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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!

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    $\begingroup$ you should conclude by saying that the Frechet derivative of $F$ at $\varphi$ is the operator $\psi \to (A \psi, \varphi) +(\varphi, A \psi)$ (and check as Calvin wrote that it is the differential of $F$ at $\varphi$) $\endgroup$ – reuns May 6 '16 at 21:12
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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.

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  • $\begingroup$ I have one more question. What happens when $A$ is unbounded? E.g. take the Hilbert space of real square integrable functions and $A$ to be the one dimensional Laplace operator, $A x= -x''$, for $x\in \mathcal{H}$. $\endgroup$ – AG1123 May 6 '16 at 22:40
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    $\begingroup$ If $A$ is unbounded then we can't get the $o(‖x‖)$ control needed so it wouldn't be Frechet differentiable. The Frechet derivative is also defined as a bounded linear map so even in the case that $F(x)=Ax$ is linear, it isn't differentiable. $\endgroup$ – Calvin Khor May 6 '16 at 23:02
  • $\begingroup$ Thanks, that's what I thought. Is there anything that can be done in this case to get something like a Frechet derivative? In any case, what I (unintentionally) derived in my original post is what is called the Gateaux differential if I am not mistaken, and the way I understand it, the unboundedness of $A$ does not cause any problem there. Do you agree with such a statement? $\endgroup$ – AG1123 May 7 '16 at 0:06
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    $\begingroup$ Yeah, thats right. It of course still wouldn't be continuous/bounded though! $\endgroup$ – Calvin Khor May 7 '16 at 9:44

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