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?


1 Answer 1


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 $\mathsf{ZF+AD}$.

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    $\begingroup$ If OP knew only half of the technical terms here he wouldn't have asked this question, and especially not in this way. $\endgroup$
    – Martin
    Jan 27, 2013 at 14:24
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    $\begingroup$ @Martin: What technical terms? linear? transformation? Banach space? $\endgroup$
    – Asaf Karagila
    Jan 27, 2013 at 14:25
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    $\begingroup$ @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. $\endgroup$
    – Asaf Karagila
    Jan 27, 2013 at 14:35
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    $\begingroup$ 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? $\endgroup$
    – Spook
    Jan 27, 2013 at 18:58
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    $\begingroup$ @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. $\endgroup$
    – Asaf Karagila
    Jul 8, 2015 at 16:18

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