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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?

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    $\begingroup$ I have a question about the formulation of the problem: Shouldn't the equation imply that $HQ - QH^* = 0$ for all $H \in \mathcal{L}(\mathcal{H})?$ (Say, take the imaginary part of the equation.) I would guess this implies that $Q$ is zero. $\endgroup$
    – J. E. Pascoe
    Commented Apr 11, 2017 at 6:02
  • $\begingroup$ If $Q$ is necessarily zero I guess the problem becomes trivial since $X=0$ would satisfy the equation. I can see that $HQ$ must be self-adjoint for all $H$ (specifically $Q$ must be self-adjoint) which is basically what you said. I guess this sub-question is equivalent to asking whether there exists a nonzero left-ideal on the multiplicative set of self-adjoint operators on a Hilbert space. I agree with your intuition that this can't always be the case, but I'll look into it tomorrow. $\endgroup$
    – Conner DiPaolo
    Commented Apr 11, 2017 at 7:43
  • $\begingroup$ Okay in general this can't hold. Consider $\mathcal{H}=\mathbb{C}^n$ and $H = iI$. Then $HQ - QH^* = 2iQ = 0$ so $Q=0$. I guess that does provoke an interesting question of whether $Q$ must be $0$ for any Hilbert space $\mathcal{H}$. $\endgroup$
    – Conner DiPaolo
    Commented Apr 11, 2017 at 7:57
  • $\begingroup$ It's not totally clear how you used $\mathcal{H} = \mathbb{C}^n,$ so the fact should probably hold for any Hilbert space. I guess my question was, is there some way the question needs to be changed to get a meaningful answer? $\endgroup$
    – J. E. Pascoe
    Commented Apr 11, 2017 at 12:16
  • $\begingroup$ Ah I think I see an issue. This question came from looking at the function $f(X) = X P X^* + XQ + R$ and trying to find an $X$ such that $f(X) \preceq_K f(Y)$ for all $Y\in\mathcal{L}(\mathcal{H})$ where $K = \mathcal{L}_+(\mathcal{H})$ is the positive cone. Since the $XQ$ term in general won't be positive, we would likely need to restrict the domain $X$ to be positive and require $Q$ to be positive as well. The resulting equation would force $Q$ to commute with all positive $H$ and would look more like the Lyapunov equation. $\endgroup$
    – Conner DiPaolo
    Commented Apr 11, 2017 at 16:30

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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.

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].

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.

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