Normal Operators (Obsolete!)

Question

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.

Disintegrating By Parts · Answer 1 · 2015-04-18T17:30:05.020

1

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.

edited Apr 18 '15 at 17:30

answered Apr 18 '15 at 11:13

Disintegrating By Parts

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Oh I should have mentioned I'm trying to construct the spectral measure by this construction... – freishahiri Apr 18 '15 at 11:17
@Freeze_S : I added a different argument, but forgot to ping you. – Disintegrating By Parts Apr 18 '15 at 19:48
No worry I'd have come by anyway... ;) – freishahiri Apr 19 '15 at 12:53
Im struggling to prove closedness; can you elaborate on it, please? – freishahiri Apr 20 '15 at 17:38
If ${ x_{n} }$ converges to $x$, and ${ N(I+N^{\star}N)^{-1}x_n }$ converges to $y$, then ${ z_n=(I+N^{\star}N)^{-1}x_n}$ converges to $(I+N^{\star}N)^{-1}x$ (why?) and ${ Nz_n }$ converges to $y$. Because $N$ is closed, then $y = N\lim_{n}z_n = N(I+N^{\star}N)^{-1}x$. – Disintegrating By Parts Apr 20 '15 at 19:45
Ah because we have $ST$ with $S$ closed and $T$ bounded. Thanks!!! – freishahiri Apr 21 '15 at 14:18
I'm so sorry that I need to change the construction... :/ Do you mind adapting your answer? – freishahiri Apr 21 '15 at 16:57
@Freeze_S : If you want a new construction, please post a new problem. – Disintegrating By Parts Apr 21 '15 at 17:56
Ok I will rename this one here since outdated, that's fine? – freishahiri Apr 21 '15 at 18:26

score 0 · Accepted Answer · answered Apr 22 '15 at 10:28

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.

Normal Operators (Obsolete!)

2 Answers2