# Is any differentiable function $f : (0,1)\rightarrow [0,1]$ is uniformly continuous

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... :(

• What is the derivative of a function of the form $f(1/x)$? – deinst Oct 7 '13 at 13:41
• $f'(\frac{1}{x}).(\frac{-1}{x^2})$.. I am not sure how does this help.... – user87543 Oct 7 '13 at 13:43
• How does $\frac{1}{x^2}$ behave for $x \to 0$? – Daniel Fischer Oct 7 '13 at 13:47
• Who told you $\sqrt{x}$ is universally continuous? – Arthur Oct 7 '13 at 13:48
• You get a factor $-\frac{1}{x^2}$ when differentiating $f(\frac1x)$. So if $f \colon (1,\infty) \to [0,1]$ is differentiable, $g(x) = f(\frac1x)$ can have unbounded derivative. That makes it a candidate for not being uniformly continuous. Choose $f$ right, and $g$ is indeed not uniformly continuous. – Daniel Fischer Oct 7 '13 at 13:55

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 $n$).
Consider the function $f:(0,1) \to [0, 1]$ be defined by $f(x)=(1+\cos(1/x))/2$ on $(0,1)$. clearly it's differentiable and $f'(x)=-\cos(1/x)/2(x^2)$ for $0< x <1$.
since, Any continuous function $f:(a,b) → \mathbb R$ is uniformly continuous on bounded open interval $(a,b)$ iff $f(a+)$ and $f(b-)$ both exist.
Here $f(1/(2n−1) \pi)=0$ and $f(1/2n \pi)=1$ for $n=1,2,3, \ldots$. so $f(0+)$ does not exist.
Hence $f(x)$ is not uniformly continuous on $(0,1)$ .