For a Lipschitz function $f : \mathbb{R}^n \rightarrow \mathbb{R}$, define the Lipschitz norm acting on the set of Lipschitz functions as $\| f \|_{Lip(\mathbb{R^n})} = \sup_{x \neq y}\frac{|f(x) - f(y)|}{\| x - y \|}.$
In this paper, the author quoted that $\| f \|_{Lip(\mathbb{R^n})} = \| \nabla f \|_{L^{\infty}{\mathbb{(R^n)}}},$ where $\nabla f$ is a distribution gradient in $L^{\infty}(\mathbb{R}^n).$
I have trouble showing the equality $\| f \|_{Lip(\mathbb{R^n})} = \| \nabla f \|_{L^{\infty}{\mathbb{(R^n)}}}.$ Any hint would be much appreciated.
The part where I have trouble with is the definition of $\| \nabla f \|_{L^{\infty}{\mathbb{(R^n)}}}.$
By definition $\Vert \nabla f\Vert_{\infty}$ is the $L^\infty$ norm of the Euclidean norm of $\nabla f$. Note that when $f$ is Lipschitz, $f$ is differentiable (in the multivariable sense) almost everywhere, this is Rademacher's theorem.
Call the Lipschitz norm $L$.
Consider then that $\left|\frac{f(x+h)-f(x)}{\Vert h \Vert}\right| \leq L$, and sending $h \to 0$ gives $|\nabla f| \leq L$ at every point of differentiability of $f$, and hence almost every $x$. Thus $\Vert \nabla f\Vert_\infty \leq L$.
On the other hand, show that the supremum $L$ can be achieved by taking $x,y$ closer and closer to some fixed $z$. Then use that $\left|\left|\frac{f(x)-f(y)}{\Vert x-y\Vert}\right|-L\right| \leq \epsilon$ for $x,y$ close enough $z$. Then for every point of differentiability of $f$ close enough to $z$, you find $|\nabla f| \geq L-2\epsilon$, and hence $\Vert \nabla f \Vert_\infty \geq L-2\epsilon$. Send $\epsilon \to 0$ to finish the claim.