Can we apply quadratic optimization in complex inner product spaces?

As I said in my previous question, I am revisiting Peter J Olver's Introduction to Partial Differential Equations. I am on chapter 9, "A General Framework for Linear PDEs". The author has been discussing some results about inner product spaces having to do with self-adjoint, positive definite linear operators. He gives the following important result, on page 362 of the linked text:

Let $S:U\to U$ be a self-adjoint positive definite linear operator on a [real] inner product space $U$. Suppose that the linear system $$Su=f$$ admits a (necessarily unique) solution $u_\star$. Then $u_\star$ minimizes the value of the associated quadratic function(al) $$Qu=\frac{1}{2}\langle u,Su\rangle-\langle f,u\rangle$$ meaning that $Qu_\star < Qu$ for all admissable $u\neq u_\star$ in $U$.

(Firstly, the last line I think makes an overly strong assumption that this minimum is achieved uniquely at $u_\star$, which I can't easily find a justification for, but that is not the main point of my question. )

The author's proof is pretty hard to follow, but I managed to find one here that includes a few more steps. It goes as follows: \begin{align*} Qu&=\frac{1}{2}\langle u,Su\rangle-\langle u,f\rangle\tag{1}\\ &=\frac{1}{2}\langle u,Su\rangle-\langle u,Su_\star\rangle\tag{2}\\ &=\frac{1}{2}\langle u,Su\rangle-\frac{1}{2}\langle u,Su_\star\rangle-\frac{1}{2}\langle u,Su_\star\rangle\\ &=\frac{1}{2}\langle u,S(u-u_\star)\rangle-\frac{1}{2}\langle u_\star,Su\rangle \tag{3}\\ &=\frac{1}{2}\langle u,S(u-u_\star)\rangle-\frac{1}{2}\underbrace{\left(\langle u_\star,S(u-u_\star)\rangle+\langle u_{\star},S(u_\star-u)\rangle\right)}_{=0}-\frac{1}{2}\langle u_\star, Su\rangle\\ &=\frac{1}{2}\langle u-u_\star,S(u-u_\star)\rangle-\frac{1}{2}\langle u_\star, Su_\star\rangle \end{align*} (1) follows because, since we are in a real inner product space, the inner product is truly symmetric. (2) follows because $f=Su_\star$. (3) follows because of symmetry of the inner product, and the fact that $S$ is both self-adjoint and linear. Finally, since $S$ is positive definite, both terms in the last line are positive, and hence the minimum is attained (but not necessarily uniquely?) when the first term is zero, i.e when $u=u_\star$.

QUESTION

Both the proof, and the initial premise, are dependent on the fact that $U$ is a real inner product space. Not only does the proof rely on the symmetry of the inner product, which is clearly violated in a complex inner product space, but the initial premise that $Qu_\star\leq Qu$ for all $u\in U$ is nonsense when $Q$ is allowed to take complex values. So, what I'm interested in is, is there any way to modify this idea to allow for complex inner product spaces? Can we somehow fix $Q$ to only take real values, or somehow perhaps talk about minimizing the absolute value $|Qu|$?

I tried my best to adjust this theorem but I was unable to reach something concrete. I also searched the internet far and wide for results that generalized to complex inner product spaces but I was unable to dig up anything. Or perhaps, is there NO way to make this work in complex spaces? That would be even more interesting. Thanks in advance for any help.

Solution 1:

$\def\<{\langle}\def\>{\rangle}\def\R{\mathbb{R}}$Here is a straightfoward approach to reduce the complex case to the real one: Since for any $u \in U$, if $u$ satisfies $Su = f$, then $\<u, f\> = \<u, Su\> \in \R$, so it suffices to focus on the following subspace (over $\R$) of $U$:$$ U_f = \{u \in U \mid \<u, f\> \in \R\}. $$ ($U_f$ is indeed a linear space over $\R$ since $\<\,·\,,\,·\,\>$ is linear in the first argument).

Now suppose $Su = f$ admits a unique solution $u^* \in U_f$. For any $u \in U_f \setminus \{u^*\}$, note that $\<u, f\> \in \R$ implies that$$ \<f, u\> = \overline{\<u, f\>} = \<u, f\>,\quad \<u^*, Su\> = \<u, Su^*\> = \<u, f\>, $$ thus\begin{align*} &\mathrel{\phantom=} \<u - u^*, S(u - u^*)\>\\ &= \<u, Su\> - \<u, Su^*\> - \<u^*, Su\> + \<u^*, Su^*\>\\ &= \<u, Su\> - 2\<u, f\> + \<u^*, f\>\\ &= (\<u, Su\> - 2\<f, u\>) - (\<u^*, f\> - 2\<f, u^*\>)\\ &= (\<u, Su\> - 2\<f, u\>) - (\<u^*, Su^*\> - 2\<f, u^*\>)\\ &= 2(Qu - Qu^*), \end{align*} and$$ Qu - Qu^* = \frac{1}{2} \<u - u^*, S(u - u^*)\> > 0. $$