Unitary operator as an exponent of self-adjoint

Question

I'm trying to proof that set of unitary operators on infinite Hilbert space is path-connected. That's why i need to show that for each unitary $U$ there is a self-adjoint $B$ such that $U=e^{iB}$. Any ideas on how to show that? Maybe by using the fact that every unitary operator is unitary equivalent to operator of multiplication by $f$ in some $L_{2}(\mu)$ ?

Answer 1

Fix a Borel function $\varphi:\mathbb T=\{z\in\mathbb C:|z|=1\}\to [0,2\pi)$ such that $e^{i\varphi(z)}=z$. Put $B=\varphi(U)$ (here we're using Borel functional calculus). Then $B$ is self-adjoint, and $U=\exp(iB)$.