Definition of the Berkovich spectrum I am trying to read these notes:
http://www-personal.umich.edu/~takumim/Berkovich.pdf
Regarding the Berkovich spectrum. In definition [2.24] it says that the spectrum is the set of bounded (non-trivial) multiplicative semi-norms. But in the explicit definition, elements in the Berkovich spectrum are said to be bounded by the norm of the ring $A$, and it says that we can assume this because we can replace the norm with an equivalent norm.
Initially I did not see the importance of this fact and assumed one can simply discuss bounded semi-norms, and not just norms bounded by the ring norm. However this part of the definition becomes essential when discussing the Gelfand transform.
I do not fully understand the remark above, and how this equivalent norm should be taken. Right now this condition seems to me to make the Berkovich spectrum much less rich than simply bounded multiplicative semi-norms. 
I would appreciate it if someone could point to what I'm missing.
 A: I don't exactly recall what was meant by that remark, but here is an observation from [Berkovich, p. 12].
Let $(\mathscr{A},\lVert \cdot \rVert)$ be the Banach ring in question, and let $\lvert \cdot \rvert$ be a multiplicative seminorm on $\mathscr{A}$ such that there exists a constant $C > 0$ for which $\lvert f \rvert \le C \cdot \lVert f \rVert$ for all $f \in \mathscr{A}$.
Claim. $\lvert f \rvert \le \lVert f \rVert$ for all $f \in \mathscr{A}$.
Proof. For every integer $n \ge 1$, we have
$$\lvert f \rvert^n = \lvert f^n \rvert \le C \cdot \lVert f^n \rVert \le C \cdot \lVert f \rVert^n$$
since $\lvert \cdot \rvert$ is multiplicative and $\lVert \cdot \rVert$ is submultiplicative. Taking $n$th roots, we then have
$$\lvert f \rvert \le \sqrt[n]{C} \lVert f \rVert$$
for every integer $n \ge 1$, hence $\lvert f \rvert \le \lVert f \rVert$. $\blacksquare$
In other words, there is no replacement by an equivalent norm necessary.
EDIT. This observation has been incorporated into the newest version of the notes.
