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$\mathcal M$ is the space of real $d\times d$ matrices and $\mathcal S\subset \mathcal M$ is its subset consisting of positive semidefinite elements. We consider the distance the product space $\mathbb R^d\times \mathcal S$ : for $(x,A),(y,B)\in \mathbb R^d\times \mathcal S$, $$dist\big((x,A),(y,B)\big):=(x-y)^T(x-y) + {\rm Tr}(A + B - 2\sqrt{\sqrt{A}B\sqrt{A}}),$$ where $T$ denotes the matrix transpose and ${\rm Tr}$ is the trace operator. My question is as follows:

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

A similar question has been addressed by fedja in Does this maximisation problem admit a finite upper bound?

PS : An obvious observation is that $dist\big((x,A),(y,B)\big)=dist\big((x-y,A),(0,B)\big)$ and $L\big((x,A),(y,B)\big)=L\big((x-y,A),(0,B)\big)$. So it suffice to take $y=0$ in the desired inequality, i.e.

$$\sup_{(x,A), (0,B)\in \mathbb R^d\times\mathcal S}\frac{dist\big((x,A),(0,B)\big)}{L\big((x,A),(0,B)\big)}<\infty?$$

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    $\begingroup$ Apparently you just have a mental block: you have the sum of two non-negative quantities $U(x,y)+V(A,B)$ and are trying to bound it by $\sup_u [S(u,x,y)+T(u,A,B)]$ where $S,T$ are non-negative as well. You already know that $V(A,B)\le C(d)\sup_u T(u,A,B)$ (the hard part) and that $U(x,y)=\sup_u S(u,x,y)$ (a triviality). Out of $U$ and $V$ one is larger, so just use $u$ to dominate that one and ignore the other one losing an extra factor of $2$. $\endgroup$
    – fedja
    Commented Jun 23 at 0:17
  • $\begingroup$ @fedja Absolutely fedja. This is exactly what I argued for such a sum and the desired result follows easily. A more challenging question is posted at mathoverflow.net/questions/473193/… and instead of the maximum, how about the average? $\endgroup$
    – Fawen90
    Commented Jun 24 at 1:00
  • $\begingroup$ @fedja I still used $\mathbb R^2$ instead of $\mathbb R^n$ for ease of presentation. This is similar to say, for the Euclidean norm on $\mathbb R^2$, i.e. $|x|^2:=x_1^2+x_2^2$ for $x=(x_1,x_2)$. Is this norm equivalent to the norm $\|x\|:=\int_0^{2\pi}|x_1\cos(t)+x_2\sin(t)|dt$? Of course $\|\cdot\|$ also defined a norm (by verification) on $\mathbb R^2$ and it is therefore equivalent to $|\cdot|$ (without more explicit calculation). For the matrix case, I don't know whether it is still true $\endgroup$
    – Fawen90
    Commented Jun 24 at 1:04
  • $\begingroup$ @fedja Of course $\|\cdot\|$ also defined a norm (by verification) on $\mathbb R^2$ and it is therefore equivalent to $|\cdot|$ (without more explicit calculation). For the matrix case, I don't know whether it is still true mathoverflow.net/questions/473193/… $\endgroup$
    – Fawen90
    Commented Jun 24 at 1:10

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