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I need to evaluate the following limit which intuitively I know its equal to 0 but I can't really prove it, so I need some help:

$$\lim_{\epsilon \to 0}{\frac{F[\rho + \epsilon\rho' + \epsilon^2\rho'']-F[\rho + \epsilon\rho']}{\epsilon}}$$

where $\epsilon$ is a real number, $F$ is a functional and $\rho$, $\rho'$ and $\rho''$ are functions in some function space.

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    What is the functional $F$? Or are you supposed to prove that the limit is zero for all possible functionals?2012-09-12
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    I don't know the form of the functional, except that it is local. I started trying to prove a general identity involving functionals and I ended with this limit. Probably it needs some additional assumptions.2012-09-12
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    Is $F$ supposed to be linear?2012-09-12
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    No, linearity is not assumed.2012-09-12
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    Does $F$ obey any sublinearity properties, or are there any inequalities given for $F$?2012-09-12
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    It was a physics problem, I'm not sure which properties the functional exactly has, I have to check.2012-09-12
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    In order to say anything at all the word *functional* must be known. To many people just a functional is a function from a topological vector space into the scalar field of the space.2012-09-12
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    @ Manuel: It was a Physics problem and you do *not* know the form of the functional? How is that even possible? What are you trying to minimize? Energy?2012-09-12
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    The problem is related with the exchange correlation functional in density functional theory, at some point the term $\partial E_{xc}[\rho] / \partial P_{\mu\nu}$ arises, then it is expanded using the identity $$\frac{\partial E_{xc}[\rho]}{\partial P_{\mu\nu}} = \int{\frac{\delta E_{xc}[\rho]}{\delta \rho(\mathbf{r})} \frac{\partial \rho(\mathbf{r})}{\partial P_{\mu\nu}}}d\mathbf{r}$$ but there was no proof, trying to prove it I ended with this limit.2012-09-12
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    The exchange correlation functionals have the general form $$E_{xc}[\rho]=\int{\rho(\mathbf{r})f(\rho(\mathbf{r}),\nabla\rho(\mathbf{r}))d \mathbf{r}}$$ where $f$ is an unknown function.2012-09-12

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Here is a very, very rough sketch of a possible proof...

Let $\bar{\rho} (\epsilon) := \rho + \epsilon \rho'$. The limit can then be written in the form

$$\displaystyle\lim_{\epsilon \to 0} \frac{1}{\epsilon}\left[ F (\bar{\rho} (\epsilon) + \epsilon^2 \rho'') - F (\bar{\rho} (\epsilon))\right]$$

The "Taylor expansion" of $F$ is the following

$$F (\bar{\rho} (\epsilon) + \epsilon^2 \rho'') = F (\bar{\rho} (\epsilon)) + \langle \nabla F (\bar{\rho} (\epsilon)), \epsilon^2 \rho''\rangle + \omicron (\epsilon^4)$$

where $\nabla F (\bar{\rho} (\epsilon))$ is the "functional gradient" of $F$. Then, we have that

$$\displaystyle\lim_{\epsilon \to 0} \frac{1}{\epsilon}\left[ F (\bar{\rho} (\epsilon) + \epsilon^2 \rho'') - F (\bar{\rho} (\epsilon))\right] = \lim_{\epsilon \to 0} \left[\frac{1}{\epsilon} \langle \nabla F (\bar{\rho} (\epsilon)), \epsilon^2 \rho''\rangle + \omicron (\epsilon^3)\right]$$

If we can show that

$$\frac{1}{\epsilon} \langle \nabla F (\bar{\rho} (\epsilon)), \epsilon^2 \rho''\rangle = \langle \nabla F (\bar{\rho} (\epsilon)), \epsilon \rho''\rangle$$

then the limit becomes

$$\displaystyle\lim_{\epsilon \to 0} \frac{1}{\epsilon}\left[ F (\bar{\rho} (\epsilon) + \epsilon^2 \rho'') - F (\bar{\rho} (\epsilon))\right] = \lim_{\epsilon \to 0} \left[\langle \nabla F (\bar{\rho} (\epsilon)), \epsilon \rho''\rangle + \omicron (\epsilon^3)\right] = 0$$

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    This is ok for my purposes, I also figured some kind of Taylor expansion for functionals but I didn't know if something like that really existed. Can you point me to literature where I can find the theory behind these expansions? Thank you.2012-09-12
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    @Manuel: This is a good introductory overview of calculus of variations: http://www.math.umn.edu/~olver/am_/cvz.pdf2012-09-12