Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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... :(

share|improve this question
    
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.... –  Praphulla Koushik 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
2  
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

1 Answer 1

up vote 3 down vote accepted

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 $n$).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.