show that $\ker(Id-T) = \ker(Id-T)^{\star}$

Question

$H$ is a hilbert space and $T$ is a bounded linear operator on $H$, also $\|T\| \leq 1$

by calculating $\|Tx-x\|^2$ I have shown the following string of equivalences $$Tx = x \iff \langle\,Tx, x\rangle = \|x\|^2 \iff \langle\,x, Tx\rangle = \|x\|^2 $$

it's supposed to help me prove $$\ker(Id-T) = \ker(Id-T)^{\star}$$

but I fail to see how one must proceed.

Answer 1

You made an error in your equivalence chain.

In fact, you have:

$$Tx = x \iff \langle\,Tx, x\rangle = \|x\|^2 \iff \langle\,x, T^* x\rangle = \|x\|^2.$$

You also have $\Vert T^* \Vert \le \Vert T \Vert \le 1$. In conjonction with $\langle\,x, T^* x\rangle = \|x\|^2$ , this leads to $$\langle\,x, T^* x\rangle = \|x\|^2 \iff T^*x=x$$

and concludes the proof.