Why is the quotient map from a Banach space open with respect to the weak topologies?

Question

Suppose $Y$ is a closed subspace of a Banach space $X$ and $q$ is the usual quotient map from $X$ to $X/Y$. I want to show that $q$ is an open map with respect to the weak topologies.

So far I have thought this: it is enough to consider the canonical open sets around zero in $X$ generated by an element $f$ in $X^*$, namely $U_{f}=\{x | \ |f(x)|<\epsilon\}$ for some $\epsilon>0$. I would like to somehow look at a form of $f$ in $(X/Y)^*$ but can't see how to do this.

Answer 1

It may be easier to picture this geometrically. A canonical open set in $X$ is a set of the form $U=\{av+m: |a|<\epsilon,\ m\in M\}$ where $M$ is a closed subspace (in your notation, $\ker f$), and $M\cup \{v\}$ is a spanning set of $X$.

Under a surjective linear operator $T:X\to Z$, the set $U$ is mapped to $$TU=\{aTv+m: |a|<\epsilon,\ m\in TM\}$$ Since $T$ is surjective, $M\cup \{Tv\}$ spans $Z$. There are two cases:

1. $Tv=0$. Then $M=Z$, hence $TU=Z$.
2. $Tv\ne 0$. Then $TU$ is a canonical weakly open set for $Z$.

Answer 2

No need to invoke any crazy theorem: It suffices to show that, for each weakly basic open neighborhood $V$ of the origin $0$ in $X$, $q[V]$ is a weakly neighborhood of the origin $\hat 0$ in $X/Y$. Let $V = \bigcap_{f \in F} \{ x \in X \colon | \langle f,x\rangle |<ε\}$ be a weakly open neighborhood of $0$, where $F\subset X^*$ is finite and $ε>0$. Let $\{x^*_1,\dots, x^*_m\}$ be a basis of $Y^\bot \cap \operatorname{span}F$ and extend this basis to a basis $\{x^*_1,\dots, x^*_n\}, \ n \ge m ,$ of $\operatorname{span}F$. Since each $f \in F$ is a linear combination of $\{x_i^*\}_{i=1}^n$ we may find $\delta>0$ such that $$ \bigcap_{i=1}^n \{ x \in X \colon | \langle x_i^*,x\rangle |<\delta\} \subset V .$$ So we may assume that $V= \bigcap_{i=1}^n \{ x \in X \colon | \langle x_i^*,x\rangle |<\delta\}$ and $F = \{ x_1^*, \dots, x_n^*\}$.

Consider the linear operator $T \colon Y \to \mathbb R^{n-m}$ given by $$y \mapsto \big ( \langle x^*_i,y \rangle \big )_{i=m+1}^n.$$ Then $T$ is onto. Indeed, suppose that $a \in \mathbb R^{n-m}$ is orthogonal to $R(T)$, that is, $\sum_{i=m+1}^n a_i \langle x^*_i,y\rangle =0$ for all $y \in Y$. This then implies that $\sum_{i=m+1}^n a_i x^*_i \in Y^\bot \cap \operatorname{span}F$ and since $x^*_1,\dots, x^*_n$ are linearly independent, we get that $a=0$.

I claim that $$ q[V] = \bigcap_{i=1}^m \{\hat x \in X/Y \colon |\langle x^*_i,x \rangle | <\delta \} \tag{$\star$} . $$ Recall that the dual of $X/Y$ is $Y^\bot$ so that the RHS of $(\star)$ is weakly open in $X/Y$. To prove $(\star)$ let $x \in V$. Then $ |\langle x^*_i,qx \rangle | = |\langle x^*_i,x\rangle |<\delta$ for $i=1,\dots,m$ so that $qx $ belongs to the RHS of $(\star)$. Conversely, suppose that $\hat x \in X/Y$ is such that $|\langle x^*_i,x\rangle |<\delta$ for $i=1,\dots,m$. Since $T \colon Y \to \mathbb R^{n-m}$ is onto, there exists $y \in Y$ such that $ \langle x_i^*, y \rangle = \langle x_i^*,x \rangle $ for $i=m+1,\dots,n$. It then follows that $\hat x = q(x-y)$ and $$ | \langle{x^*_i},x-y\rangle| = \begin{cases} | \langle x^*_i,x \rangle |< \delta & 1 \le i \le m \\ | \langle{x^*_i},x-y\rangle|=0< \delta & m<i\le n \end{cases} $$ hence, $x-y \in V$.

Answer 3

To prove this, one can use theory related to Arens-Mackey theorem.

This can be found in chapter $3$ of the book Banach Space Theory: The Basis for Linear and Nonlinear Analysis by Fabian, Habala, Hajek, Montesinos and Zizler, which I will use for reference.

In what I write, $w(E, F)$ refers to weak topology on $E$ as considered to be induced from the dual pair $\langle E, F\rangle$ (which is $\sigma(E, F)$ on wikipedia).

The most relevant is this result:

Proposition $3.50.$ Let $(E, \mathcal{T})$ be a locally convex space, $\mathcal{M}$ the family of equicontinuous subsets of $E^*$, $G\subseteq E$ a closed subspace and $\widehat{\mathcal{M}} = \{A\in\mathcal{M} : A\subseteq G^\perp\}$. Let $\widehat{\mathcal{T}}$ be the quotient topology on $E/G$ and let $\mathcal{T}_{\widehat{\mathcal{M}}}$ be topology on $E/G$ defined by taking $\{A^\circ : A\in\widehat{\mathcal{M}}\}$ as the neighbourhoods of $0$. Then $\widehat{\mathcal{T}} = \mathcal{T}_\widehat{\mathcal{M}}$.

Note that in the above theorem, we identify $G^\perp$ with $(E/G)^*$. Here $A^\circ$ denotes the polar of $A$.

Corollary $3.51.$ Let $E$ be locally convex, $G\subseteq E$ be closed. Then the canonical map $q:(E, w)\to (E/G, w)$, between $E$ and $E/G$ with weak topologies, is a quotient map.

Remark. In the book, the statement of this corollary is wrong, a mistake by the authors. Its not a theorem about subspaces, but about quotients.

Proof: Let $\mathcal{T}$ be weak topology in proposition $3.50$. Then $\mathcal{M}$ consists of subsets of balanced convex $w(E^*, E)$-closures of finite subsets of $F$. The family $\widehat{\mathcal{M}}$ is then seen to consist of subsets of balanced convex $w(G^\perp, E/G)$-closures of finite subsets of $G^\perp$, so that $\mathcal{T}_\widehat{\mathcal{M}}$ is the weak topology on $E/G$. $\square$

The exact same proof applies to show that $q:(E, \mu)\to (E/G, \mu)$ is a quotient map where $\mu$ denotes the Mackey topology. The difference is that instead of balanced convex $w(E^*, E)$-closures of finite sets, we consider all balanced convex $w(E^*, E)$-compact subets of $F$.

Using this, we see that $q:X\to X/Y$ is a quotient map when interpreted as a map between $X$ and $X/Y$ with weak topology. In particular, since all quotient maps of topological groups are open, $q$ needs to be an open map.