Potential p-norm on tuples of operators

Question

Consider $\left[\begin{matrix}A \\ B\end{matrix}\right] \in B(H)^2$. One can define $$ \left\|\left[\begin{matrix}A \\ B\end{matrix}\right]\right\|_p = \| |A|^p + |B|^p\|^{1/p}. $$

Q: Is this a norm?

Consider the matrices $C = \left[\begin{matrix}1 & 0 \\ 0 & 0\end{matrix}\right]$ and $D = \left[\begin{matrix}0 & 1 \\ 0 & 0\end{matrix}\right]$. Then $$ \left\|\left[\begin{matrix}C+D \\ D+C\end{matrix}\right]\right\|_p = 2^{1/p}||C+D\| = 2^{1/p+1/2} $$ while $$ \left\|\left[\begin{matrix}C \\ D\end{matrix}\right]\right\|_p + \left\|\left[\begin{matrix}D \\ C\end{matrix}\right]\right\|_p = 2\|C|^p + |D|^p\|^{1/p} = 2\|I\|^{1/p} = 2. $$ Therefore, when $1\leq p< 2$ then $$ \left\|\left[\begin{matrix}C+D \\ D+C\end{matrix}\right]\right\|_p > \left\|\left[\begin{matrix}C \\ D\end{matrix}\right]\right\|_p + \left\|\left[\begin{matrix}D \\ C\end{matrix}\right]\right\|_p $$ and so $\|\cdot\|_p$ is not a norm.

The $p=2$ case does give a norm as $$ \| |A|^2 + |B|^2\|^{1/2} = \left\| [A^* B^*]\left[\begin{matrix}A \\ B\end{matrix}\right] \right\|^{1/2} = \left\|\left[\begin{matrix}A & 0 \\ B & 0\end{matrix}\right] \right\| $$ which allows one to use the triangle inequality for $M_2(B(H))$.

My question is what happens for all of the other cases, $p>2$?

Answer 1

No, the expression $$ \left\|\left[\begin{matrix}A \\ B\end{matrix}\right]\right\|_p = \| |A|^p + |B|^p\|^{1/p}. $$ is not a norm for any $2<p<\infty$ (so it is a norm if and only if $p=2,\infty$).

I will justify that by proving that the triangle inequality for this expression would imply that the map $t\mapsto t^{p/2}$ is operator monotone, which is well-known to be false (Loewner).

Lemma. Given $A_1,A_2$ bounded positive operators on a Hilbert space, the following are equivalent:

1. $\|A_1+B\| \leq \|A_2+B\|$ for every positive operator $B$.
2. $A_1 \leq A_2$.

Proof: one direction ($2. \implies 1.$) is obvious. For the other, let $u$ be a unit vector, $P_u$ the rank one orthogonal projection on $\mathbf{C}u$ and take $B= s P_u$ for $s>0$ (large). Then a small computation gives that $\|A_1+B\| = s + \langle A_1 u,u\rangle + O(1/s)$, so making $s \to +\infty$, we obtain that 2. implies that $\langle A_1 u,u\rangle \leq \langle A_2 u,u\rangle$. The lemma is proven.

Assume for a contradiction that your expression was a norm. Then for any $C$ of norm $<1$, we can write $C$ as the average of two unitaries (see here), and therefore (for arbitrary $A,B$) we can write $\left[\begin{matrix} CA \\ B\end{matrix}\right]$ as a convex combination of elements of the form $\left[\begin{matrix}UA \\ B\end{matrix}\right]$ for unitares $U$, which all have the same "norm" $\| |A|^p + |B|^p\|^{1/p}$, and therefore we would have $$\| |CA|^p+|B|^p\| ^{1/p}\leq \| |A|^p + |B|^p\|^{1/p}.$$ So by the Lemma we would have $$ |CA|^p \leq |A|^p$$ for any $A$ and any $C$ of norm $\leq 1$. Note that every operator $\leq |A|^2=A^* A$ can be written as $|CA|^2$ for some $C$ of norm $\leq 1$. So we have reached the desired conclusion that the map $t\mapsto t^{p/2}$ is operator monotone.