# Is there a topology that makes every basic sequence null?

Let $$E$$ be a Banach space. Let $$F$$ be the collection of all $$f\in E^*$$ such that $$\left\to 0$$, for every normalized basic sequence $$\{e_n\}$$. It is easy to see that $$F$$ is a closed subspace of $$E^*$$.

Does $$F$$ separate points of $$E$$?

Note that if $$E$$ is reflexive, then $$F=E^*$$, since every basic sequence is shrinking (which seems too strong).

In general this is not the case: if $$E=l_{\infty}(\mathbb{Z})$$, then $$F\cap l_1=\{0\}$$. Let $$f=(f_n)\in l_1$$. WLOG an infinite number of $$f_n$$ nonnegative (otherwise replace $$f$$ with $$-f$$). By rearranging the coordinates, we may assume that $$f_n\ge 0$$, when $$n>0$$.

Take the Rademacher sequence $$r_1=(...,0,0,1,-1,1,-1,...)$$, $$r_2=(...,0,0,1,1,-1,-1,1,1,...)$$, $$r_3=(...,0,0,1,1,1,1,-1,-1,-1,-1,...)$$ and so on, which is a basic sequence in $$l_\infty$$.

Let $$n$$ be such that $$f_1+...+f_n>\frac{2}{3}\sum_{n=1}^{\infty}f_n$$. Then, for any $$m$$ such that $$2^m\ge n$$, $$\left=f_1+...+f_n\pm f_{n+1}\pm f_{n+2}...>\frac{1}{3} \sum_{n=1}^{\infty}f_n\not\to 0$$.

If $$E=C[0,1]$$, taking variations of Schauder's basis shows that $$F$$ does not contain neither discrete measures, nor the Lebesgue measure.

• Normalised basic sequence = injective sequence whose image is a basis consisting of norm-1 vectors? Commented Jan 7, 2021 at 2:49
• @LSpice not sure i understand your terminology. Normalized basic sequence is a sequence of norm-1 vectors, which form a Schauder basis of its closed span.
– erz
Commented Jan 7, 2021 at 3:58

The answer is negative in every non reflexive space. If $$X$$ is non reflexive, there is a normalized basic sequence $$(z_n)$$ in X s.t. $$(z_1 - z_n)_{n=2}^\infty$$ and $$(z_1 + z_n)_{n=2}^\infty$$ are both basic sequence (necessarily semi normalized). If $$x^*$$ tends to zero along both of these basic sequences, then $$\langle x^*, z_1\rangle =0$$.

Actually, $$z_1$$ can be any unit vector, so $$F=\{0\}$$. Take a non reflexive subspace $$E_1$$ of $$E$$ that does not contain $$z_1$$ and let $$(z_n)_{n=2}^\infty$$ be a normalized type $$\ell^+$$ basic sequence in $$E_1$$.

(A basic sequence $$(y_n)$$ is type $$\ell^+$$ provided there is a constant $$\delta>0$$ s.t. whenever $$(a_n)$$ is sequence of non negative scalars, only finitely many of which are not zero, then $$\| \sum a_ny_n\| \ge \delta \sum |a_n|.$$ Non reflexivity is equivalent to containing a normalized type $$\ell^+$$ basic sequence.)

• When looking up $l^+$ basic sequences, I came across an even narrower class of $P^*$ basic sequences, whose existence is also equivalent to non-reflexivity. To be honest I don't understand why the fact that $(x_n)$ is of type $l^+$ implies that $(x_1-x_n)$ is basic, but I understand how to get there from the assumption that $(x_n)$ is of type $P^*$,
– erz
Commented Jan 8, 2021 at 1:20
• Maybe type $P^*$ is needed. I think $\ell^+$ only gives a linear functional that is bounded below away from zero on the sequence while you want it to be one at each term of the sequence. Commented Jan 8, 2021 at 13:41

This answer is supplementary to the one of Bill Johnson, to fill in some details.

A sequence $$\{e_n\}$$ in a Banach space $$E$$ is called a basic sequence of type P* if (among other equivalent definitions) $$0<\inf\|e_n\|\le\sup\|e_n\|<+\infty$$ and there is $$r>0$$ such that for any $$a_1,...,a_n\in\mathbb{R}$$ we have $$|a_1+...+a_n|\le r\|a_1e_1+...+a_ne_n \|.$$

In the paper Singer - Basic sequences and reflexivity of Banach spaces the author showed that reflexivity is equivalent to non-existence of P* basic sequences.

Let $$E$$ be non-reflexive, let $$f\in E^*$$ and let $$e\in E$$ be such that $$f(e)\ne 0$$. Since the kernel of $$f$$ is not reflexive there is a P* basic sequence $$\{e_n\}$$. Since the distance from $$e$$ to the kernel of $$f$$ is positive, both of the sequences $$\{e-e_n\}$$ and $$\{e+e_n\}$$ are bounded from above and from below.

As $$f(e)\ne 0$$, at least one of the sequences $$\left$$ and $$\left$$ does not converge to $$0$$. Hence, it is left to show that $$\{e-e_n\}$$ and $$\{e+e_n\}$$ are basic. In order to do that we need to show that there is $$K>0$$ such that for every $$a_1,...,a_n\in\mathbb{R}$$ and $$m\le n$$ we have $$\|a_1(e+e_1)+...+a_m(e+e_m)\|\le K\|a_1(e+e_1)+...+a_n(e+e_n) \|.$$ Indeed, WLOG the norm of $$E$$ is $$l_1$$ sum of $$e$$ and $$Ker(f)$$, from where

$$\|a_1(e+e_1)+...+a_m(e+e_m)\|=|a_1+...+a_m|+\|a_1e_1+...+a_me_m\|\le$$ $$\le(r+1)\|a_1e_1+...+a_me_m\|\le (r+1)L\|a_1e_1+...+a_ne_n\|\le$$ $$\le(r+1)L(\|a_1e_1+...+a_ne_n\|+|a_1+...+a_m|)=K\|a_1(e+e_1)+...+a_n(e+e_n)\|,$$

where $$K=(r+1)L$$, and $$L$$ is the basis constant of $$\{e_n\}$$.