Proof of Goldstine's theorem

On p.98 of these notes, or the first result to come up for the search “Goldstine” one finds a proof of theorem 7.24.

http://www.math.uwaterloo.ca/~lwmarcou/Preprints/LinearAnalysis.pdf

I don’t understand the step using Hahn-Banach. Specifically, where the linear functional is chosen as an evaluation functional at some point in $X^*$. Normally, he’s only guaranteed some linear functional in $X^{***}$ that is weak* continuous in the weak* topology on $X^{**}$. Is there some reason why the evaluation functionals make up all of these? If this is not what’s going on please let me know, or if the notes are wrong, please suggest an alternative proof. Thanks!

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This is a standard argument:

Let $Y$ be a normed linear space, and suppose $y$ is in the (continuous) weak* dual of $Y^*$. Then $U=\{y^* \in Y^* : |y(y^*)|<1\}$ is a weak* neighborhood of $0$ in $Y^*$. From the definition of the weak* topology, it follows that there is an $\epsilon>0$ and elements $y_1$, $y_2$, $\ldots\,$, $y_n$ in $Y$ so that $\{ y^*\in Y^* : |y_i(y^*)|<\epsilon,\ 1\le i\le n\}\subseteq U$. From this and the linearity of $y$, it follows that $\text{ker}(y)$ contains $\cap_{i=1}^n \text{ker}(y_i)$. But, using a basic result from linear algebra, this implies that $y$ is a linear combination of the $y_i$, and thus an element of $Y$.

So, to sum up: the weak* dual of $Y^*$ is $Y$, whenever $Y$ is a normed linear space.

This follows immediately from the fact, that the dual of $X^{**}$ with weak$^*$ topology is $X^*$ itself. So every weak$^*$ continuous linear functional on $X^{**}$ can be considered as an evaluation at some element of $X^*$.