Does Holder's inequality require finite measure space?

Question

I read that when the domain of functions is compact, we have the following:

$$ C[a,b] \subseteq L^{\infty}[a,b] \subseteq L^2[a,b] \subseteq L^1[a,b] $$

and which uses Holder's inequality for the proof..

However, if it is a sequence space, I think we have:

$$ l^1[a,b] \subseteq l^2[a,b] \subseteq l^{\infty}[a,b] ? $$

for example, the sequence $(1/1,1/2,1/3,1/4,...)$ is in $l^{\infty}$ but not in $l^1$...

But a sequence is a function from $\mathbb{N} \rightarrow \mathbb{R}$ right? In that case, why doesn't Holder inequality apply to sequence spaces $l^{1},l^{2},...l^{\infty}$ s.t. we have a reverse order from those spaces in $L^p$ for $p \geq 1$? Does this mean that Holder's inequality require finite measure spaces?

Answer 1

Holder's inequlaity: $|\int fg d\mu| \leq (\int |f|^{p} d\mu)^{1/p}(\int |g|^{q} d\mu)^{1/q}$ ($\frac 1 p +\frac 1 q=1$) is valid for any measure space. However if we take $g =1$ when the measure is an infinite measure the inequality becomes useless since $1$ is not in any $L^{p}$. That is why we cannot prove any inclusion between $L^{P}$ spaces for a general measure.