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I have a question which is:

If $f(x)$ is continuous and bounded on $(0,1)$ is it uniformly continuous on $(0,1)$? And how does this change if $f(x)$ is also monotonic?

I have been told the that unless it's monotonic then the it is not uniformly continuous. But I don't see why...

Can someone please help me understand this? Thanks :)

Jason
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    consider $\sin(1/x)$ for a counterexample. one characterization is (exercise) $f$ is uniformly continuous on $(0,1)$ iff it can be extended to a continuous function on $[0,1]$ – yoyo Dec 02 '11 at 20:01
  • You should be careful with the semantics difference between "you cannot say it is uniformly continuous unless it is monotonic" and "it is not uniformly continuous unless it is monotonic". – Willie Wong Dec 02 '11 at 20:25
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    In the case when $f$ is monotone, we can see that it is uniformly continuous using yoyo's characterisation above. Note that if $f$ is monotone then $f(x)$ approaches a definite value as $x$ approaches $0$ or $1$. Therefore, defining $f(1)$ and $f(0)$ appropriately, we can extend $f$ continuously to $[0,1]$. – Srivatsan Dec 02 '11 at 20:28
  • You might be interested in this similar question. –  Dec 05 '11 at 06:17

3 Answers3

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The problem is that the function may oscillate very fast between peaks and nadirs, which would spoil uniform continuity.

For an example of a continuous bounded function on $(0,1)$ which is not uniformly continuous, consider $$f(x) = \sin\left(\frac{1}{x}\right).$$ It is certainly bounded. However, it is not uniformly continuous. Given $\epsilon=\frac{1}{4}$, for any $\delta\gt 0$ we can find a large enough value of $n$ so that $$\frac{2}{(2n+1)\pi} - \frac{1}{2n\pi} = \frac{4n - (2n+1)}{2n(2n+1)\pi} = \frac{2n-1}{2n(2n+1)\pi}\lt \delta,$$ yet $$f\left(\frac{2}{(2n+1)\pi}\right) =\sin\left(\frac{(2n+1)\pi}{2}\right) = \pm 1,$$ and $$f\left(\frac{1}{2n\pi}\right) = \sin(2n\pi) = 0,$$ so letting $x=\frac{2}{(2n+1)\pi}$ and $y=\frac{1}{2n\pi}$, we have $$|x-y|\lt\delta\text{ but }|f(x)-f(y)| \geq \epsilon.$$

Added. In the case of a continuous, bounded, and monotone function, however, this kind of phenomenon cannot occur. Since the function is bounded and monotone, it has a (one sided) limit at both $0$ and $1$. Extending $f$ to all of $[0,1]$ using those limits, we get a continuous function on $[0,1]$, which is therefore uniformly continuous. The restriction to $(0,1)$ gives the original function, which is there3fore also uniformly continuous. Note how this argument fails for $f(x)$ above: there is no limit as $x\to 0$.

In fact, as noted by yoyo, it suffices (and is necessary) that the (one-sided) limits exist at $0$ and $1$ for the function to be uniformly continuous.

Arturo Magidin
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    +1. The OP also asks about the case when $f$ is monotone. Perhaps you could add a few words on that. – Srivatsan Dec 02 '11 at 20:08
  • Great thanks :) This also seems to make it clear why monotone functions will solve the problem – Jason Dec 02 '11 at 20:19
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Let me add a little to Arturo's answer.

First: how do you approach to problem?

You may have already known that a continuous function on any compact space is necessarily uniformly continuous. You may also know that a closed interval is a compact space. (Or you may just know that a continuous function on a closed interval is necessarily uniformly continuous.)

Therefore, if your continuous function $f:(0,1)\to\mathbb{R}$ admits limits

$$ f_0 = \lim_{x\to 0} f(x) \qquad f_1 = \lim_{x\to 1} f(x) $$

then defining the function

$$ \tilde{f} :[0,1] \to\mathbb{R} $$

by $\tilde{f}(0) = f_0$, $\tilde{f}(1) = f_1$ and $\tilde{f}(x) = f(x)$ otherwise you get a continuous function on $[0,1]$, and so $\tilde{f}$ must be uniformly continuous. And so $f$, being the restriction of $\tilde{f}$, must also be uniformly continuous.

This tells us that if you want to look for possible problems, you must consider the case where one of the limits, say $\lim_{x\to 0}f(x)$, fail to exist.

How can the limit fail to exist? There are basically two ways: either the $\limsup$ or the $\liminf$ do not exist (in which case the function is unbounded, and which you ruled out by assumption), or the $\limsup$ and $\liminf$ both exist but do not equal one another. Using the definition of the $\limsup$ and $\liminf$ you then "immediately see" that the oscillation that Arturo mentioned can cause problems to uniform continuity.

Note that this also explains the remark someone else made to you about monotonic functions: if you assume that $f$ is monotonic near $0$ and $1$, then you can use the fact that for bounded monotonic functions, the $\lim$ must exist, to conclude that $f$ extends to a continuous function on $[0,1]$ (like above) and therefore must be uniformly continuous.

Srivatsan
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Willie Wong
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  • Also a great addition :) Thanks – Jason Dec 02 '11 at 20:37
  • @Willie, when you say "This tells us that if you want to look for possible problems, you must consider the case where one of the limits, say $\lim_{x\to 0}f(x)$, fail to exist", are you actually referring to the following fact: "if $f$ is uniformly continuous on $(0,1)$, then those two limits exist."? –  Oct 17 '12 at 02:27
  • (cont.)What I think is that the existence those two limit, $f_0$ and $f_1$ indeed can guarantee the uniform continuity of $f$ because of the natural extension though, the non-existence failing the uniform continuity is not a consequence of this fact. –  Oct 17 '12 at 02:33
  • I like the $\limsup$ and $\liminf$ explanation. –  Oct 17 '12 at 02:34
  • @Jack: no, I took the contrapositive (whereas you took the converse). For "if the two limits exist, then by the construction above the function $f$ is uniformly continuous", the contrapositive is "if $f$ is not uniformly continuous, then at least one of the limits do not exist." – Willie Wong Oct 17 '12 at 06:24
  • @Willie, Ah, I see. I was wondering if I can use your intuition explanation to show that the converse is also true though. –  Oct 17 '12 at 15:02
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The function $$ f(x) = \frac{2x-1}{x(1-x)} $$ is continuous and strictly increasing on $(0,1)$ and is not uniformly continuous on that interval.

I'll tell you a non-standard analysis approach to that and let you think about the details of an $\varepsilon$-$\delta$ approach.

If $x$ is a real number between $0$ and $1$ and $dx$ is infinitely small, then $dy=f(x+dx)-f(x)$ is also infinitely small; hence $f$ is continuous on $(0,1)$. But there are some nonstandard reals between $0$ and $1$---namely those that are infinitely close to the endpoints---where an infinitesimal change in $x$ results in a non-infinitesimal change in $f(x)$; hence $f$ is not uniformly continuous on $(0,1)$.

For an $\varepsilon$-$\delta$ approach, think of this: If you think $\delta$ is small enough, think about whether it's still small enough near the endpoints.

Michael Hardy
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