How to prove $\|x\|_p\leqslant n^{(1/p-1/q)}\|x\|_q$

Question

I’d like to know how to prove $$\|x\|_p\leqslant n^{(1/p-1/q)}\|x\|_q$$ for $x\in \Bbb{R}^n$ and $1\leqslant p<q$, where $$\|x\|_p=\left(\sum^n_{i=1}|x_i|^p\right)^{1/p}.$$ I think it needs Hölder’s inequality, but I’m not sure how to do it exactly.

Answer 1

I am going to assume that $x\in\mathbb R^n$ and we are trying to compare its $\ell$-norms. Well, we really can use Holder's inequality.

Start with $$\sum_{i=1}^n|x_i|^p=\sum_{i=1}^n(|x_i|^p)\cdot 1$$ and then apply Holder's with exponents $\alpha=q/p$ and $\beta=q/(q-p)$ to get $$\sum_{i=1}^n|x_i|^p\le \left(\sum_{i=1}^n(|x_i|^p)^\alpha\right)^{1/\alpha}\times \left(\sum_{j=1}^n1^\beta\right)^{1/\beta}=\left(\sum_{i=1}^n|x_i|^q\right)^{p/q}n^{(q-p)/q}$$ Then take the $p$-th root on both sides to get $$\|x\|_p\le n^{(q-p)/pq}\|x\|_q=n^{1/p-1/q}\|x\|_q$$

Answer 2

As I suspected, this is just Jensen's. Indeed, \begin{align}\|x\|_p&=\left(\sum x_i^p\right)^{1/p}\\&=n^{1/p}\left(\frac1n\sum \left(x_i^q\right)^{p/q}\right)^{1/p}\\&\leq n^{1/p}\left(\left(\frac1n\sum x_i^q\right)^{p/q}\right)^{1/p}&\text{Jensen's applied to }f(x)=x^{p/q}\\&\leq \frac{n^{1/p}}{n^{1/q}}\left(\sum x_i^q\right)^{1/q}\\&=n^{1/p-1/q}\|x\|_q\end{align}