Question is to check if :
any differentiable function $f : (0,1)\rightarrow [0,1]$ is uniformly continuous.
I know that any continuous function on compact subset of $\mathbb{R}$ is uniformly continuous.
As $(0,1)$ is not compact, we can not say anything at this time.
Now, as it is given that $f$ is differentiable, if its derivative $f'$is bounded then $f$ is uniformly continuous.
So, I am trying to look for differentiable functions $f$ on $(0,1)$ such that $f'$ is unbounded.
i am not very familiar with large number of differentiable functions with unbounded derivatives.
I know $f(x)=\sqrt{x}$ has unbounded derivative, but $\sqrt{x}$ is uniformly continuous….
So, I would like someone to help me out with some hint.
P.S : I have just now saw one example
$$f(x)=x^2\sin{\frac{1}{x^2}}$$
which is differentiable but is not bounded.
I see that $\sin(\frac{1}{x^2})$ is bounded by $1$ and if $x\in (0,1)$ then so is $x^2$ and so is $x^2\sin{\frac{1}{x^2}}$
So, $f(x)=x^2\sin{\frac{1}{x^2}}$ is from $(0,1)$ to $[0,1]$ whose derivative is unbounded.
Now, the problem reduces to show that $f(x)$ is not uniformly continuous… 🙁
Best Answer
Consider the function $f(x)=\frac12(1+\sin\frac1x)$ on $(0,1)$. It's obviously differentiable. However, $f(\frac{1}{(2n-\frac12)\pi})=0$ and $f(\frac{1}{(2n+\frac12)\pi})=1$, but $\frac{1}{(2n-\frac12)\pi}$ and $\frac{1}{(2n+\frac12)\pi}$ can be arbitrary close (if you take large $n$).