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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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    @all: Jonathan's previous comments addressed horribly confused comments of mine and I removed them in order to avoid confusing others. Sorry about that.2011-08-23

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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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    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 subsequence2012-01-08