# Finding a Lipschitz Continuous function in $D=[-1,1]$ that is not differentiable at all points in D

The problem is to find a Lipschitz Continuous function in $D=[-1,1]$ that is not differentiable at all points in D.

To tackle this, I have considered functions I know to not be differentiable at certain points on $[-1,1]$, such as $f(x)=|x|$ which is non differentiable at $x=0$, and $f(x)=x^\frac{1}{3}$ whose derivative is infinite at $x=0$. However as I found that neither of these functions are Lipschitz in [-1,1].

There must be a more logical way to approach this task. Any ideas are appreciated thanks

Edit - as Uzman pointed out in the comments, the problem may infact be to find a function that is "not differentiable at all points" in $[-1,1]$, as opposed to the condition I used.

• Your first example is definitely Lipschitz. Check again. – Ben Grossmann Feb 9 '15 at 17:43
• Ah yes, thank you. I applied the triangle inequality incorrectly when attempting to prove the Lipschitz property for $f(x)=|x|$. – Trawkley Feb 9 '15 at 17:53
• Do you mean "not differentiable" at all points or not "differentiable at all points"? – Alp Uzman Feb 9 '15 at 17:56
• Uzman - Initially I was considering the function to be not "differentiable at all points", i.e. it can be differentiable at some points in the interval. But now I think I may have interpreted the concept of the problem wrong. If the function did have to be "non differentiable at all points" in the interval would this not prove very hard to find? – Trawkley Feb 9 '15 at 18:02

I think you're wrong about $f(x)=|x|$ not being Lipschitz on [-1,1]. If you take K =1 as the constant and use the reverse triangle inequality and let $epsilon$ = $delta$ in the limit computation:
$$|x|-|y| < = |x-y|$$
I think you'll find $f(x)=|x|$ is indeed Lipschitz on all the reals and not differentiable on the entire domain. In fact, in metric spaces, this function is the usual metric in R and we know that's Lipschitz. (Try and prove it!) So I think this'll satisfy your problem.
As an alternative,did you try and take a composition of a Lipschitz function and $f(x)=|x|$ ? That might work. For example, let's try $g(x)=sin(|x|)$. We know that sin x is a Lipschitz continuous function because its derivative, the cosine function, is bounded above by 1 in absolute value. Try it,it'll work.
By Rademacher's theorem, every Lipschitz function from $[-1,1]$ into the real line (or more generally, an Euclidean space) is differentiable at almost every point (with respect to the Lebesgue measure) so your question has negative answer.