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Let $X$ be a compact metric space. Show that any ring homomorphism $C(X)\to \mathbb{R}$ is an evaluation $f \to f(x_0)$ for some point $x_0 \in X$.

$\textbf{Question}$:

have the sketch of the proof as outlined below. but Not sure of from the step 3 and also have problem with the final step.

$\textbf{Sketch of the Proof}$:

1) Show that such a homomorphism is a positive linear functional.

2) Apply the Riesz representation theorem to show that it is given by integration with respect to some measure.

3) Use the homomorphism property to argue that if U, V are disjoint open sets at most one can have positive measure.

4) Deduce that the measure in question is a Dirac delta measure at some point.

$\textbf{My attempt}$:

let $\Lambda :C(X)\to \mathbb{R}$, then by assumption it a ring homomorphism and have the following properties for $f,g \in C(X)$; we have that $(i)$ $\Lambda(f+g) = \Lambda(f)+\Lambda(g)$ ,$(ii)$ $\Lambda(f.g) = \Lambda(f).\Lambda(g)$ , and $(iii)$ $\Lambda(1_{X}) = 1$ or $\Lambda(0_X) = 0$

$\textit{Step 1} :$ I need to show that if $f\geq0$ then $\Lambda(f)\geq0$.

let $f\geq0$ be any continuous function, we can approximate that by simple functions $ f = \lim_{n\to \infty}S_n$ . Applying the operator $\Lambda (f) =\lim_{n\to \infty}\Lambda (S_n) $ ;(I need to prove that $\Lambda$ is continuous.) So it suffices to show that

$A)$ $\Lambda (S)\geq0 $ for any simple function $S=\sum_{i=1}^\infty a_i\mathcal{X_{A_{i}}}$ , $a_i\geq 0$

$B)$ $\Lambda (\mathcal{X_{A}})\geq0 $

I will show $B$ then $A$.

let $\{f_n\}_{n=1}^\infty$ be a sequence of continuous functions that we approximate the characteristic function (as it is not continuous) by them. So I start with a characteristic function $\lim_{n\to \infty} f_n^A = \mathcal{X_A}$. Clearly characteristic function satisfies the above $(i)-(iii)$ and clearly since it takes zero or one it is positive.

To show A we can write that by $(i)$ , $\Lambda(S) =\sum_{i=1}^\infty a_i\Lambda(\mathcal{X_{A_{i}})} \geq 0$ as $a_i\geq0$

$\textit{Step 2} :$ Then since $\Lambda$ is a positive linear operator by Riesz representation we have that $\exists$ measure $\nu$ that we write the functional as $\Lambda(f) = \int_X fd\nu $

$\textit{Step 3} :$ Let $U,V \subset X$ be disjoint. Then we can write for any two characteristic functions (which we'd assume that are the limit of continuous functions)

$\Lambda (\mathcal{X}_U).\Lambda (\mathcal{X}_V) = 0 $ so at most one of them can be positive. Also we have that

$\Lambda (\mathcal{X}_U) = \int_X \mathcal{X}_U d\nu =\int_U d\nu = \nu(U)$ so $\nu(U).\nu(V) =0 $ which implies that at most one of them can be positive.

$\textit{Step 4} :$

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  • $\begingroup$ If a measure has a point support, then it has to be Dirac's delta. $\endgroup$
    – Jakobian
    Oct 9, 2019 at 22:29
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    $\begingroup$ Not sure how you conclude that $\Lambda(\lim S_n) = \lim \Lambda(S_n)$ $\endgroup$
    – xyzzyz
    Oct 9, 2019 at 22:46
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    $\begingroup$ You can have 1) more easily : if $f\geq 0$, it has a square root, thus so does $\Lambda (f)$, which must then be nonnegative. And actually your proof isn't currently valid as you assume that $\Lambda$ commutes with limits, i.e. is continuous, which you can't actually do if you haven't proved it. There is also an easier proof of the global statement by looking at maximal ideals of $C(X)$ $\endgroup$ Oct 9, 2019 at 22:49
  • $\begingroup$ can you give a reference for that? $\endgroup$
    – domath
    Oct 10, 2019 at 0:23
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    $\begingroup$ @stat it's not true that all linear maps from $C(X)$ to $\mathbb{R}$ are continuous (for a counterexample, use the axiom of choice to pick a basis for $C(X)$ and then map one basis vector to $1$ and the others to $0$). It is true that all ring homomorphisms are continuous, but this needs to fall out as part of the proof. $\endgroup$
    – hunter
    Oct 10, 2019 at 2:26

2 Answers 2

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About your specific solution: as was pointed out in the comments you exchange $\lim$ and $\Lambda$ without justification, which is problematic as it's not clear that they can be exchanged (especially if $\lim$ is to be taken in the sense of simple convergence of functions).

Here are some hints on how to solve the problem (not about your specific solution):

Hints : Use square roots to prove positivity.

Use positivity to prove contintuity, at least on constant functions : deduce from that that $\Lambda$ is actually an $\mathbb R$-algebra morphism. In particular, unless $X=\emptyset$, $\Lambda$ is surjective and so its kernel is a maximal ideal of $C(X)$.

Study maximal ideals of $C(X)$ as follows : Let $m$ be such an ideal,

i) Assume for all $x\in X$ there is $f\in m$ with $f(x) \neq 0$. Use compactness to find $f\in m$ that never vanishes.

ii) Deduce (using that nonvanishing implies invertible) that there is some $x_0$ with $f(x_0) = 0$ for all $f\in m$.

iii) Prove that $I=\{f \in C(X)\mid f(x_0) = 0\}$ is a proper ideal. In particular $m\subset I \implies m= I$

Once you have the shape of maximal ideals you can deduce that $f$ is of the required shape.

You can bypass the first steps if you can find some other way to prove surjectivity some other way (I can think of at least one way to do so, although in essence it's the same thing)

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  • $\begingroup$ I did not understand how to prove positivity. Can you explain that? Or any way that if I can prove the continuity of the functional. $\endgroup$
    – domath
    Oct 11, 2019 at 1:44
  • $\begingroup$ I'm not familiar with the way you proved, cause these subjects are all new to me. However, I have been also asked to prov it as the steps I wrote in the question. $\endgroup$
    – domath
    Oct 11, 2019 at 1:45
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    $\begingroup$ If $f\geq 0$, then for some $g, f= g^2$ (take $g(x) := \sqrt{f(x)}$). Therefore $\Lambda(f) = \Lambda(g)^2 \geq 0$ $\endgroup$ Oct 11, 2019 at 6:48
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Since $X$ is compact, for every $n$, there exists a finite number of closed balls $B(x^n_1,{1\over n}),...,B(x^n_{i_n},{1\over n})$ of $X$ which cover $X$, since $\nu(\cup B(x^n_{i_j},{1\over n}))\neq 0$, we deduce that there exists a ball $B_n=B(x^n_{i_j},{1\over n}))$ such that $\nu(B_n)\neq 0$.

Consider $C=\cap_nB_n$. Suppose that $C$ is empty, there existsa finite number of $B_n$, $B_{n_1},...,B_{n_p}$ whose intersection is empty. (Baire). This implies (recursively) that there exists $B_{n_i}, B_{n_j}$ whose intersection is empty contradiction with the previous step.

$C$ contains one element $x$, and if $U$ is an open subset which does not contains $x$, for every element $y$ of $U$, there exists $B_n,$ such that $x\in B_n$ and $B(y,{1\over n})\cap B_n$ is empty, and $B(y,{1\over n})\subset U$ this implies that $\nu(B(y,{1\over n})=0$ and $\nu(U)=0$. This shows that the support of $\nu$ is $x$ and $\nu=\delta_x$.

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    $\begingroup$ So what happened, you just read the first sentence and ignored the title and the rest of the body of the post, where the actual things the user was asking about are? $\endgroup$
    – rschwieb
    Oct 10, 2019 at 18:42

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