Separation of a Subspace of a Normed Vector Space

# Separation of a Subspace of a Normed Vector Space

Recall from the Locally Convex Topological Vector Spaces page that a locally convex topological vector space (LCTVS) is a vector space $X$ with a Hausdorff topology with the following properties:

**1)**Vector addition is continuous.

**2)**Scalar multiplication is continuous.

**3)**There exists a local base at $0$ that consists of convex sets.

We are about to prove a nice result which classifies a bunch of vector spaces as LCTVSs. We will first need the following definition.

Definition: Let $X$ be a normed linear space and let $W \subseteq X^*$ be a subspace (of $X^*$). Then $W$ is said to Separate the Points of $X$ if for every pair of distinct points $x, y \in X$ ($x \neq y$) there exists $\varphi \in W$ such that $\varphi(x) \neq \varphi(y)$. |

Theorem 1: Let [$ X $]] be a normed linear space and let $W \subseteq X^*$ be a subspace (of $X^*$) which separates the points of $X$. Then $X$ equipped with the $W$-weak topology is a locally convex topological vector space. |

**Proof:**We break the proof of this theorem up into several parts.

**Showing that the $W$-weak topology is Hausdorff:**Let $x, y \in X$ be such that $x \neq y$. Then there exists a $\varphi \in W$ such that $\varphi(x) \neq \varphi(y)$. Since $\varphi$ is continuous and $\varphi(x) \neq \varphi(y)$ there exists an $\epsilon > 0$ such that:

\begin{align} \quad \{ u \in X : |\varphi(x) - \varphi(u)| < \epsilon} \cap \{ u \in X : |\varphi(y) - \varphi(u)| < \epsilon \} = \emptyset \end{align}

- That is:

\begin{align} \quad V_{\epsilon, \varphi, x} \cap V_{\epsilon, \varphi, y} = \emptyset \quad (*) \end{align}

- So for every pair of distinct points in $X$ there exists $W$-weak open sets $V_{\epsilon, \varphi, x}$ of $x$ and $V_{\epsilon, \varphi, y}$ of $y$ such that $(*)$ holds, so $X$ is Hausdorff with respect to the W-weak topology.

**Showing that there is a local base of $0$ consisting of convex sets:**Consider the following collection of sets:

\begin{align} \quad \mathcal B_0 = \{ V_{\epsilon, F, 0} : \epsilon > 0, \: F \subseteq W \: \mathrm{is \: finite} \} \end{align}

- Fix $\epsilon > 0$ and $F \subseteq W$ finite. Let $x, y \in V_{\epsilon, F, 0}$ and let $0 \leq t \leq 1$. Then for each $\varphi \in F$ we have that:

\begin{align} \quad |\varphi(tx + (1-t)y) - \varphi(0)| &= |\varphi(tx + (1-t)y)| \\ &= | t \varphi(x) + (1-t) \varphi(y)| \\ & \leq t | \varphi(x)| + (1-t)|\varphi(y)| \\ & \leq t | \varphi(x) - \varphi(0)| + (1 - t)|\varphi(y) - \varphi(0)| \\ & < t \epsilon + (1-t) \epsilon \\ & < \epsilon \end{align}

- Therefore $tx + (1-t)y \in V_{\epsilon, F, 0}$, so each $V_{\epsilon, F, 0}$ is a convex subset of $X$ and hence $\mathcal B$ is a local base of $0$ consisting of convex sets.

**Showing that vector addition and scalar multiplication is continuous:**Let $x_1, x_2 \in X$, $\epsilon > 0$, and $F \subseteq W$ be finite. Let $u, v \in X$ and suppose that:

\begin{align} \quad u + v \in V_{\epsilon, F, x_1 + x_2} \end{align}

- Choose $\delta$ such that:

\begin{align} \quad 0 < \delta < \min_{\varphi \in F} \{ \epsilon - |\varphi(u + v) - \varphi(x_1 + x_2)| \} \end{align}

- (Observe that the minimum above is well-defined since $F$ is a finite set). Now let $z_1 \in V_{\frac{\delta}{2}, F, u}$ and $z_2 \in V_{\frac{\delta}{2}, F, v}$. Then:

\begin{align} \quad |\varphi(z_1 + z_2) - \varphi(x_1 + x_2)| &= | \varphi(z_1) + \varphi(z_2) - \varphi(x_1 + x_2)| \\ & = | \varphi(z_1) - \varphi(u) + \varphi(z_2) - \varphi(v) + \varphi(u) + \varphi(v) - \varphi(x_1 + x_2)| \\ & \leq | \varphi(z_1) - \varphi(u)| + |\varphi(z_2) - \varphi(v)| + |\varphi(u + v) - \varphi(x_1 + x_2)| \\ & < \frac{\delta}{2} + \frac{\delta}{2} + |\varphi(u + v) - \varphi(x_1 + x_2)| \\ & < \epsilon \end{align}

- Therefore $z_1 + z_2 \in V_{\epsilon, F, x_1 + x_2}$. Therefore $(u, v)$ is contained in $V_{\frac{\delta}{2}, F, u} \times V_{\frac{\delta}{2}, F, v}$ which is contained in the inverse image of $V_{\epsilon, F, x_1 + x_2}$ of the addition map. So the addition map is continuous.

- An analogous argument shows that scalar multiplication is also continuous.

- Therefore $X$ with the $W$-weak topology is a locally convex topological vector space. $\blacksquare$