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I really need your help in solving the following problem: Prove that a unitary operator $U$ acting on a Hilbert space $H$ is the Cayley transform of some self-adjoint operator if and only if $1$ is not an eigenvalue of $U$.

At the first direction, I have tried assuming that 1 is an eigenvalue for the eigenfunction $f$. This implies $Uf=f$ ,where $U=(L-i)(L+i)^{-1}$ for some $L$. But since $U^* U=1$ we get that also $U^* f =f $ , and thus: $<Uf,f>=<Uf,U(U^*f)>=<f,Uf>$, which implies $U$ is symmetric, and in particular $U=U^*$ and $U^2=1$. I can't figure out how to get some contradiction out of this. Maybe we know that $U$ must be one-to-one or something?

As for the other direction, I need some detailed guidance.

Hope someone will be able to help

Thanks !

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