Inverse of operator is not continuous in Banach spaces

Question

Let $X$ be a Banach space. If $A:X\to X$ is an invertible bounded operator (injective, surjective and continuous), then $A^{-1}$ is also bounded. Now can I have an example of an unbounded operator $A:D(A)\to X$ which is invertible but $A^{-1}:X\to D(A)$ is not bounded ?

Answer 1

See the MathOverflow question https://mathoverflow.net/questions/5303/basis-of-linfinity to learn why it is consistent with ZF (without the axiom of choice) that there is no such operator.

Discontinuous everywhere defined operators can be defined using Hamel bases. In particular, as Daniel Fischer points out, one can use a Hamel basis for $X$ to define discontinuous linear bijections $X\to X$ that answer your question. All you really need is to define a discontinuous injective linear map $B$ on $X$, then take $A=B^{-1}$ on $B(X)=D(A)$.