This thread was only Q&A.
Given Hilbert spaces $\mathcal{H}$ and $\mathcal{K}$.
Consider an operator: $$J\in\mathcal{B}(\mathcal{H},\mathcal{K}):\quad P:=J^*J$$
By a previous thread:* $$\mathcal{R}:=\mathcal{N}^\perp:\quad\mathcal{N}:=\mathcal{N}P=\mathcal{N}J$$
Then one has: $$\|J\varphi\|=\|\varphi\|\quad(\varphi\in\mathcal{R})\iff P\varphi=\varphi\quad(\varphi\in\mathcal{R})$$
How can I prove this?
*See the thread: Kernel
Note: This answer is community wiki.
Suppose partial isometry: $$\|J\varphi\|=\|\varphi\|\quad(\varphi\in\mathcal{R})$$
By polarization one gets: $$\langle J\varphi,J\psi\rangle=\langle\varphi,\psi\rangle\quad(\varphi,\psi\in\mathcal{R})$$
Note that one has: $$\mathcal{N}=\overline{\mathcal{N}}:\quad\mathcal{H}=\mathcal{N}\oplus\mathcal{R}$$
So one obtains: $$\langle P\varphi,\chi\rangle=\langle J\varphi,J\chi\rangle=\langle\varphi,\chi\rangle\quad(\chi\in\mathcal{R})$$ $$\langle P\varphi,\chi\rangle=\langle\varphi,P\chi\rangle=0=\langle\varphi,\chi\rangle\quad(\chi\in\mathcal{N})$$
Conversely one has: $$\|J\varphi\|^2=\langle P\varphi,\varphi\rangle=\langle\varphi,\varphi\rangle=\|\varphi\|^2\quad(\varphi\in\mathcal{R})$$
Concluding equivalence.
