convex analysis - "Norm of norms" is another norm?

Suppose that, for some finite-dimensional real vector space $\Bbb R^n$, that $n_1(v)$, $n_2(v)$, ..., $n_k(v)$ are a set of norms on the space.

Given some $v$, then, we can look at the "vector of norms", which I will denote $v_n = (n_1(v), n_2(v), ..., n_k(v))$.

We can then look at norms on this vector of norms. For example, we could take the $\ell_1$ norm of the vector, which would be the sum of norms. It is easy to see that this will also be a norm.

• Question 1: Is any norm on this "vector of norms" also a norm?


• Question 2: Likewise, if we replace with seminorms, is any seminorm on the "vector of seminorms" also a seminorm?


• Question 3: If not, for what norms do these things hold? (Do they at least hold for $\ell_p$ norms on the vector of norms?)

It is easy to see that you get homogeneity and positive-semidefiniteness, so the question is really about convexity. Does taking a "norm of norms" preserve convexity? Equivalently, does taking a norm of convex functions preserve convexity, or does taking a strictly increasing multivariate convex function of multiple convex functions preserve convexity?

EDIT - as per the answer from "mihaild" below, this isn't true for general norms, but would still like to know when it is true (in particular if it's true for $\ell_p$ norms without changing the basis).

Answer

At least for the first (and so for the second) question the answer is "no".

Take two norms $\|\cdot\|_1$ and $\|\cdot\|_2$ and two vectors $x$, $y$ such that $\|x\|_1 = \|x\|_2 = \|y\|_1 = \|y\|_2 = 1$, $\|x + y\|_1 \approx 2$, $\|x + y\|_2 \approx 0$.

Let $n(a, b) = \max (\frac{1}{3} |a + b|, |a - b|)$ (equal to $l_\infty$ norm in some scaled and rotated basis).

Then $n(\|x + y\|_1, \|x + y\|_2) \approx n(2, 0) = 4$, but $n(\|x\|_1, \|x\|_2) + n(\|y\|_1, \|y\|_2) = 2\cdot n(1, 1) = \frac{4}{3} < 4$.

For when it holds - at least if $n$ is such that for any $a_1 > 0, a_2 > 0, \ldots a_k > 0$ and $q_i \in [-a_i, a_i]$ we have $n(q_1, \ldots, q_n) \leqslant n(a_1, \ldots, a_n)$, then it holds: $$n(\|x + y\|_1, \ldots, \|x + y\|_n) = \\
n(\|x\|_1 + (\|x + y\|_1 - \|x\|_1), \ldots, \|x\|_n + (\|x + y\|_n - \|x\|_n)) \leqslant\\
n(\|x\|_1, \ldots, \|x\|_n) + n(\|x + y\|_1 - \|x\|_1, \ldots, \|x + y\|_n - \|x\|_n)
$$
If $a_i = \|y\|_i$ and $q_i = \|x + y\|_i - \|x\|_i$, then we have
$$n(\|x\|_1, \ldots, \|x\|_n) + n(\|x + y\|_1 - \|x\|_1, \ldots, \|x + y\|_n - \|x\|_n) \leqslant\\
n(\|x\|_1, \ldots, \|x\|_n) + n(\|y\|_1, \ldots, \|y\|_n)
$$

It holds at least for all $l_p$ norms. I think it is equal to unit ball defined by $n$ to be contained in hypercube bounded by hyperplanes $x_i = \pm p_i$, where $p_i$ is such that $n(0, 0, \ldots, p_i, \ldots, 0) = 1$.

This condition if definitely not necessary: for example, if all $\|\cdot\|_i$ coincide, then any $n$ will work.