I was given this exercise in class: prove that the condition $||Tx||=||x||$ for all $x \in H$ is equivalent to $T^*T=Id$, where $T:H \rightarrow H$ is a bounded operator, $H$ is a complex Hilbert space with the scalar product $(.,.)$ and $T^*$ is the adjoint operator of $T$.
I think I managed to prove the indirect sense of the equivalence:
If $T^*T=Id$, then $|(T^* Tx,x)|=|(x,x)|=||x||^2$. On the other hand, $|(T^*Tx,x)|=|(Tx,Tx)|=||Tx||^2$, and thus $||x||^2=||Tx||^2 \Leftrightarrow ||x||=||Tx||$ .
However, I can't seem to find the way to prove the direct sense of the equivalence. There was a hint in the exercise that said to use the poralisation identity, which states that for $(H,(.,.))$ a complex pre-Hilbert space and $x,y \in H$, we have $(x,y)=\frac{1}{4}(||x+y||^2+i||(x+iy)||^2-||x-y||^2-i||(x-iy)||^2)$
How do you prove the direct sense of the equivalence?
We should use the polarization identity in the following form $$\langle Su,v\rangle ={1\over 4}\sum_{k=1}^4i^k\langle S(u+i^kv),u+i^kv\rangle $$ Thus if $\langle Sx,x\rangle = \langle Rx,x\rangle$ for every $x,$ then $\langle Ru,v\rangle =\langle Su,v\rangle$ for all $u,v.$ Hence $R=S.$ Therefore if $\langle T^*Tx,x\rangle =\langle x,x\rangle $ for every $x$ then $T^*T=I, $ by taking $R=T^*T$ and $S=I.$