Coordinate free computation of the second derivative of a functional [closed]

Question

Let $F(g(f))$ be a functional that sends functions of a vector variable (from $n$-dimensional vector space) to $\mathbb R$.

$g$ is some function of scalar valued functions $f$.

I'm interested in a coordinate free calculation of the first and second derivatives of functionals on the spaces of real valued functions of a vector variable, where the vector space dimension can be more than 3.

It can be shown using vector calculus that the first (order) functional derivative $\partial F$ is equal to the partial derivative of $g$ with respect to $f$ minus the divergence of the gradient of $g$.

\begin{gather*} F=\int {f(x)} ^{4} dx^{4} \\ \partial F= \frac{\partial{g}}{\partial{f}} - \nabla \cdot\nabla g(f). \end{gather*}

Divergence of the gradient should be equal to $4$, and if I plug in the above functional into the formula on Wikipedia it should work just as well for 4d and I get the result: $ \partial F=−12f^{3}$.

Can I just plug this into the formula again and get the second derivative of the functional?

Answer 1

For $$ F=\int d^4 z \ f^4 (z) \ , $$ the first functional derivative is $$ \frac{\delta F}{\delta f(x)} = \int d^4 z \ 4 f^3 (z)\ \delta^4 (x-z) =4 f^3 (x) $$ and the second functional derivatives are $$ \frac{\delta^{2} F}{\delta f(x) \delta f(y)} = 12 f^2 (x) \ \delta^4 (x-y) $$