An operator $T$ is an isometry if $||Tf||=||f||$ for all $f$ in a Hilbert space $H$.
(a) Show that if $T$ is an isometry, then $<Tf,Tg>=<f,g>$ for every $f,g \in H$. Prove as result that $T^{\ast}T=I$
(b) If $T$ is an isometry and $T$ surjective, then $T$ is unitary and $TT^{\ast}=I$
(c) Give an example of an isometry that is not unitary
(d) Show that if $T^{\ast}T=I$ is unitary then $T$ is an isometry.
I have prove that (a) and $T^{\ast}T=I$.
$<f,g>=<Tf,Tg>=<f,T^{\ast}Tg> then T^{\ast}T=I.$
(b) For b, i have prove that $T$ is unitary, but not $TT^{\ast}=I.$
How prove $TT^{\ast}=I?$ I tried something similar to $T^{\ast}T=I$ but I do not get to the result.
(d) If $T^{\ast}T$ unitary I have this:
$||T^{\ast}Tf||=||f||$ then $<T^{\ast}Tf,T^{\ast}Tf>=<f,f>$
then $<Tf,(T^{\ast})^{\ast}T^{\ast}Tf>=<f,f>$ then $<f,T^{\ast}TT^{\ast}Tf>=<f,f>$
This implies ¿$T^{\ast}TT^{\ast}T=I$ ?
How prove $T$ isometry?
Best Answer
(b): If $T$ is an isometry, then it is injective, and by (a) $T^*T=I$. If $T$ is a surjective isometry then it is invertible, and since $T^*$ is a right inverse it is the inverse of $T$, i.e., $T^*T=TT^*=I$.
(d): If $T^*T$ is a unitary, then since it is also positive, we have $\sigma(T^*T)\subset S^1\cap[0,\infty)=\{1\}$ (here $S^1$ is the unit circle $\{|z|=1\}$ in $\mathbb C$). By the continuous functional calculus, it follows that $T^*T=I$. From here, we have $$\|f\|^2=\langle f,f\rangle=\langle T^*Tf,f\rangle=\langle Tf,Tf\rangle=\|Tf\|^2$$ for all $f\in H$, and therefore $T$ is an isometry.