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