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?
 A: 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.
