Let $T$ be a bounded normal operator in Hilbert space such that the spectrum $\sigma(T)$ is contained in the real axis. By the Gelfand-Naimark theorem for commutative $C^*$-algebras the $C^*$-algebra generated by $T$ and $I$ is isometrically isomorphic to $C(\sigma(T)),$ the algebra of complex valued continuous functions on $\sigma(T)\subset \mathbb{R}.$ The operator $T$ corresponds to multiplication by $x$ in $C(\sigma(T)),$ therefore $T$ is self-adjoint.
I would like to prove that fact in a straightforward way, but I could not come up with any idea.
When $T$ is a compact operator the proof is relatively easy. Assume by contradiction that $T^*-T\neq 0.$ Then one of the numbers $\lambda:=\pm{1\over 2}\|T^*-T\|\neq 0$ is the eigenvalue of the self-adjoint operator $B:={i\over 2}(T^*-T).$ Let $V_\lambda$ denote the eigenspace of the operator $B$ corresponding to $\lambda.$ As $T$ and $T^*$ commute with $B,$ the subspace $V_\lambda$ is invariant for $A={1\over 2}(T+T^*).$ The operator $T=A+iB$ restricted to $V_\lambda$ is of the form $A+i\lambda I.$ Therefore $\sigma(T)\subsetneq \mathbb{R},$ which gives a contradiction.
Best Answer
I know three ways, none really elementary.
By the Spectral Theorem. You have $$ T=\int_{\sigma(T)}\lambda\,dE(\lambda). $$ Then, using the $\sigma(T)\subset\mathbb R$, $$ T^*=\int_{\sigma(T)}\overline\lambda\,dE(\lambda)=\int_{\sigma(T)}\lambda\,dE(\lambda). $$
If you know that for normal $T$ you have $$\tag1\overline{W(T)}=\overline{\operatorname{conv}}\sigma(T),$$ where $W(T)$ is the numerical range, you get easily that $T=T^*$. The problem is that the only proof of $(1)$ that I know uses the Spectral Theorem, so it is easier to do the argument directly as in the previous case.
If you have the Spectral Mapping Theorem (but again the only proof I know depends on the Spectral Theorem), you get that $T-T^*$ has real spectrum. So the selfadjoint operator $i(T-T^*)$ has imaginary spectrum, which implies that it is zero.