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Let $(X,\|\quad\|)$ be a normed vector space, and let $X^\prime$ be the set of all bounded linear maps on $X$. I need help to clarify these questions.

  1. Is it correct that the weak topology on $X$ is the topology $\tau_w$ such that for each $f \in X^\prime$ and every open interval $I=(a,b)$, $f^{-1}(I) \in \tau_w$?

In other words, $\tau_w $ is the topology generated by $\{f^{-1}(I): I = (a,b), a,b \in \mathbb{R} \}$?

  1. The definition of $x_n \to \tilde{x}$ in $\tau_w$ means that for every $U \in \tau_w$ with $\tilde{x} \in U$ there exists $M$ such that $x_n \in U$ for $n \geq M$. Why is this definition equivalent to the definition $$ f(x_n) \to f(\tilde{x})? $$ as a real number for every $f\in X^{\prime}$?
tgtt
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2 Answers2

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In 1) 'is the topology ...' is is wrong. There are many topologies with this property and weak topology is the smallest topology for which $f^{-1}(I) \in \tau_w$ for every open interval $I$. It is the topology for which sets of the form $f^{-1}(I)$ form a subbase. (Every open set in $\tau_w$ is a union of finite intersections of sets of this type).

Suppose $x_n \to x$ in $\tau_w$ and $f\in X'$. Let $I$ be an open interval containing $f(x)$. Then $f^{-1}(I)$ is an open set in $\tau_w$ containing $x$ so it contains $x_n$ for $n$ sufficiently large. Hence $f(x_n) \in I$ for such $n$ proving that $f(x_n) \to f(x)$. Conversely, suppose $f(x_n) \to f(x)$ for every $f \in X'$. Let $V$ be a neighborhood of $x$ in $\tau_w$. Then there exist $f_1,f_2,...f_k \in X'$ and an open intervals $I_1,I_2,...,I_k$ such that $x \in \cap_j f_j^{-1}(I_j) \subset V$. Hence $f_j(x) \in I_j$ for each $j \leq k$ which implies $f_j(x_n) \in I_j$ for $n$ sufficiently large. It follows that $x_n \in f_j^{-1}(I_j) \subset V$ for $n$ sufficiently large. Hence $x_n \to x$ in $\tau_w$.

Kavi Rama Murthy
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Yes, $f^{-1}(I)$ is always weakly open. It must be, because the weak topology is the coarsest topology that makes (norm) continuous functionals continuous. Since $I$ is open in $\mathbb{R}$, the inverse image $f^{-1}(I)$ is open in the weak topology.

To ask whether such sets generate the weak topology is a slightly different question, but yes, these sets form a sub-basis for the weak topology.

As for your other question about why $x_n \rightharpoonup x$ is equivalent to $f(x_n) \to f(x)$ for all $f \in X^*$, one direction is not hard to prove. If $x_n \rightharpoonup x$, $f \in X^*$, and $\varepsilon > 0$, then $f^{-1}(f(x) - \varepsilon, f(x) + \varepsilon)$ is weakly open (as above), contains $x$, and so $x_n$ must eventually lie in this set. That is, there must be some $N$ such that \begin{align*} n > N &\implies x_n \in f^{-1}(f(x) - \varepsilon, f(x) + \varepsilon) \\ &\implies f(x_n) \in (f(x) - \varepsilon, f(x) + \varepsilon) \\ &\implies |f(x_n) - f(x)| < \varepsilon, \end{align*} as required.

On the other hand, let's suppose $f(x_n) \to f(x)$ for all $f$. We need to show that, given any weakly open neighbourhood $U$ of $x$, the sequence $(x_n)$ eventually lies inside this neighbourhood.

Recall that $U$ is a union of sub-basic sets. In particular, $x$ must be contained in a set of the form $$V = \bigcap_{k=1}^m f_k^{-1}(I_k)$$ where $I_k$ is an open interval, and $V \subseteq U$. Then, since $f_k(x_n) \to x$ as $n \to \infty$, for all $k$, there must exist some $N_k$ such that $$n > N_k \implies f_k(x_n) \in I_k \implies x_n \in f_k^{-1}(I_k).$$ Taking $N = \max\{N_1, \ldots, N_k\}$, we get $$n > N \implies x_n \in \bigcap_{k=1}^m f_k^{-1}(I_k) \subseteq U.$$

Theo Bendit
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