---
tag: 27BDDC590S
kind: definition
name: Reflexive space
msc: [46B10]
defines: NN8RWPC8WG
uses: [03J5RAKB30, QKR3D65TTH]
sources:
  - url: https://en.wikipedia.org/wiki/Reflexive_space
---

## Definition

Let $X$ be a [[normed vector space|03J5RAKB30]] over $\R$ or $\C$, and let $J \colon X \to X^{**}$ be the [[canonical embedding into the bidual|QKR3D65TTH]], $(Jx)(f) = f(x)$.
The space $X$ is *reflexive* if $J$ is surjective.

## Remarks

Since $J$ is a linear isometry, $X$ is reflexive exactly when $J$ is an isometric isomorphism of $X$ onto $X^{**}$.
A reflexive space is complete, because $X^{**}$ is complete.
Example: for $1 < p < \infty$ and $1/p + 1/q = 1$, the duals satisfy $(\ell^p)^* = \ell^q$ and $(\ell^q)^* = \ell^p$; under these identifications $J$ is the identity of $\ell^p$, so $\ell^p$ is reflexive.
Every [[Hilbert space|HM2HHE3J51]] is reflexive by the [[Riesz representation theorem|TTC40N49M6]].
Non-example: $c_0$ is not reflexive, since $J$ is the inclusion $c_0 \subset \ell^\infty$, which misses the constant sequence $(1, 1, \ldots)$.
The spaces $\ell^1$, $\ell^\infty$ and $C[0, 1]$ are not reflexive either.
The definition asks for the specific map $J$ to be onto: the [[James space|DYAWGXE2GV]] is isometrically isomorphic to its bidual by some other map, but it is not reflexive.