characterization of continuous functionals in weak-star topology - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T16:34:00Z http://mathoverflow.net/feeds/question/15643 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/15643/characterization-of-continuous-functionals-in-weak-star-topology characterization of continuous functionals in weak-star topology Alex Gittens 2010-02-17T23:43:00Z 2010-02-18T00:21:14Z <p>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 $X$.</p> <p>To show the $\subseteq$ part, he says let $\varphi$ be any linear functional on $X^\star$ continuous in $\sigma(X^\star, X)$. Then <code>$\{x^\star \in X^\star : |\varphi(x^\star)| &lt; 1\} \supset \{x^\star \in X^\star : |x_j(x^\star)| &lt; \epsilon, j=1,\ldots,n\}$</code> 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?)</p> <p>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 <code>$\varphi_0 \in \text{span}\{\varphi_j\}_{j=1}^n$</code> iff $\text{ker}\varphi_0 \supset \cap_{j=1}^n \text{ker} \varphi_j$.</p> <p>How is this fact relevant?</p> http://mathoverflow.net/questions/15643/characterization-of-continuous-functionals-in-weak-star-topology/15649#15649 Answer by Bill Johnson for characterization of continuous functionals in weak-star topology Bill Johnson 2010-02-17T23:55:40Z 2010-02-18T00:21:14Z <p>Yes to your first question. As for the second, regard the $x_j$-s as linear functionals on <code>$X^*$</code>. If you have <code>$x_j(x^*)=0$</code> for all $j$, then every multiple of <code>$x^*$</code> is in the the first set you have in your second paragraph; i.e., <code>$|\phi(tx^*)| &lt;1$</code> for all $t$ and hence <code>$\phi(x^*)=0$</code>. Thus $\phi$ is a linear combination of the $x_j$-s and hence is continuous.</p> <p>Tell Joel hello for me.</p>