Separation of a Subspace by a Continuous Linear Function on LCTVS

Recall from The Separation Theorems page that if $X$ is a locally convex topological vector space and $K_1$, $K_2$ are disjoint closed convex subsets of $X$ and $K_1$ is compact then there exists a continuous linear function $f$ on $X$ and a constant $c \in \mathbb{R}$ such that for all $k_1 \in K_1$ and for all $k_2 \in K_2$:

(1)

\begin{align} \quad f(k_1) < c < f(k_2) \end{align}

We noted that as a special case of the above theorem $X$ is a locally convex topological vector space and $K$ is a closed convex subset of $X$ and $x_0 \in X \setminus K$ then there exists a continuous linear function $f$ on $X$ such that:

(2)

\begin{align} \quad f(x_0) < \inf_{k \in K} f(k) \end{align}

And if $X$ is a normed linear space and $K$ is a closed convex subset of $X$ with $x_0 \in X \setminus K$ then there exists a bounded linear functional $f \in X^*$ such that:

(3)

\begin{align} \quad f(x_0) < \inf_{k \in K} f(x) \end{align}

We will now use the above theorems to prove a very important result.

Proposition 1: Let $X$ be a locally convex topological vector space. If $Y$ is a proper closed subspace of $X$ then for every $x_0 \in X \setminus Y$ there exists a continuous linear function $f$ on $X$ such that $f(x_0) \neq 0$ and $f |_Y = 0$.

• Proof: If $Y$ is a proper closed subspace of $X$ then $Y$ is closed, and convex. By the theorem referenced above, for $x_0 \in X \setminus Y$ there exists a continuous linear function $f$ on $X$ such that:

(4)

\begin{align} \quad f(x_0) < \inf_{y \in Y} f(y) \end{align}

• Suppose that there exists a $y_0 \in Y$ such that $f(y_0) \neq 0$. Suppose that $f(y_0) = a$ where $a \neq 0$. Since $Y$ is a subspace, $ty_0 \in Y$ for every $t \in \mathbb{R}$. So $f(ty_0) = tf(y_0) = ta$. But then $f$ cannot be bounded and is hence not continuous, a contradiction. So $f(y) = 0$ for all $y \in Y$ and hence $\inf_{y \in Y} f(y) = 0$. Thus $f(x_0) \neq 0$ and $f |_Y = 0$. $\blacksquare$