$\ell^q$ stronger than $\ell^p$

Question

I have to show the following:

Let $1\le p<q\le \infty$. Then $$\|x\|_q\le\|x\|_p,\; \forall x \in \ell^p$$ with

$$\ell^p:=\{x \in \mathbb{K}^\mathbb{N}: \|x\|_p<\infty\}$$ and $$\|x\|_p:=\left(\sum_{n \in \mathbb{N}}|x_n|^p\right)^{\frac{1}{p}}$$

I know a theorem that helps me here. I need to find a $c>0$ so $\|x\|_p\le c\|x\|_q, \forall x\in\ell^p$. This is equivalent to showing the above, but no matter what I try, I can't find such a $c$. Can someone help me?

Answer 1

You can do it directly.

First show that $\|x\|_\infty\le\|x\|_p$. Then, assuming $q<\infty$, write \begin{align*} \|x\|_q &= \left(\sum_{i=1}^\infty|x_i|^q\right)^{1/q} \\ &= \left(\sum_{i=1}^\infty|x_i|^{q-p}|x_i|^p\right)^{1/q} \\ &\le \left(\sum_{i=1}^\infty\|x\|_\infty^{q-p}|x_i|^p\right)^{1/q} \\ &= \|x\|_\infty^{(q-p)/q}\left(\sum_{i=1}^\infty|x_i|^p\right)^{1/q} \\ &= \|x\|_\infty^{(q-p)/q}\|x\|_p^{p/q} \\ &\le \|x\|_p^{(q-p)/q}\|x\|_p^{p/q} \\ &= \|x\|_p. \end{align*}

Answer 2

The following trick will be very useful.

Note that $||x||_p \leq ||x||_q \iff||tx||_p \leq ||tx||_q$ for some $t > 0$. Hence, we can assume WLOG that $||x||_p = 1$, otherwise we can scale. Furthermore, we can assume that not all entries of $x$ are zero, otherwise the inequality is trivial.

Now, $\left(\sum |x_n|^p\right)^ \frac 1p = 1$, so $\sum |x_n|^p= 1$, so this means that $|x_n| \leq 1$ for each $n$, and therefore $|x_n|^q \leq |x_n|^p$ since $q > p$. Therefore, it follows that $(\sum |x_n|^q) \leq 1$ and therefore that $\left(\sum |x_n|^q\right)^ \frac 1q \leq 1$.

Hence, we conclude that $||x||_q \leq ||x||_p$ for all $q > p$, and therefore the conclusion follows.