
Best Answer
$\newcommand{\g}{\nabla f}\newcommand{\tg}{\widetilde{\nabla f}}\newcommand{\R}{\mathbb R} \newcommand{\vpi}{\varphi}\newcommand{\Om}{\Omega}$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.