Example of a normal operator which has no eigenvalues Is there a normal operator which has no eigenvalues?
If your answer is yes, give an example.
Thanks.

The answer is yes, if you allow the normal operator to be non-compact. The spectral theorem for compact normal and self-adjoint operators guarantees eigenvalues, but in general, for general bounded (or even unbounded) operators, the "big" spectral theorem is much more complicated and does not guarantee eigenvalues.
To find a counterexample, pick your Banach space to be an appropriate function space, and make your operator a multiplication operator. e.g. make your operator $T(f(x)) = g(x) f(x)$, for some fixed function $g(x)$.
