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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:

  1. Are the two definitions equivalent, and if so, how? If they are not equivalent, then what is the common definition of the weak topology?

  2. 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?

  3. For the second definition, by ‘open subsets of $ \mathbb{K} $’, do we mean the open sets belonging to the usual topology on $ \mathbb{K} $?

  4. 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!

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In what follows, $ \mathbb{F} $ shall denote either $ \mathbb{R} $ or $ \mathbb{C} $. The standard Euclidean topology on $ \mathbb{F} $ shall be denoted by $ \tau_{\mathbb{F}} $.

It is important to know that the weak topology on a vector space $ V $ over $ \mathbb{F} $ can only be defined after we have already endowed $ V $ with a linear topology, i.e., a topology on $ V $ that makes vector addition and scalar multiplication continuous operations. Hence, when one speaks of ‘weak topology’, it is always with respect to an existing linear topology.


What Semi-Norms Give Us

Let $ V $ be an abstract vector space over $ \mathbb{F} $ and $ \mathcal{P} $ a family of semi-norms on $ V $. We can define a topology on $ V $ by declaring its base to consist of subsets of $ V $ of the form $$ \{ v \in V ~|~ {p_{1}}(v - x) < \epsilon ~ \land ~ \ldots ~ \land ~ {p_{n}}(v - x) < \epsilon \}, $$ where

  • $ x \in V $,

  • $ \epsilon > 0 $ and

  • $ n \in \mathbb{N} $ and $ (p_{1},\ldots,p_{n}) $ is an $ n $-tuple of semi-norms coming from $ \mathcal{P} $.

This topology is called the semi-norm topology corresponding to $ \mathcal{P} $, and it can be shown to be a linear topology on $ V $. When we equip $ V $ with a semi-norm topology, we call $ V $ a locally convex space.

As you can see, semi-norms (whether there is only one or there are many) allow us to create a linear topology for $ V $ when there might have been none before. Its construction in no way depends on the weak topology, which comes only after.


Constructing the Weak Topology from an Existing Linear Topology

The weak topology on a vector space over $ \mathbb{F} $ can only be defined after we have equipped the vector space with some linear topology, for example the locally convex topology that was defined in the previous section.

Suppose that $ V $ is a vector space equipped with a linear topology $ \tau $. Let $ V^{*} $ denote the set of $ \tau $-continuous linear functionals on $ V $, i.e., continuous linear functions $ \varphi: (V,\tau) \to (\mathbb{F},\tau_{\mathbb{F}}) $. The weak topology on $ V $ with respect to $ \tau $ is defined as the smallest topology $ \tau_{\text{wk}} $ on $ V $ such that the mapping $ \varphi: (V,\tau_{\text{wk}}) \to (\mathbb{F},\tau_{\mathbb{F}}) $ is still continuous for all $ \varphi \in V^{*} $.

We can find a sub-basis of $ \tau_{\text{wk}} $ that consists of subsets of $ V $ of the form $ {\varphi^{\leftarrow}}[U] $, where $ \varphi \in V^{*} $ and $ U \in \tau_{\mathbb{F}} $.

Note that $ \tau_{\text{wk}} $ is a weaker topology than $ \tau $, i.e., $ \tau_{\text{wk}} \subseteq \tau $. Note also that $ \tau_{\text{wk}} $ is a linear topology on $ V $.


Wait! The Weak Topology Is Actually a Semi-Norm Topology!

I mentioned in the previous section that the weak topology arrives only after we have a linear topology. Here comes the tricky part. We can actually construct the weak topology as the semi-norm topology corresponding to some family $ \mathcal{P} $ of semi-norms. However, $ \mathcal{P} $ is not a rabbit pulled out of an empty hat. To define $ \mathcal{P} $, we need to have a linear topology in the first place. Herein lies the source of your confusion:

The weak topology is indeed a semi-norm topology, but you cannot begin to describe the semi-norms until you have put a linear topology on the vector space.

As before, let $ V $ be a vector space over $ \mathbb{F} $ equipped with a linear topology $ \tau $. For each $ \varphi \in V^{*} $, define a semi-norm $ p_{\varphi}: V \to [0,\infty) $ as follows: $$ \forall v \in V: \quad {p_{\varphi}}(v) \stackrel{\text{def}}{=} |\varphi(v)|. $$ Then $ \tau_{\text{wk}} $ as constructed in the previous section is actually the semi-norm topology corresponding to the family $ \{ p_{\varphi} ~|~ \varphi \in V^{*} \} $.


Motivation for the Weak Topology

As we already have $ \tau $, then why is there a need to create $ \tau_{\text{wk}} $? Think of it this way. The original topology $ \tau $ may contain more open sets than is actually necessary to make each $ \varphi \in V^{*} $ continuous. We can thus afford to throw out some open sets from $ \tau $ without affecting the continuity of the $ \varphi $’s. By discarding these unnecessary open sets and making the topology smaller, we can somehow endow $ V $ with certain nice topological properties. For example, with less open sets, some subsets of $ V $ automatically become compact subsets.

To give a further illustration, let us consider Kakutani’s Theorem, which states that a Banach space is reflexive if and only if its closed unit ball is compact with respect to the weak topology. For infinite-dimensional Banach spaces, the closed unit ball is never compact with respect to the norm-topology, which is a consequence of Riesz’s Lemma. However, by switching to the weak topology, the compactness of the closed unit ball may be regained, which is certainly the case precisely when the Banach space is reflexive, according to Kakutani’s result. Therefore, the weak topology gives us a complete characterization of reflexivity, and this is clear evidence of its utility.


Conclusion — $ V $ By Design

We saw a naked abstract vector space $ V $ who needed a topological outfit. We covered her shame by giving her a linear topology. $ V $ begged us, “Please respect my semi-norms if you wish to topologize me!” After we had dressed her, $ V $ examined herself in the mirror and delightfully exclaimed, “I look good with all my linear functionals!” We looked at her from top to bottom, shook our heads and told her, “No. You still look kinda baggy. But don’t worry! You’ll get to keep your linear functionals.” After removing her excess open sets, all of us concluded that $ V $’s minimalistic weak topology was stunning!

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  • $\begingroup$ ...underneath your closures... (c) $\endgroup$ – Norbert Feb 17 '13 at 3:31
  • $\begingroup$ I need to teach Shakira some topology before she’s allowed to perform that during the Grammys. $\endgroup$ – Haskell Curry Feb 17 '13 at 8:11
  • $\begingroup$ It is not true that "the weak topology on a vector space can only be defined after we have already endowed the space with a linear topology" and that "one speaks of ‘weak topology’ always with respect to an existing linear topology". In general, to define weak topology we need only a collection of mappings. Of course, if we have a topology then we have the needed collection of mappings as a byproduct (namely, the topological dual). $\endgroup$ – Pedro Sep 23 '16 at 2:00
  • $\begingroup$ @Pedro - I think the weak topology he's referring to is often called "the" weak topology on a linear space X. Strictly speaking though only a notion of bounded linear operators seems to be needed. For this you would need a norm, but not the norm induced topology. $\endgroup$ – TSU Feb 5 '17 at 17:30
  • $\begingroup$ @Pedro: In most texts on functional analysis, the weak topology is defined with respect to the continuous dual space of a topological vector space. There is therefore nothing wrong with Haskell Curry’s post. Any disagreements that you may have with him are purely nomenclatural in nature. $\endgroup$ – Transcendental Jul 12 '17 at 23:08
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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.

  1. 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 $p_{x^*}(x)=|x^*(x)|$.

  2. 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.

  1. 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.
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