Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $H$ be a Hilbert space. An operator $T\in\mathcal{B}(H)$ is called coisometric if it maps open unit ball of $H$ onto open unit ball of $H$.

Please tell me how to prove that condition $T$ is coisometric implies equality $TT^*=1_H$?

share|improve this question

1 Answer 1

up vote 3 down vote accepted

If it is interesting to someone... I found an elegant solution. Here it is $$ \Vert T^*x\Vert=\sup\limits_{\Vert y\Vert\leq 1}|\langle T^*x,y\rangle|=\sup\limits_{\Vert y\Vert\leq 1}|\langle x,Ty\rangle| $$ Since T is a coisometry $\{Ty: \Vert y\Vert\leq 1\}=\{z: \Vert z\Vert\leq 1\}$ and as the consequence $$ \Vert T^*x\Vert=\sup\limits_{\Vert y\Vert\leq 1}|\langle x,Ty\rangle|=\sup\limits_{\Vert z\Vert\leq 1}|\langle x,z\rangle|=\Vert x\Vert $$ Since $T^*$ is an isometry we have for all $x,y\in H$ an equality $\langle T^*x,T^*y\rangle=\langle x,y\rangle$. Hence for all $x,y\in H$ we have $\langle TT^*x,y\rangle=\langle x,y\rangle$. So $TT^*=1_H$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.