Lagrange multipliers on Banach spaces (in Dirac notation)

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Main Question or Discussion Point

I want to prove Cauchy–Schwarz' inequality, in Dirac notation, $\left<\psi\middle|\psi\right> \left<\phi\middle|\phi\right> \geq \left|\left<\psi\middle|\phi\right>\right|^2$, using the Lagrange multiplier method, minimizing $\left|\left<\psi\middle|\phi\right>\right|^2$ subject to the constraint $\left<\phi\middle|\phi\right> - c = 0$, where $c$ is a constant.

I'm completely new to Lagrange multipliers (although the idea is perfectly clear in simpler cases like e.g. $f : \mathbf R^2 \to \mathbf R$), and the Fréchet derivative etc., and have tried to consult https://en.wikipedia.org/wiki/Lagrange_multipliers_on_Banach_spaces, but am still quite confused, conceptually.

This is my sketchy thinking thus far (trying to follow the Wikipedia exposition, adapted to my problem):

We have a Banach space $B_\phi$. We then let $f = \left|\left<\psi\middle|\phi\right>\right|^2 : B_\phi \to \mathbf C$, which we want to minimize. The constraint is given by $g = \left<\phi\middle|\phi\right> - c : B_\phi \to \mathbf C$, which is set to zero. The Wikipedia article goes on to suppose that "$u_0$" (would "$\left|\phi_0\right>$" be a logical label in my case?) is a constrained extremum of $f$, i.e. an extremum of $f$ on $g^{-1}(0) = \big\{\left|\phi\right> \in B_\phi$ $|$ $g(\left|\phi\right>) = 0 \in \mathbf C \big\} \subseteq B_\phi$. The problem is then formulated as $$Df(u_0) = \lambda \circ Dg(u_0)$$ where $\lambda$ is the Lagrange multiplier, and $D$ the Fréchet derivative. Is it a complete misconception if I write this as (given $f$ and $g$ above, and my assumption that $u_0 = \left|\phi_0\right>$) $$D \left|\left<\psi|\phi_0\right>\right|^2 = \lambda \circ D\big(\left<\phi_0|\phi_0\right> -c\big)$$?

My main questions at the moment are:

1. What are the conceptual errors above? (I guess there are plenty)
2. How do I evaluate the Fréchet derivative, e.g. $D \left|\left<\psi|\phi_0\right>\right|^2$?

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S.G. Janssens
2. How do I evaluate the Fréchet derivative, e.g. $D \left|\left<\psi|\phi_0\right>\right|^2$?
As an example, suppose that you work in the complex Hilbert space $H = \mathbb{C}$ and let $z_0 \in H$ be fixed. If I read you correctly, you would be interested in differentiability of $z \mapsto \left|\left<z |z_0 \right>\right|^2$. However, unless $z_0 = 0$ (which makes everything trivial), this map is not differentiable except at $z = 0$. (You can check this using the Cauchy-Riemann equations.)

For the case of a real Hilbert space, it would be different. It is my impression that - since complex differentiability is such a strong requirement - Fréchet derivatives of operators and functionals are usually discussed in the context of real normed spaces, although I think that the basic definitions work fine in either case.

As an example, suppose that you work in the complex Hilbert space $H = \mathbb{C}$ and let $z_0 \in H$ be fixed. If I read you correctly, you would be interested in differentiability of $z \mapsto \left|\left<z |z_0 \right>\right|^2$.
Rather $z \mapsto \left|\left<z_0 |z \right>\right|^2$ if I use your example. That is, going back to my example again, for some arbitrarily chosen fixed element $\left|\psi\right>$ in the Banach space $B_\phi$, I'm interested in the map $\left|\phi\right> \mapsto \left|\left<\psi\middle|\phi\right>\right|^2$. This is what I wanted to minimize, by means of the Lagrange multiplier method (with constraint $g = \left<\phi\middle|\phi\right> - c = 0$). What I called $\left|\phi_0\right>$ was meant as "the element that minimizes $\left|\phi\right> \mapsto \left|\left<\psi\middle|\phi\right>\right|^2$ given the constraint".

Thanks for your reply anyway! I will look into it deeper when I have more time.