Let $G(z)$ be an $n\times m$ rational matrix-valued function of full column rank on the unit circle. Further, let $P(z)$ be an $m\times m$ rational matrix-valued function positive definite on the unit circle and let $A$, $B$ be $m\times n$ complex (constant) matrices. Consider the following inner-product condition (on $\mathcal{L}_2^{m\times m}[-\pi,\pi]$)
$$\tag{1}\label{eq:cond}
\langle G^*XG,G^* \Delta G \rangle_2 = \mathrm{tr}\int_{-\pi}^{\pi} G(e^{i\theta})^* X G(e^{i\theta}) G(e^{i\theta})^* \Delta(e^{i\theta}) G(e^{i\theta}) \frac{\mathrm{d}\theta}{2\pi}=0,\quad \forall X\in\mathcal{H}_n,
$$
where $(\cdot)^*$ denotes Hermitian transposition, $\mathcal{H}_n$ denotes the space of Hermitian $n\times n$ matrices, and 
$$
\Delta(e^{i\theta}) := A^* P(e^{i\theta}) B + B^* P(e^{i\theta}) A.
$$

> **My question.** Does condition \eqref{eq:cond} necessarily implies $G(e^{i\theta})^*\Delta(e^{i\theta})G(e^{i\theta})\equiv 0$?


----------


**Some comments.** A simple observation is that the answer is in the affirmative if $P(e^{i\theta})=p(e^{i\theta})M$ where $p(e^{i\theta})$ is scalar and $M$ is a constant $m\times m$ positive definite matrix. Indeed, in this case, the thesis readily follows from the fact that, picking $X=A^* M B + B^* M A$ in \eqref{eq:cond}, yields
$$
\left\|p^{1/2} G^*(A^* M B + B^* M A)G\right\|_2=0.
$$
I couldn't however provide an answer for the case of general $P(e^{i\theta})$.