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.



Why is the Berkovich spectrum of a C*-Algebra the same as the Gelfand spectrum?
Observe that every commutative C\*-algebra $A \not\simeq \mathbb{C}$ has a non-trivial zero-divisor.
To see this view $A$ as continuous functions over its spectrum which contains at least two points (and is Hausdorff) and use Urysohn lemma to construct two continuous functions supported at separating open sets of these two points. 
It follows that every closed$^1$ prime ideal is the kernel of a homomorphism to $\mathbb{C}$.


Now, given a bounded multiplicative seminorm $s$ on $A$, observe that its kernel $p$ is a closed prime ideal and it defines a multiplicative norm on $A/p\simeq \mathbb{C}$. This norm must coincide with the usual absolute-value $|\cdot|$.
Denoting the map $A\to\mathbb{C}$ by $a\mapsto a(x)$ we conclude that for every $a\in A$, $s(a)=|a(x)|$.


Note that the commutativity assumption was inessential in all of the above.




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1: Thanks to Robert Furber for correcting my omission here. 


