Continuity of Linear Functions on LCTVS

# Continuity of Linear Functions on LCTVS

Proposition 1: Let $X$ be a locally convex topological vector space. If $f : X \to \mathbb{R}$ is linear then $f$ is continuous on $X$ if and only if there is an open neighbourhood $N_0$ of the origin such that $f$ is bounded on $N_0$. |

*If $X$ is a topological space and $f : X \to \mathbb{R}$ then $f$ is continuous at a point $x_0$ if for all $\epsilon > 0$ there exists an open neighbourhood $N_{x_0}$ of $x_0$ such that if $x \in N_{x_0}$ then $|f(x) - f(x_0)| < \epsilon$. We use this in the proof below.*

**Proof:**$\Rightarrow$ Suppose that $f$ is continuous on $X$. Then in particular, $f$ is continuous at the origin $0$. So if $\epsilon = 1 > 0$ there exists a neighbourhood $N_0$ of $0$ such that if $x \in N_0$ then $|f(x) - f(0)| < \epsilon = 1$. But observe that $f(0) = 0$ since $f$ is linear. So $|f(x)| < 1$ for all $x \in N_0$, that is, $f$ is bounded on $N_0$.

- $\Leftarrow$ Suppose that there exists an open neighbourhood $N_0$ of the origin such that $|f(x)| \leq M$ for all $x \in N_0$. Since $N_0$ is open, so is $tN_0$ for all $t > 0$. Furthermore, since $f$ is linear, then $|f(x)| \leq tM$ for all $x \in tN_0$.

- Let $x_0 \in X$ and let $\epsilon > 0$ be given. Choose $t > 0$ such that $tM < \epsilon$. Since $X$ is a locally convex topological vector space, $x_0 + tN_0$ is an open neighbourhood of $x_0$. Note that if $x \in x_0 + tN_0$ then $x - x_0 \in tN_0$ and so:

\begin{align} \quad |f(x) - f(x_0) = |f(x - x_0)| < tM < \epsilon \end{align}

- Therefore $f$ is continuous at $x_0$, and since $x_0 \in X$ was arbitrary, $f$ is continuous on $X$. $\blacksquare$

Proposition 2: Let $X$ be a locally convex topological vector space. If $f : X \to \mathbb{R}$ is linear then $f$ is continuous on $X$ if and only if $f$ is continuous at the origin. |

*Recall that if $X$ and $Y$ are topological spaces and $f : X \to Y$ then $f$ is continuous at $x \in X$ if for every open neighbourhood $V$ of $f(x)$ there exists an open neighbourhood $U$ of $x$ such that $f(U) \subseteq V$.*

**Proof:**$\Rightarrow$ Suppose that $f$ is continuous on $X$. Then trivially since $0 \in X$ we have that $f$ is continuous at $0$.

- $\Leftarrow$ Suppose that $f$ is continuous at the origin $0$. Then for every open neighbourhood $V_{f(0)}$ of $f(0)$ there exists an open neighbourhood $U_0$ of $0$ such that $f(U_0) \subseteq V_{f(0)}$.

- Let $x \in X$ and let $V_{f(x)}$ be an open neighbourhood of $f(x)$. Consider the set $V_{f(x)} - f(x)$. Note that since $\mathbb{R}$ itself a topological vector space we have that $V_{f(x)} - f(x)$ is an open set. Furthermore, since $f(x) \in V_{f(x)}$ we see that $0 \in V_{f(x)} - f(x)$ and so $V_{f(x)} - f(x)$ is an open neighbourhood of $0 = f(0)$.

- Since $f$ is continuous at the origin $0$ there exists an open neighbourhood $U_0 $] of [[$ 0$ such that:

\begin{align} \quad f(U_0) \subseteq V_{f(x)} - f(x) \end{align}

- Consider the set $U_0 + x \subseteq X$. Since $U_0$ is open and $X$ is a topological vector space, $U_0 + x$ is open. Furthermore, since $f$ is linear:

\begin{align} \quad f(U_0 + x) = f(U_0) + f(x) \subseteq V_{f(x)} \end{align}

- Therefore $f$ is continuous at $x$. Since $x \in X$ is arbitrary, $f$ is continuous on all of $X$.

Proposition 3: Let $X$ be a locally convex topological vector space. If $f : X \to \mathbb{R}$ is linear then $f$ is continuous on $X$ if and only if there exists exists a convex open neighbourhood $O$ of the origin $0$ such that $f(O) \neq \mathbb{R}$. |

**Proof:**$\Rightarrow$ Suppose that $f$ is continuous on $X$. Let $V \neq \mathbb{R}$ be an open neighbourhood of $0 = f(0) \in \mathbb{R}$. Then by definition there exists an open neighbourhood $O$ of the origin $0 \in X$ such that $f(O) \subseteq V \neq \mathbb{R}$.

- $\Leftarrow$ Suppose that there exists an open neighbourhood $O$ of the origin $0$ such that $f(O) \neq \mathbb{R}$. We may assume that $O$ is symmetric, i.e., if $x \in O$ then $-x \in O$ (if not, take a convex subset of $O$ that is symmetric about $0$). Let $\alpha \in \mathbb{R} \setminus f(O)$. Note that $\alpha \neq 0$ since $0 = f(0) \in f(O)$.

- Suppose instead that $f(O)$ is unbounded. Since $f(O)$ is symmetric convex and contains $0$ and unbounded there exists an $o_n \in O$ such that $|f(o_n)| > |\alpha| > 0$. But then $f(O)$ cannot be convex since $f(O)$ does not contain the line segment joining $0$ and $f(o_n)$. Thus $f(O)$ must be bounded. By proposition 2 we have that $f$ is continuous. $\blacksquare$