2
$\begingroup$

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.

$\endgroup$
6
  • 3
    $\begingroup$ What do you mean by $X \subset Y$? The existence of a continuous linear injection from $X$ to $Y$? $\endgroup$ Commented Feb 14, 2023 at 13:28
  • 2
    $\begingroup$ 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 $\endgroup$ Commented Feb 14, 2023 at 15:06
  • 1
    $\begingroup$ The given example $L^\infty(0,1) \subset L^p(0,1)$ seems to indicate that one wants a continuous inclusion. $\endgroup$ Commented Feb 15, 2023 at 8:58
  • 3
    $\begingroup$ 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
    Commented Feb 15, 2023 at 15:51
  • $\begingroup$ @JochenWengenroth Yes, I am talking about continuous linear injections. $\endgroup$
    – user92646
    Commented Feb 16, 2023 at 10:21

1 Answer 1

5
$\begingroup$

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. $Y$.

$\endgroup$
11
  • $\begingroup$ A separable metric space can have cardinality the continuum, and the continuum is larger than $\aleph_1$ in some models of set theory. $\endgroup$ Commented Feb 16, 2023 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
    Commented Feb 16, 2023 at 23:14
  • $\begingroup$ @OnurOktay: regarding Bill's comment, your proof works if you just replace every instance of $\aleph_1$ by $\mathfrak{c}$. $\endgroup$ Commented Feb 17, 2023 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.) $\endgroup$ Commented Feb 17, 2023 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
    Commented Feb 17, 2023 at 14:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.