Is the CPT-norm always positive?

We give an explicit example of an exactly solvable PT-symmetric Hamiltonian with the unbroken PT symmetry which has one eigenfunction with the zero PT-norm. The set of its eigenfunctions is not complete in corresponding Hilbert space and it is non-diagonalizable. In the case of a regular Sturm-Liouville problem any diagonalizable PT-symmetric Hamiltonian with the unbroken PT symmetry has a complete set of positive CPT-normalazable eigenfunctions. For non-diagonalizable Hamiltonians a complete set of CPT-normalazable functions is possible but the functions belonging to the root subspace corresponding to multiple zeros of the characteristic determinant are not eigenfunctions of the Hamiltonian anymore.