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Definition of uniform continuity of a function $f:D\to\mathbb{R}$: $$\forall\epsilon>0,\exists\delta>0:x,y\in D \wedge |x-y|<\delta\implies|f(x)-f(y)|<\epsilon.$$

Consider the function $f(x)=\sqrt x$ on the interval $[0,1]$. This is uniformly continuous, since our domain is compact. But for a function to be uniformly continuous, don't we need a small enough $\delta$ so that it can be used on our entire domain? If we were to approach $0$ from the left, wouldn't we need our $\delta$ smaller and smaller to no end? It seems that we would not ever find a small enough $\delta$ that can be used on our domain.

I do not care for a proof that $f$ is uniformly continuous; rather, I'm curious as to how we are able to find a small enough delta (unless, of course, I am misinterpreting the definition). Thanks!

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Are you trying to find the same $\delta$ irrespective of $\epsilon$? –  Aryabhata Dec 2 '13 at 22:54
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That definition doesn't make sense. It should be $$\forall\epsilon>0,\exists\delta>0\forall x,y\in D\colon |x-y|<\delta\implies|f(x)-f(y)|<\epsilon.$$ –  Git Gud Dec 2 '13 at 22:56
    
@GitGud Can you explain why the definition I gave does not make sense? –  Bonnaduck Dec 4 '13 at 1:04
    
@Bonnaduck Because it doesn't quantify $x$ and $y$. –  Git Gud Dec 4 '13 at 7:07

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up vote 3 down vote accepted

It's a good question. For $x \mapsto \sqrt{x}$, you'd typically choose something like $\delta = \min(1, \frac{1}{2} \epsilon^2)$. (I could be off by a small constant). The point is that this choice of $\delta$ is small enough to work near $0$ (which, as you correctly observe, is the tricky area of this function), but also suffices everywhere else.

You may be used to the idea that "the way to choose $\delta$ is to divide $\epsilon$ by the derivative of $f$ at your point, perhaps with an extra factor of $2$ to be on the safe side," which is a decent first-cut strategy. But here, we've got a more clever way to pick $\delta$ -- in particular, it's not linear in $\epsilon$ -- and the wonderful thing is that it works both at the origin and elsewhere.

The "divide delta by the derivative" approach works pretty well for many functions. For instance, if your function happens to satisfy $0 < |f'(x)| < M$ for every $x$, then something like $\delta = \epsilon / M$ will probably work to show uniform continuity. But as the example above shows, other techniques can be used even in cases where there's no similar bound on the derivative.

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Interesting. Does this mean that any continuous function $f$ with the properties: $c\in D$, $f(c)$ exists, and $lim_{x\to c} f'(x)=\pm \infty$ is uniformly continuous (assuming for all other points in D, $f'$ exists)? –  Bonnaduck Dec 4 '13 at 1:10
    
I think you're asking this: Suppose $f$ is continuous and differentiable on a closed interval $D$, except at a single point $c\in D$, at which $f$ is continuous, but $\lim_{x \rightarrow c} f'(x) = \infty$. Is $f$ necessarily uniformly continuous on $D$? I'm fairly certain the answer is "no." Consider the interval $[-2, 2]$, and $f(x) = \sqrt{x} + (x-1) \sin(\frac{1}{x-1})$. At $0$, the derivative goes to infinity, because the function looks like sqrt there. At $1$, the function wiggles up and down a lot, making it not uniformly continuous. I leave the verification of that to you. –  John Hughes Dec 6 '13 at 18:49
    
Not quite what I was trying to ask. What I mean to ask is would a point $c$ that does take a value at $f(c)$ and has an infinite slope ever cause a "problem" for uniform continuity? In the above example, would the point $x=0$ of $\sqrt{x} + (x-1)\sin\left(\frac{1}{x-1}\right)$ cause a problem for uniform continuity? –  Bonnaduck Dec 10 '13 at 3:17
    
I think I understand what you're asking, but it's a little vague. :) A function continuous on a closed interval is always uniformly continuous, so "infinite slopes" never cause a problem in this case. Beyond that, I don't know what to say. (But I can say that my previous comment is surely wrong, as the function I describe is continuous on the closed interval, hence uniformly continuous there!) –  John Hughes Dec 10 '13 at 14:56

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