1. ## Uniform Continuity

Hello there. I am supposed to show whether or not the function

$\displaystyle f0,1) \rightarrow \mathbb{R}$ given by $\displaystyle f(x) = \frac{1}{x^2}$

is uniformly continuous.

I suspect it is not so I'm trying to prove the statement:

$\displaystyle \exists \epsilon > 0$ such that $\displaystyle \forall \delta > 0 \exists x,a \in (0,1)$ such that $\displaystyle |x-a| < \delta$and $\displaystyle | \frac{1}{x^2} - \frac{1}{a^2} | < \epsilon$.

i.e. our choice of delta in the definition of continuity does not depend on a. Any help with this would be appreciated.

If we have a function f continuous on [a,b] is it true that f is uniformly continuous on (a,b) assuming f(a) and f(b) are defined? If so why?

What about if f(a) is not defined? For example f(x) = 1/(x^2) is continuous on [0,1] (and so is uniformly continuous on [0,1]) but is funiformly continuous on (0,1) considering f(0) is not defined? If so why?

3. Originally Posted by slevvio

If we have a function f continuous on [a,b] is it true that f is uniformly continuous on (a,b) assuming f(a) and f(b) are defined? If so why?
Yes, remember that uniform continuity is a set property and it's trivial to show from the definition that if $\displaystyle f$ is unif. cont. on $\displaystyle A$ then it is so on every $\displaystyle B \subseteq A$

Originally Posted by slevvio
What about if f(a) is not defined? For example f(x) = 1/(x^2) is continuous on [0,1] (and so is uniformly continuous on [0,1]) but is funiformly continuous on (0,1) considering f(0) is not defined? If so why?
How is $\displaystyle f$ going to be unif. cont. on $\displaystyle [a,b]$ when it's not even defined in $\displaystyle a$?

4. thanks a very helpful post

5. how can I show that f in my original question is not uniformly continuous though? since it is actually continuous on (0,1) ?

6. http://home.iitk.ac.in/~psraj/mth101...es/uniform.pdf

on page 3 there's an useful fact.