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I have a potentially silly confusion over continuous linear functionals over topological vector spaces. Many sources say a linear functional $f: X\to \mathbb{F}$, where $X$ is a general topological vector space over $\mathbb{F}$, is continuous iff $f$ is bounded on some neighborhood of $0_X$. See, eg., Theorem 1.18 of Rudin's Functional analysis (1991).

My question is: shouldn't $f$ be bounded on every neighborhood of $0_X$, for it to be continuous? Why only some is enough?

I know, when a subset $A\subset X$ of a TVS is defined as bounded, it is defined such that, for every open neighborhood $U\ni0$ of the origin, there exists $\epsilon_U>0$ s.t. $\epsilon_U A\subset U$. Here every open neighborhood is needed. So why can we get away with only some neighborhoods, when it comes to the continuity of linear functionals?

Does the same conclusion hold for linear operators?

Hope it makes sense. Thanks in advance.

Update (self-answer):

After some searching/thinking, it appears I inherently confused neighborhoods with bounded neighborhoods when asking this question. For general TVS, neighborhoods are never bounded, unless the space is (pseudo)metrizable. So, a linear operator/functional being bounded over some neighborhood $U$ of the origin already guarantees it is bounded over all bounded sets containing the origin. Because, bounded sets can be absorbed by every neighborhood of the origin, in particular, by $U$. It is also easy to prove such a linear operator/functional is cts.

For linear functionals, the converse is also true. If $f: X\to \mathbb{F}$ is continuous, then $\text{Ker}(f)$ is closed. If $f(X)=\{0\}$. It is obviously bounded everywhere. If $f(x)\neq 0$ for some $x\in X\setminus \{0_X\}$, then $\text{Ker}(f)$ is a closed proper subspace. Hence it cannot be dense. Then there exists an open neighborhood $V_x\ni x$ such that $V_x\cap \text{Ker}(f)=\emptyset$. In a TVS, every neighborhood of the origin contains a balanced neighborhood. So there is a balanced $V_x'-x\subset V_x-x$ containing $0_X$. So the image $f(V_x'-x)\subset \mathbb{F}$ is balanced and cannot be $\mathbb{F}$. Hence it is bounded.

But the converse doesn't always hold for linear operators between TVS. CTS linear operators can be unbounded over every neighborhood of the origin.

user760
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  • @AnneBauval Yes and no, as my confusion is about the difference between neighborhoods and bounded sets. – user760 Apr 04 '23 at 11:26
  • You seem to be perfectly aware of this difference, so what point precisely don't you understand when reading the proposed duplicate? – Anne Bauval Apr 04 '23 at 11:31
  • @AnneBauval I didn't realize it's wrong to assume neighborhoods were bounded when posting this question or reading through the linked answers. I didn't even realize I was assuming it. So I didn't at all realize continuous operators are "bounded on some neighborhood" and "map every bounded set to a bounded set" are two different propositions. – user760 Apr 04 '23 at 11:58
  • But now, the proposed duplicate answers your question, doesn't it? – Anne Bauval Apr 04 '23 at 12:01
  • @AnneBauval I'll wait for a few days. If no one comes and points out there are more misunderstandings in my question/answer, I'll close it. – user760 Apr 04 '23 at 12:02
  • Technically, I believe you cannot "close" it. You can only agree (in comments, or by voting if you have enough reputation) that it should be closed as a duplicate, your "Update self-answer" being but a reformulation of the accepted answer over there. – Anne Bauval Apr 04 '23 at 12:07

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