Consider $\mathbb{R}^n$ with Lebesgue measure, $n\geq2,\ 1<p<n$ and $p^*=\frac{np}{n-p}$. If $u\in L^{p^*}(\mathbb{R}^n$), show that $$\lim _ {R \rightarrow \infty} R^{-p} \int_{R<\left|x\right|<2R}\left|u(x)\right|^p dx=0$$
I have tried to solve this question but, I couldn't conclued it, let me share my solution:
It is enough to show that the integral is finite. Since $p^*=\frac{np}{n-p}$, after some calculations $p=\frac{p^*n}{p^*+n}$. Then $\int_{R<\left|x\right|<2R}\left|u(x)\right|^p dx=\int\left|u(x)\right|^{\frac{p^*n}{p^*+n}} dx\leq (\int \left|u(x)\right|^{p^*}dx )^\frac{n}{n+p^*}({\int 1dx})^\frac{p^*}{n+p^*}$ by Hölder's İnequality. We know that the first integral is finite since $u\in L^{P^*}(\mathbb{R}^n$). I think the second itegral is $R^\frac{p*}{n+p*}$. But then what is the limit?
You are almost done. So in total you get $$\lim _ {R \rightarrow \infty} R^{-p} \int_{R<\left|x\right|<2R}\left|u(x)\right|^p dx \le c\cdot\lim _ {R \rightarrow \infty}R^{\frac{p*}{n+p*}-p}$$ with $$c = \left(\int_\Bbb R \left|u(x)\right|^{p^*}dx \right)^\frac{n}{n+p^*}$$
But $\frac{p^*}{n+p^*} - p < 0$ hence $$\lim _ {R \rightarrow \infty}R^{\frac{p*}{n+p*}-p} = 0$$ and you are done.