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A related question was asked by P. Holický, M. Šmídek, and L. Zajíček in this paper (Question 3). I have learnt from P. Koszmider and P. Hájek that the answer to this (and your) question is negative. …

Sorry for digging this up but there are way easier examples. Let $X=C[0,1]$ and let $Y$ be an infinite-dimensional separable space whose no copy is complemented in $X$ (any reflexive space would do). …

In this particular case, you can proceed similarly as in Example 2.1 here. For more general results please consult a fantastic survey on unique preduals by G. Godefroy: G. Godefroy. Existence and uni …

Edit 26.04.2022. The problem is now solved; please see my other answer. This is an open problem due to Vladimir Kadets. I would not expect an easy answer here. The problem with statistical convergenc …

Edit 26.04.2022. The problem has been recently solved in ZFC. arxiv.org/abs/2203.15123. In a recent note with J. Swaczyna (arXiv:2005.04873), we proved that assuming the existence of certain large ca …

No, the space of Lipschitz functions on an infinite metric space is non-separable so it can't have a Schauder basis.

Note that in order for $C_b(X)$ to have a Schauder basis, $X$ has to be compact. Indeed, $C_b(X)$ is naturally isomorphic to $C(\beta X)$ and the latter is non-separable (because $\beta X$ is non-metr …

No. $A = \ell_\infty$ is a dual Banach algebra. Every separable complemented subspace of $A$ is finite-dimensional, so there is no way to exhaust $A$ by nicely complemented separable subspaces.

I am inclined to say yes. This is because reflexive Banach spaces have projectional resolutions of the identity, which are increasing ordinal-indexed nets of contractive, commuting projections that ex …

Yes, as we have this theorem of Nigel Kalton: Let $K$ be a compact metric space. Then $C(K)$ is an absolute 2-Lipschitz retract. Please see [1] for details. [1] Kalton, Nigel J. "Extending Lipschitz …

Sobczyk's theorem: if $Z$ is a subspace of a separable Banach space $X$ that is isomorphic to $c_0$, then $Z$ is complemented in $X$, fails for many non-separable spaces such as $X=\ell_\infty\cong C( …

The spaces you are interested in are abstractly AL-spaces and by Kakutani's representation theorem, they can be represented as $L_1(\mu)$ for some measure. In particular, they have the RNP if and only …

Ben, $$\|x\|^\prime = \inf\Big\{ \big(\sum_{i=1}^n \|x_i\|^p\big)^{1/p}\colon \sum_{i=1}^n x_i = x, x_i\in X, n\in \mathbb N \Big\}\qquad (x\in X)$$ is the standard $p$-convex renorming. The hardish …

No, for cardinality reasons. The range of a compact operator is norm-separable hence has cardinality continuum (if non-zero). It is then enough to take $X$ to have bigger cardinality, for example, $X …

Piotr Hajłasz' answer nails the problem, however, let me point out that there are easier examples of such pairs of spaces among spaces that have the same density. Suppose that $X$ fails to have a str …