# Uniform Continuity

• Jan 20th 2011, 02:28 PM
I-Think
Uniform Continuity
There are 2 ways of saying a function,$\displaystyle f$, is uniformly continuous

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

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

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
• Jan 20th 2011, 08:09 PM
Drexel28
Quote:

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

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

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

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 $\displaystyle f$ satisfies A. then it clearly satisfies B. Conversely, try for a given $\displaystyle \varepsilon$ using $\displaystyle \frac{\varepsilon}{2}$ in B. and taking half of the guaranteed $\displaystyle \delta$
• Jan 21st 2011, 12:01 AM
FernandoRevilla
As a curiosity:

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

Fernando Revilla