Let $\psi \in C^\infty_c(\mathbb R^N)$ be a test function with support iN $B(0,R)$. Is it true that the following inequality holds $$\int_{B(0,R)} \psi^2 u^{\frac{4}{1+\beta}} dx \le R^{1+\beta} \int_{B(0,R)} \psi^2 \left(\int_{\mathbb R^N} \frac{|u(x)-u(y)|^2}{|x-y|^{N+2\alpha}} dy \right) dx$$ for at least some values of $1 \le N \le 3$, $\beta \in (0,1)$, $\alpha \in (0,1)$?
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1$\begingroup$ For all test functions $\psi$? That would simply mean that $u^{4/(1+\beta)}(x) \leqslant R^{1+\beta} \int |u(x)-u(y)|^2/|x-y|^{N+2\alpha}dy$ for all $x \in B(0,R)$, which is clearly not the case. $\endgroup$– Mateusz KwaśnickiCommented Feb 27, 2021 at 12:21
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$\begingroup$ @MateuszKwaśnicki Is there a variation of this inequality which is true? $\endgroup$– user173196Commented Feb 27, 2021 at 16:33
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$\begingroup$ No idea. I would start by looking at the local case: when $\int_{B(0,R)} \psi^2 u^p \leqslant R^q \int_{B(0,R)} \psi^2 |\nabla u|^2$? This looks a bit like the Poincaré inequality, but not quite. Also, both sides scale differently if $u$ is replaced by $\lambda u$. $\endgroup$– Mateusz KwaśnickiCommented Feb 27, 2021 at 19:53
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