I know that continuously differentiable $\implies$ lipschitz continuous on compact set.wikipedia talks about differentiable functions on compact set that are not locally lipschitz.Assume that I know nothing about what locally lipschitz being means.I am not concerned anything with locally lipschitz continuous.As far as I know I couldn't find below statement being conradicted or proven.
If function $f:A \rightarrow \mathbb{R} $ is differentible on A ,where $A$ is closed interval on $\mathbb{R}$.Then f is lipschitz continuous on A
can someone provide counterexample/proof for this. I think statement is true
Best Answer
Function $f:[0,1]\to \Bbb{R}$, defined in the following way: $$f(0)=0$$ $$f(x)=x^{\frac{3}{2}}\sin(\frac{1}{x}),\ x\in (0,1]$$ is differentiable on $[0,1]$ and $[0,1]$ is compact. However, this function is neither locally nor globally Lipschitz continuous on $[0,1]$ because its derivative isn't bounded. (Function $f$ is locally Lipschitz continuous on $A$ iff every point in $A$ has a neighborhood on which $f$ is Lipschitz continuous.)
This example is given in Wikipedia article on Lipschitz continuity: https://en.m.wikipedia.org/wiki/Lipschitz_continuity
In one of the comments above I've mentioned that we could assume (for the sake of discussion) differentiability on any set is defined as differentiability on its interior. Later we've established that we'll assume $A$ is a closed interval. Differentiability on a closed interval is, as far as I'm aware, usually defined as differentiability on its interior plus the existence of "right" derivative in one end of the interval and the existence of "left" derivative in the other end.