Is $\ell_q\subset\ell_p$? What's the "biggest" and "smallest" sequence space?

Question

(By smaller and larger, I mean as subsets/supersets of complex sequence spaces.)

I have a couple of questions about sequence spaces (over $\Bbb{C}$). I was proving that $(e^{(m)})$ (the constant $0$ sequence with $1$ in the $m$-th position) was a Schauder basis for $(\ell_2,\|\cdot\|_2)$ but not for $(\ell_\infty,\|\cdot\|_\infty)$, and to prove the latter I used that $(1,1,\ldots)\in\ell_\infty$ (it isn't in any $\ell_p$ spaces).

It makes sense that $\ell_\infty$ is "larger" than $\ell_p$, in that if $x\in\ell_p$ then it is necessarily bounded. I would imagine that, more generally, $\ell_p\subset\ell_q$ if $1\le p<q\le\infty$: reasoning that if $\sum |x_n|$ converges then $\sum|x_n|^p$ converges, and then(?) $\sum|x_n|^q$ converges. It can be proved that $\|x\|_q\le\|x\|_p$.

Are there any conditions on this property, though, maybe analagous to the $L_p$ case?

Moreover, if it does hold that for $1\le p<q\le\infty$, $$\ell_1\subset\ell_p\subset\ell_q\subset\ell_\infty,$$ can we "go further"? Is there a non-trivial normed sequence space smaller than $\ell_1$ or larger than $\ell_\infty$? We could perhaps think of trivial examples, e.g. the space of the zero sequence with "a norm" $(0,\| \cdot \|)$. I cannot think of any: but is there a way to prove they don't exist?

Answer 1

Yes, this is easy to see. Let $N>>0$ be such that $(x_n)\in \ell^p$ has that $|x_n|<1$ for all $n>N$.

Then we see that iff $\infty>q\ge p\ge 1$

$$\sum_{i=N+1}^\infty |x_i|^q\le \sum_{i=N+1}^\infty |x_i|^p<\infty$$

So $\ell^p\subseteq\ell^q$ when $q\ge p$. The case $q=\infty$ is already handled, as you noted.

Answer 2

A frequently used example for a smaller sequence space is the space of null-sequences $$c_0 = \{x : \mathbb N \to \mathbb C : x(n) \to 0 \text{ as } n \to \infty\}.$$ This space is typically equipped with the $\ell^\infty$ norm and is complete w.r.t. this norm.