Suppose $$c_{n,s}\int_{\mathbb R^n}\int_{\mathbb R^n} \frac{(u(x+y+z)-u(x+y)-u(x+z)-u(x))^2}{|y|^{n+2s}|z|^{n+2s}} dydz = C$$ for some constants $c_{n,s}$, $C$, and $s \in (0,1)$.
Is it true that $$u = Cc_{n,s}(A - |x|^2)^s$$ holds for some $A>0$?
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$\begingroup$ Of course not! Take any smooth $u_0$ with compact support. Then the integral is finite, and so $u = \lambda u_0$ for an appropriate $\lambda$ has the integral equal to $C$, but it is not of the desired form. $\endgroup$– Mateusz KwaśnickiCommented Mar 24, 2021 at 12:09
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$\begingroup$ @MateuszKwaśnicki Thank you! What if I add another assumption, for example that $(-\Delta)^s u = 1$? $\endgroup$– ZacCommented Mar 24, 2021 at 12:20
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$\begingroup$ Ah, I realise I did not read your question correctly. Let me think for a minute. $\endgroup$– Mateusz KwaśnickiCommented Mar 24, 2021 at 13:03
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$\begingroup$ @MateuszKwaśnicki Ok, thanks! Please, let me know $\endgroup$– ZacCommented Mar 25, 2021 at 6:22
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$\begingroup$ Sorry for the delay. I tried to understand what the double integral really means, and I failed. If we integrate it over $x$, this seems to give the quadratic form corresponding to $(-\Delta)^{2s}$ (with $2s$ rather than $s$ in the exponent), which seems slightly odd. I'm afraid I cannot help much here, but if I may ask, out of curiosity: how did you come up with this expression? And why do you expect the claim to be true? I think I must be missing something obvious here. $\endgroup$– Mateusz KwaśnickiCommented Mar 27, 2021 at 21:13
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