This problem arises when minimizing the operator equation $X P X^\* + X Q + R$ over positive $X$ with respect to the positive cone on a Hilbert space $\mathcal{H}$.


**The (reduced) task**:


Given $P$ and $Q$ are positive (bounded) linear operators on $\mathcal{H}$, respectively, find a positive $X\in\mathcal{L}\_+(\mathcal{H})$ so that
$$H P X + X P H + HQ = 0$$ or equivalently $$H P X + X (HP)^\* + HQ = 0$$ for all (bounded) operators $H\in\mathcal{L}\_+(\mathcal{H})$.


Note that this implies $Q$ must commute with all positive $H$, so we might assume $Q=aI$ for some $a$ in the base field (if $H$ was allowed to be *any* operator this would be necessary. Since $H$ must be positive, are multiples of the identity still the only operators that always commute?)


**My questions**:


* under what (nontrivial) conditions does the above equation have a closed form solution?
* this seems related to the Lyapunov equation $A^\*X + XA = W$ (if we let $H=I$). Is there some significance to this?


**Update**:


I tried the naive solution on $\mathbb{Z}^n,\mathbb{R}^n,$ and $\mathbb{C}^n$ of setting $H=I$, solving the resulting Lyapunov equation for $X$, and then plugging $X$ back in to the derivative. Interestingly enough, in all my experiments this naive $X$ satisfies $HPX + XPH + HQ = 0$ for all symmetric $H$. Is this some weird scale-invariance for the Lyapunov Equation?



Solving Matrix/Operator Equation $H P X + X P H + HQ = 0$
The following was resolved through personal communication, but I thought I'd post it here in case anyone has a similar problem in the future.




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I'm going to assume $P$ is nonsingular (strictly positive) so that $P^{-1}$ exists since $P$ is bounded.


As seen above $Q$ must commute with every *positive* operator. One can show that this implies $Q$ must commute with *every* arbitrary operator on $\mathcal{H}$. Therefore we must require $Q=2aI$ where $a\geq 0$ since $Q$ is positive [[1]](https://math.stackexchange.com/questions/27808/a-linear-operator-commuting-with-all-such-operators-is-a-scalar-multiple-of-the).


Then if we let $X = cP^{-1}$, the solution to the Lyapunov equation given when $H=I$, we can see that
$$
 XPH + HPX + HQ = 0
$$
for all $H$. This explains why in my numerical experiments just solving when $H=I$ works in all cases.


