1

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?

Link L
  • 717

1 Answers1

3

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.

Kavi Rama Murthy
  • 311,013
  • thanks, just to add, does $ l^1[a,b] \subseteq l^2[a,b] \subseteq l^{\infty}[a,b]$ hold for infinite sequence spaces? – Link L Nov 07 '19 at 09:38
  • 1
    @LinkL By $l^{p}[a,b]$ do you men the space of functions $f:[a,b] \to \mathbb R$ with $\sum_{t \in [a,b]} |f(t)|^{p} <\infty$. If so then YES,your inclusions are correct. – Kavi Rama Murthy Nov 07 '19 at 09:41