# $L^p$ and $\mathscr{l}^p$ embeddings and norm inequalities

I'm trying to summarize what I know about $L^p$ spaces at the moment and have a few questions. I think that for $1\le p\le q\le \infty$ one has $$\mathscr{l}^p(\mathbb{N})\subseteq\mathscr{l}^q(\mathbb{N}) \text{ and for all }x\in\mathscr{l}^p(\mathbb{N}): \| x\|_q\le\| x \|_p$$

while for a finite measure space $(X,\mu)$ $$L^q\subseteq L^p \text{ and for all }f\in L^q:\| f\|_p\le\mu(X)^{\frac{1}{p}-\frac{1}{q}}\| f\|_q$$

My questions: 1. Are the above statements correct?

1. Are the two statements about sequence spaces also true if I consider $\mathbb{Z}$ instead of $\mathbb{N}$?

2. Are the space-inclusions 'proper' for $p<q$?

3. For the non-sequential $L^p$ spaces: What if my measure isn't finite. Is there still something I can say about embeddings and norm-inequalities? I tried Wikipedia but couldn't make much sense of what I read.

4. Is there anything else important I should know about $L^p$ spaces? I'm talking very basic stuff that may be useful for an undergraduate writing a functional analysis exam. Save inequalities, I know most of these already.

I know that's quite a lot but I tried to find answers to these questions for quite a while without much success.

• $\ell^p(\Bbb Z)\cong\ell^p(\Bbb N)$, just look at a bijection between the indexing sets. Absolute convergence means this is valid. – Adam Hughes Sep 25 '15 at 17:54
• Note: you can use \| instead of \lvert, \rvert, etc. for proper spacing – Math1000 Sep 25 '15 at 18:25
• @AdamHughes Yeah I see it now, thank you. – azureai Sep 26 '15 at 11:47
• @Math1000 Where do you think that the spacing could be improved, for example? – azureai Sep 26 '15 at 11:47
• Compare e.g. $||x||$ to $\|x\|$. – Math1000 Sep 26 '15 at 22:52

Suppose $f(x)=x^\alpha, x\in (1,\infty)$
It is easy to show that, $$f\in L^p(1,\infty)\Leftrightarrow \alpha< -\frac{1}{p}$$
Here $\mu(X)=\infty$. Take $p=1$ and $q=2$ and take $\alpha$ such that $-1<\alpha <-\frac{1}{2},$ then
$\ f \in L^2(\mu)$ but $f \notin L^1(\mu)$. So above result will not be true if measure of the space is not finite.