Normal operator + only real eigenvalues implies self-adjoint operator?

Question

Let say we are in a complex vector space, is there an example of a normal operator with only real eigenvalues(or without eigenvalues) that is not a self-adjoint operator? Cause of the spectral theorem it is impossible for the finite dimensional case. I have no idea in the infinite case. I would appreciate any help! Thanks!

Answer 1

To elaborate on fedja's comment: Let $(X,\mu)$ be a measure space, let $h$ be a bounded measurable complex-valued function on $X$, and let $T$ be the multiplication operator on $L^2(X,\mu)$ defined by $Tf = hf$. Show that $T$ is normal, and is self-adjoint iff $h$ is real-valued almost everywhere. Now show that $\lambda$ is an eigenvalue of $T$ iff $\mu(\{h= \lambda\}) > 0$. Taking as an example $X = [0,1]$ with Lebesgue measure, you should be able to use this to construct a normal, non-self-adjoint operator with only real eigenvalues, or with no eigenvalues at all.

The correct statement would be if a normal operator has only real spectrum (i.e. $\sigma(T) \subset \mathbb{R}$) then it is self-adjoint. But in infinite dimensions, an operator's spectrum may be more than just its eigenvalues.