If $T$ is compact there is some $n$ such that $\operatorname{range}(T-\alpha I)^n=\operatorname{range}(T-\alpha I)^{n+1}$

Question

This is a part of the exercise 11 in the page 323 of the free digital book Measure, Integration & Real Analysis of Sheldon Axler:

If $\alpha \neq 0$ and $T$ is a compact operator in a Hilbert space $V$ then there is some $n$ such that $\operatorname{range}(T-\alpha I)^n=\operatorname{range}(T-\alpha I)^{n+1}$.

The statement is clear if $\alpha $ is not in the spectrum of $T$ (because in this case $T-\alpha I$ would be invertible) so we can assume WLOG that $\alpha $ is an eigenvalue of $T$.

In this case, naming $K_n:=\ker(T-\alpha I)^n$, $K^*_n:=\ker(T^*-\bar\alpha I)^n$ and $R_n:=\operatorname{range}(T-\alpha I)^n$ then using the following theorems:

$$ \begin{align*} \dim K_n=\dim [\ker(T^*-\bar \alpha I)^n]<\infty ,\, \text{ for compact }T\tag1\\ V=\operatorname{range}(T-\alpha I)^n\oplus \ker(T^*-\bar \alpha I)^n,\, \text{ for compact }T\tag2 \end{align*} $$

I find the following equivalence: $$ \forall n \in \Bbb N_{> 0} :R_{n+1}\subsetneq R_n\iff \forall n \in \Bbb N_{> 0} :K^*_n \subsetneq K^*_{n+1}\\ \iff \forall n \in \Bbb N_{> 0} :K_n \subsetneq K_{n+1}\tag3 $$ My idea then is to find a contradiction assuming that $\mathrm{(3)} $ holds for all $n\in \Bbb N $. Assuming it then we can construct bounded sequences $(g_n)$ or $(h_n)$ such that $g_n\in R_n\setminus R_{n+1}$ or $h_n\in K_{n+1}\setminus K_n$. These sequences can be orthonormal.

Then my idea was to find a way to contradict the compactness of $T$ using some sequence as the ones described above. To do this we can try to show that the image of the sequence, under some compact operator derived from $T$, cannot have a convergent subsequence. However I dont find a way to accomplish this strategy.

I need some help or hint.

Answer 1

I get a simple solution using an strategy already used in the cited textbook. This is my solution:

Using the notation of the question suppose that $K_n\subsetneq K_{n+1}$ for all $n\in \Bbb N $. Then there is some $e_n\in K_n \cap (K_{n-1}^\perp)$ with $\|e_n\|=1$ for all $n\in \Bbb N $, therefore for $m<n$ we find that $$ \|Te_n-Te_m\|=\|\underbrace{\overbrace{(T-\alpha I)e_n}^{\in K_{n-1}}-\overbrace{(T-\alpha I)e_m}^{\in K_{m-1}}-\overbrace{\alpha e_m}^{\in K_m}}_{\in K_{n-1}}+\underbrace{\alpha e_n}_{\in K_{n-1}^{\perp }}\|\geqslant \|\alpha e_n\|=|\alpha |>0 $$ where we used the fact that $K_{m-1}\subsetneq K_m\subset K_{n-1}$ for $m<n$. Thus the sequence $(Te_n)$ cannot have any convergent subsequence, contradicting the fact that $T$ is compact, and so our assumption that $K_n\subsetneq K_{n+1}$ for all $n\in \Bbb N $ cannot be true.$\Box$

Answer 2

Suppose that $K_n \subsetneq K_{n+1}$ for all $n \in \mathbb N.$ The Lemma of Riesz gives then:

to each $n \in \mathbb N$ there is $x_n \in K_{n+1}$ such that $||x_n||=1$ and $||x_n-x|| \ge 1/2$ for all $x \in K_n.$

Now show that $||Tx_{n+1}-Tx_m|| \ge 1/2$ for all $n$ and all $m \in \{1,2,...,n\}.$

This is a contradiction, since $(Tx_n)$ contains a convergent subsequence.