I've found the following inequality $$\int_{B_r}\vert u\vert^q\leq C \bigg(\int_{B_r}\vert\nabla u\vert^2\bigg)^{a}\bigg(\int_{B_r}\vert u\vert ^2\bigg)^{\frac{q}{2}a}+\frac{c}{r^{2a}}\bigg(\int_{B_r}\vert u\vert ^2\bigg) ^{\frac{q}{2}}$$ for $u\in H^1(\mathbb{R}^3)$,$a=\frac{3}{4}(q2)$ and $q\in [2,6]$. Some hints to prove it? I've started using interpolation and Sobolev's inequality $$\int_{B_r}\vert u\vert^q\leq \Vert u\Vert_{L^2}^{q(1\theta)}\Vert u\Vert_{L^6}^{q \theta}\leq C \Vert u\Vert_{L^2}^{q(1\theta)}\Vert u\Vert_{H^1}^{q\theta}$$ with $\theta=\frac{3}{2}\frac{q2}{q}$. How can I go on?
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$\begingroup$ Where or how did you find it? $\endgroup$– Nate EldredgeCommented Jan 21, 2014 at 17:07

$\begingroup$ In a paper about NavierStokes equation. $\endgroup$– user45822Commented Jan 21, 2014 at 17:12

$\begingroup$ Prove it on a sphere of radius 1 and then rescale $\endgroup$– Piero D'AnconaCommented Jan 21, 2014 at 20:45
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1 Answer
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Hint. For every $r>0$, and $q\le 6$, $$ W^{1,2}(B_r)\subset L^6(B_r)\subset L^q(B_r), $$ due to Sobolev Imbedding Theorem.
In particular, there is a $c>0$, such that $$ \u\_{L^q(B_r)}^2 \le c_1 \big(\\nabla u\^2_{L^2(B_r)}+\u\^2_{L^2(B_r)}\big), $$ for all $u\in W^{1,2}(B_r)$.