Equivalent formulations: pure contraction I want to prove the following equivalence: let $T$ be a bounded self-adjoint operator on a Hilbert space $H$. TFAE:


*

*$\|Tx\|<\|x\|$ for each $x\in H\setminus\{0\}$

*$\|T\|\leq1$ and $\pm1\notin\sigma_p(T)$


where $\sigma_p(T)$ is the point spectrum (the set of all eigenvalues of $T$).
One implication has to be easy from my point of view. Can someone help me with this proof?
Thanks a lot.
 A: Suppose $T$ is selfadjoint with $\|T\| =1$. Then $|(Tx,x)| \le \|T\|\|x\|=1$ for all unit vectors $x$. If $|(Tx,x)|=1$ for some unit vector $x$, then the above gives
$$
    1=|(Tx,x)|\le \|Tx\|\|x\|\le \|T\|\|x\|^{2}\le 1
$$
Therefore, $|(Tx,x)|=\|Tx\|\|x\|$ with $x\ne 0$ which, by Cauchy-Schwarz, forces $Tx=\alpha x$ for some scalar $\alpha$ and unit vector $x$. Hence $|\alpha|=1$. So $\pm 1\notin\sigma_{p}(T)$ iff $|(Tx,x)| < 1$ for all unit vectors $x$.
Assume $|(Tx,x)| < 1$ for all unit vectors $x$. Equivalently, assume $|(Tx,x)| < \|x\|^{2}$ for all $x \ne 0$. Let $x,y\in H$ and choose real $\alpha$ so that $e^{i\alpha}(Tx,y)=|(Tx,y)|$. Then
$$
     (Tx,e^{-i\alpha}y)=\Re\frac{1}{4}\sum_{n=0}^{3}i^{n}(T(x+i^{n}e^{-i\alpha}y),(x+i^{n}e^{-i\alpha}y)) \\
    =\frac{1}{4}(T(x+e^{-i\alpha}y),x+e^{-i\alpha}y)-\frac{1}{4}(T(x-e^{-i\alpha}y),x-e^{-i\alpha}y).
$$
If $x$ and $y$ are not both $0$,
$$
          |(Tx,y)| < \frac{1}{4}\|x+e^{-i\alpha}y\|^{2}+\frac{1}{4}\|x-e^{-i\alpha}y\|^{2} \\
      = \frac{1}{2}\|x\|^{2}+\frac{1}{2}\|y\|^{2}.
$$
For $x \ne 0$, let $y=Tx$ in the above in order to obtain:
$$
          \|Tx\|^{2} < \frac{1}{2}\|x\|^{2}+\frac{1}{2}\|Tx\|^{2} \\
            \|Tx\| < \|x\|.
$$
Conversely, if $\|Tx\| < \|x\|$ for all $x \ne 0$, then $|(Tx,x)| < \|x\|^{2}$.
