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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?

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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 $\mathsf{ZF+AD}$.

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If OP knew only half of the technical terms here he wouldn't have asked this question, and especially not in this way. –  Martin Jan 27 '13 at 14:24
    
@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
    
model, $\mathsf{ZF + \lnot AC}$, Solovay's model, $\mathsf{ZF+AD}$. –  Martin Jan 27 '13 at 14:30
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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? –  Montez Jan 27 '13 at 18:58
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