Weakly* continuous function

Question

I was struggling to understand a proof in the Brezis's book on functional analysis.The statement is as follows : let $E$ be a normed space over $\mathbb{R}$ and $\phi : E^* \to \mathbb{R}$ be weakly*-continuous. There exist $x_0 \in E$ such that $\phi(f) = <f,x_0>$, $\forall f \in E^*$. The idea is as follows.

The idea is to use the following : if $X$ is a normed space, and $g,f_1,\dots,f_k$ satisfy, $\forall x \in X$ $$[f_j(x) = 0, \forall j =1,\dots, k] \implies g(x) = 0$$ Then $g \in \mathrm{span}\{f_1,\dots,f_k\}$.

Then, to get back to the original problem, we take a neighborhood $V$ of $0_{X*}$, such that $|\phi(f)|< 1$ if $f \in V$, and V is taken as $$ V = \{f \in E^* | \; |<f,x_i>| < \epsilon, \, \forall i = 1,\dots,k\}$$ for arbitrarly choosen $x_1,\dots,x_k$ in $E$ and $\epsilon >0$. The book claims that $$[<f,x_i> = 0, \forall i =1,\dots, k] \implies \phi(f) = 0$$ but I don't see why. No other assumptions is made on $E$.

Thank you fellows for your help!

Giovanni D

Answer 1

Your last line comes straight from the definition of $\varphi(f)$ and the definition of continuity. Since $\varphi(f)$ is continuous, for every small number (here it is 1), we can find a neighborhood $V$ of $0$ with radius $\varepsilon$ such that for every $f \in V$ we have $\varphi (f) <$ than our small number. In particular, you see that if $f(x_i) = 0$ for every $i$ then the functional $f$ belongs to every neighborhood of the origin, for any arbitrary radius. Hence $\varphi (f) < \delta$ for any $\delta > 0$. Since $\delta$ is arbitrary, this implies that $\varphi (f)=0$.

Answer 2

Some background: Suppose X is a vector space and $\Phi $ is a family of linear functionals on $X$. This family defines a topology on $X$, denoted by $\mathcal T_\Phi$ such that $(X,\mathcal T_\Phi)$ is a topological vector space. The set
$$\mathcal S = \{ f^{-1} [(-ε,ε)] : ε>0, f \in \Phi \}.$$ is a sub-basis for $0 \in X$ and thus, a basic open set containing $0$ is of the form $ \bigcap_{i=1}^n f_i^{-1} [(-ε,ε)]$.

If $g: (X,\mathcal T_\Phi) \to \mathbb R$ is linear and continuous then $g \in <\Phi>= \operatorname{span} (\Phi)$.

To see this, let $V= g^{-1} [(-1,1)].$ By the continuity of $g$, $V \in \mathcal T_\Phi$ and so we can find $ε>0,n \in \mathbb N$ and $f_1,f_2,\dots, f_n \in \Phi$ such that $$ \bigcap_{i=1}^n f_i^{-1} [(-ε,ε)] \subset V.$$ In particular, $Y:= \bigcap_{i=1}^n \ker f_i \subset V$. Notice that $Y$ is a (non trivial) linear subspace of $X$. Furthermore, $Y \subset \ker G$ and thus we infer that $g \in <f_1,f_2, \dots , f_n> \subset <\Phi>$. Indeed, let $x \in Y$. For all $k \in \mathbb N$ one has that $kx \in Y \subset V$ and so $|g(kx) | <1$. In other words, $g(x) =0$.

The weak-star topology is just a special case, if we consider the space $X^*$ and $\Phi = j(X)$, where $j : X \to X^{**}$ is the canonical isometric embedding.