# Properties of $\bigcap_{p > 1} \ell_p$

Consider the following space of sequences $$\left\{a=(a_n)_{n\in\mathbb{N}}:a\in\bigcap_{p>1}\ell_p, a_n\in\mathbb{R}\right\}$$ What are some of its properties? What is its relation to $\ell_1$ and $\ell_\infty$?

-
which norm do you use? –  Norbert Aug 14 '13 at 5:15
I don't know! I suppose being "normable" and with which norms is one additional question. –  user90288 Aug 14 '13 at 5:21
no it it important, otherwise your qestion makes no scense –  Norbert Aug 14 '13 at 5:23
I don't think that matters, the question seems to be asking firstly what type of sequences lie in the intersection. can worry about norming it later. For starters, $l^\infty$ is the largest space, so it can't be that one. A sequence in $l^p$ for all $p$ except 1 certainly exists though... also note these spaces are nested, so should focus near p=1 –  Evan Aug 14 '13 at 6:11
Possible motivation ... there are some Banach space theorems that hold for all $\ell_p$ with $p>1$ but not for $\ell_1$. As a result, there are some Orlicz sequence spaces between $\bigcap_{p>1}\ell_p$ and $\ell_1$ that are sometimes studied. In a similar vein, we may be interested in $\bigcup_{p>1} L_p[0,1] \subset ?? \subset L_1[0,1]$. –  GEdgar Aug 14 '13 at 13:28

Obviously, $\ell_\infty\supset\ell_p\supset\ell_1$ ($p>1$), and $\bigcap_{p>1}\ell_p\supset\ell_1$, but $\bigcap_{p>1}\ell_p\ne\ell_1$, because the sequence $x_k=\frac{1}{k}$ belongs to each $\ell_p$ ($p>1$), but not to $\ell_1$.

-
Another fun game would be to come up with a sequence $(x_k)$ which is in $c_0$ (i.e. $\lim_{k \to \infty} x_k = 0$) but is not in $\ell_p$ for any $p < \infty$. The first thing I thought of was $x_k = \frac{1}{k^{1/k}}$, but that has limit $1$. Probably something else works though. –  Mike F Aug 15 '13 at 7:00
$x_k = 1/log k$ works –  Evan Aug 15 '13 at 7:13
It's strange for me that this innocent question about the intersection of $\ell_p$ generated so deep discussion. :) –  Sergei Akbarov Aug 15 '13 at 7:18
@Evan: Whoops my last (now deleted) comment was a blooper. Yes that works. Interesting to note that $\sum_{n=1}^\infty (1/\log(n))^n < \infty$ even though $\sum_{n=1}^\infty (1/\log(n))^p = \infty$ for every fixed $p$. –  Mike F Aug 15 '13 at 7:25

Sergei Akbarov showed that $E:=\bigcap_{p>1}\ell_p$ strictly contains $\ell_1$. Actually, we can associate to each element $x$ of $E$ the non-decreasing sequence $s(x):=\left(\lVert x\rVert_{\ell_{1+n^{-1}}},n\geqslant 1\right)$.

An element of $E$ is in $\ell_1$ if and only if $s(x)$ is bounded.

-
Perhaps, it is easier to see this if you replace the sequence by the function $s(p)=||x||_{\ell_p}$, where $p>1$ (or $1<p<2$). –  Sergei Akbarov Aug 14 '13 at 17:20

Another way to look is to consider weak $l^p$ spaces http://en.wikipedia.org/wiki/Lp_space#Weak_Lp

Note $\#\{i:\ |x(i)| > t\} \leq \frac{\|x\|_p^p}{t^p}$ by Markov's inequality. In this case $\#$ denotes cardinality. Then if we define an alternative norm by $\|x\|_{p,w} \leq \sup_t t (\#\{|x(i)| > t\})^{1/p}$, then $l^p$ is contained in weak $l^p$, let's call it $l^{p,w}$. The reverse is not true: the sequence $x_n = 1/n$ is in weak $l^1$ but not $l^1$.

So now there is another question to ask: If $l^1$ is not in the intersection, what about weak $l^1$?

Edit 1: Can show weak $l^1$ is in the intersection: Suppose $\#\{i: |x(i)|>t\} \leq C/t$.

$\sum_i |x_i|^p < \|x\|_\infty^p |\#\{|x_i| > \|x\|_\infty/2 \}| + \frac{\|x\|_\infty^p}{2^p} |\#\{|x_i| > \|x\|_\infty/4\}| + \ldots \leq 2\|x\|_\infty^{p-1} \sum_{k\geq 0} 2^{-k(p-1)} < \infty$ for $p>1$. The idea is that weak $l^1$ sequences are necessarily bounded (check definition), and we can group the sum into pieces that cut the sequence values in dyadic ranges (powers of 2).

So now, is the intersection equal to weak $l^1$?

Edit 2: Wellp, this is negative. Consider the sequence where $2^{-k}$ appears $k 2^k$ times. This won't be in weak $l^1$, but is in all the $l^p$ spaces. Fun...

Edit 3: For some reason this problem is just fun to play around with. Maybe this is a possibility: Suppose as Davide suggests, we look at the norms as $p$ tends to $1$, let's use $s_x(p) = \|x\|_p$. For our common example $x_n = 1/n$, we note (roughly) that $s_x(p) \sim 1/(p-1)$ (motivation is comparison to corresponding integral). Then does this characterize the intersection, that $(p-1) s_x(p)$ be bounded as $p \to 1$?

-