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Let $C(e^{i\theta})$ be an $m\times n$ ($m\ge n$) matrix-valued continuous function of $\theta\in[-\pi,\pi]$. Let $A_1(e^{i\theta})$ and $A_2(e^{i\theta})$ be two $n\times n$ positive definite matrix-valued continuous function of $\theta\in[-\pi,\pi]$ (i.e. $A_1(e^{i\theta}),\,A_2(e^{i\theta})>0$, $\forall\,\theta\in[-\pi,\pi]$) satisfying $A_1(e^{i\theta})\neq A_2(e^{i\theta})$. Suppose that it holds $$\tag{1}\label{1} \int_{-\pi}^{\pi} C(e^{i\theta})A_1(e^{i\theta})C^*(e^{i\theta})\,\mathrm{d}\theta\neq \int_{-\pi}^{\pi} C(e^{i\theta})A_2(e^{i\theta})C^*(e^{i\theta})\,\mathrm{d}\theta, $$ where the superscript $*$ denotes the Hermitian transpose.

My Question: Let $P(e^{i\theta})$ be any positive definite $n\times n$ continuous function of $\theta\in[-\pi,\pi]$, does \eqref{1} imply that $$\tag{2}\label{2} \int_{-\pi}^{\pi} CA_1^{1/2}P A_1^{1/2}C^*\,\mathrm{d}\theta\neq \int_{-\pi}^{\pi} CA_2^{1/2}P A_2^{1/2}C^*\,\mathrm{d}\theta\ \ \text{?} $$

N.B. In Eq. \eqref{2} I omitted the dependence on $\theta$ for readability. Moreover, the superscript ${1/2}$ denotes the principal matrix square root.

Note. The answer is negative, as a counterexample posted in a comment by Pietro Majer shows. However, it is not still clear if the answer is negative if $A_1$ and $A_2$ are taken of the form $$ A_1(e^{i\theta})=(C^*(e^{i\theta}) X_1 C(e^{i\theta}))^{-1},\quad A_2(e^{i\theta})=(C^*(e^{i\theta}) X_2 C(e^{i\theta}))^{-1} $$ where $X_1$ and $X_2$ are constant $n\times n$ positive matrices and $C$ is of full column rank a.e..

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    $\begingroup$ I think it does not, in general. Take e.g. $C=I$, and $A_1>0$ and $A_2>0$ two diagonal matrices, such that (1) holds, and each element in the diagonal of $A_1-A_2$ changes sign. Then there exists a diagonal matrix $P(\theta)>0$ such that $P(A_1-A_2)$ has zero mean, so in (2) there is equality. $\endgroup$ Commented Jan 25, 2017 at 20:38
  • $\begingroup$ @PietroMajer: I completely agree, thanks! Actually in my problem $A_1$ and $A_2$ are not general but they have a particular form, namely (assuming that $C$ is of full column rank a.e. and omitting the dependence on $\theta$) $A_1=(C^*X_1 C)^{-1}$ and $A_2=(C^*X_2 C)^{-1}$ with $X_1$, $X_2>0$ constant positive definite matrices. I thought however that the more general result I described in this question could have been true, but this is not the case as you correctly observed. $\endgroup$
    – Ludwig
    Commented Jan 26, 2017 at 8:04
  • $\begingroup$ I wonder whether you have some further insights w.r.t. the particular case which I just described above. (Indeed, it seems to me that your counterexample does not work in this case). $\endgroup$
    – Ludwig
    Commented Jan 26, 2017 at 8:05

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