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:

---

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 $0_E$.

---

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 $f=\tilde f\circ q$.

Answer 2

I am just a student so take my proof carefully.
The main problem that I see in the proof of Jochen, is as far that I can understand it, it implies that $E$ is of finite dimension (if not how can he conclude that the topology quotient is $E/H$ is of finite dimension ?).

1 - W.l.o.g we suppose $ \exists e \in E $ s.t. $T(e)=1$ and lets remember that in order for a linear mapping to be continuous it is enough to prove that it is continuous at $0$.
Now we look at all the sequence $(x_n)_{\mathbb{N}} \in E$ s.t. $x_n \overset{n \to \infty }{\rightarrow} 0$. By the sequence criterium of continuity if $ \forall (x_n)_{\mathbb{N}} \in E$ s.t. $x_n \overset{n \to \infty }{\rightarrow} 0$ we have too that $T(x_n ) \overset{n \to \infty }{\rightarrow} 0$ then $T$ will indeed be continuous at $0$.

2 - Now by absurd we suppose that it exists at least one sequence $(u_n)_{\mathbb{N}} \in E$ s.t. $u_n \overset{n \to \infty }{\rightarrow} 0$ but $ T(u_n ) $ does not converge to $0$.
Then it exists a sub sequence $T(x_{n_k})$ that in particular does not converge to $0$ that means to say that $\exists \epsilon ' > 0$ such that $|T(x_{n_k})|> \epsilon '$.

3 - Now we define the sequence $y_n = e - \frac{x_{n_k}}{T(x_{n_k})}$ by choice of $e$ we have necessarily that $ T(y_n)=0 $ (hence $y_n \in Ker(T)$ ).
Moreover we have that $ || y_n - e || = || \frac{x_{n_k}}{T(x_{n_k})} || \leq \frac{|| x_{n_k} ||}{\epsilon '} $ and because $x_{n_k}$ goes to $0$ we have by the sandwich theorem that $y_n$ converges to $e$ while in the same time $y_n \in Ker(T)$.

4 - By hypothesis $ker(T)$ is closed so $e$ must belong to $Ker(T)$ and that obviously contradict the choice of $e$ (because we choosed $T(e)=1$).

I hope this is correct and that it will help futur user.

Q.E.D.