I have some question regarding normal operators. I have been given the following definition for the case of bounded linear operators $A\in\mathcal{A}(H)$: $A^*A=AA^*$. My question is: what happens if we have an unbounded linear operator? From what I have found on the web, normal operators also impose some criteria on the domains $D(A)$ and $D(A^*)$.
I wondered as such if, for instance, the following (unbounded) linear operator $T$ that acts on an (infinite-dimensional) orthonormal basis $\{ e_n\}_{n=1}^{\infty}$ of a Hilbert space $H$ as follows:
$T(e_n) = n e_n,$
is normal.
It seems to me that
$T^*(e_n) = n e_n,$
and apparently then $T T^* = T^*T$.
I would also say that one could choose $D(T) = D(T^*)$ so, could one conclude that $T$ is a normal operator?
I am a bit puzzled since then such an operator seems a symmetric one. Then, wouldn't the Hellinger-Toeplitz theorem would imply that it is bounded? Obviously there must be some errors in the above reasoning...
Just an addendum, normality and self-adjointnes (and in fact closedness) depends very much on the domain on which your operator is considered. If you consider the operator you state above just on the subspace generate by the standard basis; your operator is not normal (becasue it is not closed). If, on the other hand, you consider it on the domain $$ \mathcal{D}(A):=\left\{x=(x_n)_{n \in \mathbb{N}} \in \ell^2(\mathbb{N})\mid \sum_{n=1}^\infty n^2|x_n|^2<\infty \right\}, $$ then $A$ is in fact self-adjoint and thus also normal.