I can’t seem to work out the inequality $(\sum |x_n|^q)^{1/q} \leq (\sum |x_n|^p)^{1/p}$ for $p \leq q$ (which I’m assuming is the way to go about it).

**Answer**

You are right @user1736.

If $0<a\leq1$ then

$$\left(\sum |a_n|\right)^a\leq \sum|a_n|^a.\tag{1}$$

Hence for $p\leq q$ we have $p/q\leq1$, and

$$\left(\sum_n |x_n|^q\right)^{1/q}=\left(\sum_n |x_n|^q\right)^{p/qp}\leq \left(\sum_n |x_n|^{q(p/q)}\right)^{1/p}=\left(\sum|x_n|^p\right)^{1/p}$$

*Edit:* How do we prove (1) (for $0<a<1$)?

Step 1. It is sufficient to prove this for finite sequences because then we may take limits.

Step 2. To prove the statement for finite sequences it is sufficient to prove

$$(x+y)^a\leq x^a+ y^a,\quad\text{ for $x,y>0$}\tag{2}$$

because the finite case is just iterations of (2).

Step 3. To prove (2) it suffice to prove $$(1+t)^a\leq 1+t^a,\quad\text{ where $0<t<1$} \tag{3}$$

Now, the derivative of the function $f(t)=1+t^a-(1+t)^a$ is given by $f'(t) = a(t^{a-1} – (1+t)^{a-1})$ and it is positive since $a>0$ and $t\mapsto t^b$ is decreasing for negative $b$. Hence, $f(t)\geq f(0)=0$ for $0<t<1$, which proves (3).

**Attribution***Source : Link , Question Author : user1736 , Answer Author : AD.*