Does a Lipschitz continuous function have one-sided derivatives everywhere? Let $f\colon\mathbb R\to\mathbb R$ satisfy the Lipschitz condition: there exists $K\geq 0$ such that for all $x,y\in\mathbb R$, we have $|f(x)-f(y)|\leq K\cdot |x-y|$. Is it true that $f$ has one-sided derivatives everywhere? I.e., that the limits
$$\lim_{h\nearrow 0}\frac{f(x+h)-f(x)}{h}\quad\text{and}\quad\lim_{h\searrow 0}\frac{f(x+h)-f(x)}{h}$$
exist for every $x$?
I know that both exist a.e. since the Lipschitz condition implies that $f$ is differentiable a.e., but I would like to know if it is not only true a.e., but everywhere. After working with this for some time, I got the feeling that it might be.

Let 
$$ f(x)=\begin{cases}x\sin\frac1x&0<|x|<\frac1\pi\\0&\text{otherwise}\end{cases}$$
Then $f$ is Lipschitz, but none of the one-sided derivatives exists at $x=0$.
