To show an operator is an isomorphism

Question

Given that $X$ is a Hilbert space. $T$ is a bounded linear operator where $T:X\rightarrow X$. It is also given that there exists $\theta$ where $0<\theta<1$ such that $|<Tx,x>|\leq\theta$ for all $x\in X$ and $\|x\|\leq 1$.

I am supposed to show that $I-T$ is an isomorphism from $X$ onto itself.

My idea is to show that there exists constants $c,C$ such that $0<c\leq C<\infty$ and $c\|x\|\leq \|(I-T)x\|\leq C\|x\|$ for all $x\in X$ and then $I-T$ is an isomorphism.

Now, $\|(I-T)x\|=\|x-Tx\|\leq\|x\|+\|Tx\|\leq\|x\|+C'\|x\|=(1+C')\|x\|$

The existence of the $C'$ is due to the fact that $T$ is bounded linear. So the right side is more or less done. But how to prove the left side? How do I make use of the $\theta$ given in the question?

Another attempt...

$\|x-Tx\|^{2}=<x-Tx><x-Tx>=<x,x>-2<Tx,x>+<Tx,Tx>$

Since $<x,x>$ and $<Tx,Tx>$ are both non-negative.

$\|x-Tx\|^{2}\geq <x,x>-2<Tx,x>$ ...

Answer 1

Hints:

Let's check that $I-T$ is bounded below. For $||x|| = 1$, $$ ||x - T x|| \ge |\langle x - Tx , x \rangle| \ge ||x||^2 - |\langle Tx , x \rangle| \ge 1 - \theta. $$

Since $I - T$ is bounded below, conclude that it has closed range.

At this point, you still need to show that $I - T$ has dense range, and hence is surjective.

Answer 2

In your second attempt you have $||x-Tx||^{2} \geq \theta ||x||^{2}$ if $||x|| \leq 1$. Just apply this with x changed to $x/||x||$ to see that the same inequality holds for all x.