TITLE: Smoothness of a non-local functional
QUESTION [6 upvotes]: While studying a nonlinear PDE arising from quantum mechanics, I met a statement that I cannot prove easily. Let us write $E=W^{1,2}(\mathbb{R}^3)$ for the usual Sobolev space, and define the functional 
$$
\mathcal{D}(u)= \int_{\mathbb{R}^3} \int_{\mathbb{R}^3} \frac{|u(x)|^2 |u(y)|^2}{|x-y|} \, dx \, dy \quad \text{for $u \in E$}.
$$
It is claimed without proof that $\mathcal{D} \in C^2(E)$. I think I can prove the continuity of the second derivative at zero, but I can't switch to the continuity at different points.
I would be grateful for any hint.

REPLY [2 votes]: If I'm not wrong, the derivatives of $\mathcal{D}(u)$ are given by
$$
\mathcal{D}_u(u)[\delta u] = 4\int_{\mathbb{R}^3} \int_{\mathbb{R}^3} \frac{\delta u(x) u(x) |u(y)|^2}{|x-y|} \, dx \, dy
$$
$$
\mathcal{D}_{u,u}(u)[\delta u,\delta v] = 4\int_{\mathbb{R}^3} \int_{\mathbb{R}^3} \frac{\delta u(x) \delta v(x) |u(y)|^2}{|x-y|} \, dx \, dy + \\\qquad\qquad\qquad  8\int_{\mathbb{R}^3} \int_{\mathbb{R}^3} \frac{\delta u(x) u(x) \delta v(y) u(y)}{|x-y|} \, dx \, dy
$$
From here, using the fact that $W^{1,2}(\mathbb{R}^3)\hookrightarrow L^4(\mathbb{R}^3)$ and using Cauchy-Schwarz it should follow that $\mathcal{D}_{u+h,u+h}[\delta u, \delta v]=\mathcal{D}_{u,u}[\delta u, \delta v]+o(||h||_{W^{1,2}})$, which is what you want.