# characterization of continuous functionals in weak-star topology

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?

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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.
Bill, the * is interpreted as a command to begin an italicised sequence. Enclosing it in backticks (  `) should solve the problem. (This example may all go awry --sub-comments like this seem to be formatted differently from 'main' comments.) – L Spice Feb 18 '10 at 0:14