A Banach space is reflexive if and only if its dual is reflexive

How to show that a Banach space $X$ is reflexive if and only if its dual $X'$ is reflexive?

• First show that if $X'$ is reflexive then $X''$ is reflexive. Then what can you say about $\hat X$?
– matt
Commented Jun 1, 2012 at 6:26
• @matt: what's $\hat{X}$?
– t.b.
Commented Jun 1, 2012 at 6:27
• @t.b. $\hat X$ is canonical embedding of $X$ in $X''$. Sorry I should mentioned this.
– matt
Commented Jun 1, 2012 at 6:30

I'm assuming that you have two Theorems at your disposal, easily proven:

Theorem 1: If a Banach space $X$ is reflexive then its dual space $X'$ is reflexive.

Theorem 2: A closed subspace of a reflexive Banach space is reflexive.

Claim: Let $X$ be a Banach space. If $X'$ is reflexive then $X$ is reflexive.

Proof: Suppose $X'$ is reflexive. By Theorem 1 it follows that $X''$ is reflexive. If we consider the Canonical mapping $J : X \to X''$ it follows that $J(X)$ is a subspace of $X''$. Since $X \cong J(X)$ and $X$ is a Banach space, then $J(X)$ is a Banach space and hence closed. By Theorem 2 we can conclude that $J(X)$ is reflexive. But since $X \cong J(X)$ conclude that $X$ is reflexive.

Moreover a consequence of this is that a Banach space is reflexive if and only if its dual space is reflexive.

• Why is $J(X) =X$?
– D1X
Commented Jun 26, 2016 at 18:31
• @D1X because $J$ is an isometric injection, so $X$ is isomorphic to its image under $J$. Commented Dec 4, 2016 at 21:45

It can be shown directly:

Let $J : X \to X^{**}$ and $J_*: X^* \to X^{***}$ be the canonical injections. Suppose by contradiction that $JX \subsetneq X^{**}$; using Hahn-Banach theorem, there exists $\zeta \in X^{***}$ such that $\zeta \neq 0$ and $\zeta \equiv 0$ on $JX$.

Because $X^*$ is reflexive, there exists $\theta \in X^*$ such that $\zeta = J_*\theta$. For all $x \in X$:

$$0= \langle \zeta,Jx \rangle= \langle J_*\theta,Jx \rangle = \langle Jx,\theta \rangle= \langle \theta,x \rangle$$

You deduce that $\theta=0$ and therefore $\zeta=0$: a contradiction.

• Seirios, $\zeta \in X^{***}$ and $Jx \in X^{**}$, so how exactly can an inner product $\langle \zeta ,Jx \rangle$ be defined? Secondly, there is no assumption of inner products defined on $X, X^{**}$ , etc. $X$ is merely a Banach space by hypothesis. Indeed, if an inner product were defined on $X$, then we would immediately have that $X$ is reflexive, as every Hilbert space is reflexive. Commented Oct 4, 2013 at 1:18
• $\langle \cdot, \cdot \rangle$ is not an inner product, but a duality bracket, that is: $\langle \zeta,x \rangle$ is the evaluation of $\zeta \in X^*$ at $x \in X$. Isn't a usual notation? Commented Oct 4, 2013 at 7:17
• This is a nice proof. At the beginning shouldn’t you be saying canonical injection instead of surjection? Subjectivity is what you proved here.
– gen
Commented Jun 5, 2018 at 10:05
• Where do we use that $X$ is Banach in this? Commented Jan 16 at 11:02

Really a Banach space $X$ is reflexive if and only if $X'$ is reflexive.

$$X\textrm{ is reflexive}\Longrightarrow X'\textrm{ is reflexive.}\tag{1}$$

Proof. By Banach-Alaoglu-Bourbaki theorem the closed ball $B_{X'}$ is closed w.r.t. the weak-* topology $\sigma(X',X)$. By the reflexivity of $X$ we have $\sigma(X',X'')=\sigma(X',X).$ So $B_{X'}$ is closed w.r.t. the weak topology $\sigma(X',X),$ that is $X'$ is reflexive.$\square$

$$X'\textrm{ is reflexive}\Longrightarrow X\textrm{ is reflexive.}\tag{2}$$

Proof. By hypothesis and by (1) we get that $X''$ is reflexive, and therefore even its closed vector subspace $J(X)$ is reflexive. But the canonical injection $J:X\to X''$ is an isometry so $X$ is reflexive.$\square$

• Appealing to Alaoglu is a bit of an overkill: if $X$ is reflexive then the canonical map $\iota_{X}\colon X \to X^{\ast\ast}$ is an isomorphism. Hence $\iota_{X^\ast} = ((\iota_{X})^{\ast})^{-1} = ((\iota_{X})^{-1})^\ast$ is an isomorphism.
– t.b.
Commented Jun 1, 2012 at 7:00
• Dear Theo, I haven't thought this argument, it is simpler than the one I have wrote. Thank you.
– agt
Commented Jun 1, 2012 at 7:08
• @t.b. could you expand that thought? I don't understand your argument. Commented Nov 9, 2012 at 3:21
• @Parakee: 1. Let $\iota_X\colon X \to X^{\ast\ast}$ and $\iota_{X^\ast}\colon X^\ast \to X^{\ast\ast\ast}$ be the canonical inclusions. Then we have that $(\iota_X)^\ast \circ (\iota_{X^\ast}) = \operatorname{id}_{X^\ast}$ by a direct verification. 2. If an operator is invertible then so is its adjoint $(\iota_X)^\ast$. 3. If an operator $S$ is right inverse to an invertible operator $T$ then $S$ is itself invertible and equal to the inverse: $S = T^{-1}$. 4. Apply this to $T = (\iota_X)^\ast$ and $S = \iota_{X^\ast}$.
– t.b.
Commented Nov 29, 2012 at 14:41

Here's a different, more geometric approach that comes from Folland's book, exercise 5.24

Let $$\widehat X$$, $$\widehat{X^*}$$ be the natural images of $$X$$ and $$X^*$$ in $$X^{**}$$ and $$X^{***}$$.

Define $$\widehat X^0 = \{F\in X^{***}: F(\widehat x) = 0 \text{ for all } \widehat x \in \widehat X\}$$

1) It isn't hard to show that $$\widehat{X^*} \bigcap \widehat X^0 = \{0\}$$.

2) Furthermore, $$\widehat{X^*} + \widehat X^0 = X^{***}$$. To show this, let $$f\in X^{***}$$, and define $$l \in X^*$$ by $$l(x) = f(\widehat x)$$ for all $$x\in X$$.

Then $$f(\phi) = \widehat l(\phi) + [f(\phi) - \widehat l(\phi)]$$.

Clearly $$\widehat l \in \widehat{X^*}$$, and we claim $$f - \widehat l \in \widehat X^0$$. Let $$\widehat x \in \widehat X$$. Then $$f(\widehat x) - \widehat l ( \widehat x) = f(\widehat x) - \widehat x (l) = f(\widehat x) - l(x) = 0$$

Now that 1) and 2) are verified, we prove the claim:

If $$X$$ is reflexive, then $$\widehat X^0 = \{0\}$$, and so $$X^{***} = \widehat{X^*}$$, so $$X^*$$ is reflexive.

If $$X^*$$ is reflexive, then $$X^{***} = \widehat{X^*}$$, so $$\widehat X^0 = \{0\}$$. Since $$\widehat X$$ is a closed subspace of $$X^{**}$$ (on assumption $$X$$ is Banach), if $$\widehat X$$ were a proper subspace of $$X^{**}$$, we would be able to use Hahn-Banach to construct an $$F \in X^{***}$$ such that $$F$$ is zero on $$\widehat X$$ and has ||F|| = 1. This, however, would contradict $$\widehat X^0 = \{0\}$$. So we conclude $$\widehat X = X^{**}$$.