Isometry <=> Adjoint left inverse

Question

Is it true that: $$T\text{ isometric}\iff T^*\text{ left inverse}$$ Obviously: $$\text{"}\Rightarrow\text{": }\langle x,\tilde{x}\rangle=\langle Tx,T\tilde{x}\rangle=\langle x,T^*T\tilde{x}\rangle$$ $$\text{"}\Leftarrow\text{": }\langle Tx,T\tilde{x}\rangle=\langle x,T^*T\tilde{x}\rangle=\langle x,\tilde{x}\rangle$$ My problem is that I'm not sure about if the adjoint is left inverse then the operator necessarily is bounded so that domain issues might come into play...

Answer 1

If it is an isometry then it has a left inverse that is: $$\langle x,LRz\rangle=\langle x,z\rangle=\langle Rx,Rz\rangle$$ So the adjoint must act as a left inverse too.

On the other hand if the adjoint acts as a left inverse then: $$\langle Rx,Ry\rangle=\langle x,R^*Ry\rangle=\langle x,y\rangle$$ Thus it is an isometry.

(Note, there's no need of completeness nor identification with dual space.)

Answer 2

$T$ must be densely defined for $T^\ast$ to be defined. That $T^\ast$ is a left inverse of $T$ implies that $\mathcal{R}(T) \subset \mathcal{D}(T^\ast)$. And hence for every $x \in \mathcal{D}(T)$ we have

$$\langle x,x\rangle = \langle x, T^\ast T x\rangle = \langle Tx,Tx\rangle,$$

or in other words $\lVert Tx\rVert = \lVert x\rVert$, i.e. $T$ is an isometry (and can hence be extended continuously to the entire space, still as an isometry, if it was not originally globally defined).