GSM 229 (A First Course in Fractional Sobolev Spaces) Exercise 2.28

Question

Let $1 \leq p_2<p_1<\infty$ (i) Given a sequence $\left\{a_n\right\}_n$ of nonnegative numbers (the book does not indicate whether the series formed by this sequence converges), prove that $$ \left(\sum_{n=1}^{\infty} a_n^{p_1}\right)^{\frac{1}{p_1}} \leq\left(\sum_{n=1}^{\infty} a_n^{p_2}\right)^{\frac{1}{p_2}} $$

Answer 1

If the series on the right diverges there is nothing to prove, so suppose it does converge and call the sequence $a=(a_n)$ so that $a\in l^{p_2}$. Write $a=||a||_{p_2}u$, with $u\in l^{p_2}, ||u||_{p_2}=1$. Note that $|x_n|\leq ||x||_{p_2}$ for all $x\in l^{p_2}$ and $n\in\mathbb{N}$, and so $|u_n|^{p_1}=|u_n|^{p_1-p_2}|u_n|^{p_2}\leq |u_n|^{p_2}$. By comparison, $u\in l^{p_1}, ||u||_{p_1}\leq1$, so $||a||_{p_1}=||a||_{p_2}||u||_{p_1}\leq||a||_{p_2}$, which is your inequality. As someone said in the comments, this gives you that, if $x\in l^{p_2}$, then $x\in l^{p_1}$.