Conditions on a Lipschitz function $f:U\subset \Bbb R^m\to \Bbb R^n$ which guarantees differentiability at a point $a\in U$

Question

Let $f:U\to \Bbb R^n$ be Lipschitz in the open $U\subset \Bbb R^m$. Given $a\in U$, suppose that, for all $v\in \Bbb R^m$, there exists the directional derivative $\dfrac{\partial f}{\partial v}(a)$ and it depends linearly on $v$. Prove that, for all path $g:(-\epsilon,\epsilon)\to U$, with $g(0)=a$, differentiable in $t=0$, there exists the velocity vector $(f\circ g)'(0)$. Conclude that $f$ is differentiable in the point $a$.

I was able to show the existence of such $(f\circ g)'(0)$'s (in the proof I didn't use the linearity of the directional derivatives), however, I couldn't do the final conclusion. I can define $T:\Bbb R^m\to \Bbb R^n$ as $$T(v)=\dfrac{\partial f}{\partial v}(a)$$ which is linear, by hypothesis. I wished that $T=f'(a)$, that is, I need to show that $$\lim_{h\to 0} \dfrac{r(h)}{|h|}=0$$ with $r(h)=f(a+h)-f(a)-T\cdot h=f(a+h)-f(a)-\dfrac{\partial f}{\partial h}(a)$, but I don't know how to work with this $\dfrac{}{\partial h}$ when $h\to 0$...

Answer 1

I think the intended proof is to use the result in Differentiability of composition with every path implies differentiability. However, it might be interesting to note that the Lipschitz condition allows a direct proof without use of contradiction.

By subtracting $x\mapsto f(a)+T(x-a)$ we can assume that $f(a)=0$ and that $T$ is zero. Fix $\epsilon>0.$ Let $c$ be the Lipschitz constant of $f.$ Take a finite $\epsilon/2c$-covering $X$ of the unit sphere in $\mathbb R^m$ (just apply the Heine-Borel compactness property of the unit sphere to the covering given by $\epsilon/2c$-balls). By assumption, the directional derivatives exist and are zero, so for each $x\in X$ there is a $\delta_x>0$ such that $\|f(a+\lambda x)\|\leq \epsilon\lambda/2$ for $0\leq \lambda<\delta_x.$ Define $\delta=\min_{x\in X}\delta_x.$

I claim that for any $0\leq \lambda<\delta$ and any unit vector $x$ we have $\|f(a+\lambda x)\|\leq \epsilon\lambda;$ since $\epsilon$ was arbitrary this means $f$ is differentiable at $a.$ To show this, pick a unit vector $x'\in X$ with $\|x-x'\|\leq \epsilon/2c.$ By definition of $\delta$ we know $\|f(a+\lambda x')\|\leq \epsilon\lambda/2,$ and by the Lipschitz property $|f(a+\lambda x)-f(a+\lambda x')|\leq \epsilon\lambda/2.$ By the triangle inequality, $\|f(a+\lambda x)\|\leq \epsilon\lambda.$

Answer 2

I've tried (without success) to prove it by contradiction, as follows:

We know that, if $f$ is differentiable at $a$, then necessarily $f'(a)\cdot v=T\cdot v:=\dfrac{\partial f}{\partial v}(a)$. Therefore, if we suppose that $f$ is not differentiable at $a$, then \begin{align*} \lim_{h\to 0}\frac{r(h)}{|h|}\not\to 0, \end{align*} with $r(h)=f(a+h)-f(a)-T\cdot h$. Let's suppose that. Then there is an $\epsilon>0$ such that, for all $\delta>0$, some $h_\delta$ exists with $0<|h_\delta|<\delta$ but $\dfrac{|r(h_\delta)|}{|h_\delta|}>\epsilon$. By choosing $\delta_n=\dfrac{1}{n}$, $n\in \Bbb N$, we obtain a sequence $\{h_n\}_{n\in \Bbb N}\subset \Bbb R^m$ such that $h_n\to 0$ but $\dfrac{|r(h_n)|}{|h_n|}>\epsilon$. Now, we show that $\dfrac{r(h)}{|h|}$ is bounded, for all $h\in \Bbb R^m-\{0\}$. Using that $f$ is Lipschitz, we get \begin{align*} \left|\frac{r(h)}{|h|}\right|&=\left|\frac{f(a+h)-f(a)}{|h|}-T\cdot \frac{h}{|h|}\right|\leq \frac{1}{|h|}|f(a+h)-f(a)|+\left|T\cdot\frac{h}{|h|}\right|\leq \frac{1}{|h|}c|h|+|T|\\ &\leq c+|T|. \end{align*}

In particular, the sequence given by $\dfrac{r(h_n)}{|h_n|}$ is bounded. By the Bolzano--Weierstrass theorem, up to a subsequence we can suppose that $\dfrac{r(h_n)}{|h_n|}\to v$, for some $v\in \Bbb R^n$.

What then?