Is $p(x) = \max(p_1(x), p_2(x))$ a seminorm when $p_1, p_2$ are seminorms? To start: I've shown that yes, $p(x) = \max(p_1(x), p_2(x))$ is a seminorm if $p_1, p_2$ are seminorms. But I did that by proving absolute homogeneity and the triangle inequality directly. What I'd like to know: Is there some broader theoretical result that could be used instead?
By way of comparison: if $\pi_i(x_1, x_2) = x_i$ are projection operators and $T$ is a linear transformation, then I know that $q(x) = p_1(\pi_1(Tx)) + p_2(\pi_2(Tx)))$ is a seminorm without directly proving the seminorm axioms because each $p_i(\pi_1(Tx))$ is a seminorm (as the composition of a seminorm and a linear operator), so that $q$ is the sum of seminorms. Is there some similar theorem(s) I could use for $p$?
 A: We can ask the following more general question: if $p_1, \dots p_n$ are seminorms on some real vector space $V$, what conditions does a function $f : \mathbb{R}_{\ge 0}^n \to \mathbb{R}_{\ge 0}$ need to satisfy so that $f(p_1, \dots p_n)$ is always a seminorm?
(Edit: throughout I assume that $f$ is nondecreasing, see the comments.)
Well, if we take $V = \mathbb{R}^n$ and $p_i(x_1, \dots x_n) = |x_i|$, we see that (edit: assuming that $f$ is nondecreasing, see the comments) $f$ needs to be absolutely homogeneous and subadditive. Conversely if $f$ is absolutely homogeneous and subadditive then $f(p_1, \dots p_n)$ is clearly also a seminorm.
So the answer is quite nice: $f$ itself needs to be a seminorm! The only detail that needs to be noted here is that $f$ is defined on $\mathbb{R}_{\ge 0}^n$ rather than a real vector space but this isn't a big deal. There is a generalized definition of a seminorm lurking here which has the very nice property that a composition of two generalized seminorms is a generalized seminorm.
