5
$\begingroup$

I have seen that the continuous functions $$ \sin(x^2) \quad\text{and}\quad x\sin{x} $$ are not uniformly continuous on $[0,\infty)$.

Their non uniform continuity is easy to prove with sequences, but I want to capture which properties lead to their non uniform continuity.

They "oscillate faster and faster" as $x$ grows large, and $\displaystyle\lim_{x\to\infty}f(x)$ does not exist (if exists then uniformly continuous).

I don't know how to rigorously say "oscillating faster and faster", I have tried $\displaystyle\limsup_{x\to\infty} |f'(x)| = +\infty$ and not of BV, but they do not imply non uniform continuity alone.

So my question is, are there examples of continuous functions satisfying "osciliate faster and faster" and $\displaystyle\lim_{x\to\infty} f(x)$ does not exist, but is uniformly continuous on $[0,\infty)$?

$\endgroup$
2
  • $\begingroup$ How about defining "oscillating faster and faster" as: $f(x) = a(x)\sin\bigl(\varphi(x) + \varphi_0\bigr)$ with $\lim_{x \to \infty} \frac{a(x)\varphi(x)}{x} = +\infty$? $\endgroup$ Commented Jun 1 at 15:21
  • $\begingroup$ Try variance - it measures total change of function within a segment. $\endgroup$ Commented Jun 1 at 15:44

1 Answer 1

Reset to default
5
$\begingroup$

What about $f(x)=x^{-1} \sin(x^4)+\sin(x)$?

The first addend goes to zero at infinity, but the derivative is unbounded. If you asked me whether it oscillates faster and faster I would say “yes, depending on what you mean”. The second addend is there just so that $f$ does not have limits at infinity.

$\endgroup$
5
  • $\begingroup$ Great example, so this is basically translating the idea of $g(x) = x^{-1}\sin(x^4)$ "sticking near x-axis" to this $f(x)$ "sticking near $\sin{x}$", which is uniformly continuous on $\mathbb{R}$. $\endgroup$ Commented Jun 1 at 15:51
  • $\begingroup$ And so I can say, the non uniform continuity of $\sin(x^2)$ and $x\sin{x}$ is not just due to being "oscillating faster and faster" and "$\lim_{x\to\infty} f(x)$ DNE", we need more properties to capture their essence. $\endgroup$ Commented Jun 1 at 15:56
  • $\begingroup$ Yes. What you actually need is that the function “oscillates faster and faster, with oscillation amplitude not converging to zero”. This means that you want a threshold $\varepsilon_0>0$ so that you find faster and faster oscillations of amplitude greater than $\varepsilon_0$ (this is more precise, and is essentially the opposite of the definition of uniform continuity, if you replace “oscillations” by “jumps”). $\endgroup$ Commented Jun 1 at 16:01
  • $\begingroup$ I see, I wrong identified "oscillation amplitude not converging to zero" by "limit not converging to zero". But is it true if $f(x)$ is bounded on $[a,\infty)$ for some $a$ and $f$ is not uniformly continuous, it must be "oscillates faster and faster, with oscillation amplitude not converging to zero"? Or should I have a new post for that? $\endgroup$ Commented Jun 1 at 16:06
  • 1
    $\begingroup$ Maybe you can think yourself and make a new post if needed. In any case, let $g(x)$ defined as $0$ for $x<0$, $1$ for $x>1$, and $x$ between $0$ and $1$. Let $f(x):=\sum_{n=1}^\infty (-1)^n g(n(x-n))$. This one will have nearly equally spaced jumps that are steeper and steeper, and alternate (go up and down). This one, I would say, does not look like it oscillates faster and faster (at least not in the same sense as in the counterexample), but it definitely jumps faster and faster (it is bounded and continuous, but not uniformly continuous). $\endgroup$ Commented Jun 1 at 17:05

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.