Continuous function on $[0,1]\to [0,1]$ that is not Lipschitz Continuous?

Question

Continuous function on $[0,1]\to [0,1]$ that is not Lipschitz Continuous?

One example I could perhaps think of is $f(x)=sin(\frac{1}{x})$ where we define $f(0)=0$.

Then this function has the required domain and range. Now, I was wondering by defining $f(0)=0$, would we have continuity at $x=0$?

Also, I guess this is also Lipschitz continuous since $f'(x)=-\frac{1}{x^2}f(x)$ is unbounded?

How could I improve my argument? Thanks!

Answer 1

Your function is not continuous at $0$ so does not work. It also cannot made continuous since the limit $x \to 0$ does not exist.

Instead, consider

$$f: [0,1] \to \mathbb{R}: x \mapsto \sqrt{x}$$

This is continuous but not Lipschitz-continuous since the derivative is unbounded near $0$.

Answer 2

You function is not continuous at $0$. There is a way easier counter-example : the square-root function. Show it cannot be Lipschitz continuous on a neighborhood of $0$.

Answer 3

Your function $f$ is not continuous at $0$: let $x_n:=(\pi/2+2n \pi)^{-1},$ then $x_n \to 0$ but $f(x_n)=1 \to 1 \ne 0=f(0).$

But $f(x)= \sqrt{x}$ will work.