Fuglede's theorem Let $T$ be a normal operator and let $A$ commute with $T$. Then $A$ commutes with the $T^*$. I'm having a few difficulties namely:
1)  Show that $||e^{itB}||$ is less than or equal to 1 where $B$ is a self adjoint operator.
2) Put $U=\frac{T+T^*}{2}$ and $V=\frac{T-T^*}{2i}$ and ${\lambda}=a+ib$. Show $||e^{2i(bU-aV)}||$ is precisely equal to 1. 
3) From this I can deduced that $||e^{{\lambda}T^*}Ae^{-{\lambda}T^*}||$ is less than or equal to $||A||$ so from liouvilles theorem its independent of ${\lambda}$. Can I just plug in 0 to get that it is equal to $A$.
Thanks 
 A: 1) $\|e^{itB}\|=1$ if $t\in\mathbb R$ and $B=B^*$. This is simply because $e^{itB}$ is a unitary.
2) The expression $bU-aV$ is the real part of $\lambda T$: that is, $2(bU-aV)=\lambda T+(\lambda T)^*$. So it is selfadjoint, and $e^{2i(bU-aV)}$ is a unitary as in part 1. 
3) Note that up to here you haven't used that $AT=TA$. This forces $AT^n=T^nA$ for all $n$, thus for all polynomials, and then for any limit of them. So $Ae^{-\bar\lambda T}=e^{-\bar\lambda T}A$. Then
$$
e^{\lambda T^*}Ae^{-\lambda T^*}=e^{\lambda T^*}Ae^{-\bar\lambda T}e^{\bar\lambda T}e^{-\lambda T^*}
=e^{\lambda T^*}e^{-\bar\lambda T}Ae^{\bar\lambda T}e^{-\lambda T^*}=e^{\lambda T^*-\bar\lambda T}Ae^{-(\lambda T^*-\bar\lambda T)}=e^{2i(bU-aV)}Ae^{-2i(bU-aV)}.
$$
By part 2, 
$$
\|e^{\lambda T^*}Ae^{-\lambda T^*}\|=\|e^{2i(bU-aV)}Ae^{-2i(bU-aV)}\|\leq\|e^{2i(bU-aV)}\|\,\|A\|\,\|e^{-2i(bU-aV)}\|=\|A\|.
$$
Now Liouville's Theorem guarantees that $e^{\lambda T^*}Ae^{-\lambda T^*}$ is constant, and so $e^{\lambda T^*}Ae^{-\lambda T^*}=A$, i.e. $e^{\lambda T^*}A=Ae^{\lambda T^*}$ for all $\lambda$.
Comparing the degree one terms, we get $T^*A=AT^*$. 
