norm equivalent and isometry

Question

Let $H_1$ and $H_2$ be two Hilbert spaces, $T:H_1\longrightarrow H_2$ be an one to one, onto linear map, and there exist constant $C_1, C_2>0$ such that $C_1\lVert u\rVert _1\leq\lVert Tu\rVert_2\leq C_2\lVert u\rVert_1$. Then is it possible to construct an onto isometry from $H_1$ to $H_2$ using $T?$

Answer 1

You can use polar decomposition and take

$$ U = T (\sqrt{T^{*}T})^{-1}. $$

Answer 2

Strengthening

Given Hilbert spaces $X$ and $Y$.

Suppose bounded operator: $$T\in\mathcal{B}(X,Y):\quad\mathcal{N}T=\{0\}\quad\overline{\mathcal{R}T}=Y$$

Regard modulus operator: $$|T|:X\to X:\quad|T|:=\sqrt{T^*T}\implies|T|^2=T^*T$$

Remind square root lemma: \begin{gather}\||T|x\|^2=\langle|T|x,|T|x\rangle=\langle|T|^2x,x\rangle\\ =\langle T^*Tx,x\rangle=\langle Tx,Tx\rangle=\|Tx\|^2\end{gather}

Deduce range and kernel: $$\mathcal{R}|T|^\perp=\mathcal{N}|T|^*=\mathcal{N}|T|=\mathcal{N}T=\{0\}$$

Construct unitary operator: $$U:X\leftrightarrow Y:\quad U:=\overline{T|T|^{-1}}\implies U^*=U=U^{-1}$$

Concluding unitary equivalence $X\cong Y$.

(Note how the original operator needn't
be boundedly invertible nor have closed range!)