# Non-uniform property of sequences

Let us say a sequence $$(x_n)_{n=1}^\infty$$ in some Banach space $$X$$ has $$S_C$$ if there exist $$k_1 such that for any $$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 $$C$$-$$HSP$$?

• 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? Commented May 21, 2019 at 18:17
• 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 $\sum_{n=1}^\infty 1/\sqrt{C_n}<1$.
– user78375
Commented May 21, 2019 at 18:48
• 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.
– user78375
Commented May 21, 2019 at 18:54
• 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. Commented May 21, 2019 at 19:06
• 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" Commented May 21, 2019 at 19:38

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

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