If weak topology and weak* topology on $X^*$ agree, must $X$ be reflexive?

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?

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.