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$(\Leftarrow)$ If $T\circ T^{*}=I$ we had $$\langle x,x \rangle=\langle (T\circ T^{*}) (x),x \rangle=\langle T(x),T(x) \rangle, \forall x\in H.$$ Hence, $\Vert T(x)\Vert=\Vert x \Vert$, $\forall x\in H$.

$(\Rightarrow)$ Suppose that $\Vert T(x)\Vert=\Vert x \Vert$, we had $$\langle T(x),T(x) \rangle=\langle x,x \rangle \Leftrightarrow \langle x,T(x) \rangle=\langle x,T^{*}(x) \rangle \Leftrightarrow \langle x,T(x) -T^{*}(x)\rangle=0,$$ for all $x\in H$. I do not know what to do now...

Rodolfo Cardoso
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1 Answers1

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You are almost there. Note that $$\langle Tx, Tx\rangle = \langle x,x \rangle \iff \langle (T^*T-I)x, x\rangle=0$$

Then invoke the following lemma:

Lemma: If $S$ is an operator on a complex Hilbert space $H$ such that $\langle Sx, x\rangle = 0$ for all $x \in H$, then $S=0$.

Hint: Use the polarization identity. The idea is to use the polarization identity to show that $\langle Sx,y\rangle = 0$ for all $x,y \in H$. Then take $y = Sx$ and conclude that $Sx= 0$ for all $x \in H$, i.e. $S=0.$

J. De Ro
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  • I undertand, if im showing the lemma the question is solved. But, $\langle Tx, Tx\rangle=\langle x,x \rangle\Leftrightarrow \overline{\langle (T^{*}T-I)x,x \rangle}=0$? Because i can suppose that $X$ is a complex ineer product space. And other think this Lemma not is valid for real ineer product space, because: if we are let $T(x,y)=(-y,x)$ we had $T\neq 0$ and $\langle T(x,y),(x,y)\rangle=0$ for all $(x,y)$ – Rodolfo Cardoso Apr 03 '21 at 16:14
  • @RodolfoCardoso You are right. The lemma fails over the field of real numbers, as your example shows. In practise, we always consider Hilbert spaces over the complex numbers though. – J. De Ro Apr 03 '21 at 17:10
  • @RodolfoCardoso See here for more info: https://math.stackexchange.com/questions/1307747/prove-that-if-langle-tx-x-rangle-0-for-all-x-in-x-then-t-0 – J. De Ro Apr 03 '21 at 17:12
  • Thanks! I understand now! – Rodolfo Cardoso Apr 03 '21 at 21:39
  • @RodolfoCardoso Glad to help! – J. De Ro Apr 03 '21 at 21:40