A Problem about the Differentiability and Directional Derivative I have the following problem: 

Assume that $f:\Bbb R^n\to\Bbb R$ satisfies $$|f(x)-f(y)|\le M\cdot|x-y|$$for some $M\in \Bbb R$ and $f(0)=0$. Suppose that all the directional derivatives vanish at the origin. Prove that  $f$ is differentiable at 0. 

The following was what I tried: 
In order to show that $f$ is differentiable at $0$, we need to find a linear map $A$ such that $$\lim_{h\to0}{{|f(h)-f(0)-A\cdot h|}\over{|h|}}=0.$$ Since all the directional derivatives vanish at 0, which says $A=0$ (Is this right???).$0\le{{|f(h)-f(0)|}\over{|h|}}\le M|h|/|h|=M$, which I cannot get the desired result. Any help, please. 
 A: The given assumptions indeed imply that $f$ is differentiable at $0$ and that $df(0)=0$. 
Proof. We have to prove that
$$\lim_{x\to 0}{|f(x)|\over|x|}=0\ .$$
Assume that this is not the case. Then there exists an $\epsilon>0$ and a sequence $(x_k)_{k\geq0}$ with $\lim_{k\to\infty} x_k=0$ such that
$$|f(x_k)|>\epsilon |x_k|\qquad\forall k\geq0\ .$$
The unit sphere $S^{n-1}$ is compact. After passing to a subsequence we can therefore assume that there is an $u\in S^{n-1}$ with
$${x_k\over|x_k|}\to u\qquad(k\to\infty)\ .$$
It follows that there is a $k_0$ such that
$$\left|{x_k\over|x_k|}-u\right|<{\epsilon\over 2M}\qquad(k>k_0)\ ,$$
or
$$\biggl|x_k-|x_k|u\biggr|<{\epsilon|x_k|\over 2M}\qquad(k\to\infty)\ .$$
The Lipschitz condition then implies that
$$\biggl|f\bigl(|x_k|u\bigr)\biggr|\geq|f(x_k)|-M\biggl|x_k-|x_k|u\biggr|>\epsilon|x_k|-{\epsilon|x_k|\over 2}={\epsilon|x_k|\over2}\qquad(k>k_0)\ .$$
It follows that
$${\bigl|f\bigl(|x_k|u\bigr)\bigr|\over|x_k|}\geq{\epsilon\over2}$$
for all $k>k_0$, which contradicts the assumption $D_uf(0)=0$.
A: As you found out, the Lipschitz condition applied here is useless. The essence of the problem lies in the difference between the directional derivative definition and this stronger full differentiation definition. 
Edit: In consideration of the contradiction proof from Christian Blatter, and a few attempts, it appears what I had before probably won't work. Anyway, for a direct proof suggestion, use an $\epsilon$-$\delta$ proof, and apply the definition of directional derivative to many different directions. Then for a given direction $h$, you can compare to the result to the nearest one for which you have applied the definition. Probably in essence will be the same as Christian's proof.
