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It could be a naive question. Probably, it is not true. However, this question makes sense in the setting of function spaces. For example, for $L_\infty (0,1)$, we have $L_p(0,1)\supset L_\infty (0,1)$ and $L_p(0,1)$ is reflexive when $p>1$. On the other hand, $L_1(0,1)$ itself is weakly sequentially complete and any rearrangement-invariant function spaces on $(0,1)$ is a subset of $L_1(0,1)$.

I am wondering whether we have such a result in the setting of general Banach spaces.

BTW: Davis, Figiel, Johnson and Pełczyński's construction gives us a smaller reflexive space.

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    $\begingroup$ What do you mean by $X \subset Y$? The existence of a continuous linear injection from $X$ to $Y$? $\endgroup$
    – Jochen Glueck
    Feb 14 at 13:28
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    $\begingroup$ If you mean that X is a (topological) subspace of Y, then this is true iff X itself is reflexive, for any closed subspace of a reflexive Banach space is reflexive $\endgroup$
    – Pietro Majer
    Feb 14 at 15:06
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    $\begingroup$ The given example $L^\infty(0,1) \subset L^p(0,1)$ seems to indicate that one wants a continuous inclusion. $\endgroup$
    – Jochen Wengenroth
    Feb 15 at 8:58
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    $\begingroup$ Just a trivial remark which might be helpful. If $X$ is separable, then it embeds isometrically into $C([0 1])$ and so the result holds (take the Lebesgue $L^2$-space). This can clearly be extended but Ì haven't checked whether to any Banach space (after embedding into a $C(K)$). $\endgroup$
    – terceira
    Feb 15 at 15:51
  • $\begingroup$ @JochenWengenroth Yes, I am talking about continuous linear injections. $\endgroup$
    – user92646
    Feb 16 at 10:21

1 Answer 1

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Let $K$ be a compact scattered space and $X=C(K)$ the space of continuous functions on $K$. We want to show that there is no injective bounded linear $T:X\to Y$ into a weakly sequentially complete (w.s.c.) Banach space $Y$ if $K$ has large(see below) cardinality.

Let $T:X\to Y$ be as above. $C(K)$ has Pelczynski property (V) and $Y$ is w.s.c., so $T$ is weakly compact. Since $C(K)$ has Dunford-Pettis property, then $T$ is completely continuous. Lastly $C(K)$ contains no copy of $\ell^1$ since $K$ is scattered. Thus, $T$ is compact.

Since $T$ is compact, its range $T(X)$ is separable, so $card(T(X))\leq\mathfrak{c}=card(\mathbb{R})$. $T$ is injective, so $card(X)=card(T(X))$. Together, we have $$ card(K)\leq card(X)=card(T(X))\leq\mathfrak{c}.$$ However, this leads to a contradiction when $card(K)>\mathfrak{c}$. For example, take $K=[0,\alpha]$ for some ordinal $\alpha$ with the order topology, where the cardinality of $\alpha$ is strictly greater than $\mathfrak{c}$.

Note (2023-02-18): Following the footsteps in this post, we can actually make a slightly more general statement. If $X$ is a Banach space with property (V), $X^*$ has Schur property, and $card(X)>\mathfrak{c}$, then there exists no injective bounded $T:X\to Y$ into a w.s.c. $Y$.

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  • $\begingroup$ A separable metric space can have cardinality the continuum, and the continuum is larger than $\aleph_1$ in some models of set theory. $\endgroup$
    – Bill Johnson
    Feb 16 at 22:28
  • $\begingroup$ @BillJohnson Professor Johnson, thank you truly for your comment. My limited knowledge of the set theory is mostly bounded by an undergraduate course from 20+ years ago. I think I must have assumed the continuum hypothesis with ZFC for $\aleph_1=card(\mathbb{R})$. I edit my post to fix my oversight. $\endgroup$
    – Onur Oktay
    Feb 16 at 23:14
  • $\begingroup$ @OnurOktay: regarding Bill's comment, your proof works if you just replace every instance of $\aleph_1$ by $\mathfrak{c}$. $\endgroup$
    – Philip Brooker
    Feb 17 at 12:54
  • $\begingroup$ (However, it is actually true that there is no continuous linear injective map from $c_0(\aleph_1)$ into a wsc space.) $\endgroup$
    – Philip Brooker
    Feb 17 at 13:03
  • $\begingroup$ @PhilipBrooker Professor Brooker, at the time I edited the post, I didn't know (until recently) if there existed an ordinal $\alpha$ (I use for the example above) with cardinality strictly greater than $\mathfrak{c}$ without CH in ZF. Actually, aside from $K=[0,\alpha]$, I could have picked any other compact scattered $K$ with $card(K)>\mathfrak{c}$. $\endgroup$
    – Onur Oktay
    Feb 17 at 14:32

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