Extracting convergent subsequence Let $(x_n)$ be a real sequence in $l^p$ for a fixed $p>1$ but not in $l^1$.
Is there a way to extract a sequence $(x_{\phi(n)})$ which is $l^{1+\varepsilon}$ for every $\varepsilon > 0$ and still not $l^{1}$?
 A: One approach is to take at most $2^m$ elements $x_n$ in each dyadic range $2^{-m-1}\leq |x_n|\leq 2^{-m},$ for each $m\geq 0.$ The resulting $\ell^p$ norm will be at most $(\sum_{m\geq 0}2^{m(1-p)})^{1/p}$ which is finite for $p>1.$
If there are infinitely many $m$ for which we have chosen $2^m$ elements in the range $2^{-m-1}\leq |x_n|\leq 2^{-m},$ then the $\ell^1$ norm will be infinite because each range contributes at least $\tfrac12.$ And if not, then we have only thrown out a finite number of terms, which only changes the $\ell^1$ norm by a finite amount. Note this last step relies on $x_n\to 0$ which follows from the sequence being in some $\ell^p$ space.
A: Sketch: WLOG, $0\le x_k <1$ for all $k.$ Let $\epsilon>0.$ Then there is $m$ such that $\sum_{k=m}^{\infty}x_k^p < \epsilon.$ Because $(x_k)\notin l^1$ and each $x_k \in [0,1),$ there exists $n>m$ such that
$$1 < \sum_{k=m}^{n}x_k < 2.$$
It follows by Cauchy-Schwartz that
$$\sum_{k=m}^{n}x_k^{(1+p)/2} \le (\sum_{k=m}^{n}x_k)^{1/2}(\sum_{k=m}^{n}x_k^p)^{1/2} < 2^{1/2}\epsilon^{1/2}.$$
Now let $p_1 = p, p_j=(1+p_{j-1})/2, n>1.$ Note $p_j \to 1.$ By the above, we can find disjoint blocks $[m_j,n_j]$ of integers such that the $l^{p_j}$ norm over each $[m_j,n_j]$ is as small as we like, while the $l^1$ norm over each $[m_j,n_j]$ is at least $1.$ This leads to a subsequence of the form $(x_k: k \in \cup_j [m_j,n_j])$ having the desired properties.
