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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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    I am worried by the continuity of the second derivative. For local terms, I often use the theory of superposition operators, but I do not know if I can adapt this strategy, since I cannot find a precise growth condition for the convolution term.2012-09-23

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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) 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.

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    Scratch what I said. It was orthogonal to your point. =) You might be right, it could be big-O. Which is actually enough for what we need to prove. Anyway, if the modulus of continuity is o(h), I think you only get to say that the function is sublinear. Not sure though...I should think about it.2012-10-01