Topological Vector Spaces

Let’s discuss our final application (for now!) of topology—topological vector spaces. However, we can investigate a more general idea—topological groups—of which topological vector spaces are a subset. That’s true without the “topological” bit: vector spaces are (Abelian) groups under vector addition, equipped with scalar multiplication.

Definition 1. Let G be a \mathrm T_1-space and a group with binary operation \cdot : G \times G \to G and identity element e. For any g \in G and V \subseteq G, define the translates by

g \cdot V := \{g \cdot x : x \in V\},\quad V \cdot g := \{x \cdot g : x \in V\}.

Furthermore, for any U, V \subseteq G, define

U \cdot V := \{x \cdot y : x \in G, y \in G\}.

Definition 2. A \mathrm T_1-space G is a topological group if it has a continuous group operation \cdot : G \times G \to G, as well as a continuous induced inverse map (\cdot)^{-1} : G \to G, x \mapsto x^{-1}.

Lemma 1. A \mathrm T_1-space with a group operation is a topological group if and only if the map (x, y) \mapsto x \cdot y^{-1} is continuous.

Proof. The direction (\Rightarrow) follows from the composition (x, y) \mapsto (x, y^{-1}) \mapsto x \cdot y^{-1}. The direction (\Leftarrow) follows from the compositions x \mapsto (e, x) \mapsto e \cdot x^{-1} = x^{-1} and (x, y) \mapsto (x, y^{-1}) \mapsto x \cdot (y^{-1})^{-1} = x \cdot y.

Theorem 1. G is Hausdorff.

Proof. We recall that G is Hausdorff if and only if its diagonal

\Delta_G := \{(x, x) : x \in G\}

is closed. The map f(x, y) = x \cdot y^{-1} is continuous. Since G is \mathrm T_1, \Delta_G = f^{-1}(\{e\}) is closed, so we have our result.

Example 1. (\mathbb R, +) and (\mathbb R^+, \cdot) form topological groups. Furthermore, the map x \mapsto e^x is a group isomorphism that is a homeomorphism.

Proof. (\mathbb R, +) is a topological group since the maps + and -(\cdot) are continuous, so that the map (x,y) \mapsto (x, -y) \mapsto x + (-y) is continuous.

The second claim follows from the continuity of the exponential map x \mapsto e^x, since the map

(x, y) \mapsto (\ln(x), \ln(y)) \mapsto \ln(x) + (-\ln(y)) \mapsto e^{\ln(x) + (-\ln(y))} = x \cdot y^{-1}

is continuous.

Definition 3. A topological vector space V over a field \mathbb K = \mathbb R or \mathbb C is topological space with continuous vector space operations continuous vector addition + : V \times V \to V and scalar multiplication \cdot : \mathbb K \times V \to V. Denote the additive identity by \mathbf 0_V.

Let V, W be topological vector spaces over \mathbb K and f : V \to W be a map.

Theorem 2. V forms a topological group under +. In particular, any convergent sequence in V converges to a unique limit.

Example 2. Normed spaces are topological vector spaces. In particular, \mathcal C([a, b], \mathbb R) equipped with the supremum norm forms a topological vector space. More generally, W^V = \mathcal F(V, W) forms a topological vector space when given the product topology.

Definition 4. For any f \in \mathcal F(V, W), \mathbf v_0 \in V, and \mathbf w_0 \in W, write

\displaystyle \lim_{\mathbf v \to \mathbf v_0}f(\mathbf v) = \mathbf w_0

to mean that for any neighbourhood U_W of \mathbf w_0, there exists a neighbourhood U_V of \mathbf v_0 such that f(U_V) \subseteq U_W.

Lemma 2. The set

o(1) := \left\{f \in \mathcal F(V, W) : \displaystyle \lim_{\mathbf v \to \mathbf 0_V}f(\mathbf v) = \mathbf 0_W \right\}

forms a topological vector space, when equipped with the subspace topology. Furthermore, for any \alpha \in \mathbb K,

o(1) + o(1) = o(1),\quad \alpha \cdot o(1) = o(1).

Proof. For the first result, f = f + 0 establishes (\supseteq). For the second result, fix f, g \in o(1). We need to prove that f +g \in o(1).

Fix a neighbourhood U_W of \mathbf 0_W. Since + is continuous, there exists neighbourhoods U_f, U_g of \mathbf 0_W such that U_f + U_g \subseteq U_W. Since f \in o(1), find a neighbourhood U_f' of \mathbf 0_V such that f(U_f') \subseteq U_f. Repeat likewise to obtain g(U_g') \subseteq U_g. Define U_V := U_f' \cap U_g'. Then

(f+g)(U_V) = f(U_V) + g(U_V) \subseteq U_f + U_g \subseteq U_W.

The proof is similar (obvious, even) for the scalar multiplication case.

Lemma 3. For any f \in \mathcal F(V, W), \mathbf v_0 \in V, and \mathbf w_0 \in W,

\displaystyle \lim_{\mathbf v \to \mathbf v_0}f(\mathbf v) = \mathbf w_0 \quad \iff \quad f(\mathbf v_0 + \cdot) - \mathbf w_0 \in o(1).

In particular, \displaystyle \lim_{\mathbf v \to \mathbf v_0}f(\mathbf v) = f(\mathbf v_0) if f \in \mathcal C(V, W).

Proof. Denote f_{\mathbf v_0,\mathbf w_0} := f(\mathbf v_0 + \cdot) - \mathbf w_0 for brevity, so that the right hand side simplifies to f_{\mathbf v_0,\mathbf w_0} \in o(1).

Proving the direction (\Rightarrow), fix any neighbourhood U_W of \mathbf 0_W. Then \mathbf w_0 + U_W is a neighbourhood of \mathbf w_0, where U_W is a neighbourhood of \mathbf 0_W. By hypothesis, find a neighbourhood \mathbf v_0 + U_V of \mathbf v_0, where U_V is a neighbourhood of \mathbf 0_V such that

f(\mathbf v_0 + U_V) \subseteq \mathbf w_0 + U_W \quad \iff \quad f_{\mathbf v_0,\mathbf w_0}(U_V) = f(\mathbf v_0 + U_V) - \mathbf w_0 \subseteq U_W.

Hence, f_{\mathbf v_0,\mathbf w_0} \in o(1). In the direction (\Leftarrow), fix any neighbourhood \mathbf w_0 + U_W of \mathbf w_0 where U_W is a neighbourhood of \mathbf 0_W. Since f_{\mathbf v_0,\mathbf w_0} \in o(1), there exists a neighbourhood U_V of \mathbf 0_V such that

f(\mathbf v_0 + U_V) - \mathbf w_0 = f_{\mathbf v_0,\mathbf w_0}(U_V) \subseteq W_V.

But recall that \mathbf v_0 + U_V is a neighbourhood of \mathbf v_0, so that we obtain

\displaystyle f(\mathbf v_0 + U_V)  \subseteq \mathbf w_0 + W_V \quad \Rightarrow \quad \lim_{\mathbf v \to \mathbf v_0}f(\mathbf v) = \mathbf w_0.

Theorem 3. For any f \in \mathcal C(V, W), \mathbf v_0 \in V, and \alpha \in \mathbb K,

\begin{aligned} \lim_{\mathbf v \to \mathbf v_0}(f + g)(\mathbf v) &= \lim_{\mathbf v \to \mathbf v_0}f(\mathbf v) + \lim_{\mathbf v \to \mathbf v_0}g(\mathbf v), \\ \lim_{\mathbf v \to \mathbf v_0}(\alpha \cdot f)(\mathbf v) &= \alpha \cdot \lim_{\mathbf v \to \mathbf v_0}f(\mathbf v). \end{aligned}

Proof. Denote

\displaystyle \lim_{\mathbf v \to \mathbf v_0}f(\mathbf v) = \mathbf w_f,\quad  \lim_{\mathbf v \to \mathbf v_0}g(\mathbf v) = \mathbf w_g.

By Lemma 3, f_{\mathbf v_0, \mathbf w_f}, g_{\mathbf v_0, \mathbf w_g} \in o(1). Since o(1) is a topological vector space, for any \alpha \in \mathbb K,

\begin{aligned} (f+g)_{\mathbf v_0, \mathbf w_f + \mathbf w_g} &= f_{\mathbf v_0, \mathbf w_f} + g_{\mathbf v_0, \mathbf w_g} \in o(1), \\  (\alpha \cdot f)_{\mathbf v_0, \alpha \cdot\mathbf w_f} &= \alpha \cdot f_{\mathbf v_0, \mathbf w_f} \in o(1). \end{aligned}

By Lemma 3 again,

\begin{aligned} \lim_{\mathbf v \to \mathbf v_0}(f + g)(\mathbf v) &= \lim_{\mathbf v \to \mathbf v_0}f(\mathbf v) + \lim_{\mathbf v \to \mathbf v_0}g(\mathbf v), \\ \lim_{\mathbf v \to \mathbf v_0}(\alpha \cdot f)(\mathbf v) &= \alpha \cdot \lim_{\mathbf v \to \mathbf v_0}f(\mathbf v). \end{aligned}

We conclude our discussion on topology using topological vector spaces for two reasons. Firstly, we illustrate the connection between topological vector spaces in Theorem 3 to the (linear) limit laws in classical real analysis. We could develop the notion of topological algebras and even topological fields in order to recover the multiplication and division limit laws.

Secondly, we look forward. Topological vector spaces forms the basis for distribution theory that formalises “generalised functions” that are often used in the study of Fourier analysis—in particular, the Dirac delta “function” is more properly described as the Dirac delta distribution. Distributions then help us formally discuss the Dirac delta without hand-waving.

Furthermore, normed spaces are topological vector spaces, and the latter forms the foundation for functional analysis, which is applied in measure theory to establish the notion of weak convergence, and eventually help us prove useful convergence theorems in probability theory.

But for now, we are done with topology!

—Joel Kindiak, 19 Jun 25, 1132H

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