Proof related to normal operators on an infinite-dimensional Hilbert spaces 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).
 A: 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.
A: I find the following proof:
Lemma: Let $S,T \in \mathcal{L}(E)$. Then, $ST=TS$ if and only if $e^{zS}T=Te^{zS}$ for all $z \in \mathbb{C}$.

Fuglede's Theorem:  Let $S,T \in \mathcal{L}(E)$ with $S$ normal. If $ST=TS$, then $S^*T=TS^*$.

Proof: We need to show $e^{zS^*}Te^{-zS^*}=T$ for all $z \in \mathbb{C}$. We have
$$F(z):=e^{zS^*}Te^{-zS^*} = e^{zS^*} e^{-\bar{z} S} T e^{\bar{z} S} e^{-zS^*}=e^{zS^*-\bar{z}S} T e^{-zS^* + \bar{z}S}.$$
Note that $e^{zS^*-\bar{z}S}$ (and its inverse) is unitary (since $zS^*-\bar{z}S$ is anti-self-adjoint), so $\|F(z)\|\leq \|T\|$. Moreover, $F$ is holomorphic
so by Liouville's theorem $F$ is constant. Hence $F(z)=T$ which yields 
$$e^{zS^*}Te^{-zS^*}=T.$$
Remark: One can see Lemma 6.1 in this lecture note.
