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I am looking for an infinite-dimensional, non-reflexive Banach space $X$ that does not contain an embedding of $\ell^1$ in the following sense:

There do not exist constants $K, C > 0$ and a linear operator $T\colon \ell^1 \to X$ such that:

$$ K \|x\|_1 \leq \|T x\| \leq C \|x\|_1 \quad \text{for all } x \in \ell^1$$

I suspect that $\ell^\infty$ might be such a space. I know that every separable Banach space is isomorphic to a subspace of $\ell^\infty$. However, to confirm that $\ell^\infty$ does not contain an embedded copy of $\ell^1$, I would need to show that no such $T(\ell^1)$ is not closed in $\ell^\infty$.

Does anyone have suggestions on how to approach this? Are there other well-known examples of non-reflexive, infinite-dimensional Banach spaces that do not contain $\ell^1$? Any insights or references would be greatly appreciated!

Dean Miller
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Gleberson Antunes
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1 Answers1

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The space $\ell^{\infty}$ won't work because every separable Banach space embeds isometrically into $\ell^{\infty}$. See here for more details. Some more examples of spaces that $\ell^{1}$ embeds (even isometrically) into are $C[0,1]$ and $L^{1}(\Omega , \mu )$ whenever $L^{1}(\Omega , \mu )$ is infinite-dimensional. See here and here.

However, if $X$ is a Banach space such that $X^{*}$ is separable then $\ell^{1}$ does not embed isomorphically into $X$. This is because each subspace of $X$ has a separable dual [see here] and $\ell^{1}$ does not have a separable dual. One such example of an infinite-dimensional non-reflexive Banach space is $c_{0}$. In the discussion preceding Theorem V.5.1 of A Course in Functional Analysis by Conway, it is mentioned that there are few examples of Banach spaces with a separable dual, so it is not surprising that it is difficult to find an explicit example of such a space.

Also note that Rosenthal's $\ell^{1}$ theorem provides a characterisation of such spaces. It states that $\ell^{1}$ fails to embed isomorphically into a Banach space $X$ if and only if every bounded sequence in $X$ contains a subsequence that is Cauchy in the weak topology.

Below is the original proof in this answer. It is not as elementary as the current answer because it relies on a result [Pitt's theorem] whose proof follows from the theory of basic sequences.

An example of such a space is $c_{0}$. If there was a closed subspace $X$ of $c_{0}$ and an isomorphism $T\colon \ell^{1}\to X$, then $T^{-1}\colon X\to \ell^{1}$ would be an isomorphism. However, this is not possible because by Pitt's theorem [see here] we have that every bounded operator from a closed subspace of $c_{0}$ into $\ell^{1}$ is compact. Hence no closed subspace of $c_{0}$ is isomorphic to $\ell^{1}$. Corollary 2.1.6 of Topics in Banach Space Theory by Albiac and Kalton contains some more information in this direction, namely the following result.

Corollary 2.1.6. The spaces of the set $\{c_{0}\} \cup \{\ell_{p} : 1 \leq p < \infty \}$ are mutually nonisomorphic. In fact, if $X$ is an infinite-dimensional subspace of one of the spaces $\{c_{0}\} \cup \{\ell_{p} : 1 \leq p < \infty \}$, then $X$ is not isomorphic to a subspace of any other.

Dean Miller
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