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This is a very short question, I hope it is not too broad, if so I shall try and make it more specific. I would like to start as it stands below, though, because it really points down the essence of the question:

What do people mean when they write that a map is extended by continuity ?

Thanks !

Gerry Myerson
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harlekin
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  • Generally I see this when you have a continuous function $f$ defined on a subset $Y$ of a metric space $X$ (or sufficiently nice topological space) and want a continuous function $g$ on $X$ such that $f(x)=g(x)$ for all $x\in Y$. One then says $g$ extends $f$. It is a theorem that $g$ exists if $Y$ is closed. – Alex Becker Apr 19 '12 at 22:56
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    @AlexBecker: that's extending continuously, but not extending by continuity. I'd say $f$ extends by continuity if $Y$ is a dense subset of $X$ and for each $x \in X$, $\displaystyle \lim_{y \to x, y \in Y} f(y)$ exists. – Robert Israel Apr 19 '12 at 23:07
  • @Alex: Hm .. ok, though sometimes a map is "extended by continuity" to the dual space - and this is not a subspace. I realize this concept might depend on the context, and your comment makes sense of some of the instances I hace struggled with, so thanks ! – harlekin Apr 19 '12 at 23:10
  • @RobertIsrael: many thanks for the helpful comment! May I ask, what is meant if people define a map on the dual space by "extension by continuity" ? – harlekin Apr 19 '12 at 23:12
  • @harlekin: it would be helpful if you could provide more context (such as a relevant quote from a book). What kind of map is being extended? From where? – Martin Wanvik Apr 19 '12 at 23:21
  • @MartinWanvik: I am trying to find an example that has not been answered by the comments of Robert and Alex above - though at the moment it seems I might have mixed things up in my head because I cannot find one. I'll edit the post in case I find one ! – harlekin Apr 19 '12 at 23:26

1 Answers1

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People usually meant the following:

Thm. Let $X$ and $Y$ be metric spaces with $Z$ dense in $X$. If $f\in C(Z\to Y)$ is uniformly continuous, then there is a function $F\in C(X\to Y)$ with $F|_Z=f$.
Proof. It suffices to show that, for each $x\in X$, the limit $\lim_{z\to x}{f(z)}$ is well-defined.

For any $\epsilon\in\mathbb{R}^+$, let $\delta(\epsilon)$ be such that $$d(x,y)<\delta(\epsilon)\Rightarrow d(f(x),f(y))<\epsilon$$

Existence: Suppose $\{x_n\}_n\in Z^{\omega}$ converges to $x$. Then $$\limsup_{n,m\to\infty}{d(x_n,x_m)}\leq\left(\limsup_{n\to\infty}{d(x_n,x)}\right)+\limsup_{m\to\infty}{d(x,x_m)}=0<\delta(\epsilon)$$ and so $$\limsup_{n\to\infty}{d(f(x_n),f(x_m))}<\epsilon$$ Thus $\{f(x_n)\}_n$ is Cauchy.

Uniqueness: Suppose $\{x_n\}_n,\{y_n\}_n\in Z^{\omega}$ converge to the same point, with $f(x_n)\to a$ and $f(y_n)\to b$. Fix $\epsilon$. $$\limsup_{n\to\infty}{d(x_n,y_n)}=0<\delta(\epsilon)$$ and so $$\limsup_{n\to\infty}{d(f(x_n),f(y_n))}\leq\epsilon$$ Now send $\epsilon\to0^+$. █

I leave the following exercises to you:

  1. The $F$ so defined is unique and
  2. The condition on $f$ cannot be weakened to just continuity. (Hint: Pick a function $f\in\mathbb{Q}\to\mathbb{R}$ discontinuous at $\pi$.)
Jacob Manaker
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  • The post is ten years old, but +1 all the same – FShrike Aug 07 '22 at 11:50
  • @FShrike: I was going to post a self-answered question about whether uniform continuity sufficed (vs. Hölder continuity), and then I realized the ideas would work equally well as an answer here. – Jacob Manaker Aug 07 '22 at 19:27