functional derivative of square function I have a function that looks like
$J[f]=\int L[x,f(x)]dx$, so its derivative is
$\frac{\delta J}{\delta f(x)}=\frac{\partial L}{\partial f}$.
Then how about the derivative of $J^2[f]=\{\int L[x,f(x)]dx\}^2$? 
Is it simply $\frac{\delta (J^2)}{\delta f(x)}=2J\frac{\delta J}{\delta f(x)}$?
Is there a proof of this?
Thanks.
 A: Consider a function $J:U\to\mathbb C$ where $U\subset V$ is an open subset in a complete normed vector space $V$.
$J$ is called differentiable at a point $v\in U$ if there exists linear function $dJ_v:V\to\mathbb C$ so that
$$\lim_{x\to v}\frac{J(v+x)-J(v)-dJ_v(x)}{\|x\|}=0$$
As an example consider $V=C([0,1])$, continuous functions from $[0,1]$ to $\mathbb C$ with supremum norm. $J: C([0,1])\to \mathbb C$, with $f\mapsto \int L[x,f(x)]\,dx$ where $L$ is differentiable in the second component, then
$$F(h):=h\mapsto \int(\partial_2L)[x,f(x)] h(x)\,dx$$
is a linear map and
$$\frac{J(f+h)-J(h)-F(h)}{\|h\|}=\int \left(\frac{L[x,f(x)+h(x)]-L[x,f(x)]}{h(x)}\frac{h(x)}{\|h\|}-(\partial_2L)[x,f(x)]\frac{h(x)}{\|h\|}\right)\,dx$$
Provided $h(x)$ has no zeros (this detail is not so important, but the way to go around it would be inelegant here). Taking the limi $h\to0$ in sup-norm is the same as $h\to0$ uniformly so you can pull it into the integral and you find the term inside the integral will become:
$$\lim_{h\to0}\left(\frac{L[x,f(x)+h(x)]-L[x,f(x)]}{h(x)}-(\partial_2L)[x,f(x)]\right)\frac{h(x)}{\|h\|}\to0$$

Why did I do this? I did it in order to illustrate that the functional derivative physicists always talk about is actually nothing else than the integral kernel of the normal derivative one sees in every analysis class. Since for that derivative the chain rule is well known, just apply it here to get that if $dJ_v$ is the derivative of $J:U\to\mathbb C$ at point $v$, then $2J(v)dJ_v$ is the derivative of $J^2$ at $v$.
Here $J(f)=\int L[x,f(x)]\,dx$ and $dJ_f(h)=\int (\partial_2L)[x,f(x)]h(x)\,dx$. So
$$d_f(J^2)(h)=2\int L[y,f(y)]\cdot(\partial_2L)[x,f(x)]h(x)\,dxdy$$
In physicists notation:
$$\frac{\delta J^2}{\delta f}=2\int dy\ L[y,f(y)]\ \ \frac{\partial L}{\partial f}[x,f(x)]$$
