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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?

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  • $\begingroup$ What are you allowed to use? If you know that Sobolev inequality holds on the ball of radius 1, you can use it directly (with interpolation), and then rescale to get it for all $r> 0$. $\endgroup$ Commented Jan 20, 2014 at 11:51
  • $\begingroup$ I have to use interpolation, Sobolev and Poincaré inequality $\endgroup$
    – Sue
    Commented Jan 20, 2014 at 11:54
  • $\begingroup$ There are many, many versions of the three inequalities you stated. Please edit your original post to quote exactly what inequalities are "known" and can be used. $\endgroup$ Commented Jan 20, 2014 at 14:58
  • $\begingroup$ en.wikipedia.org/wiki/Sobolev_inequality en.wikipedia.org/wiki/Poincar%C3%A9_inequality $\endgroup$
    – Sue
    Commented Jan 20, 2014 at 15:29

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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.

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