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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 $TT^*=1_H$?

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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 $TT^*=1_H$.

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