When Can We Extend Function From Dense Set by Continuity
Continuous extensions of continuous functions on dense subspaces
Uniform continuity ensures that the Cauchy sequence $(q_n)$ in $\mathbb Q$ is mapped to a Cauchy (and hence convergent) sequence $\bigl(f(q_n)\bigr)$ in $\mathbb R$. If $f$ is just continuous, $\bigl(f(q_n)\bigr)$ needn't converge (remember: $f$ is just continuous on $\mathbb Q$, so you can't argue, that convergent sequences $q_n \to r \not\in\mathbb Q$ are mapped onto convergent sequences by $f$). For example $x\mapsto \frac 1{x - \sqrt 2}$ is continuous (not uniformly!) on $\mathbb Q$ and can't be extendend.
In fact, you only need the property, that Cauchy sequences are mapped to Cauchy sequences, which is a little weaker than uniform continuity, and a little stronger than continuity (for some properties of such functions, see here for start).
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Comments
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I thought that if I have a function $f: \mathbb Q \to \mathbb R$ that is continuous then I can (uniquely) extend it to a continuous function $F: \mathbb R \to \mathbb R$ as follows: for $r \in \mathbb R \setminus \mathbb Q$ pick a sequence $q_n$ converging to $r$ and then define $F(r) = \lim_{n \to \infty} f(q_n)$.
So I thought there must be a theorem saying that given a continuous function $f: D \to Y$ where $D$ is a dense subset of a metric space $X$ one can uniquely extend it to $F: X \to Y$.
Instead I found a theorem stating this but with the additional requirement that $f$ has to be uniformly continuous. Now I'm confused: is my example above wrong? Where does uniform continuity come in here?
Thanks.
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Well, here's an easy counterexample: try extending $x \mapsto 1/x$ defined on $\mathbb{R} \setminus \{ 0 \}$ to all of $\mathbb{R}$...
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Regarding the last comment, e.g. uniform continuity on bounded sets suffices.
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Mapping Cauchy sequences to Cauchy sequences would mean that $f$ is Lipschitz? I mean the implication the other way around: for Lipschitz functions we have that Cauchy sequences are mapped to Cauchy sequences?
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(I was thinking that $f$ would have to be like an isometry times a constant. Which is, I think, the same as saying $f$ is Lipschitz continuous.)
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@MattN: No, it is weaker than uniform continuity, and uniform continuity doesn't imply Lipschitz. E.g.: $\sqrt[3]{x}$ is uniformly continuous but not Lipschitz, and $x^2$ maps Cauchy sequences to Cauchy sequences but is not uniformly continuous.
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@MattN.: No, Lipschitz is not the same as constant times isometry. E.g., $\sin(x)$ is Lipschitz but not an isometry times a constant.
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@JonasMeyer Right. Thanks! @ your penultimate comment: I am trying to weaken the uniform continuity requirement to get something more general. But less general than mapping C sequences to C sequences.
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What about the property mentioned in the first comment, that the restriction to any bounded subset is uniformly continuous? That would suffice.
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@MattN. But perserving C-sequences is, what you need in the prove of the extension theorem: If $X$, $Y$ are metric spaces, $Y$ complete, $D \subseteq X$ is dense, $f \colon D \to Y$ is a Cauchy continuous map, then there is a unique extension to a continuous $F \colon X \to Y$.
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@martini Yes, exactly, that's why I said "less general" than that, meaning that I am looking for a condition in between Cauchy-continuity and uniform continuity but not as strong as uniform continuity and not as general as Cauchy continuity. So is there a function that is Lipschitz but not Cauchy continuous?
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@JonasMeyer Yes but I was looking for something I'm more familiar with (see my previous comment).
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@MattN. No. As Lipschitz is even stronger than uniform continuity.
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Oops. ${}{}{}{}$
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and thank you @JonasMeyer
Recents
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Source: https://9to5science.com/continuous-extensions-of-continuous-functions-on-dense-subspaces
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