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?
1 Answer
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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.