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Suppose $X$ is a normal space and $f:X \rightarrow X$ is continous on X, and also uniformly continuous on a subset $A \subseteq X$. In this setting, can one conclude that f is uniformly continuous on the closure of $A$?

If not, what about a metric space? Other families of spaces?

I ask this question for the following case:

Montel's Theorem in complex analysis states that a bounded sequence of holomorphic functions on an open set has a converging subsequence (i.e, the set of bounded holomorphic functions is sequentially compact and thus compact).

In the proof, one first shows that there is a subsequence that is uniformly continous on a dense subset, and than you extend it to the whole set. The proof of the last assertion heavily uses properties of holomorphic functions (cauchy's theorems and so on). However, it seems to me that it can be proved in a (much) more general case.

Any ideas?

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  • $\begingroup$ You will need more conditions on $f$ for the question posed at the top. After all, $f$ needn't even be continuous on the closure of $A$, let alone uniformly so. $\endgroup$ Commented Dec 15, 2014 at 18:07
  • $\begingroup$ By “closer” you mean “closure”, right? $\endgroup$ Commented Dec 15, 2014 at 18:09
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    $\begingroup$ Are you assuming that $f$ is already defined and continuous on $X$? $\endgroup$ Commented Dec 15, 2014 at 18:16
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    $\begingroup$ I don't think uniform contuinity makes sense in a general topological space. Don't you need a metric space or at least a uniform space? $\endgroup$ Commented Dec 15, 2014 at 18:29
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    $\begingroup$ Nobody should need to read comments to get the correct question. Edit your question, don't just leave a comment that requires quite a lot of sifting to figure out the point. @Mike $\endgroup$ Commented Dec 17, 2014 at 20:15

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I'll assume that $X$ is a metric space.

Fix $\epsilon > 0$. Let $\delta > 0$ be a constant that satisfies $d(f(a), f(b)) < \epsilon$ for all $a, b \in A$ when $d(a, b) < \delta$. This is possible since $f$ is uniformly continuous on $A$.

Let $x, y \in \overline A$ such that $d(x, y) < \delta / 3$. Since $f$ is continuous at $x$ and $y$, we can choose $\hat x, \hat y \in A$ such that $d(x, \hat x) < \delta / 3$, $d(y, \hat y) < \delta / 3$, $d(f(x), f(\hat x)) < \epsilon/3$ and $d(f(y), f(\hat y)) < \epsilon / 3$.

We have $$ d(f(x), f(y)) \le d(f(x), f(\hat x)) + d(f(\hat x), f(\hat y)) + d(f(y), f(\hat y)) < \epsilon $$ as desired.

It is worth noting that if the codomain of $f$ is complete (which seems to be true in the special case you're considering), then it is sufficient to assume that $f$ is defined and uniformly continuous on $A$. In this case, $f$ has a unique extension to $\overline A$ and the extension is uniformly continuous. See for example this question.

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  • $\begingroup$ Thanks a lot!!! Question - why can you conclude from $d(x, \hat x) < \delta / 3$ that $d(f(x), f(\hat x)) < \epsilon/3$? Isn't this just using excatly what we need to prove? or is $\hat x$ being constant? $\endgroup$ Commented Dec 15, 2014 at 18:52
  • $\begingroup$ @Mike As to your question, that follows from the continuity of $f$ at $x$. I'm not using uniform continuity here. And yes, $\hat x$ is fixed for this $x$. $\endgroup$ Commented Dec 15, 2014 at 18:58
  • $\begingroup$ Sorry for the bother, but I think something in the proof is wrong. Continuity means that given an epsilon you can find delta; You have said that the specific delta satisfies the condition. Shouldn't it be another delta, and then take the minimum of them? $\endgroup$ Commented Dec 17, 2014 at 20:29
  • $\begingroup$ @Mike I didn't spell out all of the details. This is indeed what you need to do. To find $\hat x$, you have two deltas, you take the minimum and then pick $\hat x$ within the this minimum. The same applies to $\hat y$. $\endgroup$ Commented Dec 17, 2014 at 21:04

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