Let $E$ be a TVS and $f:E \to \mathbb R^n$ linear. If $\ker f$ is closed, then $f$ is continuous at $0_E$

Question

Previously, I showed that

Theorem Let $E$ be a (not necessarily Hausdorff) real TVS and $f:E \to \mathbb R$ linear. If $\ker f$ is closed, then $f$ is continuous at $0_E$.

The proof is as follows:

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Assume that $H := \ker f$ is closed and thus $H^c$ is open. WLOG, we assume $f$ is not constant. Then there exist $a \in H^c$ and an open neighborhood (nbh) $U$ of $0_E$ such that $U +a \subseteq H^c$.

Let $T:\mathbb R \times E \to E, (t, x) \mapsto tx$ be the scalar multiplication. Then $T$ is continuous. Then there are open nbh $V_1$ of $0_{\mathbb R}$ and open nbh $U_1$ of $0_E$ such that $T (V_1 \times U_1) \subseteq U$. There is $t_1 >0$ such that $I := (-t_1, t_1) \subseteq V_1$. Let $U_2 := \bigcup_{t \in I} t U_1 \subseteq U$.

We claim that $f(U_1)$ is bounded. If not, for each $m \in \mathbb N$, there is $x_m \in U_1$ such that $|f(x_m) |> m$. Then $m t_1(-1, 1) \subseteq f(Ix_m)$ for all $m \in \mathbb N$. Then $f(U_2) = \mathbb R$. So there is $b \in U$ such that $f(b + a) = 0_\mathbb R$, which is a contradiction. So $f(U_1)$ is bounded. Hence there is $t_2 >0$ such that $f(U_1) \subseteq (-t_2, t_2)$.

If $N = (-t_3, t_3)$ is a nbh of $0_\mathbb R$, then $f \left (\frac{t_3}{t_2}U_1 \right) \subseteq N$. It follows that $f$ is continuous at $0_E$.

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I tried but could not adapt above proof to the case $f:E \to \mathbb R^n$. Could you elaborate on how to do so?

Answer 1

Not an adaption but a different proof. For $H=$ker$(f)$ let $q:E\to E/H$ be the quotient map. Since $H$ is closed the quotient topology on $E/H$ is Hausdorff, and since finite dimensional spaces have a unique Hausdorff TVS topology the induced linear map $\tilde f: E/H\to \mathbb R^n$ (mapping the equivalence class of $x$ to $f(x)$) is continuous. This implies the continuity of $f=\tilde f\circ q$.