Normal Operators: Transform

Given a Hilbert space $\mathcal{H}$.

Consider a normal operator: $$N:\mathcal{D}(N)\to\mathcal{H}:\quad N^*N=NN^*$$

Construct the operator: $$Q:=(1+N^*N)^{-1}:\quad Z:=N\sqrt{Q}$$

Then it is normal: $$Z\in\mathcal{B}(\mathcal{H}):\quad Z^*Z=ZZ^*$$

How can I prove this?

Solution 1:

Meanwhile I got it...

Square Root

It is a well-know result: $$Q\in\mathcal{B}(\mathcal{H}):\quad Q=Q^*$$

The range includes: $$\mathcal{R}Q=\mathcal{D}(1+N^*N)=\mathcal{D}N^*N$$

One has the bound: $$\langle Q^{-1}\psi,\psi\rangle=\langle(1+N^*N)\psi,\psi\rangle=\|\psi\|^2+\|N\psi\|^2\geq0$$

So the numerical range: $$\mathcal{W}(Q^{-1})\geq0\implies\mathcal{W}(Q)\geq0$$

For bounded operators: $$Q\in\mathcal{B}(\mathcal{H}):\quad\sigma(Q)\subseteq\overline{\mathcal{W}(Q)}$$

Thus the square root exists: $$\sqrt{Q}\in\mathcal{B}(\mathcal{H}):\quad\sqrt{Q}=\sqrt{Q}^*$$

Especially range and kernel: $$\mathcal{H}\supseteq\overline{\mathcal{R}\sqrt{Q}}\supseteq\overline{\mathcal{R}Q}=\mathcal{H}$$ $$(0)\subseteq\mathcal{N}\sqrt{Q}\subseteq\mathcal{N}Q=(0)$$

(This is basic so far.)

Operator

One has the bound: $$\|NQ\varphi\|^2=\langle N^*NQ\varphi,Q\varphi\rangle\leq\langle N^*NQ\varphi,Q\varphi\rangle+\langle Q\varphi,Q\varphi\rangle=\langle\varphi,Q\varphi\rangle=\|\sqrt{Q}\varphi\|^2$$

The operator is closed: $$\sqrt{Q}\in\mathcal{B}(\mathcal{H}):\quad\quad N=\overline{N}\implies Z=\overline{Z}$$

By uniform extension: $$\|Z\restriction_{\mathcal{R}\sqrt{Q}}\|\leq1\implies Z\in\mathcal{B}(\mathcal{H})$$

Especially the range lies in: $$\mathcal{D}Z=\mathcal{H}\implies\mathcal{R}\sqrt{Q}\subseteq\mathcal{D}N$$ (Domain is unclear a priori!)

Polynomials

Here regard elements: $$\varphi,\chi\in\mathcal{D}N=\mathcal{D}N^*$$

Exploit the selfadjoint inverse: $$\langle NQ\varphi,\chi\rangle=\langle Q\varphi,N^*(1+N^*N)Q\chi\rangle=\langle Q\varphi,(1+N^*N)N^*Q\chi\rangle=\langle QN\varphi,\chi\rangle$$

Thus a calculation reveals: $$\langle\sqrt{Q}N\varphi,\chi\rangle=\lim_n\langle Q_nN\varphi,\chi\rangle=\lim_n\langle NQ_n\varphi,\chi\rangle=\lim_n\langle Q_n\varphi,N^*\chi\rangle=\langle N\sqrt{Q}\varphi,\chi\rangle$$

Hence on operator level: $$N1=1N\quad NQ\supseteq QN\quad NQ^k\supseteq Q^kN\quad N\sqrt{Q}\supseteq\sqrt{Q}N$$

(The analogous hold for adjoint.)

Normality

Formally the adjoint acts as: $$\langle Z\varphi,\psi\rangle=\langle\varphi,\sqrt{Q}N^*\psi\rangle\quad(\psi\in\mathcal{D}N^*)$$

Regard dense elements: $$\varphi\in\mathcal{R}\sqrt{Q}\subseteq\mathcal{D}N=\mathcal{D}N^*$$

Remind also that: $$\sqrt{Q}\varphi\in\mathcal{R}Q=\mathcal{D}N^*N=\mathcal{D}NN^*$$

Thus one derives: $$Z^*Z\varphi=\sqrt{Q}N^*N\sqrt{Q}\varphi=\sqrt{Q}NN^*\sqrt{Q}\varphi\\ =N\sqrt{Q}N^*\sqrt{Q}\varphi=N\sqrt{Q}\sqrt{Q}N^*\varphi=ZZ^*\varphi$$

Concluding normality.