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Let $X$ be a Banach space and suppose that the weak topology on $X^*$ agrees with the weak* topology on $X^*$. Must $X$ be reflexive?

To prove the contrapositive, it will suffice to assume that $X$ is not reflexive and construct a sequence $\phi_n \in X^*$ such that $\phi_n(x) \rightarrow \phi(x)$ for each $x\in X$, but $\lambda(\phi_n)\not\rightarrow \lambda(\phi)$ for some bounded linear fucntional $\lambda$ on $X^*$. However, I have been unable to do so. Does anyone have any ideas?

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    $\begingroup$ This is a part of Theorem 4.2 in A Course in Functional Analysis by Conway. The key ingredient of the proof is that the closed unit ball of $X$ (denoted $B_1(X)$) is weak*-dense in $B_1(X^{**})$, for any Banach space. $\endgroup$ – user53153 Feb 4 '13 at 4:43
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    $\begingroup$ @user15464: The theorem that 5PM is referring you to is known as Goldstine’s Theorem. More precisely, if $ J: X \to X^{\ast \ast} $ is the canonical embedding, and $ \mathbb{B}(X) $ and $ \mathbb{B}(X^{\ast \ast}) $ denote the closed unit balls of $ X $ and $ X^{\ast \ast} $ respectively, then $ J[\mathbb{B}(X)] $ is a $ \sigma(X^{\ast \ast},X^{\ast}) $-dense subset of $ \mathbb{B}(X^{\ast \ast}) $. $\endgroup$ – Haskell Curry Feb 4 '13 at 6:26
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5PM and Haskell Curry pointed out that this is a corollary of Goldstine's theorem.

  1. A Banach space $X$ is reflexive if and only if its closed unit ball $B$ is weakly compact.

    Proof: Suppose $B$ is weakly compact. The canonical embedding $I\colon X \to X^{\ast\ast}$ is a homeomorphism from $X$ with the weak topology to $I(X)$ with the relative weak*-topology. By Goldstine's theorem $I(B)$ is weak*-dense in $B^{\ast\ast}$ and it is compact since $I$ is continuous. Since the weak*-topology is Hausdorff, $I(B)$ is therefore closed and thus it is all of $B^{\ast\ast}$. It follows that $I\colon X \to X^{\ast\ast}$ is surjective. The other direction is a consequence of Alaoglu's theorem.

  2. Suppose the weak and weak*-topologies on $X^\ast$ coincide. By Alaoglu's theorem the unit ball in $X^\ast$ is weak$^\ast$-compact and hence it is weakly compact, so $X^\ast$ is reflexive by 1.

  3. A Banach space $X$ is reflexive if and only if $X^\ast$ is reflexive.

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    $\begingroup$ Is that true when X is not a Banach space? In Folland's "Real Analysis" book he argues "The two coincide percisely when X is reflexive" and there's no assumption that X is complete (p. 169) $\endgroup$ – Shirly Geffen Feb 21 '16 at 10:56
  • $\begingroup$ @ShirlyGeffen: Of course an incomplete normed space cannot be reflexive (because $X^{**}$ is always complete). For a normed space $X$, the statement would be that if the weak and weak-* topologies on $X^*$ agree, then the completion of $X$ is reflexive. $\endgroup$ – Nate Eldredge Feb 6 at 19:16
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Let $J : X \longrightarrow X^{**}$ be the natural embedding of $X$ into $X^{**}.$ Suppose that weak-topology and weak$^{*}$-topology on $X^*$ coincides i.e. $\sigma (X^*, J(X)) = \sigma (X^*, X^{**}).$ Then each $\xi \in X^{**}$ being continuous with respect to the weak topology $\sigma (X^*,X^{**})$ on $X^*$ they will be continuous with respect to the weak$^{*}$-topology $\sigma(X^*,J(X))$ on $X^*.$ That means $\xi \in \text {span}\ (J(X)) = J(X).$ So we have $X^{**} \subseteq J(X),$ proving that $J(X) = X^{**}$ i.e. $X$ is a reflexive normed linear space, as required.

Note that completeness of $X$ is superfluous here. The above argument works fine for any arbitrary normed linear space. Also the above argument doesn't involve Banach-Alaoglu's theorem.

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