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Crossposted from MSE

How do the maximal ideals of $C_0(X)$ look like where $X$ is a locally compact Hausdorff space?

I know that if $X$ is a compact Hausdorff space then the maximal ideals of $C(X)$ are of the form $I_{c} : = \left \{f \in C(X)\ |\ f(c) = 0 \right \},$ for some $c \in X.$ I can also able to show that if $X$ is a locally compact Hausdorff space then $I_{c}$ is a maximal ideal of $C_0 (X)$ but can't able to prove the converse of that. Could anyone please help me in this regard?

Thanks a bunch.

EDIT $:$ Let $X^+$ denote the one point compactification of $X.$ Let $\varphi : C_0(X) \longrightarrow \mathbb C$ be a multiplicative linear functional on $C_0(X).$ Then it can be extended to a multiplicative linear functional $\overline {\varphi} : C(X^+) \longrightarrow \mathbb C$ defined by $\overline {\varphi} (f) = \varphi \left (f\ \big |_{X} \right ).$ Also given any $f \in C_0(X)$ there is an unique way to extend it to a function $\overline {f} \in C(X^+)$ as follows $:$ $$\overline {f} (x) = \begin{cases} f(x) & x \in X \\ 0 & x = \infty \end{cases}$$ Now take any $f \in C_0(X).$ Then since $X^+$ is compact we have $\varphi (f) = \overline {\varphi} (\overline {f}) = \overline {f} (c),$ for some $c \in X^+.$ If $c = \infty$ we have $\varphi \equiv 0$ i.e. $\varphi$ is a zero linear functional on $C_0(X).$ Otherwise there always exixts some $c \in X$ such that $\varphi (f) = f(c),$ for all $f \in C_0(X).$ So for any non-zero multiplicative linear functional $\varphi$ on $C_0(X)$ we have $\text {Ker}\ (\varphi) = \{f \in C_0(X)\ |\ f(c) = 0 \} = I_{c},$ for some $c \in X.$ Since we know that maximal ideals in any Banach algebra $A$ are precisely the kernels of non-zero multiplicative linear functionals on $A$ and $C_0(X)$ is a Banach algebra (in fact a $C^{\ast}$-algebra) we are through.

Possible Mistake $:$ I think where my above argument goes wrong is the stage where I assume $f\ \big |_{X} \in C_0 (X)$ whenever $f \in C(X^{+}).$ Instead when $f \in C_0 (X^+)$ this happens. Now we need to search for maximal ideals of $\{f \in C(X^+)\ |\ f(\infty) = 0 \}.$ So the above is not a valid argument. What my above argument shows is the following $:$

Multiplicative linear functionals on $C(X)$ are precisely evaluations even if $X$ is locally compact.

So the original question ultimately boils down to the following question $:$

Can we always extend every multiplicative linear functional on $C_0(X)$ to a multiplicative linear functional on $C(X)$ if $X$ is a locally compact Hausdorff space?

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    $\begingroup$ What is $C_0(X)$? Functions with compact support or functions going to zero at infinity? $\endgroup$
    – Denis Nardin
    Commented Oct 17, 2021 at 19:36
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    $\begingroup$ If $\varphi$ is a multiplicative linear functional on $C_0(X)$, then isn't $\varphi^+ : c + f \mapsto c + \varphi(f)$ a multiplicative linear functional on $C(X^+)$, where we write $f$ also for the extension of $f$ to $X^+$ that vanishes at $\infty$, and $c$ also for the constant function on $X^+$ with value $c$? Then $\varphi^+$ is the evaluation at some point $x \in X^+$ ($x = \infty$ if and only if $\varphi = 0$), and $\ker(\varphi) = \ker(\varphi^+) \cap C_0(X)$. $\endgroup$
    – LSpice
    Commented Oct 17, 2021 at 20:26
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    $\begingroup$ Although the question has received answers, or outlines of answers, here, it belongs more properly on MSE. It has only been on MSE for less than 24 hours; in future, perhaps try waiting a bit longer before crossposting here. $\endgroup$
    – Yemon Choi
    Commented Oct 18, 2021 at 2:18
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    $\begingroup$ Oh right. Since kernels of non-zero multiplicative linear functionals are necessarily maximal ideals of the underlying Banach algebra we are through. But how do I guarantee that such an evaluation is always non-zero? In other words given $c \in X$ can we always find some $f \in C_0(X)$ such that $f(c) \neq 0$ @PietroMajer? For specific $X ( = (0,1)$ (say)) it is easy to see. But for arbitrary $X$ (LCH) is it always true? $\endgroup$
    – RKC
    Commented Oct 19, 2021 at 8:24
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    $\begingroup$ Because X was assumed to be a locally compact Hausdorff space, hence completely regular (so for instance, for any c and any nbd U of c, there are functions with f(c)=1 and vanishing outside U). en.wikipedia.org/wiki/Tychonoff_space $\endgroup$
    – Pietro Majer
    Commented Oct 19, 2021 at 8:35

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The maximal ideals are exactly the kernels of point evaluation. This is proved in Chapter II, Prop. 7.4.5 on page 120 of

  • Z. Semadeni: Banach spaces of continuous functions. Warsaw 1971.
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  • $\begingroup$ Is the pdf of this book available online? I have searched for it but unfortunately couldn't able to find it anywhere. $\endgroup$
    – RKC
    Commented Oct 18, 2021 at 16:54

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