Let $\mathcal{H}$ be a Hilbert space and $A$ a self-adjoint operator.
With $(\, ,\, )$ denoting the inner product and $\psi\in \mathcal{H}$, I want to formally show that the Frechet derivative of the expression $F[\psi]=\left(\psi, \, A\psi \right)$ at $\varphi$ is equal to $d_\varphi F[\psi]=\left(\varphi, \, A\psi \right) + \left(A \psi, \, \varphi \right) $.
The way I show it is the following. For some $\epsilon\in \mathbb{R}$, I write
\begin{align}
F[\psi+\epsilon\varphi]=\left(\psi+\epsilon \varphi, \, A(\psi+\epsilon\varphi) \right)= \left(\psi, \, A\psi \right) +\epsilon \left(A\psi, \, \varphi \right)+\epsilon \left(\varphi, \, A\psi \right)+\epsilon^2 \left(\varphi, \, A\varphi \right).
\end{align}
Finally,
\begin{equation}
\lim_{\epsilon\rightarrow 0} \frac{F[\psi+\epsilon\varphi]-F[\psi]}{\epsilon}=\left(A\psi, \, \varphi \right)+ \left(\varphi, \, A\psi \right)= d_\varphi F[\psi].
\end{equation}
Is this sufficient for my claim or have I missed anything?
Moreover, assuming this is correct, how would I extend the above argument to something like $f(F[\psi])$, where $f$ is some function from $\mathbb{R}$ to $\mathbb{R}$. Essentially, I would like to know what would be the chain rule in this case, and what are the restrictions(if any) on the function $f$, for the Frechet derivative to exist?
Thanks in advance!
Best Answer
Its not quite correct; you have computed a directional derivative. You need that $$ |F(\phi + x) - F(\phi) - (A\phi,x)-(\phi,Ax)| = o(‖x‖_{H}) $$ Which is indeed true by Cauchy-Schwarz, $$|F(\phi + x) - F(\phi) - (A\phi,x)-(\phi,Ax)| = |(x,Ax)| \leq ‖A‖_{\text{op}} ‖x‖_H^2$$
If $f:\Bbb R→\Bbb R $ is a differentiable function, there is indeed a chain rule, and the result is
$$ \text{d}_\phi (f(F))[h]=f'(F(\phi))\text{d}_\phi F[h]$$
In fact if $F:X→ Y$ and $G:Y→ Z$ are Frechet differentiable Banach space valued functions, then their composition $GF$ satisfies the chain rule,
$$ \text{d}_\phi (GF)[h] = (\text{d}_{F(\phi)}G)[\text{d}_\phi F[h]]$$
Its proof is very similar to the one-dimensional proof.