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?



linear bijective operator
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](http://www.rac.es/ficheros/doc/00213.pdf). 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](https://mathoverflow.net/questions/5303/basis-of-linfinity/5313#5313).) 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.


