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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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What did you find as second derivative? –  Davide Giraudo Sep 23 '12 at 11:31
    
The formula for $D^2 \mathcal{D}(u)$ is trivially deduced by formal rules, since $\mathcal{D}$ contains only powers of $u$. My problem is to show that the formula is rigorous, and that it is a continuous function of $u$. This functional is used in Buffoni's paper (library.epfl.ch/en/theses/?nr=1035), where the smoothness is stated but not proved. –  Siminore Sep 23 '12 at 11:36
    
First, how do you see it's well-defined? –  Davide Giraudo Sep 23 '12 at 11:53
    
Buffoni shows that $\int \frac{|u(y)|^2}{|x-y|}dy$ is bounded (in $x$). But it can also be seen as a consequence of some Young inequality for convolutions. –  Siminore Sep 23 '12 at 12:00
    
In this case, we can rigorously compute the first derivative, expanding $(u(x)+h(x))^2$ and the same for $y$. Then I guess from that we can prove the formula for second derivatives. –  Davide Giraudo Sep 23 '12 at 12:04
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1 Answer

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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I don't see why it's $o(\cdots)$ instead of $O(\dots)$. –  user31373 Oct 1 '12 at 22:56
    
When it would be big-O of 2nd order terms, which means little-O of 1st order terms. Just like for functions of real variables. The differential is the linear part of the incremental quotient (as the norm of the increment goes to zero), so whatever is left has to be a little-O of the norm of the increment. –  bartgol Oct 1 '12 at 23:16
    
My point is, if a function on a normed linear space has a modulus of continuity that's $o(h)$, then it has to be constant. –  user31373 Oct 1 '12 at 23:18
    
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. –  bartgol Oct 1 '12 at 23:27
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