Prove that if $1 \leq p \leq q \leq \infty$, then $||x||_q \leq ||x||_p$, for every $x \in\mathbb{R}^{n}$.

Question

(Hint) Consider the case $||x||_{p} = 1$.

So far I have that $1= |x_{1}|^p+|x_2|^{p}+..+|x_{n}|^{p}$, because I used the fact that $||x||_{p}=1$. Why do I have to use the positive homogeneity property for norms? So now all of the $x_{i}$ elements are all less than $1$ for it to equal $1$. So that means $P$ has to be greater than $1$ for this to make sense right? Then, from here, I don't know where to go...

Answer 1

Here is how to apply the hint: if $\|x\|_p = 1$ then each component $x_i$ is less than or equal to $1$. In particular $|x_i|^p \ge |x_i|^q$ for each $i$. This means $$ 1 = \sum_{i=1}^n |x_i|^p \ge \sum_{i=1}^n |x_i|^q = \|x\|_q^q.$$ Take the $q$ root on each side to get $\|x\|_q \le 1 = \|x\|_p$.

In general use $\|ax\|_p = a\|x\|_p$ whenever $a > 0$, and choose $a$ carefully.

Answer 2

$L_{P}$s are comparable when they are in finite spaces.

Answer 3

More generally given that $ 1 \leq p < q < \infty $ you can prove that $||a||_q \leq ||a||_p$ for a sequence of real numbers $a = (a_n)$ verifying $a \in l_p $.

1- Indeed $ \forall n \in \mathbb{N}$ we have that because of the sum of positive numbers:
$$||a||_p^{q-p}=(|a_1|^p+|a_2|^p+...+|a_n|^p+...)^{\frac{q-p}{p}} \geq |a_n|^{\frac{q-p}{p}}=|a_n|^{q-p} \Rightarrow \forall n , |a_n|^q \leq ||a||^{q-p}_p |a_n|^p$$

2- According to "1-" we can write. $$||a||_q = (|a_1|^q+|a_2|^q+...+|a_n|^q+...)^{\frac{1}{q}} \leq (||a||^{q-p}_p |a_1|^p+ ||a||^{q-p}_p |a_2|^p + ... + ||a||^{q-p}_p |a_n|^p +...)^{\frac{1}{q}} = ||a||_p^{\frac{q-p}{q}}||a||_p^{\frac{p}{q}} = ||a||_p^{\frac{q-p}{q} + \frac{p}{q}} = ||a||_p$$

And this finish the prove.