A normed space $E$ is reflexive if and only if the weak- and weak*-topology coincide on $E'$ Let $E$ be a normed space. It is claimed in this answer that $E$ is reflexive (which is to say that the canonical embedding of $E$ into $E''$ is surjective) if and only if "the weak- and weak*-topology on $E'$ coincide".
I didn't found this result in the literature on functional analysis I'm usually consulting. So, could anyone recommend a good reference?
But more importantly: How do we need to understand this result at all? The "weak*-topology" is a topology on $E'$. So, that's fine. But the "weak-topology" is a topology on $E$. So, what does it mean that both topologies coincide on $E'$ when the latter is not even a topology on $E'$?
 A: I don't know a good reference to a proof of the equivalence you are looking but I think the proof is elementary.
Let $\mathcal{T} \subset 2^{E'}$ denote the weak topology on $E'$ and let $\mathcal{U} \subset 2^{E'}$ denote the weak$^{\ast}$ topology.
Assume first that $E$ is reflexive and aim to show that $\mathcal{U}= \mathcal{T}$. By definition, $\mathcal{T}$ is the weakest (smallest w.r.t inclusion) topology on $E'$ making all norm-continuous functionals on $E'$ continuous. Also by definition, $\mathcal{U}$ is the smallest topology making all point-evaluation functionals $\text{ev}_x:E' \rightarrow \mathbb{K}$, $x \in E$ on $E'$ continuous. Since $E$ is reflexive these sets of linear functionals on $E'$ are the same, so $\mathcal{T}= \mathcal{U}$.
Assume now that $\mathcal{T}= \mathcal{U}$ and aim to show that $E$ is reflexive. Let $\ell: E' \rightarrow \mathbb{K}$ be norm continuous. Then it is continuous w.r.t $\mathcal{T}$ and hence with respect to $\mathcal{U}$. The latter implies that the set $\{x' \in E'\,:\, |\ell(x')| < 1\}$ is weak$^{\ast}$ open.  It follows that there are $x_1, \dots, x_n \in E$ and $\varepsilon_1,\dots, \varepsilon_n > 0$ so that
$$
\bigcap_{i=1}^n{\{x' \in E' \,:\, |x'(x_i)| < \varepsilon_i\}} \subset \{x' \in E'\,:\, |\ell(x')| < 1\}.
$$
Hence
$$
\bigcap_{i=1}^n{\ker{(\text{ev}_{x_i})}} \subset \ker{\ell}.
$$
This implies that $\ell = \text{ev}_x$ for some $x$ in the linear span of the $x_i$'s . What I've just explained in the last paragraph is probably very standard: Every weak$^{\ast}$ continuous linear functional is a point-evaluation functional.
To answer your last paragraph: The weak topology can be defined on any normed vector space, in particular on dual spaces.
