Reading Wojtaszczyk's Banach spaces for analysts, I'm trying to understand his proof that the space of all continuous linear functionals on $(X^\star,\sigma(X^\star, X))$ is $X$.


To show the $ \subseteq$ part, he says let $\varphi$ be any linear functional on $X^\star$ continuous in $\sigma(X^\star, X)$. Then $\{x^\star \in X^\star : |\varphi(x^\star)| < 1\} \supset \{x^\star \in X^\star : |x\_j(x^\star)| < \epsilon, j=1,\ldots,n\}$ for some $\epsilon >0$ and some $x\_1, \ldots, x\_n \in X$. (Isn't this just saying since $\varphi^{-1}((-1,1))$ is open, it contains a neighborhood of 0?)


Then--- this is where I get lost--- he says that the result follows from the fact that if $\varphi\_0, \ldots, \varphi\_n$ are linear forms on a linear space $X$ (without any topology), then $\varphi\_0 \in \text{span}\{\varphi\_j\}\_{j=1}^n$ iff $\text{ker}\varphi\_0 \supset \cap\_{j=1}^n \text{ker} \varphi\_j$.


How is this fact relevant?



characterization of continuous functionals in weak-star topology
Yes to your first question. As for the second, regard the $x\_j$-s as linear functionals on $X^\*$. If you have $x\_j(x^\*)=0$ for all $j$, then every multiple of $x^\*$ is in the the first set you have in your second paragraph; i.e., $|\phi(tx^\*)| <1$ for all $t$ and hence $\phi(x^\*)=0$. Thus $\phi$ is a linear combination of the $x\_j$-s and hence is continuous.


Tell Joel hello for me.


