How to show a normed space $X$ is reflexive if the weak topology and weak* topology on $X^*$ are equal?

Let $X$ be a normed space such that the weak topology and weak* topology on $X^*$ are the same. I want to show $X$ is reflexive.

My attempt is: Since we can use $\sigma(X^*,\widehat{X})$ to denote the weak* topology on $X^*$, $\sigma(X^*,\widehat{X})=\sigma(X^*,X^{**})$. But to get $X^{**}=\widehat{X}$ is not as easy as I think. Since $\widehat{X}\subseteq X^{**}$, we only need to show $X^{**}\subseteq\widehat{X}$. I know I can use this lemma:

Lemma: Let $X$ be a vector space and suppose that Y is a subspace of $X^*$. Then $f \in X^*$ is $\sigma(X,Y)$-continuous if and only if $f \in Y$.

But I don't know how to prove the Lemma and I am not familiar with the notation $\sigma(X,Y)$. Is there a better way or can anyone help me with the lemma? Thank you so much!

• The $\sigma(X,Y)$ topology is the weakest topology such that each $y\in Y$ is continuous on $X$. For example, $\sigma(X,X^*)$ is just the usual weak topology on $X$, while $\sigma(X^*,X)$ denotes the weak$^*$-topology on $X^*$. – Aweygan Jun 18 '18 at 20:21

Proof of Lemma: For ease of notation, I'll give a proof in the case $Y = \widehat{X}$ so that $\sigma(X^*,Y)$ is the weak$^*$-topology. It is clear that we can assume that $X$ is infinite dimensional without loss of generality.

The non-trivial point is to show that if $\phi \in X^{**} \setminus \widehat{X}$ then $\phi$ is not weak$^*$-continuous.

Suppose that $\phi \in X^{**}$ is continuous (for the weak$^*$-topology on $X^*$). Then, in particular, $U = \{f \in X^* : |\phi(f)| < 1\}$ is an open set. As a result, it contains a weak$^*$-basic open neighbourhood.

This means that there is a $g \in X^*$ and $x_1, \dots, x_k \in X$ and $\varepsilon > 0$ such that $$V = \{f \in X^*: |f(x_n)| < \varepsilon \text{ for } 1 \leq n \leq k\}$$ satisfies $g + V \subseteq U$. Notice that $\cap_{n=1}^k \ker J_X(x_n) \subseteq V$ where $J_X: X \to X^{**}$ is the canonical embedding. Since $\cap_{n=1}^k \ker J_X(x_n)$ has finite codimension, it is a nontrivial subspace of $X^*$.

Also, since $g + V \subseteq U$, if $f \neq 0$ and $f \in \cap_{n=1}^k \ker J_X(x_n)$ then

$$|\phi(g) + \phi( \lambda f)| = |\phi(g) + \lambda \phi(f)| < 1$$ for every $\lambda \in \mathbb{F}$. In particular, $f \in \ker \phi$ (otherwise we can make the inequality false by sending $\lambda \to \infty$). Since $f$ was arbitrary, $$\cap_{n=1}^k \ker J_X(x_n) \subseteq \ker \phi$$ and then it follows that $\phi \in \operatorname{span} \{J_X(x_1), \dots, J_X(x_k)\} \subseteq \widehat{X}$ which completes the proof.

Proof of Main Result: Suppose that the weak and weak$^*$ topology on $X^*$ coincide. Then if $\phi \in X^{**}$ we have that $\phi$ is continuous when $X^*$ is given the weak topology by definition of that topology.

But this then gives that $\phi$ is weak$^*$-continuous since the weak and weak$^*$-topology coincide. By the Lemma, we then have that $\phi \in \widehat{X}$ so that $X^{**} \subseteq \widehat{X}$ as desired.