Normal operator close in norm to projection, what about the spectrum?

Question

Let $T$ be a normal operator on a Hilbert space and suppose $T$ is close in norm to a projection $P$. Can I say that the spectrum of $T$ is contained in small balls around $0$ and $1$?

Answer 1

Even with $P$ just an idempotent and not requiring $T$ to be normal.

We have \begin{align} \|T-T^2\|&\leq\|T-P\|+\|P-P^2\|+\|P^2-T^2\|\\ \ \\ &=\|T-P\|+\|P^2-T^2\|\\ \ \\ &\leq\|T-P\|+\|P^2-PT\|+\|PT-T^2\|\\ \ \\ &\leq (1+\|P\|+\|T\|)\,\|T-P\|. \end{align} Now, using $\rho$ for the spectral radius, $$\tag{1} \rho(T-T^2)\leq\|T-T^2\|\leq k\|T-P\| $$ for a certain $k$ (we can take $k=1+\|P\|+\|T\|$ as above). Let $r=k\|T-P\|$. Since $$ \sigma(T-T^2)=\{\lambda-\lambda^2:\ \lambda\in\sigma(T)\} $$ we get from $(1)$ that $|\lambda-\lambda^2|\leq r$ for all $\lambda\in\sigma(T)$. Thus, either $$ |\lambda|\leq \sqrt r,\ \ \ \text{ or } \ \ \ |1-\lambda|\leq\sqrt r. $$ In other words, if $B_\delta(\mu)$ denotes the ball of radius $\delta$ around $\mu$, then $$\sigma(T)\subset B_{\sqrt r}(0)\cup B_{\sqrt r}(1).$$

Answer 2

First of all $\rho(T)=||T||$ because $T$ is normal ($\rho$ is the spectral radius). So by the triangle inequality $\rho(T)\leq1+c$, given $||T-P||\leq c$. Now the formula in the comment $$(T-\lambda I)^{-1}=((T-P)(P-\lambda I)^{-1}+I)^{-1}(P-\lambda I)^{-1}$$ makes sense, using the Von Neumann series, if $(T-P)(P-\lambda I)^{-1}$ is less than $1$ in norm. If we let $A$ be the closed ball centered at $0$ of radius $1+c$, with two small open balls removed around $0$ and $1$, we have $\sup_A ||(P-\lambda I)^{-1}||<\infty$ by compactness. So if $c$ is small enough we have the result. The smaller the open balls, the smaller the $c$.