If one also enforces non-triviality and continuity, then there is a one-to-one correspondence between multiplicative seminorms and points. Namely:

Proposition. Let $X$ be compact Hausdorff, and let $\| \|: C(X \to {\bf R}) \to {\bf R}$ be a continuous map that obeys the homogeneity property $\|cf\| = |c| \|f\|$, the multiplicative property $\|fg\|=\|f\|\|g\|$, and the triangle inequality $\|f+g\| \leq \|f\| + \|g\|$, and such that $\| \|$ is not identically zero. Then there exists a unique $x_0 \in X$ such that $\|f\| = |f(x_0)|$.

Proof. Since $\|f\| \neq 0$ for at least one $f$, and $f \cdot 1 = f$, we conclude from multiplicativity that $\|1\| = 1$.

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 $\|f\| \neq 0$. 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 $\|0\| = \| f_1 \dots f_k \| = \|f_1\| \dots \|f_k\| \neq 0$, which is absurd. Thus there must be a point $x_0$ which is not typical; thus $\|f\|=0$ whenever $f$ vanishes near $x_0$, and hence by the triangle inequality $\|f\|=\|g\|$ 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 $\|f\| = \|f_\varepsilon\|$; taking limits as $\varepsilon \to 0$, we conclude that $\|f\| = \| f(x_0) \| = |f(x_0)|$, 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 $\| f \| := |f(0)| \exp( \alpha \frac{f'(0)}{f(0)} )$, 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.]

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

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 $\Abs f = \abs{f(x_0)}$.

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.]

If one also enforces non-triviality and continuity, then there is a one-to-one correspondence between multiplicative seminorms and points. Namely:

Proposition. Let $X$ be compact Hausdorff, and let $\| \|: C(X \to {\bf R}) \to {\bf R}$ be a continuous map that obeys the homogeneity property $\|cf\| = |c| \|f\|$, the multiplicative property $\|fg\|=\|f\|\|g\|$, and the triangle inequality $\|f+g\| \leq \|f\| + \|g\|$, and such that $\| \|$ is not identically zero. Then there exists a unique $x_0 \in X$ such that $\|f\| = |f(x_0)|$.

Proof. Since $\|f\| \neq 0$ for at least one $f$, and $f \cdot 1 = f$, we conclude from multiplicativity that $\|1\| = 1$.

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 $\|f\| \neq 0$. 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 $\|0\| = \| f_1 \dots f_k \| = \|f_1\| \dots \|f_k\| \neq 0$, which is absurd. Thus there must be a point $x_0$ which is not typical; thus $\|f\|=0$ whenever $f$ vanishes near $x_0$, and hence by the triangle inequality $\|f\|=\|g\|$ 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 $\|f\| = \|f_\varepsilon\|$; taking limits as $\varepsilon \to 0$, we conclude that $\|f\| = \| f(x_0) \| = |f(x_0)|$, 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 $\| f \| := |f(0)| \exp( \alpha \frac{f'(0)}{f(0)} )$, 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.

If one also enforces non-triviality and continuity, then there is a one-to-one correspondence between multiplicative seminorms and points. Namely:

Proposition. Let $X$ be compact Hausdorff, and let $\| \|: C(X \to {\bf R}) \to {\bf R}$ be a continuous map that obeys the homogeneity property $\|cf\| = |c| \|f\|$, the multiplicative property $\|fg\|=\|f\|\|g\|$, and the triangle inequality $\|f+g\| \leq \|f\| + \|g\|$, and such that $\| \|$ is not identically zero. Then there exists a unique $x_0 \in X$ such that $\|f\| = |f(x_0)|$.

Proof. Since $\|f\| \neq 0$ for at least one $f$, and $f \cdot 1 = f$, we conclude from multiplicativity that $\|1\| = 1$.

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 $\|f\| \neq 0$. 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 $\|0\| = \| f_1 \dots f_k \| = \|f_1\| \dots \|f_k\| \neq 0$, which is absurd. Thus there must be a point $x_0$ which is not typical; thus $\|f\|=0$ whenever $f$ vanishes near $x_0$, and hence by the triangle inequality $\|f\|=\|g\|$ 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 $\|f\| = \|f_\varepsilon\|$; taking limits as $\varepsilon \to 0$, we conclude that $\|f\| = \| f(x_0) \| = |f(x_0)|$, 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 $\| f \| := |f(0)| \exp( \alpha \frac{f'(0)}{f(0)} )$, 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.]