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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:

Set $u_{\theta}:=\big(\cos({\theta}),\sin(\theta)\big)^T$ for ${\theta}\in [0,2\pi]$. Define $$L(A,B):=\int_0^{2\pi} \Big\{ u_{\theta}^TAu_{\theta}+u_{\theta}^TBu_{\theta}-2\sqrt{u_{\theta}^TAu_{\theta} u_{\theta}^TBu_{\theta}}\Big\} d\theta.$$ Does the maximisation problem below admit a finite upper bound? $$\sup_{A, B\in \mathcal S_2}\frac{d(A,B)}{L(A,B)}<\infty?$$

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  • $\begingroup$ For $2\times2$ matrices don't we have $\operatorname{tr}(A^2) = \operatorname{tr}(A)^2 - 2\det(A)$? You might be able to do stuff with that $\endgroup$
    – Daniel Weber
    Commented Jun 13 at 19:24
  • $\begingroup$ @DanielWeber I don't see how your comment helps estimate the ratio $d/L$. Do you mind explaining more? $\endgroup$
    – Fawen90
    Commented Jun 13 at 19:29
  • $\begingroup$ Why? mathoverflow.net/questions/471794/… $\endgroup$
    – Fabian Wirth
    Commented Jun 13 at 19:38
  • $\begingroup$ @FabianWirth I don't understand your concern. This post is a continuation. Roughly speaking, these are two metrics where the former is to take the maximum among all projections along $u$, and the later is to take the average over all projections $u$ $\endgroup$
    – Fawen90
    Commented Jun 13 at 19:59
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    $\begingroup$ Now the answer is negative: take $A=\begin{bmatrix}1&0\\0&0\end{bmatrix}$ and $B=\begin{bmatrix}1&0\\0&\varepsilon\end{bmatrix}$ with $\varepsilon\to 0+$. $\endgroup$
    – fedja
    Commented Jun 24 at 2:15

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