I'm having trouble to understand the proof of the following Lemma where $r(T)$ is the spectral radius of an operator $T$.
If $T\in\mathcal{B}(\mathcal{X})$ is an operator on a complex Banach space $\mathcal{X}$, then the following assertions are pairwise equivalent
(a) $T^{n}\stackrel{u}{\rightarrow} O$
(b) $r(T)<1$
(c) $||T^{n}||\leq \beta \alpha ^{n}$ for every $n\geq0$, for some $\beta \geq 1$ and some $\alpha \in (0,1)$
$\mathbf{(b)\Rightarrow (c) \Rightarrow (a)}$ Suppose $r(T)<1$ and take an arbitrary $\alpha$ in $(r(T),1)$. Since $r(T)=\lim||T^{n}||^{\frac{1}{n}}$, there is an interger $n_{\alpha} \geq 1$ for which $||T^{n}||\leq \alpha^{n}$ whenever $n\geq n_{\alpha}$. Thus $||T^{n}||\leq \beta \alpha^{n}$ for every $n \geq 0$ with $\beta=\max_{0 \leq n \leq n_{\alpha}}||T^{n}||\alpha^{-n_{\alpha}}$.
What I do understand is the following:
Since $1>r(T)=\lim\limits_{n \to \infty}||T^{n}||^{\frac{1}{n}}$ and $\alpha \in (r(T),1)$ the existence of $n_{\alpha}$ with $||T^{n}|| ^{\frac{1}{n}}\leq \alpha$ for all $n\geq n_{\alpha}$ follows. This is equivalent to $||T^{n}|| \leq \alpha^{n}$ for all $n\geq n_{\alpha}$. It is also clear that $\alpha^{n_{\alpha}} \geq \alpha^{n}$ for $n \geq n_{\alpha}$ since $0\leq \alpha < 1$.
Furthermore it is obvious that $||T^{n}|| \leq \max_{0 \leq k \leq n_{\alpha}}||T^{k}||$ for $n \leq n_{\alpha}$. Thus $\max_{0 \leq k \leq n_{\alpha}}||T^{k}||$ is an upper bound for the first $n_{\alpha}$ elements of the sequence $\{||T^{n}||\}_{n=1}^{\infty}$ and $\alpha^{n_{\alpha}}$ an upper bound for the infinite sequence starting at $n_{\alpha}$.
What I do not understand is:
- How does this yields the inequality $||T^{n}||\leq \beta \alpha^{n}$?
- Why does $\beta \geq 1$ hold?
Update I guess I've understood why $\beta \geq 1$ holds. $\beta \geq 1$ is equivalent to $\max_{0 \leq k \leq n_{\alpha}}||T^{k}||\geq \alpha^{n_\alpha}$. Assume that this inequality is not satisfied, then $||T^{n}|| \leq \max_{0 \leq k \leq n_{\alpha}}||T^{k}||< \alpha^{n_\alpha}\leq \alpha^{n}$ for all $n\leq n_{\alpha}$ (since $0\leq \alpha <1$). This is a contradiction since $n_{\alpha}$ was choosen to be the smallest integer to satisfy this inequality.
I cannot understand the exact way the proof is written, but here is my take.
For $n\geq n_\alpha$, you have $$\tag1\|T^n\|\leq\alpha^n.$$ For $n<n_\alpha$, we have $$\tag2 \|T^n\|\leq\beta', $$ where $\beta'=\max\{\|T^n\|:\ n=0,\ldots,n_\alpha-1\}$. Now we take $$ \beta''=\max\{\|T^n\|:\ n=0,\ldots,n_\alpha-1\}\,\alpha^{-n_\alpha}. $$ When $n\leq n_\alpha$, since $0<\alpha<1$ we get $\alpha^{n_\alpha}\leq\alpha^n$. From $(2)$ we get $$\tag3 \|T^n\|\leq \beta'= \beta''\,\alpha^{n_\alpha}\leq\beta''\,\alpha^n,\qquad\qquad 0\leq n<n_\alpha. $$ Finally, take $$\beta=\max\{\beta'',1\}.$$ Then $\beta\geq1$ and from $(1)$ we get $$\tag4 \|T^n\|\leq\alpha^n\leq\beta\alpha^n,\qquad\qquad n\geq n_\alpha. $$ Combining $(3)$ and $(4)$, we have $$\tag5 \|T^n\|\leq\beta\,\alpha^n $$ for all $n$.