# Prove $Tx=x$, for $x\in H$, if and only if $(Tx,x)=\|x\|^2$ and $\ker(I-T)=\ker(I-T^*)$

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.

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.