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Let $f:S\subseteq\mathbb{R}^m\to\mathbb{R}^n$ be a function defined on an open set $S$ in $\mathbb{R}^m$ with the property that the image of every sequence in $S$ which is convergent in $\mathbb{R}^m$ is also a convergent sequence. Does this condition imply that $f$ is uniformly continuous?

As proved here the condition implies the continuity of $f$. The function $f$ can be extended, by continuity, to $\bar{S}$ in an obvious manner. So, if $S$ is bounded then the extension is continuous on a compact set, hence uniformly continuous. So the discussion is about the case when $S$ is not bounded...

aly
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  • What if $f$ is continuous, but not uniformly continuous? – Angina Seng Sep 05 '20 at 15:09
  • It can't always be extended to $\overline S$, for example if $S$ is bounded and $f$ is not. – Matt Samuel Sep 05 '20 at 15:21
  • @Angina Seng I am thinking about the sequences which are in S, while their limits are not in S. – aly Sep 05 '20 at 15:27
  • @Matt Samuel Such an $f$ verifies the condition about convergence? – aly Sep 05 '20 at 15:30
  • @aly It can. Your condition is equivalent to mere continuity. Let $S=(0, 1)$, $f(x) =1/x$. – Matt Samuel Sep 05 '20 at 15:31
  • @Matt Samuel Maybe I am missing something. In your example, take the sequence $(\frac{1}{n})$ which converges to $0$, while $(f(\frac{1}{n}))$ diverges. – aly Sep 05 '20 at 15:53
  • @aly There is no $0$ in $S$, so $\left(\frac 1n\right)$ is not a convergent sequence in $S$. – Matt Samuel Sep 05 '20 at 15:54
  • @Matt Samuel No, I mean sequences in S convergent in $\mathbb{R}^m$. – aly Sep 05 '20 at 15:56
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    @aly I see. Well that does imply uniform continuity on a bounded set, but for example you could define $f(x)=x^2$ on $(0,\infty)$ and this sends convergent sequences to convergent sequences and is not uniformly continuous. – Matt Samuel Sep 05 '20 at 15:58
  • @Matt Samuel Ok, thanks. I was not checking the most obvious examples. – aly Sep 05 '20 at 16:02

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