Let $A = \mathcal{C}(X)$ be a commutative (unital) C*-Algebra. Let $Spec(A)$ denote its Gelfand spectrum $$ Spec(A) = \{A \rightarrow \mathbb{C} : \text{non-zero *-homomorphism} \} \simeq X. $$ Now view $A$ as a normed algebra over $\mathbb{C}$ with respect to the supremum norm $\| a\|_{sup} = \sup_{x \in X} (\|a(x)\|)$. Its Berkovich spectrum $Sp^B(A)$ is then given by $$Sp^B (A) = \{|\cdot|: A \rightarrow \mathbb{R}_+ \text{ multiplicative seminorm bounded by }\| \|_{sup} \}.$$

Why do we have $Spec(A) = Sp^B(A)$? (as for example stated in the wikipedia article on Berkovich spectra https://en.wikipedia.org/wiki/Berkovich_space#Berkovich_spectrum)

It is clear that every homomorphism $A \rightarrow \mathbb{C}$ composed with the absolute value gives a seminorm of the required form. However I struggle to see the converse.

Given a multiplicative seminorm $|\cdot| : A \rightarrow \mathbb{R}_+$ bounded by $\| \|_{sup}$, why must there be an $x \in X$ such that for all $a \in A:$ $|a| = \| a(x) \|$?

Using the boundedness and multiplicativness of $|\cdot|$, I can show that for each $a \in A$ there must be an $x$ such that $|a| = \| a(x) \|$. But a priori these $x$ can be different for different $a$ - I simply fail to see why this cannot happen.