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Let $X, Y$ be topological spaces and let $D \subset X$ be dense. Suppose $f : D \to Y$ is continuous. Then $f : X \to Y$ is continuous on all of $X$.

The above stated assertion does clearly not hold in general. However, I am curious if the assertion holds under certain conditions on the spaces $X, Y$ or function $f$, or whether there are similar results.

Any comment is appreciated.

  • Do you want to extend the function from $D$ to $X$ such that it remains continuous, or do you want a function which is defined on $X$, and continuous only on $D$? – Sarvesh Ravichandran Iyer Sep 13 '16 at 09:23
  • I am curious when given a function $f : X \to Y$, it is sufficient for the continuity on $X$ to check continuity on $D \subset X$ only. So if you could provide me with a function defined on $X$, which is only continuous on $D$, then this clearly shows that checking continuity on $D$ is not sufficient. Note, however, that I am curious if there are conditions on $X,Y$ when this is sufficient. –  Sep 13 '16 at 09:25

2 Answers2

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As a counterexample you can consider $D=(0,1]$, $X=[0,1]$, and $Y=\mathbb{R}$ with the Euclidean topology. Take $f:D\to\mathbb{R}$ as $$f(x)=\frac 1x.$$ There is no way to extend continuously this function on the whole interval $[0,1]$.

However there is a requirement on $f$ that makes it possible (if for example $X,Y$ are metric spaces and $Y$ is complete), namely uniform continuity of $f$.

Del
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  • Thanks. Do you have a reference for the fact regarding metric spaces? –  Sep 13 '16 at 09:46
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    You need $Y$ to be complete as well; otherwise you can take $X=\Bbb R$, $D=Y=\Bbb Q$, and $f$ the identity map. – Brian M. Scott Sep 13 '16 at 09:50
  • I didn't checked the answers there but you can have a look at http://math.stackexchange.com/questions/245237/extension-of-a-uniformly-continuous-function-between-metric-spaces. I forgot you need completeness of $Y$ (I'll edit my answer). – Del Sep 13 '16 at 09:51
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    @jvnv: With the emendation in my previous comment it’s Theorem $\mathbf{4.3.17}$ in Engelking’s General Topology. – Brian M. Scott Sep 13 '16 at 09:51
  • @BrianM.Scott. Thanks for the reference. I have difficulties with the term extendable. In my above question, I have a function $f : X \to Y$ and know that $f|_D : D \to Y$ is uniformly continuous. Can I then deduce that $f$ is uniformly continuous on all of $X$, i.e., is the extension $\tilde{f}$ of $f|_D$ equal to $f$? –  Sep 13 '16 at 09:57
  • @jvnv: Yes, if you know that $f$ is continuous. If $f\upharpoonright D$ is uniformly continuous, then it has a unique continuous extension to $X$; since the continuous extension is unique, it must be $f$ if $f$ is continuous (and obviously won’t be $f$ otherwise). – Brian M. Scott Sep 13 '16 at 10:00
  • @BrianM.Scott. True. However, I do not know whether $f$ is continuous, right? That was the main point of my question. I know only that $f|_D : X \to Y$ is (uniformly) continuous, and would like to conclude that $f$ is continuous. –  Sep 13 '16 at 10:01
  • @jvnv: If you don’t know that $f$ is continuous, you can’t say anything: the restriction of $f$ to $X\setminus D$ can be anything at all, no matter how nice $f\upharpoonright D$ is. – Brian M. Scott Sep 13 '16 at 10:05
  • @BrianM.Scott. Thanks. I think that answers my question. One final question: I have seen the assertion that given a linear operator on a Banach space $X$ such that $|T x |_X \leq C |x|_X$ for all $x \in D$, then $T$ is bounded. This can then not be true, right? –  Sep 13 '16 at 10:08
  • The proposition is true because the condition $|Tx|_X\leq C|x|_X$ is precisely a uniform continuity condition (in fact a bit more: Lipschitz continuity) – Del Sep 13 '16 at 10:11
  • @Del. I see how it follows from your answer when it is assumed that $T$ is bounded, but I only assume that $T$ is linear. –  Sep 13 '16 at 10:12
  • Well, okay. If the question is: "does $T$ admit a continuous linear extension, given the condition on the norm?" then the answer is yes. If the question is: "given $T$ linear such that $T|D$ is bounded, is $T$ bounded?" then the answer in general is no in infinite dimension, because you can make it behave wildly (unboundedly) outside $D$ – Del Sep 13 '16 at 10:18
  • @jvnv: You’re right; see the accepted answer to this MathOverflow question. – Brian M. Scott Sep 13 '16 at 10:19
  • @BrianM.Scott: thanks for the reference, that's a simple and beautiful proof – Del Sep 13 '16 at 10:25
  • @BrianM.Scott. Thanks; this is a good counterexample. I think my question is solved now as this shows that it is not even possible for linear, uniformly continuous functions on Banach spaces. –  Sep 13 '16 at 11:51
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Consider $f:R\rightarrow R$ defined by $f(x)=x,x\neq 0, f(0)=1$.$f$ is continuous on $R-\{0\}$ but not on $R$ and $R-\{0\}$ is dense in $R$.

Tsemo Aristide
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