Consider a normal operator on a complex Hilbert space with its spectrum contained in the real line. Show that the operator is hermitian without using the spectral theorem.
What I have tried- Since the spectrum is inside the real line, the boundary of the spectrum as a subset of the complex numbers is the spectrum itself. Hence the spectrum is equal to the approximate point spectrum. I have been unable to proceed further.
If the spectrum $\sigma(N)$ of $N$ is contained in $\mathbb{R}$, then you want to show that $N=N^*$. So assume $\sigma(N)\subseteq\mathbb{R}$ and consider $M=N-N^*$. The spectrum of $M$ lies on the real axis by the spectral mapping theorem. But $iM$ is selfadjoint, which means that the spectrum of $M$ lies on the imaginary axis. So $\sigma(M)=\{0\}$. Because the norm and spectral radius are the same for a normal operator, then $M=0$, which proves that $N=N^*$.