# For a Banach space $X$, can we find a reflexive (or weakly sequentially complete) space $Y$ such that $X\subset Y$?

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.

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

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. $$Y$$.

• A separable metric space can have cardinality the continuum, and the continuum is larger than $\aleph_1$ in some models of set theory. Feb 16 at 22:28
• @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. Feb 16 at 23:14
• @OnurOktay: regarding Bill's comment, your proof works if you just replace every instance of $\aleph_1$ by $\mathfrak{c}$. Feb 17 at 12:54
• (However, it is actually true that there is no continuous linear injective map from $c_0(\aleph_1)$ into a wsc space.) Feb 17 at 13:03
• @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 $card(K)>\mathfrak{c}$. Feb 17 at 14:32