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A character on $C\left( \mathbb{R}^n\right)$ (the set of all complex-valued continuous functions on $\mathbb{R}^n$) is a continuous $^*$-algebra homomorphism into $\mathbb{C}$. For any fixed $x_0\in \mathbb{R}^n$, the function $\widetilde{x_0}:C\left( \mathbb{R}^n\right) \rightarrow \mathbb{C}$ defiend by $\widetilde{x_0}(f)=f(x_0)$ is a character.

Are there any characters of $C\left( \mathbb{R}^n\right)$ not of this form?

Thanks much in advance!

EDIT: Just to clarify, a $^*$-algebra homomorphism preserves addition, scalar multiplication, multiplication, the involution, and sends $1$ to $1$.

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What topology do you put on $C(\mathbb{R}^n)$ (since you speak of continuous $\ast$-algebra homomorphisms). –  t.b. Aug 23 '11 at 17:35
    
Good question. Let $\{ K_m|\, m\in \mathbb{N}\}$ be any sequence of increasing compact sets whose union is all of $\mathbb{R}^n$ and define the seminorm $p_m(f)=\sup \left\{ \left| f(x)\right| |\, x\in K_m\right\}$. The topology on $C\left( \mathbb{R}^n\right)$ is the topology generated by this collection of seminorms. –  Jonathan Gleason Aug 23 '11 at 17:39
    
Huh? Do you mean the inclusion from $\widehat{C(K_n)}$ into $\widehat{C\left( K_{n+1}\right)}$? This should be unital . . . –  Jonathan Gleason Aug 23 '11 at 17:55
    
It does? There is no inclusion from $C(K_m)$ into $C\left( \mathbb{R}^n\right)$, so you can't just restrict the functional. $C(-)$ is a *contra*variant functor. –  Jonathan Gleason Aug 23 '11 at 18:05
    
@all: Jonathan's previous comments addressed horribly confused comments of mine and I removed them in order to avoid confusing others. Sorry about that. –  t.b. Aug 23 '11 at 19:37
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1 Answer

up vote 2 down vote accepted

Since $\mathbb C[x_1,\ldots,x_n]$ is dense in $C(\mathbb R^n)$, it is sufficient to show that every character of $\mathbb C[x_1,\ldots,x_n]$ is of the form $\widetilde{x_0}$. But this is easy to see: if $\phi$ is any character, let $a_i = \phi(x_i)$. Then $\phi(f) = f(a_1,\ldots,a_n)$ since $\phi$ is a homomorphism.

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What does $\phi(x_i)$ stand for? –  anon Aug 23 '11 at 18:31
    
$x_i$ is the function sending a point $(p_1,\ldots,p_n)$ to $p_i$ and $\phi(x_i)$ is the character applied to the function $x_i$. –  Eric O. Korman Aug 23 '11 at 18:37
    
Just seen this. I'm being a bit slow here, but why is ${\mathbb C}[x]$ dense in $C({\mathbb R})$ for the topology of uniform convergence on compact sets? The natural attempt would be to take an exhaustion by compact subsets (WLOG, closed intervals) and then approximate on each one by (Stone-)Weierstrass, but I'm having trouble seeing how one can extract a diagonal subsequence –  user16299 Jan 8 '12 at 5:14
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