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Davis, Figiel, Johnson and Pełczyński's Factorization Theorem states that each weakly compact operator $T:X \to Y$ between Banach spaces $X$ and $Y$ factors through a reflexive Banach space $Z$. In addition, they prove that $Z$ can be constructed to have a Schauder basis if $Y$ has a shrinking basis or an unconditional basis. It was later shown that if $Y=C[0,1]$ then $Z$ can be constructed with a basis.

I am interested generally in positive and negative results regarding when $Z$ can be constructed to have a basis. For example:

Q1: If $Y=L_1[0,1]$, can $Z$ be constructed to have a basis for any weakly compact $T:X \to Y$?

Q2: Is there an example of a space $Y$ with basis and weakly compact $T:X \to Y$ so that every reflexive $Z$ which $T$ factors through does not have a basis?

I'm also interested to know if anyone is aware of other results similar to the those stated above.

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    $\begingroup$ I never thought about this nice question. My guess is that Q2 has an affirmative answer when $Y^*$ has the bounded approximation property but a negative answer in general. $\endgroup$
    – Bill Johnson
    Jun 5, 2016 at 2:25
  • $\begingroup$ Who proved the positive result when $Y=C[0,1]$? $\endgroup$
    – Bill Johnson
    Jun 5, 2016 at 2:26
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    $\begingroup$ The $Y=C[0,1]$ theorem was proved by Ghoussoub, Maurey and Schachermayer in "Slicings, Selections and their applications." (Can. J. Math). They use it to give an alternative proof of Zippin's Theorem (i.e. a space with a separable dual embedds into a space with a shrinking basis). $\endgroup$
    – Kevin Beanland
    Jun 5, 2016 at 16:34

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I think Q1 has a positive answer. To see this, review how the results are proved in DFJP. You have a weakly compact subset $W$ of $Y$ (the closure of the image of the unit ball of $X$ under a weakly compact operator) and you apply the factorization technique (which we now know is just a real interpolation method applied to the pair $(W, B_Y)$) to $W$ to obtain a larger weakly compact symmetric subset $C$ of $Y$. Consider the normed space that has $C$ as its unit ball. It is proved that $C$ is weakly compact in in this normed space and hence this normed space is reflexive.

When $Y$ has a (always Schauder) basis, instead of applying the interpolation technique to the original weakly compact set $W$, we enlarged $W$ to a bigger weakly compact set $W_1$ that is invariant for the partial sum projections associated with the basis for $Y$. Then the interpolation technique, applied to $W_1$, produces a still larger weakly compact set $C$ that is invariant for the partial sum projections associated with the basis for $Y$. From this it is easy to see that the basis for $Y$ is also a basis for the space that has $C$ as its unit ball.

For $W_1$ we used the closed convex hull of the union over $n$ of $P_n W$, where $(P_n)$ are the partial sum projections for the basis for $Y$. When the basis is shrinking, we proved that $W_1$ is weakly compact. Suppose that $Y=L_1$ and you use the Haar basis for $Y$. Even though the Haar basis is far from being shrinking, the same construction produces a weakly compact set $W_1$. To see this, recall that a weakly closed set $D$ in $L_1$ is weakly compact iff for all $\epsilon > 0$ there is $M=M_\epsilon$ s.t. $D\subset M B_{L_2} + \epsilon B_{L_1}$ (``uniform integrability") and use the fact that both $B_{L_2}$ and $B_{L_1}$ are invariant under the Haar basis projections.

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    $\begingroup$ Thanks for the answer. Combing this with what I proved with Dan a few years ago it follows that if A is an analytic (with respect to the strong operator topology) collection of weakly compact operators into $L_1$ there is a separable reflexive space $X$ that every operator in A factors through $X$. $\endgroup$
    – Kevin Beanland
    Jun 6, 2016 at 14:30
  • $\begingroup$ To clarify: for Q2 you guess that if $Y$ has a basis, $Y^*$ has the BAP and $T:X \to Y$ is weakly compact then there is a $Z$ with a basis so that $T$ factors though $Z$. $\endgroup$
    – Kevin Beanland
    Jun 6, 2016 at 14:32
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    $\begingroup$ Yes, that was one of my guesses. But thinking more about why I guessed that, I now conjecture that not both of my guesses are correct. In other words, I haven't a clue about the general case. $\endgroup$
    – Bill Johnson
    Jun 6, 2016 at 19:26

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