Prove that $f$ is uniformly continous

I have to prove this:

Suppose $f:(a,b)\to \mathbb{R}$ is differentiable and $|f´(x)| \leq M$ for all $x\in (a,b)$.Prove that $f$ is uniformly continous on $(a,b)$.Give an example of a function $f:(0,1) \to \mathbb{R}$ that is differentiable and uniformly continous on $(0,1)$, but such that $f´$ is unbounded.

My attempt:

Since $f$ is differentiable we know that for all $\epsilon>0$ there exists a $\delta>0$ such that if $|x-y|<\delta$ then:

$$|\frac{f(x)-f(y)}{x-y}-f´(x)|< \epsilon$$

then we choose $\delta=\frac{\epsilon}{\epsilon+M}$ and we have that:

$$|\frac{f(x)-f(y)}{x-y}-| \epsilon+f´(x) \leq \epsilon +M$$

$$\Rightarrow |f(x)-f(y)|<(\epsilon +M)|x-y|<(\epsilon +M) \delta= \epsilon$$

Therefore $f$ is uniformly continous.

Now let´s consider the function $f: \mathbb{R_{\geq 0}} \to \mathbb{R}$ such that $f(x)= \sqrt{x}$. To see that $f$ is uniformly continous we note that it is continous at all its dominion, therefore $f$ is continous at $[0,1]$ then we have that $f$ is uniformly continous at $[0,1]$, this means that the same $\delta$ Works for $x,y \in (0,1)$, and therefore $f$ is uniformly continous at $(0,1)$ , but $f´(x)=1/2 \sqrt{x}$ tat is clearly unbounded.

Can you tell me if I am right please, if not how can I fix the problems please, thank you a lot.

• Correct, in the first part you essentially prove that if $|f'|\leq M$ then $f$ is $M$-Lipschitz, which implies uniform continuity. Second part correct. – Milly Oct 19 '14 at 3:23
• Thanks Milly then I don't have to fix something, right? :) – user162343 Oct 19 '14 at 3:26
• It is all right, apart from the typo in formula after "then we choose $\delta$ ..." (I think meant $\frac{|f(x)-f(y)|}{|x-y|}\leq \left|\frac{f(x)-f(y)}{x-y}-f'(x)\right|+|f'(x)|\leq \varepsilon+M$). – Milly Oct 19 '14 at 3:29

Example is $f(x)=x^{\frac{1}{n}}$, basically you have to construct a function whose limit exists at end points and derivative is unbounded in $(0,1)$