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I'm trying to prove that

Let $E$ be a topological vector space and $f:E \to \mathbb R$ linear. Then $f$ is continuous if and only if $f$ is continuous at $0$.

Could you verify if my below proof is fine?

My attempt: The direction $\implies$ is clear. Let's prove the reverse. Let $(x_d)_{d \in D}$ be a net in $E$ such that $x_d \to a \in E$. Let $x'_d := x_d -a$. Let $U$ be an open neighborhood of $0 \in E$.

Consider a map $T: E \to E, x \mapsto x-a$. Then $T$ is continuous by definition of a t.v.s., so $T^{-1} (U) =U+a$ is an open neighborhood of $a$. By net convergence, there is $n_U \in \mathbb N$ such that $x_d \in U+a$ and thus $x'_d \in U$ for all $d \ge n_U$. This means $x'_d \to 0$.

By continuity of $f$ at $0$, we get $f(x'_d) = f(x_d) - f(a) \to f(0) = 0 \in \mathbb R$. Just as previously, we get $f(x_d) \to f(a)$. This completes the proof.

Akira
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1 Answers1

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The proof is fine. If you're using nets anyway (which I would do too BTW), note that $t_a(x)=x-a$ is continuous for any $a$ (axioms of a TVS) so

$$(x_d-a)_d = t_a(x_d) \to t_a(a)=a-a=0$$ by continuity of $t_a$. So continuity at $0$ of $f$ (plus linearity of $f$) tells us that

$$f(x_d)-f(a) = f(x_d -a) \to f(0)=0$$ and so by continuity of $u(x)=x+f(a)$ (where $u$ is defined and continuous on $\Bbb R$) we get

$$f(x_d)= u((f(x_d) - f(a)) \to u(0)=f(a)$$ as required.

So it's basically a consequence of translations (in $\Bbb R$ and $E$) being continuous. "Anything at any point happens at all points" (homogeneity).

Henno Brandsma
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