Non-differentiability implies non-Lipschitz continuity

Question

In a simple one-dimensional framework, it is known that the differentiability of a function (with bounded derivative) on an interval implies its Lipschitz continuity on that interval. However, non-differentiability does not implies non-Lipschitz continuity as shown by the function $f:x\to |x|$. Still, there are functions that are not differentiable at a point and this argument is used to say that this same function is not Lipschitz-continuous at that point as, for example, $f:x\to \sqrt{x}$ at $x=0$ (we say that the slope of the function at that point is "vertical"). So my question is: is there a theorem telling us that for some functions (to be characterized), their non-differentiability implies their non-Lipschitz continuity? Or is this result obvious from the definition of Lipschitz continuity?

Answer 1

The following is a consequence of MVT :

If $f$ is differentiable and if $f'$ is bounded $\implies f$ is Lipschitz.

But, $f$ is Lipschitz $ \implies$ $f$ is differentiable does not hold and a counterexample for this you've already given.

Coming to your main question: When can we say "non-differentiability $\implies$ non-Lipschitz continuity"?

Intuitively speaking, if $f$ is not differentiable then either $f$ is not smooth(in the sense that it has some sharp peaks) or $f$ has infinite slope at some point. In the latter case $f$ cannot be Lipschitz, whereas in the former case if the slope of the function is finite at each point of the domain the it must be Lipschitz.

This isn't an answer to your question, but it helps build some intuition towards the problem.

Answer 2

It is worthwhile pointing out Rademacher's theorem, which states that a Lipschitz function is differentiable almost everywhere. Also, and much more trivially, the derivative will be bounded.