Normal Operators (Obsolete!) Problem
Given a Hilbert space $\mathcal{H}$.
Consider a normal operator:
$$N:\mathcal{D}(T)\to\mathcal{H}:\quad N^*N=NN^*$$
Construct the operator:
$$A:=N(1+N^*N)^{-1}\in\mathcal{B}(\mathcal{H})$$
Then it is normal:
$$A^*\in\mathcal{B}(\mathcal{H}):\quad A^*A=AA^*$$
How can I prove this?
Disclaimer
This question became obsolete due to: Transform
I kept it open for the sake of respect for TAE's answer.
Also it may serve as a manual to handling such problems.
 A: If $E_N$ is the spectral resolution for $N$, then the following is bounded and normal by the functional calculus:
$$
       N(I+N^{\star}N)^{-1} = \int \frac{\lambda}{1+|\lambda|^{2}}dE_{N}(\lambda).
$$
In line with your comment under this point that you are trying to derive the spectral measure, let's try a different approach.
For any $x$, one has $y=(I+N^{\star}N)^{-1}x \in \mathcal{D}(N^{\star}N)=\mathcal{D}(NN^{\star})$ and, hence, $Ny$ and $N^{\star}y$ are defined. So
$$
   Ax = N(I+N^{\star}N)^{-1}x,\;\;\; Bx = N^{\star}(I+N^{\star}N)^{-1}x
$$
are defined everywhere. $A$ and $B$ are bounded because they are everywhere defined, and easily verified to be closed. If $x \in \mathcal{D}((N^{\star}N)^{2})$, then
$$
                 (I+N^{\star}N)N^{\star}x = N^{\star}(I+NN^{\star})x=N^{\star}(I+N^{\star}N)x.
$$
If $y \in \mathcal{D}(N^{\star}N)$, then $x=(I+N^{\star}N)^{-1}y \in \mathcal{D}((N^{\star}N)^{2})$, and the above gives
$$
      (I+N^{\star}N)N^{\star}(I+N^{\star}N)^{-1}y = N^{\star}y \\
           N^{\star}(I+N^{\star}N)^{-1}y=(I+N^{\star}N)^{-1}N^{\star}y.
$$
Therefore, if $x \in \mathcal{D}(N^{\star}N)$, $y \in \mathcal{H}$,
\begin{align}
      (Bx,y) & = (N^{\star}(I+N^{\star}N)^{-1}x,y) \\
             & = ((I+N^{\star}N)^{-1}N^{\star}x,y) \\
             & = (N^{\star}x,(I+N^{\star}N)^{-1}y) \\
             & = (x,N(I+N^{\star}N)^{-1}y) \\
             & = (x,Ay).
\end{align}
That's enough to prove that $B^{\star}=A$ and $A^{\star}=B$ because $\mathcal{D}(N^{\star}N)$ is dense in $\mathcal{H}$. To show that $A$ is normal, keep in mind that $N$ is normal and, therefore,
\begin{align}
     \|Ax\| & =\|N(I+N^{\star}N)^{-1}x\| \\
            & =\|N^{\star}(I+N^{\star}N)^{-1}x\| \\
            & =\|A^{\star}x\|.
\end{align}
So $A$ is normal and $A^{\star}=B$ is normal.
A: By the other thread:
$$WN\subseteq NW\quad WN^*\subseteq N^*W$$
And formally it holds:
$$\psi\in\mathcal{D}(N^*):\quad A^*\psi=WN^*\psi$$
So a calculation gives:
$$AA^*\psi=NWWN^*\psi=WNN^*W\psi=WN^*NW\psi=A^*A\psi\quad(\psi\in\mathcal{D}N^*)$$
But the domain was dense:
$$\overline{\mathcal{D}(N^*)}=\mathcal{H}\implies AA^*=A^*A$$
That finishes the proof.
