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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.

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    I had problems with the same part of the proof, so thanks for your help. Can you explain why $\text{ker}(y)$ contains $\cap_{i=1}^n \text{ker}(y_i)$? It is clear that $\cap_{i=1}^n \text{ker}(y_i) \subset U$.2014-08-23
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    @J.B. If $z\in\text{ker}\, (y_i)$, then $y_i(z)=0<\epsilon$. So, the containment $\cap_{i=1}^n\text{ker}\,(y_i) \subset\{y^*\in Y^*:|y_i(y^*)|<\epsilon,1\le i\le n\}\subset U$ should be clear. For your first question: let $y^*\in\cap_{i=1}^n\text{ker}\,(y_i)$. Then $y^*\in U$ and so is every scalar multiple of $y^*$ (since every scalar multiple of $y^*$ is in $\cap_{i=1}^n\text{ker}\,(y_i)$). We then have $|y(\alpha y^*)|<1$ for all scalars $\alpha$, which implies $y^*$ is in the kernel of $y$.2014-08-23
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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 $X^*$.

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    This is precisely what I wanted to see proven. Also, is this true in general when $X^*$ is given the weak* topology from $X$, then $X^*$ with weak * topology has continuous dual that is just evaluations? Or is it important that $X$ be the continuous dual of some banach space?2012-10-06
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    @Jeff: Consider locally convex spaces $X,Y$ with a bilinear form $b:X\times Y \to \mathbb{C}$. Then the weak topology $\sigma(X,Y)$ is the weakest topology on $X$ such that $\{b(\dot\, ,y)| y \in Y\}$ are the continuous linear forms on $X$. Now apply this for your $X^{**}$ and $Y=X^*$ with the bilinear evaluation form.2012-10-06
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    @Jeff: Maybe this http://en.wikipedia.org/wiki/Dual_topology article will help you.2012-10-06
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    @Jeff: You have to be careful: On $X^*$, there is a difference between $\sigma(X^*,X) $ and $\sigma (X^* ,X^**) $ iff $X$ is not reflexive. Usually the first is taken as the weak* topology, whereas the second is named as the weak topology on $X^*$ induced by $X^{**}$.2012-10-06
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    For whatever reason, I could not enter $\sigma(X^*, X^{**})$ in the last comment.2012-10-06