Does this maximisation problem admit a finite upper bound?

Question

Let $\mathcal M_2$ be the space of real $2\times 2$ matrices and $\mathcal S_2\subset \mathcal M_2$ be its subset consisting of positive semidefinite elements, i.e. $A\in \mathcal S_2$ iff $A$ is symmetric and $u^TAu\ge 0$ for all $u\in\mathbb R^2$. We consider the distance between $A,B\in \mathcal S_2$ given by $$d(A,B):={\rm Tr}(A + B - 2\sqrt{\sqrt{A}B\sqrt{A}}),$$ where ${\rm Tr}$ denotes the trace operator. My question is as follows:

Define the unit ball $B_1 :=\{u=(u_1,u_2)\in \mathbb R^2: u_1^2+u_2^2\le 1\}$. Set $$L(A,B):=\max_{u\in B_1} \Big\{ u^TAu+u^TBu-2\sqrt{u^TAuu^TBu}\Big\}.$$ Does the maximisation problem below admit a finite upper bound? $$\sup_{A, B\in \mathcal S_2}\frac{d(A,B)}{L(A,B)}<\infty?$$

PS : It is straightforward to observe $d(A,B)=d(B,A)$ and $d(cA,cB)=cd(A,B)$ for $c\ge 0$ (where the same relations hold for $L$).

Answer 1

The answer is "yes" and it is quite a nice linear algebra problem, but let me restate it first in a less intimidating way. We'll deal with $\mathbb R^n$ for any finite $n$.

The first thing I will do is to replace $A$ with $A^2$ and $B$ with $B^2$. Next, notice that $\newcommand{\tr}{\operatorname{Tr}}$ $\tr\sqrt{AB^2A}=\sup_{u,v}\sum_k\langle BAu_k,v_k\rangle=\sup_{u,v}\sum_k\langle Au_k,Bv_k\rangle$ where $u,v$ are two orthonormal bases in $\mathbb R^n$. Indeed, if $C$ is any linear operator in $\mathbb R^n$, then writing $C=RD$ where $R$ is orthogonal and $D$ is self-adjoint positive semi-definite, we get $\sqrt{C^*C}=D$. Also the expression $\sup_{u,v}\sum_k\langle Cu_k,v_k\rangle$ is invariant under multiplications of $C$ by orthogonal operators from either side, so the supremum for $C$ is the same as for $D$. However, for $D$, we have $$ \sum_k\langle Du_k,v_k\rangle\le\frac 12\left[\sum_k\langle Du_k,u_k\rangle+\sum_k\langle Dv_k,v_k\rangle\right]=\tr D $$ with equality when $u=v$.

Since $\tr A^2=\sum_k |Au_k|^2$ and $\tr B^2=\sum_k |Bu_k|^2$ regardless of the choice of $u,v$, we see that the LHS is just $$ \inf_{u,v}\sum_k|Au_k-Bv_k|^2 $$ Let $R$ be the orthogonal mapping defined by $Ru_k=v_k$. Then the infimum can be rewritten as $$ \inf_R\sum_k|(A-BR)u_k|^2=\inf_R\|A-BR\|_{HS}^2=\inf_R\|A-RB\|_{HS}^2\,. $$ (the last equality is due to the fact that $\|A-BR\|_{HS}=\|A-R^*B\|_{HS}$).

The RHS in this notation is merely $\sup_x(|Ax|-|Bx|)^2$ where $x$ runs over the unit sphere. Thus we arrive to the question if for two operators $A$ and $B$, the only reason why $\left||Ax|-|Bx|\right|<\varepsilon$ for all unit vectors $x$ is that $\|A-RB\|\le K\varepsilon$ for some orthogonal $R$ with some $K=K(n)\in[1,+\infty)$ (obviously, if the latter condition holds with $K=1$, then the former one is satisfied as well). What norm to use doesn't matter in the finite dimension, so I'll switch to the operator norm now.

What I suspect is that $K$ should be independent of the dimension in this setting. However, I can do it only with $K(n)$ growing exponentially in $n$ at this moment, which is rather dismal. Still, it solves the problem as posted.

We shall proceed by induction on $n$. The base $n=1$ is trivial with $K(1)=1$.

For the step $n-1\to n$, we can always assume that $A$ is self-adjoint PSD by multiplying it by an orthogonal operator on the left. Also, since the inequality scales correctly, we can normalize to $\|A\|=1$. If $\varepsilon>\frac 13$, we can take $K=7$, so I'll assume that $0<\varepsilon\le\frac 13$.

Then $\|B\|\le\frac 43$. Denote by $e$ the top unit eigenvector of $A$, so $Ae=e$. Then $\mu=|Be|\in[\frac 23,\frac 43]$. Let $R'$ be the rotation such that $B'e=R'Be=\mu e$. Consider now a unit vector $y\in e^\perp$ and put $x=(\cos t)e+(\sin t)y$. On the one hand, we have $$ ||Ax|^2-|B'x|^2|=||Ax|^2-|Bx|^2|=||Ax|-|Bx||(|Ax|+|Bx|)\le \frac 73\varepsilon\,. $$ On the other hand, using that $Ae^\perp\subset e^\perp$, the left hand side can be written as $$ |J_1\cos^2 t+J_2\sin^2 t-\mu\langle e,B'y\rangle\sin 2t| $$ where $J_1$ and $J_2$ is some junk independent of $t$. Since $\sin 2t$ is orthogonal to $\sin^2 t$ and $\cos^2 t$ in $L^2(-\pi,\pi)$, we conclude that $$ \mu|\langle e,B'y\rangle|\le \frac{14}{3}\varepsilon\,, $$ so, since $\mu\ge \frac 23$, we get $$ |\langle e,B'y\rangle|\le 7\varepsilon\,. $$ Now consider the operator $B''=B'-S$ where $Sx=\langle B'P_{e^\perp}x,e\rangle e$. Note that $\|S\|\le 7\varepsilon$, so $||Ax|-|B''x||\le 8\varepsilon$ for all unit $x$. Also $B''e=B'e=\mu e$ and $e^\perp$ is invariant under both $A$ and $B''$. Thus, applying the induction assumption, we can find $R''$ that rotates only in $e^\perp$ (i.e., $R''e=e$) such that $\|A-R''B''\|_{e^\perp\to e^\perp}\le 8K(n-1)\varepsilon$. Together with $|(A-R''B'')e|=|(A-B')e|\le\varepsilon$, it yields $$ \|A-R''B''\|\le 8K(n-1)\varepsilon $$ and, finally, $$ \|A-R''R'B\|=\|A-R''B'\|\le \|A-R''B''\|+\|S\|\le [8K(n-1)+7]\varepsilon\,, $$ i.e., we can take $K(n)=8K(n-1)+7$.

As I said, this bound (growing as $8^n$) is ridiculously large and it would be nice to figure out what really happens here. Any ideas?