Polar sets and weak star topology I have to prove the following:
Let X a real normed space, and I a non-empty index set. Let $ \{ x_{i}, i \in I \}$ $\subset$ X ,   $\{  \alpha_{i},\ i \in I \}$ a set of positive real numbers.
Define $ S:= \{x^{*} \in X^{*}, ||{x^{*}}|| \leq R \ and \ |x^{*}(x_{i})| \leq \alpha_{i}, \forall i \in I \} $. Show that $\forall \  x_{0} \in X$, the set $ \{x^{*}(x_{0}), \ x^{*} \in S \} $ is a bounded closed interval in the real line.
My claim was the following:
$ \forall i \in I $, let $ x_{i}^{\circ} = \{x^{*} \in X^{*}, \ such \ that \ |x^{*}(x_{i})| \leq 1 \}$ the polar of $x_{i}$. Then we know that $\alpha_{i} x_{i}^{\circ} = (x_{i}/\alpha_{i})^{\circ}$.
Now we can characterize $S= B_{R}^{\star} \  \cap \ \cap_{i \in I} (x_{i}/\alpha_{i})^{\circ}$. Thus S is $wk^{\star}$compact because $B_{R}^{\star}$ is $wk^{\star}$compact  (Banach-Alaoglu Theorem) and $\cap_{i \in I} (x_{i}/\alpha_{i})^{\circ} = (\cup_{i \in I} (x_{i}/\alpha_{i}))^{\circ}$ which is $wk^{\star}$ closed because polars are.
Thus if we take the evaluation map $\phi_{x_{0}}:S \to \mathbb{R}$ such that $\phi_{x_{0}}(x^{*})=x^{*}(x_{0})$, this is a continuous map from a compact set so its image must be compact, therefore it is bounded and closed.
Am I missing something or it sounds good?
Thanks in advance for any help! :)
 A: First observe that $S$ is closed. It follows from the fact that if $x^*_n\in S$ and $x^*_n\to y^*$ in the weak* topology then $||y^* ||\leq R$ and $|y^* (x_i ) |\leq \alpha_i $ for all $i\in I.$ So $S$ is closed in weak* topology in $X^*.$
Moreover $S$ is convex this follows from the fact that the sets $$H^{x_i}_{\alpha_i } =\{ x^* \in X^* : |x^* (x_i )|\leq \alpha_i \}$$ are convex and $$S=\{x^*\in X:||x^* ||\leq R\}\cap\bigcap_{i\in I} H^{x_i}_{\alpha_i }$$
Moreover from the above follows that the set $S$ is also bounded.
Now observe that $$\{x^* (x_0 ): x\in S \} = k_{x_0} (S)$$ where $k_{x_0 }: X^{*} \to\mathbb{R} $ is defined by $$k_{x_0 } (x^* ) =x^* (x_0 ).$$
Moreover $k_{x_0} $ is continous when $X^*$ is equiped with  weak* topology this follows from fact topology $$k_{x_0 } (\{x\in X^* : |x^* (x_0 )|<\varepsilon \} ) \subset (-\varepsilon , \varepsilon ).$$
So we prove that $S$ is closed bounded and convex set in weak* topology and hence by Banach - Alaoglu theorem it is compact in weak* topology.
The set $\{x^* (x_0 ): x\in S \} = k_{x_0} (S)$ is a image of compact convex set by a continous linear functional and hence it is compact convex subset of real line thus it is a closed interval.
