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I have a questions regarding the definition of the functional derivative. Unfortunately a lot of text books give not a proper formal definition. Wikipedia gives the following definition \begin{align} \int \frac{\delta F}{\delta\rho}(x) \phi(x) \; dx &= \lim_{\varepsilon\to 0}\frac{F[\rho+\varepsilon \phi]-F[\rho]}{\varepsilon} \\ &= \left [ \frac{d}{d\varepsilon}F[\rho+\varepsilon \phi]\right ]_{\varepsilon=0}, \end{align} with $\phi$ an arbitrary function, $M$ be a manifold of continous functions $\rho$ and $F:M\to \mathbb{R}$

If $\phi$ is arbitrary then how do I know the left integral exists? Are there no constraints on $\phi$ like it has to be integrable and in $C_c^{\infty}$?

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  • First of all, don't be fooled by the integral on the right side of the formula reported by Wikipedia for the functional derivative: not all the functional derivatives have that structure. More information on this can be find in the links given in the notes below.
  • An answer to your question. The constraint on $\phi$ is simply the fact that the functional $F$ should be defined on all points $\rho+\varepsilon \phi$ for a sufficiently small $\varepsilon\in [0,\varepsilon_0]$ for some $\varepsilon_0>0$ (including $\varepsilon_0=+\infty$): it is the structure of $F$ that implies the structure of the variation $\delta \rho=\phi$. More precisely, if $F$ is a functional defined on a (subset of a) topological vector space $X$, then $\rho+\varepsilon \phi\in X$ for all $\varepsilon$ belonging to a suitable neighborhood of $0\in\Bbb R$. And as said in the comments below, this also implies that $X$ can be only a topological manifold, i.e. a manifold that is locally isomorphic to a topological vector space.

Notes

  • For the definition of functional derivative, perhaps it would be useful to have a look at this MathOverflow Q&A, where some commonly spread misunderstanding is corrected.
  • For an example on how $\phi$ is chosen, you could also have a look at this answer where, for formally the same functional, two different kind of function spaces are used, depending on their characteristics.
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  • $\begingroup$ Hi, thank you for this nice answer, so theoretically $\phi$ doesn't have to belong to the preimage $M$ of $F$ as long as $\rho + \varepsilon \phi$ is part of it and $F$ at these points well defined? $\endgroup$ Commented Jun 27, 2021 at 10:10
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    $\begingroup$ @Matriz exactly. This is important when the domain of $X$ has not the structure of a topological vector space, but "only" that of a topological manifold: in that case $\phi$ can belong to (an appropriate neighborhood of $0$ in) the tangent space $\operatorname{TM}_\rho(X)$. $\endgroup$ Commented Jun 27, 2021 at 10:49
  • $\begingroup$ thank you very very much! That was eye-opening, have a nice day! $\endgroup$ Commented Jun 27, 2021 at 10:56
  • $\begingroup$ @Matriz you are welcome, and have a nice day! $\endgroup$ Commented Jun 27, 2021 at 10:57

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