Uniform convergence of difference quotient function implies differentiability in $\mathbb{R}^n$ This arises in the context of the proof of the Rademacher theorem.
Suppose that $f: \mathbb{R}^n \to \mathbb{R}$ is a Lipschitz function. It therefore has a distributional gradient $L$ and suppose that $x$ is a point in the Lebesgue set of $L$.
We can define $f_r(y) = \frac{f(x+ry)-f(x)}{r}$ and the claim is that if $f_r(y) \to \langle L(x), y\rangle$ uniformly as $r \to 0$ then $f$ is differentiable at $x$. Why is this true and why is uniform convergence necessary rather than just point wise convergence?

In order for the existence of all directional derivatives to imply full differentiability, you need


*

*the directional derivative $Df|_x(y) = \lim_{r \to 0}f_r(y)$ to be linear in $y$, and

*the directional difference quotients to converge uniformly in $y$.


You can find some discussion of this on this Wikipedia page, including an example of what can go wrong when the convergence is not uniform in $y$: a multivariable function with vanishing directional derivatives at a point that is not differentiable there.
When both conditions are satisfied, the proof is fairly straightforward. We know the directional derivative $Df|_x (y) = \langle L, y\rangle$ is linear, so it is a suitable candidate to be the derivative. The uniform convergence of the quotients can be written $|f_r(y) - \langle L,y \rangle| \le\omega(r)$ for some function satisfying $\lim_{r \to 0} \omega(r) = 0$.
Thus we calculate $$f(x+y)-f(x)-\langle L,y\rangle=|y|f_{|y|}(\frac{y}{|y|})-\langle L,y\rangle;$$ so we have $$\frac{|f(x+y)-f(x)-\langle L,y\rangle|}{|y|}=\left|f_{|y|}(\frac{y}{|y|}) -
 \langle L, \frac{y}{|y|}\rangle\right| \le\omega(|y|)$$which converges to zero as $y \to 0$ as desired.
