Prove that a Lipschitz continuous function is differentiable at a point ${\bf x}_0$

Question

Consider $f: B({\bf x}_0,r)\to \Bbb R$, that apart from being Lipschitz continuous, has directional derivatives at the point $x_0$ and $\frac{\partial f}{\partial{\bf v}}({\bf x}_0)=\sum_{i=1}^n \frac{\partial f}{\partial x_i}({\bf x}_0)v_i$ for every direction $\bf v$. How do I use these facts to show that it is differentiable at $x_0$? I am pretty clueless about how to go about this problem; the hint given says we need to use contradiction by assuming it is not differentiable and use the Bolzano Weierstrass theorem.

Answer 1

Instead of using a proof by contradiction, one can also give a (more or less) constructive proof:

Let $\varepsilon>0$ be arbitrary. Since $\partial B_{1}\left(0\right)$ is compact, there are $y_{1},\dots,y_{N}\in\partial B_{1}\left(0\right)$ such that for each $y\in\partial B_{1}\left(0\right)$, we have $\left|y-y_{j}\right|<\varepsilon$ for some $j\in\left\{ 1,\dots,N\right\} $. Without loss of generality (why?), let us assume $x_{0}=0$ as well as $f\left(x_{0}\right)=0$. Set $v:=\nabla f\left(0\right)$.

Now for each $x\neq0$, there is some $j_{x}\in\left\{ 1,\dots,N\right\} $ with $\left|\frac{x}{\left|x\right|}-y_{j_{x}}\right|<\varepsilon$. Hence, $\left|x-\left|x\right|y_{j_{x}}\right|<\varepsilon\left|x\right|$. Thus, using the fact that $f$ is Lipschitz (with Lipschitz constant $L\geq 0$), we get \begin{eqnarray*} \left|f\left(x\right)-\left\langle v,x\right\rangle \right| & \leq & \left|f\left(x\right)-f\left(\left|x\right|y_{j_{x}}\right)\right|+\left|f\left(\left|x\right|y_{j_{x}}\right)-\left\langle v,\left|x\right|y_{j_{x}}\right\rangle \right|+\left|\left\langle v,\left|x\right|y_{j_{x}}\right\rangle -\left\langle v,x\right\rangle \right|\\ & \leq & L \cdot \left|x-\left|x\right|y_{j_{x}}\right|+\left|x\right|\cdot\max_{j\in\left\{ 1,\dots,N\right\} }\sup_{\left|t\right|\leq\left|x\right|}\left|\frac{f\left(ty_{j}\right)}{t}-\left\langle v,y_{j}\right\rangle \right|+\left|v\right|\left|x-\left|x\right|y_{j_{x}}\right|\\ & \leq & \left|x\right|\cdot\underbrace{\left[\varepsilon L \cdot +\max_{j\in\left\{ 1,\dots,N\right\} }\sup_{\left|t\right|\leq\left|x\right|}\left|\frac{f\left(ty_{j}\right)}{t}-\left\langle v,y_{j}\right\rangle \right|+\left|v\right|\varepsilon\right]}_{=:\gamma\left(x\right)}. \end{eqnarray*} By your assumption, we have $$ \frac{f\left(ty_{j}\right)}{t}\xrightarrow[t\to0]{}\frac{\partial f}{\partial y_{j}}\left(0\right)=\left\langle v,y_{j}\right\rangle . $$ This implies $$ \gamma\left(x\right)\xrightarrow[x\to0]{}\varepsilon\left(L+\left|v\right|\right). $$ As $\varepsilon>0$ was arbitrary, we have shown that $$ \frac{\left|f\left(x\right)-\left\langle v,x\right\rangle \right|}{\left|x\right|}\xrightarrow[x\to0]{}0. $$ This proves what you want.