# differentiability and Lipschitz continuous on compact set

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

• Differentiability cannot be defined on arbitrary compact sets. For example $\{0,1,\frac 1 2,\frac 1 3,..\}$ is a compact set. What do you mean by a differentiable function on this set? – Kavi Rama Murthy Sep 21 '19 at 12:01
• Locally Lipschitz should mean that it's Lipschitz continuous locally, which should be a weaker condition than Lipschitz continuity. We could assume that differentiability on any set is defined as differentiability on its interior. My question is: where does $A$ "live"? In $\mathbb{R}$? In a normed vector space? – blue Sep 21 '19 at 12:36
• okay assume it is closed interval .then – viru Sep 21 '19 at 15:48
• It's a good question. It's easy to see that it's true for $f\in C^1$, but what about differentiable functions whose derivative isn't continuous? Can you tell us what wiki page you were reading? Maybe there are some clues there. – blue Sep 21 '19 at 16:41
• @blue can you write an answer so that I can accept? – viru Sep 23 '19 at 14:18

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.)
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.