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Let $\mathcal A$ denote the C*-algebra of complex valued, bounded continuous functions on the reals $\mathrm{C}_\mathrm{b}(\mathbb{R})$. Furthermore, let $\mathcal A_c$ denote the C*-subalgebra of $\mathcal A$ such that for every $f \in \mathcal A_c$ the limits $\lim_{x \to \infty} f(x)$ and $\lim_{x \to -\infty}$ exist and coincide.

On the one side, it is possible to identify the C*-subalgebra $\mathcal A_c$ with the continuous functions on the one-point compactification on $\mathbb{R}$. On the other side, one can see $\mathcal A$ as continuous functions on the Stone-Cech compactification of $\mathbb{R}$.

Let $K$ be another (Hausdorff-) compactification of the reals, i.e. $K$ is compact and we have a continuous embedding $\varphi \colon \mathbb{R} \to K$ such that $\overline{\varphi(\mathbb{R})} = K$. I want to show, that I can identify the C*-Algebra $\mathcal B := \mathrm{C}(K)$ with a sub-algebra of $\mathcal A$ which contains $\mathcal A_c$.

Consider some $f \in \mathcal B$. Since $\varphi(\mathbb{R})$ is dense in $K$, I can identify $f$ with its restriction $f_{|\mathbb{R}}$, which is bounded because $K$ is compact. This should show that we can consider $\mathcal B$ a subalgebra of $\mathcal A$, correct?

What about the other "inclusion". Given $f \in \mathcal A_c$ how would one "extend" this function to $K$. Since $K$ is Hausdorff, it is clear that once a function in $\mathcal A_c$ exhibits an extension, this extension will be unique.

Do you know how to prove the existence of a continuous extension to $K$?

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  • $\begingroup$ Can you specify the question? It is true that if you have a compactification $K$ of $\Bbb R$ that then the restriction $C(K)\to C_b(\Bbb R)$ is an isometric $*$ morphism. But not every bounded continuous function can be extended to a function on $K$, ie there doesn't exist an inverse map $C_b(\Bbb R)\to C(K)$. $\endgroup$
    – s.harp
    Commented Mar 9, 2017 at 18:10
  • $\begingroup$ @s.harp: Note that I don't want to extend just bounded continuous functions. I want to extend functions lying in $\mathcal A_c$, i.e. having a limit $\lim_{|x| \to \infty}$. At least I'm hoping to use this fact to get the existence of an extension. $\endgroup$
    – el_tenedor
    Commented Mar 9, 2017 at 18:22

2 Answers 2

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Response to comment by OP: How does local compactness of $\mathbb R$ imply $\mathbb R$ is open in K? (Too long for a comment).

Lemma. Let $X$ be a dense subspace of $Y$ and let $U$ be an open subset of the space $X.$ If $Cl_X(U)$ is closed in $Y$ then $U$ is open in Y....And we also have $Cl_Y(U)=Cl_X(U)$.

Proof. Let $U=V\cap X$ where $V$ is an open subset of $Y.$ Since X is dense in $Y $ and $V$ is open in $Y$ we have $Cl_Y(V)=Cl_Y(V\cap X).$ Hence $Cl_Y(V)=Cl_Y(U).$ Now since $Cl_X(U)$ is closed in $Y,$ we have $$X\supset Cl_X(U)=Cl_Y(Cl_X(U))\supset Cl_Y(U)=Cl_Y(V)\supset V.$$ So $X\supset V,$ implying $U=V\cap X=V,$ so $U$ is open in $Y.$

And we also have $X\supset Cl_Y(V)=Cl_Y(U)$ so $Cl_X(U)=X\cap Cl_Y(U)=Cl_Y(U).$

Corollary. If $Y$ is Hausdorff and $X$ is dense in $Y,$ and if $U$ is an open subset of the space $X$ such that $Cl_X(U)$ is compact then $U$ is open in $Y$. Because $Cl_X(U)$ must be closed in $Y$ because $Y$ is Hausdorff and $Cl_X(U)$ is compact.

Now let $X$ be a locally compact Tychonoff space and $id_X:X\to Y$ be a compactification of $X$. For any $p\in X$ let $p\in U_p\subset X$ where $U_p$ is open in $X$ and $Cl_X(U_p)$ is compact. By the corollary, $U_p$ is open in $Y.$ So $X=\cup_{p\in X}U_p$ is open in $Y.$

BTW. The converse also holds: If $c:X\to K$ is a compactification of the Tychonoff space $X,$ such that $c(X)$ is open in $Y,$ then $X$ is locally compact.

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  • $\begingroup$ Thanks, for taking the time to write all this down. In the first line of the lemma, should $U$ maybe be an open subset of the space $X$ instead of $Y$? $\endgroup$
    – el_tenedor
    Commented Mar 10, 2017 at 22:17
  • $\begingroup$ Do you have a reference at hand for the converse result? $\endgroup$
    – el_tenedor
    Commented Mar 10, 2017 at 22:32
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    $\begingroup$ I was going to give you Engelking, General Topology, Ch 3, Section 3.5. It's part of Proposition 3.5.8. But when I looked at the proof of that part, it says "it is obvious."...... $K$ is compact Hausdorff so it is regular. So if $K$ \ $c(X)$ is closed in $K,$ and $ p\in X$, there exists open $V$ in $K$ with $c(p)\in V$ and $$Cl_K(V)\cap (K \backslash c(X))=\phi.$$.... So $V \cap c(X)=V$ is open in $c(X)$ and its closure in $X$ (which equals its closure in $K$) is compact (because $K$ is compact Hausdorff). So $c(p)$ has a nbhd $V$ in the space $c(X)$ whose closure in $c(X)$ is compact. $\endgroup$
    – DanielWainfleet
    Commented Mar 10, 2017 at 23:28
  • $\begingroup$ Thanks for pointing out the typo in my A. In the lemma, $U$ is indeed an open subset of the space $X$. $\endgroup$
    – DanielWainfleet
    Commented Mar 10, 2017 at 23:31
  • $\begingroup$ Thanks again, for your help. I think all my questions have been answered. $\endgroup$
    – el_tenedor
    Commented Mar 11, 2017 at 7:40
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Given $f \in \mathcal A_c$, let $L(f) = \lim_{x \to \infty} f(x)$. Then you extend $f$ to $\tilde{f} \in C(K)$ by $\tilde{f}(\varphi(x)) = f(x)$ while $\tilde{f}(k) = L(f)$ for $k \in K \backslash \varphi(\mathbb R)$. It is not hard to show that this is continuous.

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  • $\begingroup$ Thank you. I have been trying to prove that the suggested extension is continuous but have not had success yet. Apparently I don't know enough about the topology of $K$. For example, if I take some open neighbourhood $U$ of $L(f)$ and try to show that it has an open preimage under $\tilde f$ I think one could try to show, that each $y \in \tilde f^{-1}(U)$ has an open neighbourhood $O_y$ with $\tilde f(O_y) \subseteq U$. If $y = \varphi(x)$ this shouldn't be a problem. But what if $y = k \in K \subseteq \varphi(\mathbb{R})$? I could use another hint, I'm not even sure if I'm on the right track $\endgroup$
    – el_tenedor
    Commented Mar 9, 2017 at 22:30
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    $\begingroup$ WLOG $ \psi =id_{\mathbb R}.$ For $n>0$ the set $I_n=[-n,n]$ is compact, hence closed in $K$. If $ p\in K$ \ $\mathbb R$ and $V$ is a nbhd of $\bar f (p)=L(f),$ let $r>0$ with $(-r+L(f),r+L(f))\subset V.$ Take $n>0$ where $|f(x)-L(f)|<r$ for all $x\in \mathbb R$ \ $I_n$. Then $U=K$ \ $I_n$ is a nbhd of $p$ and $\bar f (U) \subset \mathbb R$ \ $ I_n \subset V.$ So $\bar f$ is continuous at $p.$..... For continuity of $\bar f$ on $\mathbb R,$ note that $\mathbb R$ is open in $K$ (as $\mathbb R$ is locally compact) and that $\bar f|_{\mathbb R}=f$ is continuous. $\endgroup$
    – DanielWainfleet
    Commented Mar 10, 2017 at 5:03
  • $\begingroup$ @user254665: Thank you for this proof, instead of $\psi$ it should be $\varphi$, though. I don't understand your last sentence. How does local compactness of $\mathbb{R}$ imply that $\mathbb{R}$ is open in $K$? $\endgroup$
    – el_tenedor
    Commented Mar 10, 2017 at 8:56
  • $\begingroup$ @user254665: Note that instead of the interval $(-r + L(f), r + L(f))$ we want the open ball $U_r(L(f)) \subset V \subseteq \mathbb{C}$. $\endgroup$
    – el_tenedor
    Commented Mar 14, 2017 at 18:52
  • $\begingroup$ Yes. I had forgotten that we have complex-valued f. $\endgroup$
    – DanielWainfleet
    Commented Mar 14, 2017 at 21:27

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