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Let $X$ be a Banach space and $X^*$ its dual. We know that the weak* topology is the least topology that makes every $x \in X$ continuous as an evaluation functional. However, this does not imply that every weak* continuous linear functional is something in $X$, even though this happens to be true.

The question is: how can we prove this?

What have I though is:

It is enough to show that $\cap_{i=1}^{k} \ker{x_i} \subset \ker{\phi}$ for some $x_i \in X, i=1,2,...,k$

I have shown this for infinitely many $x_i$s (easy, using the weak* continuity and that 0 is always in the ker) and in order to pass to finitely many I would need some kind of compactness result (probably by using Banach-Alaoglu somehow), but I do not know how to do this.

Can anyone help?

Martin Argerami
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Nick Kolliopoulos
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3 Answers3

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Since $\phi$ is weak*-continuous, the set $\{z\in X^*:|\phi(z)|<1\}$ is weak*-open. Therefore, it contains a weak*-neighborhood of $0$, which by the definition of weak* topology means there exist vectors $x_1,\dots,x_n$ such that $$ \{z\in X^*: |z(x_k)| <1,\quad k=1,\dots,n\}\subseteq \{z\in X^*:|\phi(z)|<1\} $$ By homogeneity, this implies $$ |\phi(z)| \le \max_{k=1,\dots,n} |z(x_k)| $$ and therefore $\bigcap_{k=1}^n \ker x_k \subset \ker \phi$.

It follows that $\phi$ is a linear combination of $x_1,\dots,x_n$.

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    Can explain why by homogeneity we obtain $|\phi(z)| \leq \max_{1 \leq k \leq n}{|z(x_k)|}$? – Idonknow Nov 23 '15 at 08:34
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    Let $M$ be greater than the right hand side. Then $w = z/M$ is such that $|w(x_k)|<1$ for all $k$. So $|\phi(w)|<1$. Hence $|\phi(z)|<M$. –  Nov 23 '15 at 08:36
  • Do you mean 'Let $M$ be an integer smaller than the RHS'? – Idonknow Nov 23 '15 at 09:00
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    No, I mean greater. The point is, if you prove $|\phi(z)|<M$ for any $M$ that is greater than RHS, then it follows $|\phi(z) |\le RHS$ –  Nov 23 '15 at 09:01
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    The argument in this answer has a problem: $M$ can be zero, which implies that it can not be divided by $M$. – Diego Fonseca Nov 03 '17 at 17:47
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    There is no problem with $M$, as it is enough to consider $M$ strictly bigger than the maximum. – Martin Argerami May 20 '21 at 22:56
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As $f$ is weak$^{\star}$ continuous then it is weak$^{\star}$ continuous in $0$, therefore, there exists $x_{1},\ldots,x_{n}\in X$ such that \begin{equation}\begin{aligned}V(x_{1},\ldots,x_{n})&:=\left\{z\in X^{*} \::\: |z(x_{i})|\leq1 \: i=1,\ldots,n \right\}\\ &\subseteq \left\{z\in X^{*} \: :\: |\phi(z)|<1\right\} \end{aligned}\tag{$\bigstar$}\end{equation}

We show that $$\bigcap_{i=1}^{n}\mathrm{ker}x_{i}\subseteq\ker f.$$ In fact, let $z \in \ker x_{i}$, then $\left| z(x_{i}) \right|=0$ for each $i=1,\ldots ,n$. Let $\varepsilon >0$, we consider $w=\frac{z}{\varepsilon}$, then $|w(x_{i})|=\frac{1}{\varepsilon}|z(x_{i})|=0$ for each $i=1,\ldots,n$. In particular, $w\in V(x_{1},\ldots,x_{n})$, then, by ($\bigstar$) we have $$|\phi(w)|<1 \quad \Longrightarrow \quad |\phi(z)|<\varepsilon.$$ But $\varepsilon$ is arbitrary, therefore, $\phi(z)=0$.

Dean Miller
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Diego Fonseca
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  • And I believe a similar argument can be made for the weak(initial) topology generated by a collection of continuous linear functionals in general? i.e. the continuous linear functionals are exactly the ones in the span of the collection. – Smooth Alpert Frame Sep 06 '20 at 11:06
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For any vector space $X$ over $\mathbb{F}=\mathbb{R}/\mathbb{C}$, its algebraic dual $X^\#$ is a proper subset of $\mathbb{F}^X$, the set of all functions from $X$ to $\mathbb{F}$. The product topology of $\mathbb{F}^X$ is the topology of pointwise convergence.

Lemma 1. $X^\#$ is a closed subset of $\mathbb{F}^X$ under the product topology. In addition, $(X^\#, \tau^\#)$, where $\tau^\#$ is the subspace topology, is a topological vector space. That is, both vector addition $+$ and scalar multiplication are $\tau^\#$-continuous.

Lemma 2. The continuous dual of $(X^\#, \tau^\#)$ is exactly the space of evaluation maps $X^{\#*}=\{\delta_x: x\in X \}$, where $\delta_x: f\mapsto f(x)$ is the evaluation of $f$ at point $x$.

Lemma 3. If $Y\subset X^\#$ is a subspace which separate points in $X$, meaning, for every nonzero $x\in X$, there is some $f\in Y$ s.t. $f(x)\neq 0$, then $Y$ is $\tau^\#$-dense in $(X^\#, \tau^\#)$.

Lemma 4. (Hahn-Banach) If $X$ is a Banach space, then its continuous dual $X^*$ separates points in $X$.

Lemma 5(i). For a topological space $(X, \tau_F)$, where $\tau_F$ is the initial topology of a family of functions $F:=\{f: X\to Y\}$, where $Y$ is some other topological space, let $X'\subset X$ be a dense subspace, and let $F':=\{f|_{X'}:X'\to Y\}$ be the restriction of family $F$ on $X'$. Then the initial topology on $X'$, induced by $F'$, is the same as the subspace topology $\tau_F|_{X'}$.

Lemma 5(ii). Any function $g: X'\to Y$ continuous on $X'$, under the subspace topology, admits a continuous extension $\tilde{g}$ on $X$.

From these lemmas, it follows

Proposition. For a Banach space $X$, the weak*-continuous dual of $X^*$ is exactly the the space of evaluation maps $\{\delta_x: x\in X \}$, where $\delta_x$ is defined by $f\mapsto f(x)$.

Proof. Suppose $f: X^*\to \mathbb{F}$ is a weak$^*$-continuous linear functional. Note the weak$^*$ topology of $X^*$ is the same as the subspace topology it inherits from $(X^\#, \tau^\#)$, and $X^*$ is $\tau^\#$-dense in $X^\#$. Hence $f$ admits a $\tau^\#$-continuous extension on $X^\#$. But every such linear functional is an evaluation map, and so $f$ must be an evaluation map. $\square$

user760
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