Bounded operator on Hilbert space Let $H$ is a Hilbert space. If $T\in B(H)$ show that $T+T^*\ge 0$ iff $T+I$ is invertible in $B(H)$ with $\|(T-I)(T+I)^{-1}\|\le 1$.
(Hint is  $T+T^*\ge 0$ iff  $\|(T+I)x\|\ge \|x\|\ $ and 
$\|(T+I)x\|\ge \|(T-I)x\|\ $.
 A: Suppose $T+\mathbb 1$ is invertible and $\|(T-\mathbb 1)(T+\mathbb 1)^{-1}\|≤1$. Then you have:
$$(T^\mathstrut-\mathbb 1)(T^*-\mathbb 1)=T^\mathstrut T^* - T^* - T^\mathstrut +\mathbb 1 =(T\mathstrut+\mathbb 1)(T^*+\mathbb 1)-2(T^\mathstrut+T^*)$$
And then you get
$$2(T\mathstrut+T^*)=(T^\mathstrut+\mathbb 1)(T^*+\mathbb 1)-(T^\mathstrut-\mathbb 1)(T^*-\mathbb 1)$$
Multiply both sides from the left by $(T+\mathbb 1)^{-1}$ and from the right by $(T^*+\mathbb 1)^{-1}$ (call these $B$ and $B^*$ to simplify the expression) to get 
$$2B(T^\mathstrut+T^*)B^*=\mathbb 1-B(T^\mathstrut-\mathbb 1)(T^*-\mathbb 1)B^*$$
Now note that $B(T^\mathstrut-\mathbb 1)(T^*-\mathbb 1)B^*$ is of the form $F^\mathstrut F^*$ for $F=B(T-\mathbb 1)$. This means that it is a positive operator. Furthermore, from $\|F^\mathstrut F^*\|=\|F\|^2$ you get: 
$$\|B(T^\mathstrut-\mathbb 1)(T^*-\mathbb 1)B^*\|=\|(T^*-\mathbb 1)(T^*+\mathbb 1)^{-1}\|^2≤1$$
So $2B(T^\mathstrut+T^*)B^*=\mathbb 1-A$, where $A$ is positive and has norm smaller than $1$. This means $B(T^\mathstrut+T^*)B^*$ is positive.
If you have a positive operator $P$, then $B^{-1}P(B^{-1})^*$ is also positive. If you take $P=B(T^\mathstrut+T)B^*$ and apply this you get that $T^\mathstrut + T^*$ must be positive. This would conclude the first direction.

For the second direction let $T+T^*≥0$, then $$(T^\mathstrut+T^*)+T^*T^\mathstrut=(T^*+\mathbb 1)(T^\mathstrut+\mathbb 1)-\mathbb 1$$ is a sum of positive operators, and thus positive. Remember that if a hermitian operator $P$ is positive, then $P+ \epsilon I$ is invertible for any $\epsilon >0$ (this is actually the usual definition of positive in $*$ algebras). For that reason
$$(T^*+\mathbb 1)(T^\mathstrut+\mathbb 1)=(T^\mathstrut+T^*)+T^*T^\mathstrut+\mathbb 1$$ is invertible. The same argument shows that $(T^\mathstrut+\mathbb 1)(T^*+\mathbb 1)$ is invertible.
If $AB$ and $BA$ are invertible, then $A$ and $B$ must both be invertible. To see this consider for example $A^{-1}:=B(AB)^{-1}$, then
$$A(B(AB)^{-1})=(AB)(AB)^{-1}=\mathbb 1$$ and
$$B(AB)^{-1}A=B(AB)^{-1}A \ (BA) (BA)^{-1}=B(AB)^{-1}AB A (BA)^{-1}=BA(BA)^{-1}=\mathbb 1$$
so you get that $(T+\mathbb 1)$ and $(T^*+\mathbb 1)$ are invertible.
To get the bound we have to turn the first proof around. We have from $T^\mathstrut+T^*$ being positive that $2B(T^\mathstrut+T^*)B^*$ is positive, which is equal to $\mathbb 1-B(T^\mathstrut-\mathbb 1)(T^*-\mathbb 1)B^*$. But the thing we are subtracting from $\mathbb 1$ is positive, so for the subtraction to result in a positive thing, what we are subtracting must have norm smaller than $1$.
What we are subtracting is however
$$(T+\mathbb 1)^{-1}(T-\mathbb 1) \cdot (T^*-\mathbb 1)(T^*+\mathbb 1)^{-1}$$
Which has norm $\|(T^*-\mathbb 1)(T^*+\mathbb 1)^{-1}\|^2$, for this to be smaller than one it must be true that $\|(T-\mathbb 1)(T+\mathbb 1)^{-1}\|≤1$ holds.
