0

What exactly is a uniform continuous function on a metric space?

My book says $f:X\to Y$ is uniform continuous if $\forall \epsilon\in\Bbb{R}$, for any points $x,y\in X$, there exists a constant $\delta$ such that $\rho(f(x),f(y))<\epsilon$ iff $d(x,y)<\delta$.

Is this equivalent to the condition that for every point $p\in X$, and $\epsilon\in \Bbb{R}$, there exists a common $\delta$ such that $\rho\bigl(f(p),f(x)\bigr)<\epsilon$ if $d(p,x)<\delta$?

Thanks in advance!

Cameron Buie
  • 102,994
  • What exactly does $f(p)-{f(|x-a|<\delta)}$ mean? What is $a$? What is $f(|x-a|<\delta)$? What do the ${;}$ indicate in this case? How is that subtraction working? – Cameron Buie Aug 21 '13 at 04:23
  • ${}$ are just braces put to separate one expression from another. That expressoin just means $|f(p)-f(x)|<\epsilon$ if $|x-p|<\delta$ –  Aug 21 '13 at 04:27
  • My basic question is, for two different points $p$ and $q$ in $X$, $|f(p)-f(x)|<\epsilon$ for $|x-p|<\delta_1$, and $|f(q)-f(x)|<\epsilon$ for $|x-q|<\delta_2$. If we say $\delta_1=\delta_2$ for any such two points in $X$, is this the complete definition of a uniform continuous function? –  Aug 21 '13 at 04:30
  • 1
    In your definition of uniform continuity, you don't want an "iff", but only an "if". – Martin Argerami Aug 21 '13 at 04:37
  • @MartinArgerami- I don't agree. If $|f(p)-f(x)|<\epsilon$, then $|x-p|<\delta$. If $|x-p|<\delta$, then $|f(p)-f(x)|<\epsilon$. Shouldn't "iff" work for both continuity and uniform continuity? –  Aug 21 '13 at 06:06

1 Answers1

1

What do you mean by $|f(p) - \{f(|x-a| < \delta)\}|$?.

Uniform continuity means that for every $\epsilon > 0$, $\exists \delta > 0$ such that $d(x,y) < \delta, \; x,y \in dom(f) \Rightarrow d(f(x),f(y)) < \epsilon$.

If you observer closely, the definition of continuity says that for every $x$ and every $\epsilon > 0$, $\exists \delta > 0$ (which may depend on $x$). But in case of uniform continuity, the $\delta$ does not depend on $x$.

For example, the function $f: \mathbb{R} \to \mathbb{R}$ given by $f(x) = x$ is uniformly continuous but the function $g:(0,\infty) \to \mathbb{R}$ given by $g(x) = 1/x$ is not uniformly continuous.

In case of $f$, for every $\epsilon > 0$, $\delta = \epsilon$ will "work" for all $x \in \mathbb{R}$. As $|x-y| < \delta \Rightarrow |f(x) - f(y)| < \epsilon $, no matter what $x$ and $y$ are.

Whereas, in the case of $g$, given $\epsilon$, you cannot find a $\delta$ which works for all $x$ in the domain. The delta you need to find will depend on $x$ as well. It is an instructive exercise to try to find a $\delta$ for a given $\epsilon$ in this case and see what happens.

Vishal Gupta
  • 6,946