Let $H$ be an infinite dimensional Hilbert space over $\mathbb C$, $T$ be a continuous linear operator of $H$, $r(T)=\sup_{||x||=1}|(Tx|x)|$ be the numerical radius of $T$, and $z\in \mathbb C$, such that $|z|<1$. Assume that $r(T)\leqslant 1$.
Clearly $\ker (I-zT)=\{0\}$. So, if I could show that $T$ is compact then it would follow that $I-zT$ is invertible. So my questions are :
- Is there an example of $T$ non compact ?
- Is there an example of $T$ such that $I-zT$ is non surjective?
The answer to the first question is yes, any $T = \lambda\cdot I$ with $0 < \lvert\lambda \rvert < 1$ is a non-compact operator satisfying the requirements.
The answer to the second question is no, all such $I - z\cdot T$ are invertible.
First, by
$$\lvert \langle (I - z\cdot T)x \mid x\rangle\rvert = \lvert \lVert x\rVert^2 - z \langle Tx\mid x\rangle\rvert \geqslant (1 - \lvert z\rvert r(T)) > 0$$
for $\lVert x \rVert = 1$, it follows that $\lVert (I - z\cdot T) x\rVert \geqslant (1 - \lvert z\rvert r(T))\lVert x\rVert$, hence $\mathcal{R}(I - z\cdot T)$ is closed.
So either it is all of $H$, or $\mathcal{R}(I - z\cdot T)^\perp$ is nontrivial. Suppose $y \in \mathcal{R}(I - z\cdot T)^\perp$. then, by the above computation,
$$0 = \lvert \langle (I - z\cdot T)y\mid y\rangle\rvert \geqslant \lVert y\rVert^2 (1 - \lvert z\rvert r(T))$$
and the second factor on the right is strictly positive, thus $\lVert y\rVert = 0$.