1. ## Uniform Continuity

There are 2 ways of saying a function, $f$, is uniformly continuous

A. For every $\epsilon>0$, there exists a $\delta>0$ such that
$|f(x)-f(p)|<\epsilon$ whenever $x,p\in{I}$ and $|x-p|<\delta$

B. For every $\epsilon>0$, there exists a $\delta>0$ such that
$|f(x)-f(p)|\leq\epsilon$ whenever $x,p\in{I}$ and $|x-p|\leq\delta$

Prove f satisfies A if and only if it satisfies B

This is a confusing proposition. I think if f satisfies B, then it obviously satisfies A, but I could be mistaken.
And if it satisfies A, how do I go about demonstrating B?

Thanks in advance for the help

2. Originally Posted by I-Think
There are 2 ways of saying a function, $f$, is uniformly continuous

A. For every $\epsilon>0$, there exists a $\delta>0$ such that
$|f(x)-f(p)|<\epsilon$ whenever $x,p\in{I}$ and $|x-p|<\delta$

B. For every $\epsilon>0$, there exists a $\delta>0$ such that
$|f(x)-f(p)|\leq\epsilon$ whenever $x,p\in{I}$ and $|x-p|\leq\delta$

Prove f satisfies A if and only if it satisfies B

This is a confusing proposition. I think if f satisfies B, then it obviously satisfies A, but I could be mistaken.
And if it satisfies A, how do I go about demonstrating B?

Thanks in advance for the help
As a heuristics less than and less than or equal to are indistinguishable in hard analysis. If $f$ satisfies A. then it clearly satisfies B. Conversely, try for a given $\varepsilon$ using $\frac{\varepsilon}{2}$ in B. and taking half of the guaranteed $\delta$

3. As a curiosity:

We say that $A\leq 0$ is obtained by relaxing the strict inequalyty $A<0$ .

Fernando Revilla