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.
 A: 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.  
A: 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.
