Which norms on vectors can be consistently decomposed? I need to know which permutation-invariant norms can be consistently decomposed in the sense that for any vector $v = (a,b,c)$ we have that
$$\|(a,b,c)\| = \|(\|(a,b)\|,c)\|.$$
More precisely, let $v = \sum_{i=1}^n v_ie_i$ be a finite-dimensional
vector, and $\{P_j\}_{j=1}^k$ a partition of the index set $\{i\}_{i=1}^n$
into $k$ subsets, so that $v = \sum_{j=1}^k \sum_{i \in P_j} v_ie_i$. The
question is then for which norms is it true that for all vectors and partitions
$$
\|v\| = \left\|\sum_{j=1}^k \Bigg\|\sum_{i \in P_j} v_ie_i\Bigg\|e_j\right\|.
$$
It is easy to see that this is true for every $p$-norm, and every other norm that I've tried failed to have this property, so it would be natural to conjecture that $p$-norms are the only consistently decomposable ones.  Just finding a counterexample to this conjecture would be very useful.
 A: The theorem by Bohnenblust is not exactly what I wanted, but it's what I need. Adapting its statement, we have
Let $\|\cdot\|:\mathbb R^N \to \mathbb R$ be a permutation-invariant norm for $N \ge 3$ such that


*

*$\|x+y\| = \|(\|x\|,\|y\|)\|$ for all $x,y$ with disjoint support.

*$\|(1,1)\| \neq 1 $.


Then for any rationals $|a|$ and $|b|$ we have that $\|(a,b)\| = (|a|^p + |b|^p)^{\frac1p}$ for some real number $p \ge 1$.
Proof: Let $f$ be such that $f(1)=1$ and $f(n+1) = \|(1,f(n))\|$. We want to show that $f(n+m) = \|(f(m),f(n)\|$. Assume that it holds for some $m$. Then $$f(n+1+m) = \|(f(m),f(n+1)\| = \|(f(m),1,f(n)\| = \|(f(m+1),f(n))\|.$$
Since it holds for $m=1$ by definition, by induction it holds for all $m$. From that it is easy to see that $f(nm) = f(n)f(m)$ and therefore $f(n^k) = f^k(n)$. We also need to show that $f(n)$ is monotonous. This follows from applying the triangle inequality to the identity $$2(f(n),0) = (f(n),1) + (f(n),-1),$$ which implies that $f(n) \le f(n+1)$.
Now let $m,n\ge 2$ be some fixed integers, and $h$ the integer such that for any positive integer $k$
$$m^h \le n^k < m^{h+1}.$$
Using the properties of $f(n)$, it follows that
$$h\log f(m) \le k \log f(n) < (h+1) \log f(m),$$
and elementary manipulations with $h$ and $k$ let us conclude that
$$\frac{\log f(m)}{\log m} = \frac{\log f(n)}{\log n},$$ which means that this fraction is a constant independent of $n$ and different than $0$. Calling this constant $1/p$, we conclude that
$$f(n) = n^\frac1p.$$
Now for any positive rational $m/n$ we have that
$$\|(1,m/n)\| = \frac1n\|(n,m)\| = \frac1n\|(f(n^p),f(m^p)\| = \frac1nf(n^p+m^p) = (1+(m/n)^p)^\frac1p,$$
so by homogeneity $\|(a,b)\| = (|a|^p + |b|^p)^{\frac1p}$ for any rationals $|a|$ and $|b|$. I guess I can't extend this for all reals without some continuity condition.
