Closed Preimage Criterion for a Linear Form to be Continuous in a TVS

# Closed Preimage Criterion for a Linear Form to be Continuous in a TVS

Lemma 1: Let $E$ be a vector space and let $f$ be a nonzero linear form on $E$, with $a \in E$ such that $f(a) = 1$. If $U$ is a balanced set, then $(a + U) \cap f^{-1}(0) = \emptyset$ if and only if $U \subseteq \{ x : |f(x)| < 1 \}$. |

**Proof:**$\Rightarrow$ Let $(a + U) \cap f^{-1}(0) = \emptyset$. Suppose that there exists a $u \in U$ such that $|f(u)| \geq 1$. Then observe that $\displaystyle{\left | -\frac{1}{f(u)} \right | \leq 1}$. Since $U$ is a balanced set, we have that:

\begin{align} \quad \frac{-1}{f(u)} U \subseteq U \end{align}

- In particular, $\displaystyle{-\frac{u}{f(u)} \in U}$. So $\displaystyle{a - \frac{u}{f(u)} \in a + U}$. But:

\begin{align} \quad f \left ( a - \frac{u}{f(u)} \right ) = f(a) - \frac{f(u)}{f(u)} = 0 \end{align}

- so that $\displaystyle{a - \frac{u}{f(u)} \in f^{-1}(0)}$, which implies that $(a + U) \cap f^{-1}(0) \neq \emptyset$, which is a contradiction. Hence for all $u \in U$ we must have that $|f(u)| < 1$, so $U \subseteq \{ x : |f(x)| < 1 \}$.

- $\Leftarrow$ Let $U \subseteq \{ x : f(x) < 1 \}$. Then if $y \in a + U$, write $y = a + u$ where $u \in U$. Then $|f(u)| < 1$, and so:

\begin{align} \quad f(y) = f(a + u) = f(a) + f(u) = 1 + f(u) \neq 0 \end{align}

- $y \not \in f^{-1}(0)$. Since this is true for all $y \in a + U$, we conclude that $(a + U) \cap f^{-1}(0) = \emptyset$.

Theorem 2: Let $E$ be a topological vector space and let $f$ be a linear form on $E$. Then $f$ is continuous if and only if $f^{-1}(0)$ is closed. |

**Proof:**Observe that if $f$ is a zero form, then $f$ is certainly continuous and $f^{-1}(0) = E$ is certainly closed. So we will assume that $f$ is a nonzero linear form.

- $\Rightarrow$ Let $f$ be continuous. Since $\{ 0 \}$ is a closed subset of $\mathbb{R}$, we have by the continuity of $f$ that $f^{-1}(0)$ is closed.

- $\Leftarrow$ Let $f^{-1}(0)$ be closed. Since $f$ is a nonzero linear form, there exists an $a \in E$ such that $f(a) = 1$.

- Since $f^{-1}(0)$ is closed, $E \setminus f^{-1}(0)$ is open, and so $[E \setminus f^{-1}(0)] - a$ is an open neighbourhood of the origin (it contains the origin, since $a \in E \setminus f^{-1}(0)]$, and so $o = a - a \in [E \setminus f^{-1}(0)] - a$) . Since $E$ is a topological vector space, by the proposition on the Every TVS Has a Base of Closed and Balanced Neighbourhoods of the Origin page, $E$ has a base of (closed) and balanced neighbourhoods of the origin. So there exists a balanced neighbourhood $U$ of the origin such that:

\begin{align} \quad U \subseteq [E \setminus f^{-1}(0)] - a \end{align}

- And thus;

\begin{align} \quad a + U \subseteq E \setminus f^{-1}(0) \end{align}

- Hence $U$ is a balanced neighbourhood such that $(a + U) \cap f^{-1}(0) = \emptyset$. By the previous lemma, this implies that $U \subseteq \{ x : |f(x)| < 1 \}$. But then $U$ is a neighbourhood of the origin for which $f$ is bounded on $U$. So by a proposition on the Continuous Linear Forms on a TVS and its Continuous Dual page we have that $f$ is continuous. $\blacksquare$