Criterion for a LCTVS to be Metrizable

Theorem 1: Let $E$ be a locally convex topological vector space. Then $E$ with its topology is metrizable if and only if $E$ with its topology is Hausdorff and possesses a countable base of neighbourhoods of the origin.

• Proof: $\Rightarrow$ Suppose that $E$ is metrizable. Then there exists a metric $d$ on $E$ where the topology defined by the metric $d$ coincides with the topology on $E$. Given $x, y \in E$, the open ball centered at $x$ with radius $\displaystyle{\frac{d(x, y)}{2}}$, and the open centered at $y$ with radius $\displaystyle{\frac{d(x, y)}{2}}$ are disjoint neighbourhoods of $x$ and $y$ respectively, and so $E$ is Hausdorff. Moreover, $\mathcal U := \left \{ \left \{ z : d(o, z) < \frac{1}{n} \right \} : n \in \mathbb{N} \right \}$ is a countable base of neighbourhoods of the origin.

• $\Leftarrow$ Suppose that $E$ is Hausdorff and possesses a countable base of neighbourhoods of the origin for its topology. Since $E$ is a locally convex topological vector space, by the result on the Every LCTVS Has a Base of Closed Absolutely Convex Absorbent Neighbourhoods of the Origin page we have that $E$ has a countable base of absolutely convex and absorbent neighbourhoods of the origin for its topology. Let $\{ U_1, U_2, ... \}$ be such a base of neighbourhoods of the origin.

• For each $n \in \mathbb{N}$, let $p_n := p_{U_n}$ be the gauge of $U_n$, and let $f : X \to [0, \infty)$ be defined for each $x \in X$ by:

(1)

\begin{align} \quad f(x) := \sum_{n=1}^{\infty} \frac{1}{2^n} \inf \{ p_n(x), 1 \} \end{align}

• Then observe that if $f(x) = 0$ then $p_n(x) = 0$ for all $n \in \mathbb{N}$. But since $E$ is Hausdorff and by the result on the Criterion for the Coarsest Topology Determined by a Set of Seminorms to be Hausdorff page, we have that this implies that $x = o$. Conversely, if $x = o$ then $p_n(0) = 0$ since each $p_n$ is a seminorm, and thus $f(x) = 0$.

• Also, for each $x \in E$, we have that $p_n(-x) = |-1| p_n(x) = p_n(x)$ since each $p_n$ is a seminorm, and thus $f(x) = f(-x)$.

• Lastly, if $x, y \in E$ then $p_n(x + y) \leq p_n(x) + p_n(y)$ for each $n \in \mathbb{N}$, since each $p_n$ is a seminorm, so that $\inf \{ p_n(x + y), 1 \} \leq \inf \{ p_n(x), 1 \} + \inf \{ p_n(y), 1 \}$, and:

(2)

\begin{align} f(x + y) = \sum_{n=1}^{\infty} \frac{1}{2^n} \inf \{ p_n(x + y), 1 \} \leq \sum_{n=1}^{\infty} \frac{1}{2^n} \inf \{ p_n(x), 1 \} + \sum_{n=1}^{\infty} \frac{1}{2^n} \inf \{ p_n(y), 1 \} = f(x) + f(y) \end{align}

• Let $d : E \times E \to [0, \infty)$ be defined for all $x, y \in E$ by:

(3)

\begin{align} \quad d(x, y) = f(x - y) \end{align}

• We claim that $d$ is a metric on $E$. Indeed, $d(x, y) = 0$ if and only if $f(x - y) = 0$, which happens if and only if $x - y = 0$, which happens if and only if $x = y$.

• Secondly, $d(x, y) = f(x - y) = f(-(x - y)) = f(y - x) = d(y, x)$.

• And lastly, for all $x, y, z \in E$, we have that:

(4)

\begin{align} \quad d(x, y) = f(x - y) = f((x - z) + (z - y)) \leq f(x - z) + f(z - y) = d(x, z) + d(z, y) \end{align}

• So $d$ is indeed a metric on $E$.

• Note that a countable base of metric open balls of the origin for the metric topology, $\mathcal V = \{ V_n : n \in \mathbb{N}$ are given for each $n \in \mathbb{N}$:

(5)

\begin{align} \quad V_n := \left \{ x : d(x, 0) < \frac{1}{2^n} \right \} = \left \{ x : f(x) < \frac{1}{2^n} \right \} = f^{-1} \left ( \left ( -\frac{1}{2^n}, \frac{1}{2^n} \right ) \right ) \quad (\star) \end{align}

• Note that each seminorm $p_n = p_{U_n}$ is continuous since $U_n$ is a neighbourhood of the origin (see the proposition on the Continuity of Seminorms on Vector Spaces page), and so each $\displaystyle{\frac{1}{2^n} \inf \{ p_n, 1 \}}$. Note that the sum $\displaystyle{\sum_{n=1}^{\infty} \frac{1}{2^n} \inf \{ p_n(x), 1 \}}$ is uniformly convergent, since for all $x \in E$:

(6)

\begin{align} \quad \left | \sum_{n=1}^{\infty} \frac{1}{2^n} \inf \{ p_n(x), 1 \} \right | \leq \left | \sum_{n=N}^{\infty} \frac{1}{2^n} \right | \to 0 \end{align}

• as $N \to \infty$. Thus since each $\displaystyle{\frac{1}{2^n} \inf \{ p_n(x), 1 \}}$ is continuous and since the sum of these functions is uniformly convergent, we have that $f$ is continuous on $E$ with its given topology. So from $(\star)$, each $V_n$ is open in $E$ with its given topology.

• We also have that$V_n \subseteq U_n$. To see this inclusion, take $x \in V_n$. Then $\displaystyle{f(x) < \frac{1}{2^n}}$ which implies that $p_n(x) < 1$ (if instead $p_n(x) \geq 1$ then $\displaystyle{\frac{1}{2^n} \inf \{ p_n(x), 1 \} = \frac{1}{2^n}}$ and so $f(x) \geq \frac{1}{2^n}$). So $x \in \{ y : p_n(y) < 1 \} \subseteq U_n$ and thus $V_n \subseteq U_n$.

• So the topology defined by the countable base of neighbourhoods $\{ U_1, U_2, ... \}$ of the origin is the same as the topology defined by $\mathcal V$. So $E$ with its topology defined by its countable base $\{ U_1, U_2, ... \}$ is metrizable. $\blacksquare$