Isometric <=> Left Inverse Adjoint

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

Let $H$ be a complex Hilbert space and let $T \in B(H)$ be an isometry. We claim that $T^\ast T = I$. To this end let $h \in H$ and note that by the Hilbert space Riesz representation theorem a linear functional in $H^\ast$ corresponds to an element $h \in H$ ($h \mapsto \langle \cdot, h \rangle$). Also note that if $\varphi (x) = 0$ for all $\varphi \in H^\ast$ then $x=0$.

Hence if for all $h, h' \in H$:

$$ \langle h, T^\ast T h' - h'\rangle = 0,$$

then $T^\ast T h' = h'$ for all $h' \in H$.

But since $T$ is an isometry

$$ \langle h, T^\ast T h' - h'\rangle = \langle h, T^\ast T h' \rangle-\langle h, h'\rangle = \langle Th,Th'\rangle - \langle h,h'\rangle = \langle h, h'\rangle - \langle h, h'\rangle= 0,$$

hence $T^\ast T = I$ as desired.