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.
Best Answer
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.