Let $E$ be a Banach space, let $e_{n}\in E$ and $g_{n}\in E^{*}$ be biorthogonal basic sequences (i.e. $\left<e_n,g_m\right>=\delta_{mn}$ ). Moreover, both of these sequences are weakly null. (note that existence of these sequences is equivalent to negation of Dunford-Pettis property)

Can we always find $\alpha>0$, an infinitely dimensional subspace $H\subset E$ and a weakly compact $D\subset E^{*}$, such that $\sup\limits_{d\in D} |\left<h,d\right>|\ge \alpha \|h\|$, for every $h\in H$?

(note that this implies reflexivity of $H$)


1 Answer 1


I think I've found a counterexample in the literature. I would really appreciate if somebody verified that I didn't get confused about the terminology of Orlicz spaces.

  1. Recall that a Banach space $E$ has Dunford-Pettis property if for every Banach space $F$ any weakly compact operator $T:E\to F$ operator is completely continuous, i.e. maps weakly compact sets (weakly convergent sequences) into compact sets (convergent sequences). As was mentioned in the question, existence of biorthogonal weakly null basic sequences is equivalent to negation of Dunforf-Pettis property. This is Theorem 3.8 in the paper Ghenciu, Lewis - Almost Weakly Compact Operators.

  2. The negation of the desired property is equivalent to the fact that every weakly compact operator is strictly singular, i.e. not bounded from below on any infinitely dimensional subspace of $E$. Indeed, if $T:E\to F$ is weakly compact and bounded from below on $H$, then we can take $D=T^*B_{F^*}$. Conversely, if $D$ exists make it convex and balanced (weak compactness will be preserved due to Krein), generate a seminorm $\rho_{D}$ on $E$, take the quotient $F=E/ D^\bot$ (with the induced norm), and then the corresponding quotient $Q$ map is not strictly singular (due to assumption on $D$), and weakly compact since $Q^*B_{F^*}=D$.

  3. Note that if $T$ is weakly compact and completely continuous, then it is strictly singular. Indeed, if $\left.T\right|_{H}$ is bounded from below, then $Id_H=\left.T\right|_{H}^{-1}\left.T\right|_{H}$ is weakly compact and completely continuous; then $Id_H^2=Id_H$ is compact, from where $\dim H<\infty$. Hence, Dunford-Pettis property implies the property in 2, and in order to show that they are not equivalent it is enough to present an example of a Banach space without Dunford Pettis property, but such that every weakly compact operator from it is strictly singular.

  4. Such counterexample can be found in the papers Lefèvre - When strict singularity of operators coincides with weak compactness and Lefèvre, Li, Queffèlec, Rodriguez-Piazza - Weak compactness and Orlicz spaces. Namely, every weakly compact operator from the Morse-Transue space $M^{e^{t^3}-1}$ is strictly singular (see Proposition 2.7 in the first paper), but this space does not have Dunford Pettis property (see the proof of Proposition 7 in the second paper: it works for $e^{t^3}-1\ge t^3$).

The unsatisfying thing about all of this is that nothing is revealed about biorthogonal basic sequences. Are they shrinking at least? Or perhaps have a shrinking subsequence?

  • $\begingroup$ In your proof of the "conversely" part of (2) you implicitly assume that $D$ is the unit ball of a subspace of $E^*$. $\endgroup$ May 24, 2021 at 23:51
  • $\begingroup$ @BillJohnson i think i should have specified that $\|\cdot\|_D$ is a semi-norm on $E$, rather than $E^*$. I changed the notation for it. $\endgroup$
    – erz
    May 25, 2021 at 5:34

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