TITLE: Norms as Points in $C(X)$
QUESTION [7 upvotes]: $\newcommand\abs[1]{\lvert{#1}\rvert}$Let $X$ be a compact hausdorff space, and put $C(X)$ for the $\mathbb{R}$-algebra of continuous maps from $X$ to $\mathbb{R}$.
For each point $x$, there is a multiplicative semi-norm $\abs-_x$ on $C(X)$, where $\abs f_x = \abs{f(x)} \in \mathbb{R}_{\geq 0}$. That is,

$\abs{f}_x \geq 0$.

$\abs{cf}_x = \abs c \cdot \abs f_x$.

$\abs{fg}_x = \abs{f}_x\cdot\abs{g}_x$.

$\abs{f + g}_x \leq \abs f_x + \abs g _x$.


Let $Y$ be the set of multiplicative semi-norms on $C(X)$ mod the equivalence relation where $\abs- \sim \abs-'$ when each is bounded by the other up to a constant. We have just constructed a function $\nu : X \rightarrow Y$. My questions are:


How might we put a topology on $Y$ using only the information of $C(X)$?

Is $\nu$ above a bijection? In light of the last question, is it a homeomorphism?




So in this view, multiplicative seminorms are like points or generalized points. Interestingly, the submultiplicative norm $\abs- : C(X) \rightarrow \mathbb{R}$ sending $f$ to $\operatorname{sup}_x \abs f_x$ seems like it could be the supremum of multiplicative norms in $Y$. That way, multiplicative norms are the "local" versions of submultiplicative norms, which would be "global".

REPLY [8 votes]: A slight improvement to Terry Tao's answer: It is not necessary to assume a priori that the multiplicative seminorm $\|\cdot\|$ is continuous with respect to the sup-norm on $C(X)=C(X\to \mathbb R)$ to conclude $\|f\|=|f(x_0)|$ for some $x_0\in X$.
This follows from a theorem of Mazur [Sur les anneaux linéaires, C. R. Acad. Sci. Paris, 2 0 7 (1938), 1025-1027] which says that the only real algebras with a multiplicative norm are (isometrically isomorphic to) the real numbers, the complex numbers, or the quaternions.
We will apply this to the the quotient algebra $C(X)/I$ with the ideal $I=\{f\in C(X): \|f\|=0\}$. Denoting by $\tilde f$ the equivalence class of $f\in C(X)$ one easily sees that $\tilde f \mapsto \|f\|$ is a well-defined multiplicative norm on $C(X)/I$ to which Mazur's theorem applies. Since this quotient algebra is commutative it cannot be the quaternions and we show that it neither can be ismorphic to $\mathbb C$. Indeed, assume that $\pi: C(X)\to \mathbb C$ is an algebra morphism onto $\mathbb C$ with kernel $I$. If $g\in C(X)$ satisfies $\pi(g)=i$ we get $\pi(g^2+1)=0$ so that $g^2+1 \in I$ which is absurd because $g^2+1$ is invertible in $C(X)$ and thus does not belong to the (proper) ideal $I$.
We conclude that $C(X)/I$ is isomorphic to $\mathbb R$ so that $I$ is a maximal ideal in $C(X)$ and hence of the form $I=\{f\in C(X): f(x_0)=0\}$ for some $x_0\in X$. Finally, given $f\in C(X)$ write $f=f-f(x_0)+f(x_0)$ (where the number $f(x_0)$ is identified with the constant function $f(x_0)\mathrm 1)$ to get $$\|f\|=\|f(x_0)\mathrm 1\|=|f(x_0)| \|\mathrm 1\|=|f(x_0)|.$$