PDE-Cayley Transform Of An Operator

[Combinatorics]

Homework Statement

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 $U$.


Hope someone will be able to help

Thanks !

Homework Equations

The Attempt at a Solution

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.


Thanks in advance

 

[mighty2000]

for any help and insights.

To prove that a unitary operator U acting on a Hilbert space H is the Cayley transform of some self-adjoint operator, we need to show that U=(L-i)(L+i)^{-1} for some self-adjoint operator L.

First, assume that 1 is not an eigenvalue of U. This means that Uf=f for some eigenfunction f, but f cannot be the zero vector since U is one-to-one. Thus, we can define a new operator L on H such that Lf=i(f-1).

To show that L is self-adjoint, we need to show that <Lf,g>=<f,Lg> for all f,g in H.

<Lf,g>=<i(f-1),g>=i<f-1,g>=i<f,g>-i<g,1>=<f,i(g-1)>=<f,Lg>

Thus, L is self-adjoint.

Now, we need to show that U=(L-i)(L+i)^{-1}. Since Uf=f, we have:

(L-i)(L+i)^{-1}f=(L-i)f=(i(f-1)-i)f=i(f-1)=Uf

Thus, U=(L-i)(L+i)^{-1} and we have proven one direction of the statement.

For the other direction, assume that U=(L-i)(L+i)^{-1} for some self-adjoint operator L.

If 1 is an eigenvalue of U, then there exists an eigenfunction f such that Uf=f. This implies (L-i)(L+i)^{-1}f=f, which can be rewritten as (L-i)f=(L+i)f.

Since L is self-adjoint, we have:

<Lf,f>=<f,Lf>=<f,L^*f>=<f,Lf>

This means that Lf is an eigenvector of L with eigenvalue 1. But since L is self-adjoint, it can only have real eigenvalues. This contradicts the fact that 1 is an eigenvalue of L.

Thus, 1 cannot be an eigenvalue of U and we have proven the other direction of the statement.

Therefore, a unitary operator U acting on a Hilbert space H is the Cayley transform of some self-adjoint operator