2
$\begingroup$

There are two questions.

First: is the proof underneath correct?

Let $\epsilon>0$ and let $f(x)=x^{1+\epsilon}$. I aim to show that $f$ is not uniformly continuous on $[0,\infty)$.

We will show that for any $\delta>0$ there exists $x$ and $y$ such that $|x-y|<\delta$ and $|f(x)-f(y)|\geq 1$.

Now let $\delta>0$ be given. Consider $f'(x)=(1+\epsilon)x^\epsilon$. We have that $f'\rightarrow\infty$ as $x\rightarrow \infty$ and that $f'$ is monotone increasing. Now choose an $x_0$ large enough so that $f'(x)$ is larger than $2/\delta$ for all $x\geq x_0-\delta$.

Now consider $|f(x_0)-f(y)|$ for some $y\in(x_0+\delta/2,x_0+\delta)$. The Mean Value Theorem guarantees us that there is a $\zeta\in(x_0, y)$ such that $|f(x_0)-f(y)|=|f'(\zeta)||x_0-y|$.

We thus have that $$ \begin{array}{rl} |f'(\zeta)||x_0-y|&=(1+\epsilon)\zeta^\epsilon|x_0-y|\\ &\geq(1+\epsilon)\zeta^\epsilon\delta/2\\ &\geq(1+\epsilon)x_0^\epsilon\delta/2\\ &\geq2/\delta\cdot\delta/2\\ &=1 \end{array} $$ Thus for this choice of $x_0$ and $y$, we have that $|f(x_0)-f(y)|\geq 1$ and $f$ is not uniformly continuous.

Second: If all is well and good, what other ways are there to prove this claim? I feel like the approach I took was a little bulky.

$\endgroup$
  • $\begingroup$ Kind of interesting that this is an example of how being uniformly-continuous is not an open condition; $f(x)=x=x^1$ is uniformly-continuous. $\endgroup$ – user99680 Jun 23 '14 at 23:03
  • $\begingroup$ Thank you everyone for your input. I'm glad I asked, as I probably would have learned about the more general result and about cofinal sequences much later. $\endgroup$ – Robert Wolfe Jun 23 '14 at 23:04
2
$\begingroup$

Your proof is fine, and the same proof works for any $f$ with $f' \to \infty$ as $x \to \infty$. Personally, I think your approach is excellent in that it is extremely clear. Perhaps you could have been more terse, but I like your style.

Another approach, however, if you insist on being as terse as possible, is to note that $f$ is uniformly continuous if and only if it maps Cauchy pairs of cofinal sequences to pairs of cofinal sequences. See Pedro's answer for an explanation.

(in fact, this function takes Cauchy sequences to Cauchy sequences but is not uniformly continuous. The condition of taking Cauchy sequences to Cauchy sequences is sufficient, however, when considering functions over bounded sets).

$\endgroup$
  • $\begingroup$ No, not "Cauchy", rather "Cofinal". One direction holds though. $\endgroup$ – Pedro Tamaroff Jun 23 '14 at 22:51
  • $\begingroup$ @PedroTamaroff right, good call $\endgroup$ – Omnomnomnom Jun 23 '14 at 22:54
2
$\begingroup$

Yes the proof is correct, basically you use the fact that, for every nonnegative $u$, $$ (1+u)^{1+\varepsilon}-1\geqslant(1+\varepsilon)u, $$ hence, for every nonnegative $x$ and $\delta$, applying this to $u=\delta/x$ and multiplying by $x^{1+\varepsilon}$, one gets $$ (x+\delta)^{1+\varepsilon}-x^{1+\varepsilon}\geqslant(1+\varepsilon)x^\varepsilon\delta, $$ and that, for every positive $\delta$, the RHS is unbounded when $x\to\infty$.

$\endgroup$
  • $\begingroup$ And this is from the binomial theorem, yes? $\endgroup$ – Robert Wolfe Jun 23 '14 at 22:51
  • $\begingroup$ For example, yes. $\endgroup$ – Did Jun 23 '14 at 22:52
2
$\begingroup$

An alternative solution is to find a pair of cofinal sequences $(x_n)$ and $(y_n)$ such that $(x_n^{1+\varepsilon})$ and $(y_n^{1+\varepsilon})$ are not cofinal. Can you do this?

Here cofinal means $d(x_n,y_n)\to 0$. For example, if $\varepsilon=1$; I can take $x_n=\sqrt n$ and $y_n=\sqrt{n+1}$. Then $|x_n-y_n|\to 0$ but $|x_n^2-y_n^2|=1\not\to 0$.

$\endgroup$
  • $\begingroup$ I'm unfamiliar with cofinal sequences. Can you elaborate? $\endgroup$ – Robert Wolfe Jun 23 '14 at 22:53
  • $\begingroup$ @Bryan I added something. It is a theorem that a function is uniformly continuous iff it maps cofinal sequences to cofinal sequences. $\endgroup$ – Pedro Tamaroff Jun 23 '14 at 22:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.