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I've $I = [0 ,+\infty)\,$ and $f: I \rightarrow \Bbb R.$

a. I've proved that if $f'$ is bounded on $I$ then $f$ is uniformly continuous on $I$.
b. I've proved that if $\lim f' = \infty$ (with $x \rightarrow +\infty$) then $f$ isn't uniformly continuous on $I$.

c. Now I should prove that if $f'$ is unbounded on $I,$ then isn't uniformly continuous on $I$.
Using b I've proved that if c is wrong, there is a segement $T = [0, t]$ where $f'$ is unbounded.

Added: the also known that $f'$ exists on every point on $I.$

Martin Sleziak
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RiaD
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4 Answers4

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As @Kannappan suggested, here is my previous comment, expanded into an answer.

Assertion c) is indeed not correct, not even for bounded intervals. That is, every function that is differentiable on a closed and bounded interval is continuous and hence automatically uniformly continuous there, even if its derivative is unbounded (which is of course possible). An example for the latter case (differentiable everywhere, unbounded derivative on a bounded and closed interval) is $$ f(x) = \begin{cases} x^{3/2} \sin \frac{1}{x} (0 < x \le 1) \\ 0 (x = 0) \end{cases} $$ For the case $I = [0,\infty)$, a counterexample is given by $g(x) = f(\frac{x}{1+x})$ where $f$ is defined as above.

Hans Engler
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This is a wrong claim as the answer by Professor Aguirre suggests. I give an example in the case of a neither closed nor open set. It is known to be false even in general as the other answer suggests.

You seem to claim that:

If a function $f$ has unbounded derivative, this does not mean that $f$ is not uniformly continuous.

This is a wrong claim. Here is a counter example.

Consider $f:(0,1] \to \mathbb R$ defined by $f(x)=\sqrt x$. I leave it to you to prove that $f'$, the derivative of $f$ is unbounded on $(0,1]$ but $f$ is uniformly continuous on $(0,1]$.

Conclusion:

What is true is, there are functions with unbounded derivatives nevertheless uniformly continuous.

  • Hmmm, but I have task to prove it:( Maybe $I$ matter here? For, example it is closed. – RiaD Mar 10 '12 at 22:03
  • I think it would matter, give me some time to work out the details. –  Mar 10 '12 at 22:08
  • We need a function for which $f'$ exists everywhere. – Hans Engler Mar 11 '12 at 05:00
  • $f'$ does exist everywhere on $(0,1]$ @HansEngler –  Mar 11 '12 at 05:04
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    Let $f(x) = x^{3/2} \sin \frac{1}{x}$ for $x \in (0,1]$ and $f(0) = 0$. One can calculate that $f'$ exists everywhere on $[0,1]$, in particular $f'(0) = 0$, and $f'(\frac{1}{n \pi})$ is unbounded as $n \to \infty$. Yet $f$ is of course uniformly continuous on $[0,1]$ since the interval is compact. The OP's assertion c) therefore is incorrect. @RiaD - go and give them a counterexample and you will henceforth be afforded great respect. – Hans Engler Mar 11 '12 at 05:18
  • @HansEngler You can write this as an answer but still I am puzzled why you'd leave this as a comment under my answer. Did you see that $f'$ exists everywhere? I'd happy if you told me if I am missing something. Regards, –  Mar 11 '12 at 05:22
  • My point was that $f$ should be defined on a closed and bounded interval and be differentiable on the entire interval. The function $f(x) = \sqrt{x}$ doesn't quite satisfy that since $f'(0)$ does not exist. – Hans Engler Mar 11 '12 at 05:27
  • Well, I agree about closedness which fails in my example and I have pointed it out in my answer and this is the whole point my answer is Community Wiki. :-) @HansEngler –  Mar 11 '12 at 05:31
  • Even $f(x)=\sqrt{x}$ does not satisfy the lipschitz condition but it is UC . – RAM_3R Mar 19 '18 at 14:10
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Claim c. is not true in general. I will provide an example in $[\,0,+\infty)$, as asked by the OP. Let $g\colon[\,0,+\infty)\to\mathbb{R}$ be defined as follows. For each $n\in\mathbb{N}$, $g$ is piecewise linear on the interval $$\Bigl[\,2^n-\dfrac{1}{n\,2^n},2^n+\dfrac{1}{n\,2^n}\,\Bigr]$$ and $$ g\Bigl(2^n-\frac{1}{n2^n}\Bigr)=g\Bigl(2\,^n-\dfrac{1}{n2^n}\Bigr)=0,\quad g(2^n)=n. $$ Outside those intervals, $g$ is equal to zero. Let $f(x)=\int_0^xg(t)\,dt$. Then $f$ is increasing and $\lim_{x\to+\infty}f(x)=\sum_{n=1}^\infty2^{-n}=1$. This implies that $f$ is uniformly continuous on $[\,0,+\infty)$, but $f'=g$ is unbounded.

Julián Aguirre
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  • just a question, why does the $\lim_{x\to+\infty}f(x)=1$ imply $f$ is uniformly continuous on $[0,+\infty)$? – C Squared Nov 07 '20 at 10:54
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    also, kind of nit picky, but do you mean $g(2^n\textbf{+}\frac{1}{n2^n})=0$ in the very middle of the answer? – C Squared Nov 07 '20 at 11:02
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One thing at work here is that, although the derivative might be large, if it doesn't have a sufficiently large interval on which to work, then it doesn't always have a chance to create a large increase in $y$-values.

For example, consider $x^{1/3}$ near the origin and the interval $[0,\delta]$. No matter how close a positive $x$ is to $0$, the maximum change possible is completely dependent on the value $\delta$.

Patrick
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  • Interesting example. What makes $x^{1/3}$ different from, say, $\ln(x)$? Why is the former uniformly continuous whereas the latter is not? They both have unbounded derivatives near $0$. – chharvey Feb 01 '13 at 15:16
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    @chharvey $\ln(x)$ is unbounded itself, unlike $x^{1/3}$. – Ruslan Feb 05 '15 at 15:01