Let $E$ be an infinite-dimensional complex Hilbert space.


I look for an elementary proof of the following result:



> 
> If $A,B \in\mathcal{L}(E)$ be two normal operators such that $AB=BA$. Then
>  $$A^\*B=BA^\*\;\;\text{and}\;\; AB^\*=B^\*A.$$
> 
> 
> 


If $E$ is finite dimensional, I found a rather difficult proof for this in Proposition 6 in this set of lecture notes ([1](http://www.math.mcgill.ca/makkai/MATH2472010/Notes%20on%20the%20adjoint%20and%20normal%20operators.pdf)).



Proof related to normal operators on an infinite-dimensional Hilbert spaces
A proof via the multiplication operator representation: the assumption on $A$ and $B$ ($A$ and $B$ bounded normal commuting operators on a Hilbert space $H$) are sufficient for the multiplication operator representation to hold. So up to unitary isomorphism, $A=M\_{a}:u\mapsto a(x)u(x)$ and $B=M\_{b}:u\mapsto b(x)u(x)$ are multiplication operators on some space $L^2(X,\mu)$, for some bounded Borel measurable functions $a$ and $b$. Then $A^\*=M\_{\overline a} $ and $B^\*=M\_{\overline b}$ are multiplication operators too, and they all commute.


