Prove that any eigenspace of an operator T ∈ C(X ) is closed. Definition:-
An operator $T:X\to Y$ is called closed if $Γ(T)=\{(x,T(x)):x\in X\}$ is a closed set in $X ⊕Y.$
Let $C(X)$ denotes the set of all closed linear operators from a Banach space $X$
to itself.
Problem: Prove that any eigenspace of an operator T ∈ C(X ) is closed.
(Hint:- Use the result)
My attempt:-
Let $T$ be a closed operator. Let $\lambda $ be an eigenvalue of the closed operator $T$. We need to prove that $\overline{\{x\in X: T-\lambda I(x)=0\}}=\{x\in X: T-\lambda I(x)=0\}.$
We know that $\{x\in X: T-\lambda I(x)=0\}\subseteq \overline{\{x\in X: T-\lambda I(x)=0\}}$.
Let $x\in \overline{\{x\in X: T-\lambda I(x)=0\}}\implies \exists \{x_n\}\subseteq \{x\in X: T-\lambda I(x)=0\}: \lim_{n\to \infty}x_n=x.$
We have $(T-\lambda I)(x_n)=0.$ Taking limit and linearity
We have $T(x_n)-\lambda I(x_n)=0\implies T(x_n)=\lambda x_n.$ As $n\to \infty , x_n \to x$ and $T(x_n) \to T(x)\implies T(x)=\lambda x.$
Hence, $x$ is an eigenvector of $T$. This shows that Eigen space is closed. Is my proof Correct?
 A: Fix an eigenvalue $\lambda$ . Let $\{x_{n}\}$ be a sequence such that $x_{n}\to x$ and $(T-\lambda I)x_{n}=0$
To show that the eigenspace corresponding to $\lambda$ is closed you only need to show that $(T-\lambda I)x=0$ .
To that extent you have $(T-\lambda I) x_{n}=0\implies T(x_{n})=\lambda x_{n}$ .
As $\lambda x_{n}\to \lambda x$ you have $T(x_{n})\to \lambda x$
Now what you have is that $T(x_{n})$ converges as a sequence and $x_{n}$ also converges . So the sequence $(T(x_{n}),x_{n})$ in the set $\Gamma(T)$ is a convergent sequence. And hence by the closedness of the graph, you have $(T(x_{n}),x_{n})\to (Tx,x)$ .
Hence $T(x_{n})\to T(x)$
So $T(x)= \lambda x$.
Hence $(T-\lambda I)x=0$ as required .
Point to note . If $Y$ is also Banach then by the Closed Graph Theorem we have $\Gamma(T) \,\text{is closed} \iff T\text{ is continuous}$ . So here you can directly use the fact that $T-\lambda I$ is a continuous linear map from $X\to Y$ and hence $(T-\lambda I)x_{n}=0$ implies $(T-\lambda I)x=0$
