Differentiation under the integral sign and chain rule

Question

Let $(E,\mathcal E,\mu)$ be a probability space, $I$ be a finite nonempty set, $\gamma:(E\times I)^2\to[0,\infty)$ be measurable, $g\in\mathcal L^2(\mu)$ with $g\ge0$ and $\int g\:{\rm d}\mu=0$, $$F_1(w):=\sum_{i\in I}\int\mu({\rm d}x)w_i(x)g(x)\sum_{j\in I}\int\mu({\rm d}y)w_j(y)\gamma((x,i),(y,j))(g(y)-g(x))$$ and $$F_2(w):=\sum_{i\in I}\int\mu({\rm d}x)w_i(x)\left|\sum_{j\in I}\int\mu({\rm d}y)w_j(y)\gamma((x,i),(y,j))(g(y)-g(x))\right|^2$$ as well as $$F(w):=2F_1(w)+F_2(w)$$ for $w\in\mathcal L^2(\mu)^I$.

How can we compute the Fréchet derivative of $F$?

Using differentiation under the integral sign, I'm able to obtain $${\rm D}F_1(w)h=\sum_{i\in I}\int\mu({\rm d}x)g(x)\sum_{j\in I}\int\mu({\rm d}y)\gamma((x,i),(y,j))(g(y)-g(x))(w_i(x)h_j(y)+w_j(y)h_i(x))$$ for all $h,w\in\mathcal L^2(\mu)^I$. However, I'm struggling to deal with $F_2$. Obviously, we need to apply a chain rule.

Answer 1

The Gateaux derivative ${\rm D}_hF_2(w)$ at $w\in\mathcal L^2(\mu)^I$ in direction $h\in\mathcal L^2(\mu)^I$ should be \begin{equation}\begin{split}&{\rm D}_hF_2(w)=\sum_{i\in I}\int\mu({\rm d}x)\left[h_i(x)\left|\sum_{j\in I}\int\mu({\rm d}y)w_j(y)\gamma((x,i),(y,j))(g(y)-g(x))\right|^2\right.\\&\;\;\;\;\left.+2w_i(x)\left(\sum_{j\in I}\int\mu({\rm d}y)w_j(y)\gamma((x,i),(y,j))(g(y)-g(x))\right)\sum_{j\in I}\int\mu({\rm d}y)\gamma((x,i),(y,j)((y)-g(x))h(y)\right]\end{split}\end{equation}