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Let $X$ be a locally compact Hausdorff space, such that $C_0(X)$ is an injective Banach space, i.e. a $\mathfrak{P}_\lambda$ space for some $\lambda\geq 1$.

  • Is it true that $X$ is compact?

  • If additionally $X$ is a locally compact group, is it true that $X$ is finite?

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    $\begingroup$ NO. $C_0(X\setminus\{x_0\})$ is complemented in $C(X)$. $\endgroup$
    – Narutaka OZAWA
    Commented Jul 3, 2014 at 21:51
  • $\begingroup$ @NarutakaOZAWA I agree for the 1st question, but is the 2nd one also obvious? $\endgroup$
    – Yemon Choi
    Commented Jul 3, 2014 at 21:55
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    $\begingroup$ Regarding your second question, one obvious attempt seems to be ruled out by encyclopediaofmath.org/index.php/Extremally-disconnected_space which says, essentially, that the only compact groups $G$ for which $C(G)$ is $1$-injective are the finite ones. $\endgroup$
    – Yemon Choi
    Commented Jul 3, 2014 at 22:30
  • $\begingroup$ @Yemon Choi: It's not obvious, but probably not too difficult (modulo a structure theorem of LCA groups) to show it's YES. Every LCA group $G$ has a compact open subgroup $K$ such that $G/K$ is a Lie group. One can probably show $K$ is finite and then $G=K$. $\endgroup$
    – Narutaka OZAWA
    Commented Jul 4, 2014 at 0:31
  • $\begingroup$ @NarutakaOZAWA Thanks (by the way, you meant locally compact, not LCA, right?) $\endgroup$
    – Yemon Choi
    Commented Jul 4, 2014 at 0:43

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$C_0(G)$ is not injective for infinite $G$, because it is not complemented in $L_\infty(G)$.

See Lau, Anthony To Ming; Losert, Viktor. Complementation of certain subspaces of $L_\infty(G)$ of a locally compact group. Pacific Journal of Mathematics 141 (1990), no. 2, 295--310, corollary 3.

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