Problem: If A is a bounded normal operator, the spectrum $\sigma(A)=\{s+it:s \in \sigma(B),t \in \sigma(C)\}$, where B, C are bounded self adjoint operators which commute.
Fact: A bounded normal operator A can be written $A=B+iC$, where B,C are bounded self adjoint operators which commute.
Fact: Let H be a complex Hilbert space and let $A:H \rightarrow H$ be a bounded complex linear operator, then A is normal if only if $\Vert A^*x \Vert=\Vert Ax \Vert$ for all $x \in H$. Also every self-adjoint operator is normal.
I was told there is a mistake in the problem, but have not spotted it. Thanks in advance.
Best Answer
This is not true even in the simplest of examples. For instance $$ \begin{bmatrix} 1&0\\0&i\end{bmatrix} =\begin{bmatrix} 1&0\\0&0\end{bmatrix} +i\begin{bmatrix} 0&0\\0&1\end{bmatrix}. $$ Here $\sigma(A)=\{1,i\}$ and $\sigma(B)=\sigma(C)=\{0,1\}$. But $0=0+i0\in\{s+it:\ s\in\sigma(B),\ t\in\sigma(C)\}$ is not in $\sigma(A)$.
What does hold is that $\sigma(A)\subset \{s+it:\ s\in\sigma(B),\ t\in\sigma(C)\}$. This can be proven via functional calculus, for instance. If $f(t)=g(t)+ih(t)$, then the spectrum of $M_f$ is $$ \sigma(M_f)=\operatorname{ran}f=\{g(t)+ih(t):\ t\}\subset \{s+it:\ s\in\operatorname{ran}g,\ t\in\operatorname{ran}h\}. $$