The theorem by Bohnenblust is not exactly what I wanted, but it's what I need. Adapting its statement, we have

Let $\|\cdot\|$ be a permutation-invariant norm 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 $f(n) = \|1^{(n)}\|$, where $1^{(n)}$ is the vector of $n$ ones. It is easy to see that $f(n+m) = \|(f(n),f(m)\|$, that $f(nm) = f(n)f(m)$ and therefore $f(n^k) = f^k(n)$. The crucial thing to show is that $f(n)$ is monotonous. This follows from applying the triangle inequality to $2(1^{(n)},0) = (1^{(n)},1) + (1^{(n)},-1)$, 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 rational number $m/n$ 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.

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.

The theorem by Bohnenblust is not exactly what I wanted, but it's what I need. Adapting its statement, we have

Let $\|\cdot\|$ be a permutation-invariant norm such that

• $\|x+y\| = \|(\|x\|,\|y\|)\|$ for all $x,y$ with disjoint support.
• $\|e_1\| = 1$
• $\|(1,1)\| \neq 1 $.

Then $\|(a,b)\| = (|a|^p + |b|^p)^{\frac1p}$ for any rationals $|a|$ and $|b|$.

Proof: Let $f(n) = \|1^{(n)}\|$, where $1^{(n)}$ is the vector of $n$ ones. It is easy to see that $f(n+m) = \|(f(n),f(m)\|$, that $f(nm) = f(n)f(m)$ and therefore $f(n^k) = f^k(n)$. The crucial thing to show is that $f(n)$ is monotonous. This follows from applying the triangle inequality to $2(1^{(n)},0) = (1^{(n)},1) + (1^{(n)},-1)$, 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 rational number $m/n$ 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.

The theorem by Bohnenblust is not exactly what I wanted, but it's what I need. Adapting its statement, we have

Let $\|\cdot\|$ be a permutation-invariant norm 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 $f(n) = \|1^{(n)}\|$, where $1^{(n)}$ is the vector of $n$ ones. It is easy to see that $f(n+m) = \|(f(n),f(m)\|$, that $f(nm) = f(n)f(m)$ and therefore $f(n^k) = f^k(n)$. The crucial thing to show is that $f(n)$ is monotonous. This follows from applying the triangle inequality to $2(1^{(n)},0) = (1^{(n)},1) + (1^{(n)},-1)$, 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 rational number $m/n$ 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.