Let $\lambda\in \Bbb R_{\ne 0}$ and let $E$ be a vector space and let $T:E\to E$ be a linear continuous compact operator.
Prove there exists some $n\in \Bbb N$ such that $$\ker((\lambda Id-T)^n)=\ker((\lambda Id-T)^{n+1})$$
I am trying to prove this by contradiction and some intuition (I have big doubts but here is my attempt)
For a subset $A\subset E$, denote its complement in E by $A^c$.
Denote $\forall n\in \Bbb N$
$$U_n:= (\ker((\lambda Id-T)^n))^c.$$
Suppose that $\forall n\in \Bbb N$,
$$\ker((\lambda Id-T)^n) \subsetneq \ker((\lambda Id-T)^{n+1})$$
Hence the sequence $(U_n)_{n\ge 0}$ is strictly decreasing.
We know that each $U_n$ is open (since $\ker((\lambda Id-T)^n)$ is closed by continuity of $(\lambda Id-T)^n$)
By The Baire category theorem,
$$\lim\limits_{n\to \infty}U_n=\bigcap_{n\in\Bbb N} U_n$$ is dense in $E$.
I am stuck here, I feel that this last result contradict the compacity of $T$, but I don't see how to prove it.
Best Answer
Im not sure how to do it with Baire Category Theorem but I know how to do it using Riesz Lema. First, let $V$ be a vector space and $A:V \rightarrow V$ a linear function, then one of the following must be true $$\begin{align} \{ 0 \} \subsetneq Ker(A) \subsetneq Ker(A^2) \subsetneq Ker(A^3)\subsetneq Ker(A^4) \subsetneq \cdots \\ \{ 0 \} \subsetneq Ker(A) \subsetneq \cdots \subsetneq Ker(A^m) = Ker(A^{m+1}) = \cdots \end{align} $$ This shouldn't be hard to prove. Let $E$ be a Banach space, $K:E \rightarrow E$ a linear continuous compact operator and $T:=Id_E-K$, I will prove there is an $n \in \Bbb N$ such that $$Ker(T^n)=Ker(T^{n+1})$$ What you want to prove follows directly from this (by writing $ \gamma .Id-K=\gamma .(Id-\gamma ^{-1}.K)$). We will prove it by contradiction, suppose $$\{ 0 \} \subsetneq Ker(T) \subsetneq Ker(T^2) \subsetneq Ker(T^3)\subsetneq Ker(T^4) \subsetneq \cdots$$ The inclusions are strict so, by Riesz Lema, for every $n \in \Bbb N$ there is a $y_n \in Ker(T^n)$ with $\vert \vert y_n \vert \vert = 1$ and $dist(y_n,Ker(T^{n-1}))>1/2$. Suppose $n>m$, by the definition of $T$ we have $$K(y_n)-K(y_m)=y_n-T(y_n)-y_m+T(y_m)$$ Its easy to check $$\begin{align} T(y_n) \in Ker(T^{n-1}) \\ y_m \in Ker(T^m) \subseteq Ker(T^{n-1}) \\ T(y_m) \in Ker(T^{m-1}) \subseteq Ker(T^{n-1}) \end{align}$$ So $T(y_n)+y_m-T(y_m) \in Ker(T^{n-1})$ and then
$$\vert \vert K(y_n)-K(y_m) \vert \vert = \vert \vert y_n -(T(y_n)+y_m-T(y_m)) \vert \vert \geq dist(y_n,Ker(T^{n-1})) \geq 1/2$$ So $\{ K(y_n) \}_{n \in \Bbb N}$ cant have a convergent subsequence (It wouldnt be cauchy) but $\vert \vert y_n \vert \vert =1$, this contradicts $K$ being a compact operator.