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 $\ell_\infty$?
 A: 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 $\ell_1$.
A: 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.
A: 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 $p \to 1$? 
