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

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!

• Consider the multiplication by your favorite bounded complex valued function that doesn't take any particular value on a set of positive measure in $L^2$. – fedja Feb 15 '13 at 15:34
• @fedja What does it mean, not to take any particular value on a set? – Julien Feb 15 '13 at 15:48

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.