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Let $E$ be a normed space over $\mathbb K$. For each $x \in E$, let $x̂ : E^* \to \mathbb K$ be defined by $x̂ (l) = l(x)$, $l \in E^*$. Then the map $\Lambda_E: E \to (E^*)^*, \Lambda_E(x) = x̂ $ is an isometric linear operator. $E$ is called $reflexive$ if $\Lambda_E(x)$ is surjective.

Why can only Banach spaces be reflexive?

Pazu
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1 Answers1

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The dual space of any Normed vector space is complete and therefore $(E^*)^*$ is a Banach space. In the case that the isometric map is surjective it therefore transfers the completeness to the space $E$.

  • Thanks. I see why it works when we have a surjective isometric map. Is it already sufficient if there exists a bounded + linear isomorphism from one normed space into another to transfer completeness? – Pazu Oct 14 '17 at 17:25
  • That depends on what an isomorphism defines for you. The definitions i know, require that the inverse operator is bounded. But i have seen one where they do not require that. It does not suffice to have a bijective bounded linear Operator. The inverse must be bounded (by the inverse map theorem). An example: $X=(C[0,1],|\cdot|\infty)$ and $Y=(C[0,1],|\cdot|_1)$. Then the Operator $T \colon X \to Y, f \mapsto f$ is clearly linear bijective and bounded. But $Y$ is not banach and the inverse operator is not bounded ($f_n = t^n$ leads to $|f_n|_1 = \frac{1}{n+1}$ but $|f_n|\infty = 1$). –  Oct 15 '17 at 06:29