# Potential p-norm on tuples of operators

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, $$p>2$$?

• Hi Chris, when $p>2$ do you even know if the triangle inequality works for $A \mapsto \Vert \ \vert A \vert \ \Vert^{1/p}$ (i.e. the $B=0$ case)? Jul 27, 2020 at 19:39
• @YemonChoi, if $B=0$ then you've missed a power p. In one dimension these are all just identical to the usual norm. Jul 27, 2020 at 19:43
• Thanks, yes I did intend to have $|A|^p$, and yes, I was being slow (forgetting basic facts about norms of positive operators) Jul 27, 2020 at 19:49

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 $$2 (so it is a norm if and only if $$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.