# Spectrum of non-invertible operator

By an operator on Banach space $$X$$ we mean a bounded linear map $$T:X\to X$$. The spectrum of an operator $$T$$ on a complex Banach space $$X$$ is the set

$$\begin{equation} \sigma(T)= \{\lambda\in\mathbb{C}: T-\lambda I \text{is not invertible}\} \end{equation}$$

We denote by $$\mathbb{D}$$ the open unit disc in the complex plane $$\mathbb{C}$$.

Can I say that:

1) If $$\sigma(T)\subset \mathbb{D}$$, then there is $$0 and $$C\geq 0$$ such that $$||T^n||\leq Ct^n$$ for all $$n\in\mathbb{N}$$?

2) If $$T:X\to X$$ is non-invertible, then $$\sigma(T)\cap (\mathbb{C}-\overline{\mathbb{D}})=\emptyset$$ ?

• What have you tried? Did you play around with any 2x2 matrices? – Daniel McLaury Mar 11 '19 at 21:30
• @DanielMcLaury, In my research, $X$ is a Banach space and $T:X\to X$ is an operator. My field is dynamical system and I want to study dynamic of operator on $X$ and I do not know properties of spectrum. – user479859 Mar 11 '19 at 21:37
• The Euclidean plane is a perfectly good Banach space, and every 2x2 matrix gives you a bounded linear map on it. – Daniel McLaury Mar 11 '19 at 21:41

Yes to 1. If $$\sigma(T)\subseteq\{ \lambda : |\lambda| < 1 \}$$, then, because the spectrum is closed, there exists $$0 < r < 1$$ such that $$\sigma(T)\subseteq \{ \lambda : |\lambda| \le r \}$$, and there exists $$N$$ large enough that
$$\|T^n\|^{1/n} \le \frac{1+r}{2} < 1,\;\;\; n \ge N,\\ \|T^n\| \le \left(\frac{1+r}{2}\right)^n,\;\; n \ge N.$$
There is a constant $$C > 1$$ such that $$\|T^n\| \le C\left(\frac{1+r}{2}\right)^n$$ for all $$1 \le n < N$$. So $$\|T^n\| \le C\left(\frac{1+r}{2}\right)^n$$ for all $$n \ge 1$$.
No to 2. If $$T$$ is non-invertible, then the only thing you can say is that $$0\in\sigma(T)$$. The spectrum can be any compact subset of $$\mathbb{C}$$ that includes $$0$$, which would alow $$\sigma(T)\cap(\mathbb{C}\setminus\mathbb{D})$$ to be non-empty.
Hint: For 1) if the subset is not strict consider the identity operator on $$\mathbb R^2$$. If the subset is strict then remember we know that $$\rho(T)=\lim_{n\to\infty}\|T^n\|^{1/n}$$, where $$\rho(T)$$ is the spectral radius, and we also know $$\sigma(T)$$ is compact.
For 2) consider the operator $$T_A:\mathbb R^2\to \mathbb R^2$$ given by the matrix: $$A=\begin{pmatrix}2&0\\0&0\end{pmatrix}.$$