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Let $X$, $Y$ be Banach spaces, and $T\colon X\to Y$ be linear and bijective ($D(T)=X, R(T)=Y)$. Can one infer directly that $T$ is continuous? If not, is there a counterexample?

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Of course not, using vector space bases of $X,Y$ you can construct lots of bijective linear maps as soon as the vector space dimensions agree. Usually they are not continuous. – Martin Brandenburg Oct 28 '11 at 12:23
On the other hand, there is no counterexample! More precisely, it is not possible to write down a counterexample. If you assume enough choice then you can justify the existence of a counterexample, but if you assume a different family then you can assert that all such maps are continuous. I learnt this from Greg Kuperberg on this site. Take a look at his answer here: – Loop Space Oct 28 '11 at 12:34
I don't find it obvious that in this situation it is possible to get a well-defined operator on the $\underline{\mathrm{whole}}$ domain $X$, or do I miss an important point? – Marc Oct 28 '11 at 12:40
N0te: a linear bijective non continuous operator on a Banach space can be easily built starting from a non-continuous linear form on it (yet an object with no concrete examples). Check also this question and answers:… – Pietro Majer Oct 28 '11 at 15:47
I wonder about the appropriateness of this question for MO. – Todd Trimble Oct 28 '11 at 17:16
up vote 1 down vote accepted

The question was simply answered in the comments. If you choose algebraic bases (which are also called Hamel bases) for $X$ and $Y$, then it is elementary to make a bijective linear operator which is unbounded and therefore discontinuous. You only have to check that the algebraic bases have the same cardinality, but note that the algebraic dimension of an infinite-dimensional Banach space equals its cardinality. In particular, if the Banach space is separable but infinite-dimensional, then its algebraic dimension is $c = 2^{\aleph_0}$.

On the other hand, these algebraic bases require the axiom of choice. The question of whether there are any unbounded, fully defined operators between Banach spaces is known to be independent of the axioms of set theory (ZF) without the axiom of choice. (See this previous MO question.) So there will never be an explicit example. Actually you can interpret this independence result as a proof of continuity. If you describe a fully defined linear operator $T$ (whether or not it is a bijection) between Banach spaces, and if you didn't use the axiom of choice in its definition, then theorem, it's continuous.

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I'm not sure there's any way to define "fully defined" that would work. $\;$ You can't allow "If all reals are constructible, then the first discontinuous linear bijection from $\ell^2$ to itself in whatever the canonical well-order of $L$ is, otherwise the identity map from $\ell^2$ to itself.". – Ricky Demer Oct 28 '11 at 19:40
Maybe it's better to say that explicitly define... – Greg Kuperberg Oct 29 '11 at 0:26

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