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.
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If I'm not wrong, the derivatives of $\mathcal{D}(u)$ are given by $$ \mathcal{D}_u(u)[\delta u] = 4\int_{\mathbb{R}^3} \int_{\mathbb{R}^3} \frac{\delta u(x) u(x) |u(y)|^2}{|x-y|} \, dx \, dy $$ $$ \mathcal{D}_{u,u}(u)[\delta u,\delta v] = 4\int_{\mathbb{R}^3} \int_{\mathbb{R}^3} \frac{\delta u(x) \delta v(x) |u(y)|^2}{|x-y|} \, dx \, dy + \\\qquad\qquad\qquad 8\int_{\mathbb{R}^3} \int_{\mathbb{R}^3} \frac{\delta u(x) \delta u(x) \delta v(y) \delta u(y)}{|x-y|} \, dx \, dy $$ 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. |
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