It is well-known that, assuming the axiom of choice, there are unbounded linear maps defined not only on a dense subset but on all of Hilbert space. Is it possible that such a map is invertible?



Invertible unbounded linear maps defined on a Hilbert space
Yes.
Consider the space $E=\ell^2(\mathbb N)$.
($0\notin\mathbb N$.)
It has an orthonormal Hilbert basis $A=\{e\_i;i\in\mathbb N\}$, where $e\_i(n)=\delta\_{in}$.
Extend this to a Hamel basis $B$ of $E$ (this is where the axiom of choice comes in).
Then define $L:E\to E$ by $L(e\_i)=ie\_i$ and $L(x)=x$ for $x\in B\setminus A$ (and extend linearly).
The mapping is diagonal in the Hamel basis, so bijectivity is easy to check.
Also it maps the bounded set $A$ to an unbounded one, so it is not continuous.


