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Let $(E, | \cdot|)$ be a normed linear space and $E^\star$ its topological dual. Let $\sigma(E, E^\star)$ be the weak topology on $E$, and $\sigma(E^\star, E)$ the weak$^\star$ topology on $E^\star$. Let $\overline x \in E$ and $(x_n)$ be a sequence in $E$. Let $\overline f \in E^\star$ and $(f_n)$ be a sequence in $E^\star$. Then

  • $x_n \rightharpoonup \overline x$ in $\sigma(E, E^\star)$ if and only if $\langle f, x_n \rangle \to \langle f, \overline x \rangle$ for all $f \in E^\star$.

  • $f_n \overset{\star}{\rightharpoonup} \overline f$ in $\sigma(E^\star, E)$ if and only if $\langle f_n, x \rangle \to \langle \overline f, x \rangle$ for all $x \in E$.

Above convergences look quite "symmetric/dual". Is there any result relating these topologies?

Akira
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    One could show for example that a functional $\phi:E^\to\mathbb{C}$ is weak-star continuous if and only if there exists $e\in E$ such that $\phi(f)=f(e)$ for all $f\in E^$, so the dual space of $(E^,wk^)$ is equal to ${\phi_e:e\in E}$ where $\phi_e(f)=f(e)$. As a dual space, this space has a weak-star topology. It is easily seen that a net $(\phi_{e_i})_{i\in I}$ in the dual converges to $\phi_e$ in the weak-star topology if and only if $e_i\to e$ in the weak topology of $E$. If this is what you're looking for, I will write a detailed answer with the proofs of these facts. – Just dropped in Jan 27 '22 at 09:49
  • @JustDroppedIn Please write it as an answer :v – Akira Jan 27 '22 at 09:58

1 Answers1

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Let $(E,\|\cdot\|)$ be a normed space. Consider the dual space $E^*$ and equip it with the weak star topology. Set $V:=(E^*,\text{wk}^*)^*=\{\phi:E^*\to\mathbb{C}: \phi\text{ is wk}^*\text{ continuous}\}$. It is obvious that, if $e\in E$, then $\phi_e:E^*\to\mathbb{C}$, $\phi_e(f):=f(e)$ is weak-star continuous, so $\{\phi_e:e\in E\}\subset V$. We will show that $V=\{\phi_e:e\in E\}$.

Now let $\phi\in V$. Note that the weak-star topology on $E^*$ is a locally convex topology generated by the seminorms $\{p_e\}_{e\in E}$, where $p_e(f):=|f(e)|$. By elementary theory of locally convex spaces (see for example the appendix in Murphy's book, theorem A.1), there exist $e_1,\dots,e_n\in E$ and $M>0$ such that $$|\phi(f)|\le M\cdot\max_{1\le j\le n}|p_{e_j}(f)|=M\cdot\max_{1\le j\le n}|f(e_j)|$$ Therefore, if $f(e_j)=0$ for all $j=1,\dots,n$, then $\phi(f)=0$. In other words, $$\bigcap_{j=1}^n\ker(\phi_{e_j})\subset\ker(\phi)$$ By this exercise of linear algebra this shows that $\phi$ is a linear combination of the $\phi_{e_j}$, so find $\lambda_1,\dots,\lambda_n\in\mathbb{C}$ such that $\phi=\sum_{j=1}^n\lambda_j\phi_{e_j}$. But if $e:=\sum_{j=1}^n\lambda_je_j\in E$, then $\phi_e=\sum_{j=1}^n\lambda_j\phi_{e_j}$, so $\phi=\phi_e$ as we wanted.

Finally, $V$ has its own weak-star topology, i.e. $\phi_i\to\phi$ weak-star in $V$ if and only if $\phi_i(f)\to\phi(f)$ for all $f\in E^*$. But writing $\phi_i=\phi_{e_i}$ and $\phi=\phi_e$ for $(e_i)\subset E, e\in E$, we see that $\phi_{e_i}\to\phi_e$ weak-star in $V$ if and only if $f(e_i)\to f(e)$ for all $f\in E^*$, i.e. if and only if $e_i\to e$ weakly in $E$. In other words, what we have almost just shown is that the map $(E,\text{weak})\to (V,\text{weak*})$, $e\mapsto \phi_e$ is an isomorphism of locally convex topological spaces, i.e. a linear homeomorphism. All we haven't showed yet is injectivity of this map. But indeed, if $\phi_e=0$, then $f(e)=0$ for all $f\in E^*$, and it is well-known by the Hahn-Banach theorem that this implies that $e=0$.

Just dropped in
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