prove that f is uniformly continuous..

**alice8675309** · April 24th 2010, 07:23 PM

Suppose f: (a,b) -> R is differentiable and |f'(x)|<=M for all x in (a,b). Prove that f is uniformly continuous on (a,b). Give an example of a function f: (0,1)->R that is differentiable and uniformly continuous on (0,1) but such that f' is unbounded.

I have no clue.

**maddas** · April 24th 2010, 08:22 PM

$x\sin(\frac1x)$ should do for your example.

**becko** · April 25th 2010, 09:15 AM

Let $x,t\in \left(a,b \right)$ . By the mean value theorem of derivatives:

$f(x)-f(t)=(x-t)f'(\lambda )$

where $\lambda \in (x,t)$ . Hence:

$\left| f(x)-f(t)\right|\leq M\left| x-t\right|$

Now for any $\epsilon >0$ , we can put

$\delta =\frac{\epsilon }{M}$

so that whenever $\left| x-t\right|<\delta$ , we will have $\left| f(x)-f(t)\right|<\epsilon$ . This proves that $f(x)$ is uniformly continuous.

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