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?

  • $\begingroup$ What is Lipschitz continuity ? is it Uniform continuity? $\endgroup$ – hamam_Abdallah Jul 23 '17 at 0:35
  • $\begingroup$ @Salahamam_Fatima. A function f is Lipschitz iff there exists $K>0$ such that for all $x,y\in dom(f)$ we have $|f(x)-f(y)|\leq K|x-y|.$ $\endgroup$ – DanielWainfleet Jul 23 '17 at 1:19
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    $\begingroup$ Differentiabilty on an interval does NOT imply Lipschitz continuity on that interval, because $ f'$ may be unbounded on that interval. (Use the MVT.)... For example $f(x)=1/x$ on the interval $(0,1).$.... A sharper example is $f(0)=0$ and $f(x)=x^2\sin (1/x^2)$ when $x\ne 0, $ which is differentiable for all real $x,$ but $f'$ is unbounded on any neighborhood of $0.$ $\endgroup$ – DanielWainfleet Jul 23 '17 at 1:27
  • $\begingroup$ @DanielWainfleet you are right, I have added the boundedness argument. $\endgroup$ – pluton Jul 23 '17 at 12:40

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.

  • $\begingroup$ Rademacher's theorem is primarily a theorem in several variables; it holds true for single variable functions too, but then the proof is much easier, being a standard result in integration theory. $\endgroup$ – Harald Hanche-Olsen Jul 23 '17 at 12:51

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.


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