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I'm not sure how to prove or give a counter example to this. I wasn't able to prove it but I couldn't think of any counter example. Here's what I tried: By definition there exists $M>0$ such that for every $x>0$, $(a-1)x<f(x)<(a+1)x$. $f$ is uniformly continuous in $[0,M]$ (Cantor), but I'm not sure how to prove this for $(M, \infty)$. Can someone please help?

Omer
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  • @Surb but the derivative of ln(x+1) is 1/(x+1) which is bounded in [0,infinity) so ln(x+1) is uniformly continuous. – Omer Jan 22 '19 at 10:28
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    How about $f(x)=\sin{e^{x^2}}$? – Aphelli Jan 22 '19 at 10:29
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    @Mindlack actually, I think sin(x^2) is enough. because if we take $x_n=\sqrt{2 \pi n}$ and $y_n= \sqrt{2 \pi n + \pi/2}$ then $x_n-y_n$ tends to 0 but f(x_n)-f(y_n) doesn't. – Omer Jan 22 '19 at 10:33
  • @DavideGiraudo Yes, sorry, I edited the title. – Omer Jan 22 '19 at 10:35
  • Intuitively, for nice enough functions (in this case I would say $C^1$), uniform continuity is equivalent to boundedness of the derivative, and the limit condition to boundedness of the function. So unless the counterexample is "pathological", you should be looking for a bounded function with unbounded derivative, and $\sin(x^2)$ is a fine example (or more generally $\sin$ of any fast growing function). – Mees de Vries Jan 22 '19 at 10:37

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No.

For a function to be uniformly continuous you can't have segments where the "rate of change" is arbitrarily large.

So for example, $f(x) = cos(x^2)$ is not uniformly continuous, because there are arbitrarily short segments where the function goes from $-1$ to $1$ ($[\sqrt{(2n-1)\pi},\sqrt{2n\pi}]$ for each $n$).

However, this does not constitute a counterexample because clearly $f(x)$ has no limit at infinity.

Alright, we can fix this, $f(x)$ is bounded so multiplying it by a vanishing function should do the trick, and indeed $g(x) = \frac{cos(x^2)}{x}$ vanishes at infinity.

This is no good either, because (as you can check) $g(x)$ has a bound derivative and therefore it is uniformly continuous.

The geometric meaning of this is essentially that multiplying by $1/x$ has tamed the rate of change so much that it is no longer arbitrarily fast. To overcome this, we need to make the rate of change even faster.

So we choose $h(x) = \frac{cos(x^3)}{x}$ (or, alternatively, make the decline to $0$ softer by choosing $h(x) = \frac{cos(x^2)}{\sqrt{x}}$), which, as you can readily check, is not uniformly continuous.

Shai Deshe
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