# Inequality in the Sobolev space $H^1$

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 $u\in H^1(\mathbb{R}^3)$,$a=\frac{3}{4}(q-2)$ 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 $\theta=\frac{3}{2}\frac{q-2}{q}$. How can I go on?

• Where or how did you find it? Commented Jan 21, 2014 at 17:07
• In a paper about Navier-Stokes equation.
– user45822
Commented Jan 21, 2014 at 17:12
• Prove it on a sphere of radius 1 and then rescale Commented Jan 21, 2014 at 20:45

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 $u\in W^{1,2}(B_r)$.