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Let's note $E=C([0,1],\mathbb{R})$ the Banach space of real continuous funtions from the [0,1] interval with the uniform norm.

Is it possible to show a non-continuous linear form on $E$ exists without using a basis, i.e. without AC?

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No. It's impossible.

In certain models of $\sf ZF+DC$ there is a property known as "automatic continuity" for Banach spaces, that means that every linear operator to a normed space is continuous.

Such models are, for example, Solovay's model where all sets of reals are Lebesgue measurable, and have the Baire property; and Shelah's model where all sets of reals have the Baire property. Note that Solovay's model require the existence of an inaccessible cardinal, a statement which is unprovable from the usual axioms of set theory, whereas Shelah's model does not require assumptions beyond those of $\sf ZF$ (and was in fact used to show that the consistency of the statement "all sets of reals have the Baire property" requires no additional hypothesis).


You can find a nice exposition to Solovay's model in Aki Kanamori's "The Higher Infinite", as well Solovay's original paper "A model of set-theory in which every set of reals is Lebesgue measurable".

Shelah's result is much more difficult (but more rewarding too, as it omits the requirement of an inaccessible cardinal), and can be found in the seventh section of Shelah's celebrated paper "Can you take Solovay's inaccessible away?". The paper is long and requires quite an understanding of forcing and set theory in order to full appreciate its content.

To see why either one of these results imply the answer above, see Kechris' "Classical Descriptive Set Theory", there he proves Pettis theorem stating that if a group homomorphism between Polish groups is Baire measurable then it is continuous. The proof goes through in $\sf ZF+DC$, and while it does not immediately imply the automatic continuity for all Banach spaces, the one in the question is indeed separable and so forms a Polish group.

However a nice argument suggested to me by Harvey Friedman is that in $\sf ZF+DC$ continuity and sequential continuity are equivalent. If $T$ was a discontinuous linear operator on a non-separable space, then we could have found a sequence which witnesses that, and the restriction of $T$ to the separable space generated by that sequence has to be discontinuous as well.

Other authors have dealt with the question of automatic continuity (for Banach spaces) in such models. Amongst them are Garnir, Wright and Brunner. You can also find some information in Eric Schechter's "Handbook of Analysis and Its Foundations".

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Thanks Asaf for your answer. Maybe you have a reference for the construction of such models? –  Jean-Pierre Merx Oct 14 '13 at 18:05
    
I was actually trying to come up with a good reference for those. Solovay's construction is "rather simple" (as a construction of a symmetric extension), but Shelah's model is much much more complicated. –  Asaf Karagila Oct 14 '13 at 18:12
    
I hope you will find the references, some are trickier to find than others. –  Asaf Karagila Oct 14 '13 at 18:27
    
I can see why Pettis's theorem implies automatic continuity in Shelah's model (if every set is Baire measurable, then every homomorphism is Baire measurable hence continuous). What do you do in Solovay's model? It's not clear to me how to exploit Lebesgue measurability. –  Nate Eldredge Oct 15 '13 at 15:29
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@Nate: In Solovay's model every set of reals has the Baire property as well. But one can prove that directly from measurability in a similar manner. That's what Wright and Garnir essentially do, if I recall correctly. I'll put some clarifications later on tonight (there was a minor addition I wanted to make anyway). Thanks for the comment. –  Asaf Karagila Oct 15 '13 at 15:31

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