Polarization identity in Hilbert space and some application

Question

Show that $‖Tx‖=‖x‖$ for all $x \in H$ if and only if $\langle Tx,Ty \rangle=\langle x,y\rangle$ for all $x,y \in H$. How can I use the polarization identity to prove this statement?

Polarization identity : $\ 4(Tx,y)=(T(x+y),(x+y))-(T(x-y,(x-y))+i(T(x+iy),(x+iy))-i(T(x-iy),(x-iy))$

the proof for the second direction seems clear for me but I struggle with the the first direction as I need to show that $\langle Tx,Ty \rangle=\langle x,y\rangle$

Answer 1

Assume $\|Tx\| = \|x\|$ for all $x \in H$. The polarization identity gives:

\begin{align}\langle Tx, Ty\rangle &= \frac14\left(\|Tx + Ty\|^2 + \|Tx - Ty\|^2 + i\|Tx + iTy\|^2 - i\|Tx - iTy\|^2\right)\\ &= \frac14\left(\|T(x + y)\|^2 + \|T(x - y)\|^2 + i\|T(x + iy)\|^2 - i\|T(x - iy)\|^2\right)\\ &= \frac14\left(\|x + y\|^2 + \|x - y\|^2 + i\|x + iy\|^2 - i\|x - iy\|^2\right)\\ &= \langle x, y\rangle \end{align}

Conversely, assume $\langle Tx, Ty\rangle = \langle x, y\rangle$ for all $x, y \in H$. In particular, for $y = x$ we obtain:

$$\|Tx\|^2 = \langle Tx, Tx \rangle = \langle x, x\rangle = \|x\|^2$$

Hence $\|Tx\| = \|x\|$.

Answer 2

$HINT$

You do not need the polarization identity.

We have that $$||Tx-Ty||^2=||x-y||^2 \Rightarrow Re<Tx,Ty>=Re<x,y>$$

Also $$||Tx-Tiy||^2=||x-iy||^2 \Rightarrow Im<Tx,Ty>=Im<x,y>$$

I leave to you the calculations for proving the above equalities.