Let $(E, |\cdot|)$ be a real Banach space. For a linear map $T: E \to E$, we denote by $N(T)$ its kernel and $R(T)$ its range. Let $I: E \to E$ be the identity map.
I'm trying to prove Lemma 6.2. in Brezis' Functional Analysis, i.e.,
Let $T: E \to E$ be a compact (bounded linear operator) and $\sigma (T)$ its spectrum. Let $(\lambda_n)$ be a sequence of distinct real numbers such that $(\lambda_n) \subset \sigma (T) \setminus \{0\}$ and $\lambda_n \to \lambda$. Then $\lambda = 0$.
Could you elaborate on how to prove $u=0$ in my below failed attempt?
My attempt
Because $T$ is compact, $T-\lambda_n I$ is injective IFF $T-\lambda_n I$ is bijective. Then $\lambda_n$ is also an eigenvalue of $T$. Let $E_n := N(T-\lambda_n I)$ be the eigenspace corresponding to $\lambda_n$. Then $E_n \neq \{0\}$. Then there is $e_n \in E_n$ such that $|e_n|=1$. Then
- $\{e_1, e_2, \ldots\}$ is linearly independent.
- $Te_n = \lambda_n e_n$ for all $n$.
It suffices to prove $\lambda_ne_n \to 0$. This is equivalent to proving $Te_n \to 0$. It suffices to prove each subsequence of $(Te_n)$ has a further subsequence that converges to $0$. For simplicity, we also denote this subsequence by $(Te_n)$. Because $T$ is compact, there is a subsequence $(Te_{n_k)}$ and $u\in E$ such that $Te_{n_k} \xrightarrow{k\to \infty} u$. Then $\lambda_{n_k} e_{n_k} \xrightarrow{k\to \infty} u$. We want to prove $u=0$.
As @daw mentioned in a comment, there is no "easy" workaround. Below is the proof form Brezis' textbook.
Because $T$ is compact, $T-\lambda_n I$ is injective IFF $T-\lambda_n I$ is bijective. Then $\lambda_n$ is also an eigenvalue of $T$. Let $E_n := N(T-\lambda_n I)$ be the eigenspace corresponding to $\lambda_n$. Then $E_n \neq \{0\}$. Then there is $e_n \in E_n$ such that $|e_n|=1$. Then
Let $F_n := \operatorname{span} \{e_1, e_2, \ldots, e_n\}$. Then $F_n \subsetneq F_{n+1}$ for all $n$. Then $$ F_{m-1} \subsetneq F_m \subseteq F_{n-1} \subsetneq F_n \quad \forall m <n. $$
By Riesz's lemma, for each $n\ge 1$ there is $u_n \in F_n$ such that $|u_n|=1$ and $d(u_n, F_{n-1}) := \inf_{v \in F_{n-1}} |v-u_n| \ge 1/2$. Assume the contrary that $\lambda \neq 0$. Because $T$ is compact, $(\frac{Tu_n}{\lambda_n})_n$ has a convergent subsequence. On the other hand, $$ \frac{Tu_n}{\lambda_n} - \frac{Tu_m}{\lambda_m} = \frac{(T-\lambda_n I)u_n}{\lambda_n} - \frac{(T-\lambda_m I)u_m}{\lambda_m} -u_m+u_n. $$
We have $(T-\lambda_n I) F_n \subset F_{n-1}$. Then $\frac{(T-\lambda_n I)u_n}{\lambda_n} \in F_{n-1}$ and $\frac{(T-\lambda_m I)u_m}{\lambda_m} \in F_{m-1}$. Then $$ \frac{(T-\lambda_n I)u_n}{\lambda_n} - \frac{(T-\lambda_m I)u_m}{\lambda_m} -u_m \in F_{n-1} \quad \forall m<n. $$
Then $$ \bigg | \frac{Tu_n}{\lambda_n} - \frac{Tu_m}{\lambda_m} \bigg | \ge d(u_n, F_{n-1}) \ge 1/2 \quad \forall m<n. $$
Then $(\frac{Tu_n}{\lambda_n})_n$ is not a Cauchy sequence, which is a contradiction.