Let $M$ be a smooth manifold. Let $d$ be any metric on $M$ which generates the topology of $M$. Let $f:M \to R$ be Lipschitz w.r.t the metric $d$.
Is it true that $f$ is differentiable a.e?
Note: This question is different from this one, where $d$ is assumed to come from a Riemannian metric. In that case, the induced distance is locally equivalent to the "Euclidean" distance.
I am allowing metrics which are not locally equivalent. For instance, if $d$ is a metric induced by a Riemannian metric, we can choose $\tilde d =\sqrt d$, and so $\tilde d$-Lipschitzity does not imply $d$-Lipschitzity.
I am not even sure whether the assertion is true for $f:\mathbb{R} \to \mathbb{R}$ which satisfies $|f(x)-f(y)| \le C \cdot \sqrt{|x-y|}$.
($f(x)=\sqrt x$ which is defined on $\mathbb{R}^{\ge 0}$ satisfies this with $C=1$, and it is locally Lipschitz around every point except $x=0$, hence differentiable a.e)
Using studiosus's comment:
Let $f:\mathbb{R} \to \mathbb{R}^2$ represent the Koch snowflake in the plane.
Define a metric $\tilde d$ on $\mathbb{R}$ as follows:
$\tilde d(x,y) =d(f(x),f(y))$, where $d$ is the standard Euclidean distance on $\mathbb{R}^2$.
$\tilde d$ generates the stabdard topology on $\mathbb{R}$.
Define $g:\mathbb{R} \to \mathbb{R}$ by $g(x)=\tilde d(x,0)$. Then
$$ |g(x)-g(y)| \le \tilde d(x,y), $$
so $g$ is $\tilde d$-Lipschitz, but not differentiable at every point.