Prove that a Lipschitz function is continuous. Must a Lipschitz function necessarily be differentiable?

Question

I have:

Pf. Let epsilon>0 be given. Select a delta such that delta is less than epsilon/M so that |x-y| is less than delta. Then|(f(x)-f(y)| is less than M|x-y| is less than M(delta) is less than epsilon. So |f(x)-f(y)| is less than epsilon So f is continuous.

Now I do not know how to show that Lipschitz does not imply differentiability unless i use counterexample f(x)=|x|.

Then, there is also a T/F statement: If f is Lipschitz on R, then f attains its max and min values.

Answer 1

Try the T/F statement (about attaining the max) on the example you gave: $f(x) = |x|$.

And if you want other examples of a function that is Lipschitz but fails to be differentiable at multiple points, consider the Triangle wave. Now, this one does attain its mins and maxes. However, if we add to the Triangle wave (denote it $w(x)$) the function $x$, then the resulting function, $$ w(x) + x, $$ is Lipschitz, still fails to be differentiable at multiple points, and does not attain its min or max.