$\newcommand\abs[1]{\lvert{#1}\rvert}$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,


1. $\abs{f}\_x \geq 0$.
2. $\abs{cf}\_x = \abs c \cdot \abs f\_x$.
3. $\abs{fg}\_x = \abs{f}\_x\cdot\abs{g}\_x$.
4. $\abs{f + g}\_x \leq \abs f\_x + \abs g \_x$.


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:



> 
> 1. How might we put a topology on $Y$ using only the information of $C(X)$?
> 2. Is $\nu$ above a bijection? In light of the last question, is it a homeomorphism?
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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".



Norms as Points in $C(X)$
$\newcommand\Abs[1]{\lVert{#1}\rVert}\newcommand\abs[1]{\lvert#1\rvert}$If one also enforces non-triviality and continuity, then there is a one-to-one correspondence between multiplicative seminorms and points. Namely:



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> **Proposition**. Let $X$ be compact Hausdorff, and let $\Abs{ }: C(X \to {\bf R}) \to {\bf R}$ be a continuous map that obeys the homogeneity property $\Abs{cf} = \abs c\Abs f$, the multiplicative property $\Abs{fg}=\Abs f\Abs g$, and the triangle inequality $\Abs{f+g} \leq \Abs f + \Abs g$, and such that $\Abs{ }$ is not identically zero. Then there exists a unique $x\_0 \in X$ such that $\Abs f = \abs{f(x\_0)}$.
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**Proof.** Since $\Abs f \neq 0$ for at least one $f$, and $f \cdot 1 = f$, we conclude from multiplicativity that $\Abs1 = 1$.


Call a point $x \in X$ *typical* if there is a function $f \in C(X \to {\bf R})$ that vanishes in a neighbourhood of $x$ with $\Abs f \neq 0$. If every point was typical, then by compactness one could find finitely many functions $f\_1, \dots,f\_k \in C(X \to {\bf R})$, all of nonzero norm, with each $f\_i$ vanishing on an open set $U\_i$ with the $U\_i$ covering $X$. But then $\Abs0 = \Abs{f\_1 \dotsb f\_k} = \Abs{f\_1} \dotsb \Abs{f\_k} \neq 0$, which is absurd. Thus there must be a point $x\_0$ which is not typical; thus $\Abs f=0$ whenever $f$ vanishes near $x\_0$, and hence by the triangle inequality $\Abs f=\Abs g$ whenever $f$, $g$ agree near $x\_0$.


For any $f \in C(X \to {\bf R})$ and any $\varepsilon>0$, one can use Urysohn's lemma to find a function $f\_\varepsilon$ that lies with $\varepsilon$ of $f(x\_0)$ and agrees with $f$ on a neighbourhood of $x\_0$. Then $\Abs f = \Abs{f\_\varepsilon}$; taking limits as $\varepsilon \to 0$, we conclude that $\Abs f = \Abs{f(x\_0)} = \abs{f(x\_0)}$, so we have located an $x\_0$ with the required properties, which is unique since $C(X \to {\bf R})$ separates points. $\Box$


Without the continuity axiom there appear to be additional, weirder seminorms arising from other distributions supported at points than Dirac masses. For instance, for $X = [-1,1]$, the map $\Abs f \mathrel{:=} \abs{f(0)} \exp( \alpha \frac{f'(0)}{f(0)} )$, for any fixed real $\alpha$, gives a densely defined map on $C([-1,1] \to {\bf R})$ that obeys the homogeneity, multiplicativity, and triangle inequality (the latter is a calculation using the convexity of $t \mapsto \exp(\alpha t)$), but is discontinuous. It may be possible to extend this map to the rest of $C([-1,1] \to {\bf R})$ by a Zorn's lemma argument, though I have not checked this. [EDIT: actually, the triangle inequality only seems to work for functions with $f(0)>0$. Not sure whether this example can be fixed to cover the opposite case $f(0)<0$. I'll leave it up here in case anyone else has an idea.]


