TITLE: Semi-Norms and the Definition of the Weak Topology
QUESTION [27 upvotes]: When I was searching for the definition of the weak topology, I found two different definitions. One defines the weak topology in terms of a family of semi-norms, while the other defines it in terms of the inverse image of open subsets of the base field $ \mathbb{K} $ under continuous linear functionals.
My questions are:

Are the two definitions equivalent, and if so, how? If they are not equivalent, then what is the common definition of the weak topology?
For the first definition (in terms of semi-norms), how can we define the open sets if we do not have a single fixed semi-norm?
For the second definition, by ‘open subsets of $ \mathbb{K} $’, do we mean the open sets belonging to the usual topology on $ \mathbb{K} $?
The topological dual space of $ X $ is defined to be the set of continuous linear functionals defined on $ X $. Is this continuity with respect to each topology or just a fixed one?

I am so confused. Could someone help me to understand the definition of the weak topology? Thanks!

REPLY [9 votes]: Let us remember two definitions (bold terminology from Narici & Beckenstein).

Seminorm Topology. Let $X$ be a vector space and let $\mathcal{P}=\{p_i\}_{i\in I}$ be a family of seminorms on $X$. The seminorm topology on $X$ generated by $\mathcal{P}$ is the topology generated by the sets of the form $\{x\in X\mid p_i(x-x_0)<\varepsilon\}$, where $i\in I$, $x_0\in X$ and $\varepsilon>0$.
Inverse Image Topology. Let $X$ be a nonempty set and let $\mathcal{F}=\{f_i:X\to Y_i\}_{i\in I}$ be a collection of maps, where each $Y_i$ is a topological space. The inverse image topology on $X$ generated by $\mathcal{F}$ is the topology generated by the sets of the form $f_i^{-1}(A_i)$, where $i\in I$ and $A_i$ is open in $Y_i$.

Working a little with basis we conclude that the inverse image topology on $X$ generated by $\mathcal{F}$ is, in fact, the topology generated by the sets of the form $f_i^{-1}(B_i)$, where $i\in I$ and $B_i$ runs over a basis of $Y_i$. In particular, if $Y_i=\mathbb{K}$ for all $i\in I$ ($\mathbb{K}=\mathbb{C}$ or $\mathbb{R}$), then the inverse image topology is generated by the sets of the form $f_i^{-1}(B(y;\varepsilon))$, where $i\in I$, $y\in Y_i$ and $\varepsilon>0$. Indeed we can prove more:

Proposition. If $Y_i=\mathbb{K}$ for all $i\in I$ ($\mathbb{K}=\mathbb{C}$ or $\mathbb{R}$), then a subbasis for the inverse image topology consists of all sets of the form
$$f_i^{-1}(B(f_i(x_0);\varepsilon))=\{x\in X\mid|f_i(x)-f_i(x_0)|<\varepsilon\},$$
where $i\in I$, $x_0\in X$ and $\varepsilon>0$. (See Brezis, page 57, for the proof of the case $\mathbb K=\mathbb R$.)

Now, let $X$ be a normed space, $X^*$ be the dual of $X$ and $\mathcal{P}=\{p_i\}_{i\in X^*}$ be the family of seminorms on $X$ defined by $p_i(x)=|i(x)|$. Then the seminorm topology on $X$ generated by $\mathcal{P}$ is equal to the inverse image topology on $X$ generated by $X^*$ because, from the first definition and from the above proposition, the collection of sets of the form
$$\{x\in X\mid p_i(x-x_0)<\varepsilon\}=\{x\in X\mid |i(x-x_0)|<\varepsilon\}=\{x\in X\mid |i(x)-i(x_0)|<\varepsilon\}$$
is a subbasis for both. Thus, the following definitions are equivalent.


Let $X$ be a normed space and let $X^*$ be the dual of $X$. The weak topology on $X$ is the topology generated on $X$ by the family $\{p_{x^*}\}_{{x^*}\in X^*}$ of seminorms defined by $p_{x^*}(x)=|x^*(x)|$.

Let $X$ be a normed space and let $X^*$ be the dual of $X$. The weak topology on $X$ is the topology generated by the sets of the form $f^{-1}(A)$, where $f\in X^*$ and $A$ is open in $\mathbb K$.



Since the inverse image topology generated by $\mathcal{F}$ is the weakest topology with respect to which all the members of $\mathcal{F}$ are continuous (see pages 93 and 94 in Bachman), these two definitions are also equivalent to the following.


Let $X$ be a normed space. The weak topology on $X$ is the weakest topology on $X$ with respect to which all bounded linear functionals on $X$ are continuous.