Let us say a sequence $(x_n)_{n=1}^\infty$ in some Banach space $X$ has $S_C$ if there exist $k_1<k_2<\ldots$ such that for any $t\in \mathbb{N}$ and scalars $(a_n)_{n=1}^t$, $$\|\sum_{n=1}^t a_n x_{k_n}\|\leqslant C\Bigl(\sum_{n=1}^t |a_n|^2\Bigr)^{1/2}.$$

Let us say the Banach space $X$ has $HSP$ if every normalized, weakly null sequence in $X$ has $S_C$ for some $C>0$.

For $C>0$, let us say the Banach space $X$ has $C$-$HSP$ if every normalized, weakly null sequence in $X$ has $S_C$.

Is there a reflexive Banach space $X$ such that $X$ has $HSP$ but for each $C>0$, $X$ fails $C$-$HSP$?

  • $\begingroup$ Did you try to consider the direct sum in 2-norm of reflexive spaces with the C_n - HSP , attaining C_n, with C_n tending to infinity? $\endgroup$
    – S Argyros
    May 21, 2019 at 18:17
  • $\begingroup$ Yes. Suppose that $\sup_n C_n=\infty$ and for each $n$, $(x^n_i)_{i=1}^\infty$ is a normalized weakly null sequence which has a $C_n+1$ upper $\ell_2$ estimate but has no subsequence with a $C_n$ upper $\ell_2$ estimate. Suppose also that $(x^n_i)_{i=1}^\infty$ lies in a subspace $E_n$ of $X$ such that there exist uniformly bounded projections $P_n:X\to E_n$ such that $E_m\subset \ker(P_n)$ for all $m\neq n$. Then, after passing to a subsequence to assume $(C_n)_{n=1}^\infty$ satisfies $\sum_{n=1}^\infty 1/\sqrt{C_n}<1$. $\endgroup$
    – user78375
    May 21, 2019 at 18:48
  • $\begingroup$ Then we can let $w_n=1/\sqrt{C_n}$ and $x_i=\sum_{n=1}^\infty w_n x^n_i$. If $b=\sup_n \|P_n\|$, then $(x_i)_{i=1}^\infty\subset B_X$ and for each $n$, we can show that no subsequence has a $w_nC_n/b$ upper $\ell_2$ estimate. Since $(x_i)_{i=1}^\infty\subset B_X$ is weakly null, this shows that $X$ cannot have $HSP$. At the moment, without the assumption that we can project from $(x_i)_{i=1}^\infty$ onto $(w_nx^n_i)_{i=1}^\infty$ by $P_n$, I cannot see how to show that $(x_i)_{i=1}^\infty$ does not have good $\ell_2$ upper estimates. $\endgroup$
    – user78375
    May 21, 2019 at 18:54
  • 1
    $\begingroup$ ok I agree with you. Your question reminds me of a recent result I have with Pavlos Motakis that there exists a reflexive Banach space with unique $\ell_1$ spreading model but no subspace with a uniformly unique $\ell_1$ spreading model. Probably a 2-convexification of that example could lead to a strong answer to your question. The paper is posted to arXiv. $\endgroup$
    – S Argyros
    May 21, 2019 at 19:06
  • $\begingroup$ When I say that "I agree with you" I think that it is clear. I also agree with the last remark made by "user 19871987". Hence it is not clear to me what is the answer. The paper I mentioned is that quoted by "LSpice" $\endgroup$
    – S Argyros
    May 21, 2019 at 19:38

1 Answer 1


No, and the reflexivity plays no role. This is actually a theorem of Knaust and Odell.

  • $\begingroup$ I have also found that Freeman generalized this to any normalized basic sequence. arxiv.org/pdf/0705.0218.pdf $\endgroup$
    – user78375
    May 23, 2019 at 16:19

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