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Given a colloquial definition of uniform continuity as $f(x)$ and $f(y)$ can be made to be arbitrarily close when $x$ and $y$ are sufficiently close, and the distance between $x$ and $y$ is independent of $x$ and $y$.

I'm not really sure how to picture a uniformly continuous function in my head. I showed that if the derivative of a function is bounded, then it will be uniform continuous. (I had trouble with the converse though.) Thinking along the lines that I need to bound the change in $f$.

How do you visualize uniform continuity?

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I am not sure we can get good advices about "visualizing stuff". Using the modulus of continuity ( might be a good idea here. – D. Thomine Jan 11 '12 at 18:32
If you have a bounded derivitve then the function is uniformly continuous - in other words, things can't get too steep (things being tangent lines here). Now, this is actually okay because on a compact subset of the real line we know the function is uniformly conintous so we only have to look near the ends - obviously if our function isn't differentiable we have work ahead. It should be stressed the converse is false also - a bounded derivitve doesn't imply uniform continuity. – Adam Jan 11 '12 at 19:12
I've posted below what I think may be the most "visual" answer to this question that you'll ever see. – Michael Hardy Jan 11 '12 at 19:48
up vote 2 down vote accepted

Fix $\varepsilon > 0$ and fix a $\delta > 0$ which works in the definition of uniform continuity.

The statement $|x-y| < \delta \Rightarrow |f(x)-f(y)| < \varepsilon$ tells you that you can place a rectangle of width $\delta$ and height $\varepsilon$ with its centre on any point on your graph, and the graph will always go through the middle of the rectangle, i.e. it never touches the top or bottom of the rectangle.

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Either I do not understand your "sloped lines" image, or it is false. It implies that the function is lipschitz, and not every absolutely continuous function is Lipschitz. – D. Thomine Jan 11 '12 at 19:20
@D.Thomine: You're right, it's false. Silly mistake on my part. I've edited my response. – Clive Newstead Jan 12 '12 at 1:38

Here is a visualization of uniform continuity:

Let $\gamma$ be the graph of a function $f\colon\ {\mathbb R}\to{\mathbb R}$ and let an $\epsilon>0$ be given. The $\epsilon$-tube with soul $\gamma$ is the set $T_\epsilon(\gamma):=\{(x,y)\in{\mathbb R}^2\ |\ |y- f(x)|<\epsilon\}$.

Together with $f$ we consider its translates $f_a$ defined by $f_a(x):=f(x-a)$. Let $\gamma_a$ be the graph of $f_a$.

The given function $f$ is uniformly continuous on ${\mathbb R}$ if, given an $\epsilon>0$, there is a $\delta>0$ such all graphs $\gamma_a$ for $|a|<\delta$ fit into the $\epsilon$-tube with soul $\gamma$: $$\gamma_a\subset T_\epsilon(\gamma)\qquad \bigl(|a|<\delta\bigr)\ .$$

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$f$ is continuous at a real number $x$ if $f(x+dx)-f(x)$ is infinitesimal whenever $dx$ is infinitesimal.

$f$ is continuous at all reals if the same is true at all reals $x$.

$f$ is uniformly continuous on the real line if the same is true not only when $x$ is real, but also for all infinitely large $x$ and all $x$ infinitely close to a real.

For example $x\mapsto \sin(1/x)$ is not uniformly continuous because for $x$ infinitely close to $0$, $\sin(1/x)$ can go all the way from $-1$ to $1$ when $x$ changes by an infinitely small amount.

And $x\mapsto e^x$ is not uniformly continuous because for an infinitely large value of $x$, $x$ can increase infinitesimally while $e^x$ increases by $1$.

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I guess the following is a way to make this more precise: – Jonas Meyer Jan 12 '12 at 3:22
@Jonas : I've just done some minor editing. – Michael Hardy Jan 14 '12 at 18:19

Definition of Uniform Continuity

A function $f$ is said to be uniformly continuous if to each $\epsilon >0$, $\exists~\delta >0$ such that whenever $|x-y|<\delta$, $|f(x)-f(y)|<\epsilon$.

The following is just an "uninformed" intuition:

If your function will have arbitrary long steeper chords, then for close values of $x$, the difference in the value of the function when $x$'s are close to each other. This means the condition of UC will fail for some epsilon.

So, this means, a function is uniformly continuous if it does not have steep chords or its arbitrarily steep chords are arbitrarily small.

This idea is rigorously proved here in this 1968 paper of Dwight Paine.

Also, do note that, if I drop the condition about the length of the steep chords, I can be wrong sometimes (consider $\sqrt x$) and I can be right sometimes as well ( consider $\dfrac{1}{x}$)

Incase you need the paper, do feel free to contact me!

P.S.: I welcome comments as to can there be a "iff" characterisation here!

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The link fails for me. – robjohn Jan 18 '12 at 19:34
@robjohn Thanks for the pointer. Fixed now. – user21436 Jan 18 '12 at 19:41

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