# “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).

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.

• Yes, this is correct. Do you know when this does hold? Does it hold for Lp norms (without rotating the basis)? – Mike Battaglia Apr 23 '19 at 21:57
• I wrote my thoughts about it. I don't know and could'n come up with any good criteria. – mihaild Apr 23 '19 at 22:14
• I think the outer norm (we call it $n$) needs to be increasing in a coefficient-wise way, i.e., it satisfies: $|v| \le |w|$ (coefficientwise) implies $n(v) \le n(w)$. – gerw Apr 24 '19 at 6:18
• I am not sure it's enough: my deduction uses that, for example, if $a > b > 0$ then $n(a, c) \geqslant n(-b, c)$. Though I can't think of any counterexample for your hypothesis either. – mihaild Apr 24 '19 at 9:59