Inequality for function in certain Sobolev space

Question

I have to prove the following inequality for a function $u$ in $H^1(\mathbb{R}^3)$: $$\int_{B_r}\vert u\vert^q\leq C\bigg(\int_{B_r}\vert\nabla u\vert^2\bigg)^a\bigg(\int_{B_r}\vert u\vert^2\bigg)^{q/2-a}+C/{r^{2a}}\bigg(\int_{B_r}\vert u\vert^2\bigg)^{q/2}$$ with $$q\in [2,6]\,\,\,\,\,,\,\,\, a=(3/4)(q-2)$$ I succed in proving it if I replace the ball with the whole space using interpolation and Sobolev inequality. Some hint for the other case?

Answer 1

For $\theta\in [0,1]$ write the Hölder inequality $$\int_{B_r}|u|^q\,dx=\int_{B_r}|u|^{(1-\theta)q}\!\cdot|u|^{\theta q}\,dx\leqslant \biggl(\int_{B_r}|u|^2\,dx\biggr)^{(1-\theta)\frac{q}{2}}\!\cdot\!\biggl(\int_{B_r}|u|^6\,dx\biggr)^{\theta\frac{q}{6}} $$ with the exponents $$s=\frac{2}{(1-\theta)q}\,,\quad s'=\frac{6}{\theta q}\,,\quad\frac{1}{s}+\frac{1}{s'}=1.$$ For a unit ball $B_1$ write the Sobolev inequality $$ \|u\|_{L_6(B_1)}\leqslant C\!\cdot\!\Bigl(\|\nabla u\|_{L_2(B_1)}+\|u\|_{L_2(B_1)}\Bigr),$$ which by virtue of dilations with coefficient $r$ immediately transforms into inequality $$ \|u\|_{L_6(B_r)}\leqslant C\!\cdot\!\Bigl(\|\nabla u\|_{L_2(B_r)}+\frac{1}{r}\cdot\|u\|_{L_2(B_r)}\Bigr)$$ with the same constant $C>0$ independent of $r$. The rest is trivial.