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My question is about the proof that $l^1$ is irreflexive. I have seen multiple proofs ($L^1$ and $L^{\infty}$ are not reflexive, Dual of $l^\infty$ is not $l^1$ and more), and all proofs stop after showing that there exists $f \in (l^\infty)' \cong (l^1)''$ that is not representable as \begin{equation} f(x) = \sum_{i \in \mathbb{N}} a_i x_i \quad \text{with} \quad a \in l^1 \end{equation} like the Hahn-Banach extension of $\lim_{n \to \infty} x_n$ on $c \subseteq l^\infty$.

With the isometric, linear injection $X \to X'', x \mapsto (\phi \mapsto \phi(x))$ we get that $l^1 \subsetneq (l^1)''$, but why does it follow from this that $l^1 \not\cong (l^1)''$? After all, there are spaces that are isometrically isomorphic to proper subspaces of themselves, like $l^1 \cong X := \{ x \in l^1 | x_1 = 0 \} \subsetneq l^1$ by \begin{equation} \phi: l^1 \to X, \quad (x_1, x_2, x_3, ...) \mapsto (0, x_1, x_2, ...) \end{equation}

Feanor
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1 Answers1

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By definition a normed linear space $X$ is reflexive if the canonical injection from $X$ to the second dual is surjective.

For the definition see https://en.wikipedia.org/wiki/Reflexive_space

Martin Sleziak
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Kavi Rama Murthy
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  • So there could still be a continuous, linear bijection between $l^1$ and $(l^1)''$? – Feanor Jul 26 '20 at 11:51
  • But that is not the definition of reflexivity. It does not matter if there is an isomorphism between $X$ and $X''$. Your confusion is with the definition of reflexivity. @Feanor – Kavi Rama Murthy Jul 26 '20 at 11:53