# Extracting subsequences in Banach spaces, along an ultrafilter?

There are various principles in Banach space theory that allow one to pass from a given sequence of vectors $(x_n)$, to a subsequence $(x_{n_k})$ with some desired property. I'm thinking here, in particular, of the Bessaga-Pelczynski selection principle which, given a Schauder basis $(e_n)$ for a Banach space $X$, allows the passage from a normalized sequence $(x_n)$ in $X$ which converges weakly to $0$, to a subsequence $(x_{n_k})$ which is congruent to a block basic sequence.

What I am wondering is if, given a non-principal ultrafilter $\mathcal{U}$ on $\mathbb{N}$, and a sequence $(x_n)$ of unit vectors weakly converging to 0, can we find such a subsequence $(x_{n_k})$ as above so that the index set $\{n_k:k\in\mathbb{N}\}$ is in $\mathcal{U}$?

I would also be interested any related selection principles which can be done "along an ultrafilter", or reasons why one cannot hope for results like this.

• I'm a bit confused here: is the point that you have fixed your ultrafilter beforehand, and then someone hands you a sequence, and you want to have a nice subsequence that corresponds to an element of the ultrafilter? Jul 6 '16 at 22:45
• @YemonChoi Yes, that is what I'm looking for. Jul 6 '16 at 22:47
• Possibly related: mathoverflow.net/questions/111842/… Jul 7 '16 at 13:20

A (non-trivial) ultrafilter $\mathcal{U}$ on $\mathbb{N}$ is a p-point if and only if every bounded sequence $(a_n)$ of real numbers contains a convergent subsequence $(a_{n_k})$ such that $\{n_k\}\in \mathcal{U}$.

See the Introduction of [Israel J. Math. 122 (2001), 189-206 DOI: 10.1007/BF02809899] for an application of a subclass of these ultrafilters to obtain an ultrapower version of the subsequence splitting property for the Banach space $L_1(\mu)$, some discussions related with the question, and proper references.

I'll post a partial answer (it should probably be checked) to my question here, but I welcome additional answers and examples. I also sense that selective might be overkill here, and perhaps this can be done with a p-point.

My $\mathbb{N}$ is a set theorist's $\mathbb{N}$, so $\mathbb{N}=\{0,1,2,\ldots\}$ and $1=\{0\}$, $2=\{0,1\}$, etc.

Recall that an ultrafilter $\mathcal{U}$ on $\mathbb{N}$ is selective (or Ramsey) if and only if for every $2$-coloring of the pairs of $\mathbb{N}$, $\pi:[\mathbb{N}]^2\to2$, there is a homogeneous set $U\in\mathcal{U}$ for $\pi$. In addition, such ultrafilters satisfy that whenever $(U_n)$ is a decreasing sequence of sets in $\mathcal{U}$, there is a $U=\{k_0<k_1<k_2<\cdots\}\in\mathcal{U}$ (called a diagonalization) for which $U\setminus k_n\subseteq U_n$ for all $n$.

Let's call the following a "selective Ramsey's theorem for analysts", in the spirit of Francois Dorais' answer to Applications of infinite Ramsey's Theorem (on N)?.

Lemma: Let $\mathcal{U}$ be a selective ultrafilter on $\mathbb{N}$. If $(a_{i,j})_{i,j=0}^\infty$ is an infinite matrix of real numbers such that $a_i = {\displaystyle\lim_{j\to\infty} a_{i,j}}$ exists for each $i$, and $a = {\displaystyle\lim_{i\to\infty} a_i}$ exists, then there is a set $\{k_0<k_1<k_2<\cdots\}\in\mathcal{U}$ such that $a = {\displaystyle\lim_{i<j} a_{k_i,k_j}}$.

I'll omit the proof; I believe it's the same as the one for the non-selective version, except that you take homogenous sets in $\mathcal{U}$ at each step, and then diagonalize within $\mathcal{U}$ to get the desired set.

Proposition: Let $\mathcal{U}$ be a selective ultrafilter on $\mathbb{N}$, and $(e_n)$ a Schauder basis of a Banach space $X$, with basis constant $K$. If $(x_n)$ is a normalized weakly null sequence in $X$, then there is a set $V=\{n_0<n_1<n_2<\cdots\}\in\mathcal{U}$ such that $(x_{n_k})$ is congruent to some block basic sequence $(y_k)$ of $(e_n)$. Moreover, for $\varepsilon>0$, $V$ can be chosen so that $(x_{n_k})$ has basis constant $K+\varepsilon$.

Proof. This is based on the proof of the selection principle in Albiac & Kalton. For $i,j\in\mathbb{N}$, let $$a_{i,j}=\max\{\|S_jx_i\|,\|x_i-S_jx_i\|\}$$ where $S_jx=\sum_{k=0}^j e_k^*(x)e_k$ for $x\in X$. Note that $\lim_{j\to\infty}a_{i,j}=0$ for all $i$. Let $U$ be as in the Lemma above, and take $0<\nu<1/4$.

Let $U_0=U\setminus N_0\in\mathcal{U}$, where $N_0$ is such that $$\max\{\|S_jx_i\|,\|x_i-S_jx_i\|\}<\frac{\nu}{2K}$$ for all $N_0<i<j\in U$. Inductively, we define $U_{k+1}=U_k\setminus N_{k+1}\in\mathcal{U}$ so that $$\max\{\|S_jx_i\|,\|x_i-S_jx_i\|\}<\frac{\nu^{k+2}}{2K}$$ for all $N_{k+1}<i<j\in U$.

As $\mathcal{U}$ is selective, we may choose $V=\{n_0<n_1<n_2<\cdots\}\in\mathcal{U}$ such that $U\setminus n_k\subseteq U_k$ for all $k$. Let $y_k=S_{n_{k+2}}x_{n_k}-S_{n_{k+1}}x_{n_k}$, a block sequence of $(e_n)$, thus having basis constant $\leq K$. Observe, $$\|y_k-x_{n_k}\|=\|S_{n_{k+2}}x_{n_k}-S_{n_{k+1}}x_{n_k}-x_{n_k}\|<\frac{\nu^{k+1}}{K},$$ as $n_k<n_{k+1}<n_{k+2}$ are all in $U_k$. Hence $\|y_k\|>1-\frac{\nu}{K}\geq 1-\nu$, and $$2K\sum_{k=0}^\infty\frac{\|y_k-x_{n_k}\|}{\|y_k\|}<2(1-\nu)^{-1}\sum_{k=0}^\infty\nu^{k+1}=2\nu(1-\nu)^{-2}<1.$$ The result follows by the principle of small perturbations. Taking $\nu$ smaller will control the basis constant of $(x_{n_k})$. QED.