Unitary operator in von Neumann algebra

Question

Let $R\subseteq B(H)$ be a von Neumann algebra, and $U\in R$ be unitary. Prove that there is a self adjoint operator $A\in R$ such that $||A||\leq \pi$, and $U=\exp(iA)$ . Any idea how to start! Thank you

Answer 1

Since $U$ is a normal operator in $R$, we can use the $L^\infty$ functional calculus which yields an isometric $R$-valued *-homomorphism $$ L^\infty(\sigma(U))\longrightarrow R\qquad f\longmapsto f(U). $$ The principal branch of the complex logarithm $\log$ induces a function of $L^\infty(\sigma(U))$ on the spectrum of $U$, extended by $0$ if you want at $-1$. So $$ A:=-i\log U $$ defines a bounded operator, element of $R$. Now let us use the properties of the functional calculus. Note that the spectrum of $U$ is contained in the unit circle.

1- Since $\log(\overline{z})=-\log z$ on the unit circle and $g(z)=\overline{\log \overline{z}}=\log z$ everywhere, we have $(\log U)^*=g(U^*)=\log U^*=-\log U$. Hence $A^*=(-i\log U)^*=i(\log U)^*=-i\log U=A$. That is $A$ is hermitian.

2- Since $e^{i(-i\log z)}=z$ a.e., we have $e^{iA}=U$ by composition of the functional calculus of $A$ and $U$.

3- Since $-i\log z \in (-\pi,\pi]$ for every $z$ on the unit circle, we see that $\|A\|=\sup_{\sigma(U)}|-i\log z|\leq \pi$.

Answer 2

Hint: $A = f(U)$ for a certain function $f$.