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On every infinite-dimensional Banach space there exists a discontinuous linear functional.

Assuming the axiom of choice, every vector space has a basis. With an infinite basis, I can define on a countable subset $\{e_n:n\in\mathbb{N}\}$ a function $f(e_n)=n\|e_n\|$ and let $f(x)=1$ for all other basis vectors.

Then this determines an unbounded linear functional, which is therefore discontinuous.

But this argument, a, applies to any infinite-dimensional normed spaces, b, relies on the assumption of the axiom of choice.

Is there a smart answer that does make use of the condition that the space in question is a Banach space, and even better, avoids the use of axiom of choice?

Asaf Karagila
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Spook
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1 Answers1

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No. There are models of $\mathsf{ZF+\lnot AC}$ in which every linear transformation from a Banach space to a normed space is automatically continuous. In particular this is true for linear functionals.

In such models, it follows, every linear functional has to be continuous.

An example for these models are Solovay's model, or models of $\mathsf{ZF+AD}$.

Asaf Karagila
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    @Martin: What technical terms? linear? transformation? Banach space? – Asaf Karagila Jan 27 '13 at 14:25
  • @Martin: Maybe you mean $\mathsf{ZF}$ and the symbol $\lnot$? – Asaf Karagila Jan 27 '13 at 14:27
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    model, $\mathsf{ZF + \lnot AC}$, Solovay's model, $\mathsf{ZF+AD}$. – Martin Jan 27 '13 at 14:30
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    @Martin: It's lovely that you decide what the OP may or may not know. When I was an undergrad student and learned functional analysis in my third year I knew what does the important symbols here mean. The last line is just to give an example for the interested reader. Nothing more. And sure, I'm a set theorist now, but I wasn't back then. Please let the OP post a comment and ask for further clarification, that if history is any indication, I will be more than happy to write. – Asaf Karagila Jan 27 '13 at 14:35
  • Keep cool. I didn't decide anything. The only set-theoretic term appearing in the post is "axiom of choice". I was merely trying to point out in a clumsy way that what may seem a perfectly natural way for you to phrase things involves quite a bit of guesswork for the uninitiated. – Martin Jan 27 '13 at 14:43
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    Thank you. I think I roughly know what you mean--under the ZF axioms and "not AC", linear functionals must be continuous, so the assumption of AC is necessary. Does this also mean that it is impossible to write down explicitly a discontinuous linear functional for some specific infinite-dimensional Banach space? – Spook Jan 27 '13 at 18:58
  • @Montez: Almost, in the sense that if we replace AC by its negation it still might be that there are discontinuous linear functionals. However it is possible to replace "not AC" by a stronger assumption from which we can prove that there is no such functional. So we cannot write an explicit linear functional, we can only "start the recipe" and let the axiom of choice (in one form or another) finish the work for us. – Asaf Karagila Jan 27 '13 at 19:10
  • @AsafKaragila Is it true in any of these models that there exist infinite dimensional vector space on which we can't define norm? – Sushil Jul 08 '15 at 15:50
  • @AsafKaragila what is the relationship of Axiom: ''On every infinite-dimensional Banach space there exists a discontinuous linear functional." with AC. Is it equivalent? – Sushil Jul 08 '15 at 16:01
  • @AsafKaragila what does ZF+AD stands for. I know ZF + axiom of dependent choice(but with ¬AC, right) – Sushil Jul 08 '15 at 16:02
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    @Sushil: (1) Please don't flood me with questions. I'm more than happy to answer, but try to pace yourself. It's quite terrible to come to the site and find a bunch of comments posted in succession and each asking a different question, whose answers you may or may not expect to be somewhat detailed. (2) I think that the answer for the first question is positive, I'm not 100% sure though, but it is possible in general to add these sort of spaces along with, say, many Banach spaces without discontinuous functionals. (3) If my memory serves me right, this is equivalent to the Hahn-Banach theorem. – Asaf Karagila Jul 08 '15 at 16:18
  • @Sushil: I don't fully remember if this is indeed equivalent to Hahn-Banach, and I'll have to check. (4) AD is the Axiom of Determinacy. – Asaf Karagila Jul 08 '15 at 16:18
  • Ok I'll try to ask questions in a better way. And as was suggested by you I am studying Jech book. And I am about to start model theory. So I want answers in detailed because it will help me in solving these problems when I'll be studying models.(So it is not like I am just asking questions because I want just a word from you Yes or no. But as always you give answers in much more details) – Sushil Jul 08 '15 at 16:30
  • @Sushil: Yes, in Solovay's model there exist infinite dimensional vector spaces on which there is no norm. See http://math.stackexchange.com/a/1018905/822. (I cite a model produced by Shelah but it has the same properties as the Solovay model mentioned here.) – Nate Eldredge Oct 11 '15 at 03:39