Usually, we are interested in finding the Frechet derivative of a given functional. My problem is the opposite; to find a functional satisfying a condition given in terms of the Frechet derivative of the functional.

Let the Frechet derivative of the unknown, possibly nonlinear, functional $\Xi : L_2(0,1) \rightarrow \mathbb{R} $, be defined as the linear operator $D\Xi_f : L_2(0,1) \rightarrow \mathbb{R} $ such that

\begin{align} \lim_{h\rightarrow 0} \frac{||\Xi[f+h] -\Xi[f] - D \Xi_f[h]||_\mathbb{R}}{||h||_{L_2(0,1)}} = 0, \end{align} and satisfying a given condition \begin{align} D \Xi_f[h] = \left<\eta[f],h\right> \end{align} where $\eta : L_2(0,1) \rightarrow L_2(0,1) $ is a known functional.

Is there any systematic approach to find such a functional $\Xi$, besides trial and error?

More specifically, I am interested in situations where $\eta$ takes the form \begin{align} \eta[f](x) = \int_x^1 (2-x) \phi(f(x)) dx \end{align} and where $\phi : L_2(0,1) \rightarrow \mathbb{R} $ is a low order polynomial functional. For example $\phi(f) = f$ or $\phi(f) = f^2$.

Any suggestions on

  1. Conditions $\eta$ must satisfy for a functional $\Xi$ to exist
  2. General solution strategies or tricks
  3. Specific solutions for the specific $\eta$ provided above

will be appreciated.

  • $\begingroup$ If $f$ is a linear function (between two normed spaces), then the derivative of $f$ at any point is just $f.$ $\endgroup$ – Will M. May 3 '18 at 15:57
  • $\begingroup$ $f$ is an unknown function and can not be assumed to be linear. $\endgroup$ – Haavard May 3 '18 at 20:03
  • $\begingroup$ Read again my comment and generalise. Choose $f$ (the function you want to find) to be the linear function given by the derivative you know. $\endgroup$ – Will M. May 3 '18 at 20:05
  • $\begingroup$ The unknown functional (which I called $\Xi$) will, in general, be a non-linear operator on $f$. For $\phi=1$, the Frechet derivative $D\Xi_f[h]$ will be independent of $f$, and we can do as you suggested. $\endgroup$ – Haavard May 4 '18 at 6:42
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    $\begingroup$ I thought $\eta$ was linear. In the conditions you are stating above, your question doesn't make sense. There are a few inconsistencies. For instance, is $\eta:L^2 \to \Bbb R$? $\endgroup$ – Will M. May 5 '18 at 4:31

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