Local Convex Topological Vector Spaces

In revisiting the proofs of many theorems in mathematics, we often need to use sequential compactness. However, the Riesz lemma in normed linear spaces tells us that every bounded set in an infinite-dimensional normed linear space is not sequentially compact, which brings many inconveniences. In fact, the essential reason for this phenomenon is that the topology induced by the norm in a normed linear space is “too strong.” For this reason, we need to introduce the weak topology, so that bounded sets are sequentially compact in the weak-topology sense. In this article, we mainly review local convex topological vector spaces and some main results on weak convergence.

Note: Unless otherwise specified, all normed linear spaces in this article are understood to be complex normed linear spaces.

Why do we need weak convergence?

For a normed linear space $(X, |\cdot|)$, we already have the definition of convergence with respect to the norm, namely:

Definition 1 (Norm convergence) Let $(X, |\cdot|)$ be a normed linear space, $\{x_n\}\subset X$, and $x_0\in X$. We say that $\{x_n\}$ converges to $x_0$ with respect to the norm $|\cdot|$ if
$$
\lim_{n\to\infty}|x_n-x_0| = 0.
$$

When $X$ is a finite-dimensional normed linear space, since it is linearly isomorphic and homeomorphic to $\mathbb{C}^n$ (where $n=\dim X<+\infty$), every bounded subset of $X$ is sequentially compact. However, when $X$ is an infinite-dimensional normed linear space, we have the following Riesz lemma.

Theorem 1 (Riesz lemma) Let $(X, |\cdot|)$ be a normed linear space, and let $M$ be a closed subspace of $X$. Then for any $\varepsilon\in (0,1)$, there exists $x_{\varepsilon}\in X$ such that $|x_{\varepsilon}|=1$ and
$$
\operatorname{dist}(x_{\varepsilon},M)\ge 1-\varepsilon.
$$

The Riesz lemma has the following two basic corollaries.

Corollary 1 Let $(X, |\cdot|)$ be an infinite-dimensional normed linear space. Then there exists $\{x_n\}_{n=1}^{\infty}\subset X$ such that
$$
|x_m-x_n|\ge \frac{1}{2}.
$$

Corollary 2 Let $(X, |\cdot|)$ be a normed linear space. Then the following are equivalent:

(1) $\dim X<+\infty$.

(2) Every bounded subset of $X$ is sequentially compact.

(3) The closed unit ball $D = \{x\in X \ |\ |x|\le 1\}$ in $X$ is sequentially compact.

According to Corollary 2, we know that bounded sets in infinite-dimensional normed linear spaces are never sequentially compact. Therefore, we need to give a “suitable, weaker” topology on $X$ so that bounded sets have weakly convergent subsequences in the new weak-topology sense. This “suitable, weaker” topology is precisely the topology corresponding to a local convex topological vector space.

Local Convex Topological Vector Spaces (LCTVS)

First, let us recall the definition of a seminorm.

Definition 2 (Seminorm) Let $X$ be a vector space. A mapping $p:X\to \mathbb{R}$ is called a seminorm if it satisfies

(1) Nonnegativity: $p(x)\ge 0,\ \forall x\in X$.

(2) Absolute homogeneity: $p(\lambda x) = |\lambda|p(x),\ \forall \lambda\in \mathbb{C},\ x\in X$.

(3) Triangle inequality: $p(x+y)\le p(x)+p(y),\ \forall x,y\in X$.

Definition 3 (A family of seminorms separating points) Let $X$ be a vector space, and let $\mathcal{P}$ be a family of seminorms on $X$. If $p(x) = 0,\ \forall p\in \mathcal{P}$ implies $x=0$, then $\mathcal{P}$ is called a family of seminorms separating points on $X$.

Note: The point-separating condition is equivalent to the following: if $x\neq y$, then there exists a seminorm $p\in \mathcal{P}$ such that $p(x)\neq p(y)$.

Example 1 Let $(X, |\cdot|)$ be a normed linear space, and let $X^\star$ be its dual space. For any $f\in X^\star$, define
$$
\begin{aligned}
P_f: X&\to \mathbb{R}\\\
x&\mapsto |f(x)|.
\end{aligned}
$$
Then it is easy to verify that $P_f$ is a seminorm on $X$, and if
$$
P_f(x) = 0,\ \forall f\in X^\star,
$$
that is,
$$
f(x) = 0,\ \forall f\in X^\star,
$$
then by the Hahn–Banach theorem, we must have $x=0$. Hence
$$
\mathcal{P} = \{P_f; f\in X^\star\}
$$
is a family of seminorms separating points on $X$.

Definition 4 (Local convex topological vector space, LCTVS) Let $X$ be a vector space, and let $\mathcal{P}$ be a family of seminorms on $X$ separating points. Then the topology generated by the following convex sets
$$
U(x_0,p,r): = \{x\in X\ ;\ p(x-x_0)<r\},\quad x_0\in X,p\in \mathcal{P},r>0
$$
is denoted by $\tau_{\mathcal{P}}$, and $(X,\tau_{\mathcal{P}})$ is called a local convex topological vector space (Local Convex Topological Vector Space).

Note: Let
$$
\mathcal{S} = \{U(x_0,p,r)\ ;\ x_0\in X,p\in \mathcal{P},r>0\}
$$
be a subbase for the topology, and let
$$
\mathcal{B} = \left\{\bigcap_{U\in \mathcal{G}}U\ ;\ \mathcal{G}\subset \mathcal{S}, |\mathcal{G}|<+\infty\right\}
$$
be a base for the topology. Then
$$
\tau_{\mathcal{P}} = \{A\ ;\ A = \bigcup_{\lambda\in \Gamma}B_{\lambda},\ B_{\lambda}\in \mathcal{B}\}.
$$
It is easy to verify that $\tau_{\mathcal{P}}$ is indeed a topology on $X$ (the verification that $\varnothing\in \tau_{\mathcal{P}}$ requires the point-separating condition).

In order to give the basic properties of local convex topological vector spaces and the definition of weak convergence, let us first recall the definition of a net.

Nets

A net is a generalization of a sequence. By means of the convergence of nets, several common types of limits can be unified, and nets also have some additional properties.

Definition 5 (Directed set) Let $(I,\le)$ be a partially ordered set. If for any $\alpha,\beta\in I$, there always exists $\gamma\in I$ such that $\alpha\le \gamma$ and $\beta\le \gamma$, then $(I,\le)$ is called a directed set.

Definition 6 (Net) Let $X$ be a set, and let $(I,\le)$ be a directed set. A mapping $I\to X$ is called a net on $X$, denoted by $\{x_\alpha\}_{\alpha\in I}$.

Definition 7 (Convergence of a net) Let $(X,\tau)$ be a topological space, and let $\{x_\alpha\}, \alpha\in I$ be a net in $X$. We say that the net $\{x_\alpha\}, \alpha\in I$ converges to $x_0\in X$ if for every neighborhood $U$ of $x_0$, there exists $\alpha_U\in I$ such that whenever $\alpha\ge \alpha_U$, we have $x_\alpha\in U$. We write
$$
\lim\limits_{\alpha\in I}x_\alpha = x_0.
$$

Example 2 (Limit of a sequence) The set of natural numbers $\mathbb{N}$, equipped with its usual order relation, forms a directed set. Hence any sequence $\{x_n\}_{n=1}^{\infty}\subset X$ is a net on $X$, and convergence of nets in this case is exactly convergence of sequences.

Example 3 (Limit of a function) Let $f:\mathbb{R}\to \mathbb{R}$ be a function. Then continuity of $f$ at $x_0$ is equivalent to
$$
\lim_{x\to x_0} f(x) = f(x_0).
$$
Let $I = (x_0-1,x_0+1)$, and define a partial order on $I$ as follows: for $x_1,x_2\in I$, write $x_1\ge x_2$ if one of the following holds:

(1) $x_1 = x_2$.

(2) $|x_1-x_0|<|x_2-x_0|$.

Then it is easy to verify that $(I,\le)$ is a directed set. Define the net
$$
\begin{aligned}
(I,\le )&\to \mathbb{R}\\\
x&\mapsto f(x).
\end{aligned}
$$
Then
$$
\lim_{x\to x_0}f(x) = f(x_0)\iff \lim_{x\in I}f(x) = f(x_0).
$$

Example 4 (Riemann integral) A function $f:[0,1]\to \mathbb{R}$ is Riemann integrable if the following holds: there exists $\alpha\in \mathbb{R}$ such that for every $\varepsilon>0$, there exists $\delta>0$ such that for every partition $\Delta:0=x_0<x_1<\cdots<x_n=1$ of $[0,1]$, and for every choice of tags $\xi_i\in [x_{i-1},x_i],\ i=1,\cdots,n$, whenever the mesh of the partition satisfies
$$
|\Delta| =\max_{1\le i\le n}|x_i-x_{i-1}|<\delta,
$$
we have
$$
\left|\sum_{i=1}^n f(\xi_i)(x_i-x_{i-1})-\alpha\right|<\varepsilon.
$$
Then $f:[0,1]\to \mathbb{R}$ is said to be Riemann integrable, and we write
$$
\alpha = \int_{0}^{1}f(x),dx.
$$

Now consider
$$
I = \{(\Delta,\Lambda)\ ;\ \Delta\text{ is a partition of }[0,1],\ \Lambda \text{ is a choice of tags for this partition}\}.
$$
Write $(\Delta_1,\Lambda_1)\le (\Delta_2,\Lambda_2)$ if one of the following holds:

(1) $(\Delta_1,\Lambda_1)=(\Delta_2,\Lambda_2)$.

(2) $|\Delta_1|>|\Delta_2|$.

Then it is easy to prove that $(I,\le)$ is a directed set. For a partition $\Delta:0=x_0<x_1<\cdots<x_n=1$ and a choice of tags $\Lambda = \left\{\xi_i\right\}_{i=1}^n$, define

$$
S_{(\Delta,\Lambda)} = \sum_{i=1}^n f(\xi_i)(x_i-x_{i-1}).
$$
Then it is easy to verify that the function $f:[0,1]\to \mathbb{R}$ is Riemann integrable if and only if

$$
\lim_{(\Delta,\Lambda)\in I}S_{(\Delta,\Lambda)}
$$

exists.

A net is not only a generalization of the limit of a sequence; because its condition is stronger, it also has some properties better than those of sequence limits:

Proposition 1 (Properties of net convergence) Let $(X,\tau)$, $(X_1,\tau_1)$, and $(X_2,\tau_2)$ be topological spaces, and let $A\subset X$. Then the following conclusions hold:

(1) $(X,\tau)$ is a Hausdorff space $\iff$ every convergent net in $X$ has a unique limit.

(2) $x\in \bar{A}\iff$ there exists a convergent net $\{x_\alpha\}_{\alpha\in I}$ in $A$ such that

$$
\lim\limits_{\alpha\in I} x_\alpha = x.
$$

(3) $f:(X_1,\tau_1)\to (X_2,\tau_2)$ is continuous $\iff$ for every convergent net $\{x_\alpha\}_{\alpha\in I}$ in $X_1$, if

$$
\lim\limits_{\alpha\in I} x_\alpha = x_0,
$$

then

$$
\lim\limits_{\alpha\in I} f(x_\alpha) = f(x_0).
$$

(4) $\tau_1$ is stronger than $\tau_2\iff$ for every convergent net $\{x_\alpha\}_{\alpha\in I}$ in $(X_1,\tau_1)$, if

$$
\lim\limits_{\alpha\in I} x_\alpha = x_0,
$$

then $\{x_\alpha\}_{\alpha\in I}$ is also a net converging to $x_0$ in $(X_2,\tau_2)$.

Properties of local convex topological vector spaces

Theorem 2 Let $(X,\tau_{\mathcal{P}})$ be a local convex topological vector space. Then it has the following properties:

(1) $(X,\tau_{\mathcal{P}})$ is a Hausdorff space.

(2) For $x_0\in X$,
$$
\{x_0+U_{F,r}\ ;\ F\subset \mathcal{P},|F|<+\infty,r>0\}
$$
forms a neighborhood base at $x_0$, where
$$
U_{F,r} = \bigcap_{p\in F} \{x\in X\ ;\ p(x)<r\}.
$$

(3) For a net $\{x_\alpha\}_{\alpha\in I}$ in $X$ and $x_0\in X$,

$$
\lim\limits_{\alpha\in I} x_\alpha = x_0
$$

if and only if

$$
\lim_{\alpha\in I}p(x_\alpha-x_0) = 0,\quad \forall p\in \mathcal{P}.
$$

(4) Addition and scalar multiplication on $(X,\tau_{\mathcal{P}})$ are continuous.

Weak convergence and weak-$\star$ convergence

Weak topology and weak convergence

Let $(X, |\cdot|)$ be a normed linear space, and let $X^\star$ be its dual space. According to Example 1,
$\mathcal{P} = \{P_f\ ;\ f\in X^\star\}$ is a family of seminorms on $X$ separating points, where
$$
\begin{aligned}
P_f: X&\to \mathbb{R}\\\
x&\mapsto |f(x)|
\end{aligned}
$$
is a seminorm on $X$. Therefore, the corresponding $(X,\tau_{\mathcal{P}})$ forms a local convex topological vector space, and the topology $\tau_{\mathcal{P}}$ on it is also denoted by $\tau(X,X^{\star})$, called the weak topology on $X$.

According to Theorem 2 (3), for a net $\{x_\alpha\}_{\alpha\in I}$ in $X$ and $x_0\in X$,

$$
\begin{aligned}
\lim\limits_{\alpha\in I} x_\alpha = x_0 &\iff \lim_{\alpha\in I}P_f(x_\alpha-x_0) = 0,\quad \forall f\in X^\star\\\
&\iff \lim_{\alpha\in I} |f(x_\alpha-x_0)| = 0,\quad \forall f\in X^\star\\\
&\iff \lim_{\alpha\in I} f(x_\alpha) = f(x_0),\quad \forall f\in X^\star.
\end{aligned}
$$

In particular, for any sequence $\left\{x_n\right\}$ in $X$ and $x_0\in X$, we say that the sequence $\left\{x_n\right\}$ converges weakly to $x_0$ if

$$
\lim_{n\to \infty}f(x_n) = f(x_0),\quad \forall f\in X^\star.
$$

We write
$$
x_n\stackrel{w}{\longrightarrow} x_0.
$$

Weak-$\star$ topology and weak-$\star$ convergence

Let $(X, |\cdot|)$ be a normed linear space, and let $X^\star$ be its dual space. For any $x\in X$, define
$$
\begin{aligned}
P_x: X^\star&\to \mathbb{R}\\\
f&\mapsto |f(x)|.
\end{aligned}
$$
Then it is easy to verify that $\mathcal{P}^\star = \{P_x\ ;\ x\in X\}$ is a family of seminorms on $X^\star$ separating points. Therefore, the corresponding $(X^\star,\tau_{\mathcal{P}^\star})$ forms a local convex topological vector space, and the topology $\tau_{\mathcal{P}^\star}$ on it is also denoted by $\tau(X^\star,X)$, called the weak-$\star$ topology on $X^\star$.

According to Theorem 2 (3), for a net $\{f_\alpha\}_{\alpha\in I}$ in $X^\star$ and $f_0\in X^\star$,

$$
\begin{aligned}
\lim\limits_{\alpha\in I} f_\alpha = f_0 &\iff \lim_{\alpha\in I}P_x(f_\alpha-f_0) = 0,\quad \forall x\in X\\\
&\iff \lim_{\alpha\in I} |f_\alpha(x)-f_0(x)| = 0,\quad\forall x\in X\\\
&\iff \lim_{\alpha\in I} f_\alpha(x)= f_0(x),\quad \forall x\in X.
\end{aligned}
$$

In particular, for any sequence $\{f_n\}$ in $X^\star$ and $f_0\in X^\star$, we say that the sequence $\{f_n\}$ converges weak-$\star$ to $f_0$ if
$$
\lim_{n\to \infty}f_n(x) = f_0(x),\quad \forall x\in X.
$$
We write
$$
f_n\stackrel{w^\star}{\longrightarrow} f_0.
$$

The Banach–Alaoglu theorem

Finally, let us answer the question raised at the beginning: the weak-$\star$ topology is indeed a “suitable, weaker” topology, so that bounded sets have weak-$\star$ convergent subsequences in the new weak-$\star$ topology sense. We have the very important Banach–Alaoglu theorem.

Theorem 3 (Banach-Alaoglu) Let $(X, |\cdot|)$ be a normed linear space, and let $X^\star$ be its dual space. Then the closed unit ball in $X^\star$,
$$
X^\star_1 = \{f\in X^\star\ ;\ |f|\le 1\},
$$
is compact in the weak-$\star$ topology. Moreover, when $(X, |\cdot|)$ is a separable normed linear space, $X_1^\star$ is sequentially compact in the weak-$\star$ topology; that is, for any $\{f_n\}\subset X^\star$, there exist $f_0\in X^\star$ and a subsequence $\{f_{n_k}\}$ such that
$$
f_{n_k}\stackrel{w^\star}{\longrightarrow} f_0.
$$

The cover image of this article was taken on the way to Lake Königssee, Germany.