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Let $H$ be a complex Hilbert space and $T:H\rightarrow H$ an operation such that $\|T\|\leq 1$. Show that

  1. $Tx=x$ if and only if $(Tx,x)=\|x\|^2$
  2. $\ker(I-T)=\ker(I-T^*)$.

My attempt
1. $(Tx,x)=(x,x)=\|x\|^2$ if $Tx=x$. Conversely, WTS $\|Tx-x\|=0$ $\forall x\in H$.

We get $\|Tx-x\|=(Tx-x,Tx-x)=\|Tx\|^2-(x,Tx)$. But $\|T\|\leq 1$ has not been used.

Please how do I proceed?

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For the first point, first off since $\left \langle Tx,x\right\rangle=\|x\|^2$ you also have $\left \langle x,Tx\right\rangle = \overline{\|x\|^2}=\|x\|^2$. Thus $$0 \leq\left\langle Tx-x,Tx-x\right\rangle =\|Tx\|^2+\|x\|^2-\left\langle Tx, x\right\rangle-\left\langle x,Tx\right\rangle= \|Tx\|^2-\|x\|^2 $$ Since $\|T\|\leq 1$, $$0 \leq \|Tx\|^2-\|x\|^2 \leq \|x\|^2-\|x\|^2=0 $$ and hence $Tx=x$.

For the second point, using the first point \begin{align*}x\in \ker(I-T)\iff Tx=x \iff \left \langle Tx,x\right\rangle =\|x\|^2\end{align*}

Now observe that if $\|T\|\leq 1$ then also $\|T^*\|\leq 1$. Indeed, \begin{align*}\|T^*\|&=\sup_{x\neq 0}\frac{\|T^*x\|}{\|x\|}=\sup_{x\neq 0}\frac{\left\langle T^*x,T^*x\right\rangle^{1/2}}{\|x\|}=\sup_{x\neq 0}\frac{\left\langle TT^*x,x\right\rangle^{1/2}}{\|x\|}\leq \sup_{x\neq 0}\frac{\|TT^*x\|^{1/2}}{\|x\|^{1/2}}\leq\\ &\leq\sup_{x\neq 0}\frac{\|T^*x\|^{1/2}}{\|x\|^{1/2}}=\|T^*\|^{1/2}\end{align*} i.e. $\|T^*\|\leq \|T^*\|^{1/2}$ and thus $\|T^*\|\leq 1$ (remark: this argument can be easily extended to prove that in general $\|T\|=\|T^*\|$ for any bounded linear operator $T$).

Therefore the first point may be applied to $T^*$, too.

\begin{align*} x\in \ker (I-T^*)\iff T^*x=x\iff \left \langle T^*x,x\right\rangle = \|x\|^2\iff \left \langle x,Tx\right\rangle =\|x\|^2 \end{align*} But $\left \langle Tx,x\right\rangle =\|x\|^2 $ implies $\left \langle x, Tx\right\rangle = \left \langle Tx,x\right\rangle =\|x\|^2$, as shown above, and viceversa, thus the equivalence is proved.

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