# Rudin's functional analysis, theorem 4.23 (existence of a certain sequence)

If $$X$$ is a Banach space, $$T \in \mathcal{B}(X)$$, $$T$$ is compact, and $$\lambda \neq 0$$, then $$T - \lambda I$$ has closed range.

Proof with questions below

Proof: By (d) of Theorem 4.18, $$\text{dim } \mathcal{N}(T - \lambda I) < \infty$$. By (a) of Lemma 4.21, $$X$$ is the direct sum of $$\mathcal{N}(T - \lambda I)$$ and a closed subspace $$M$$. Define an operator $$S \in \mathcal{M,X}$$ by $$Sx = Tx - \lambda x \;\;\; (1)$$ Then $$S$$ is one-to-one on $$M$$. Also, $$\mathcal{R}(S) = \mathcal{R}(T - \lambda I )$$. To show that $$\mathcal{R}(S)$$ is closed, it suffices to show the existence of an $$r > 0$$ such that $$r \lVert x \rVert \leq \lVert S x \rVert \;\;\; (2)$$ by Theorem 1.26.

First bit I don't get

If (2) fails for every $$r > 0$$, there exists $$\left\{ x_n \right\}$$ in $$M$$ such that $$\lVert x_n \rVert = 1$$, $$S x_n \to 0$$, and (after a passage to a subsequence) $$T x_n \to x_0$$ for some $$x_0 \in X$$.

Why does such $$x_n$$ exist?

If (2) fails for every $$r>0$$, then we have $$\forall r>0, \ \exists x_r \in M \ \ni r \lvert x_r \rvert > \lvert Sx_r \rvert.$$ The $$x_r$$'s clearly can't be zero, so we can normalize them to have unit norm. Take a sequence $$r_n \downarrow 0$$ and let $$x_n = x_{r_n}$$. This should give everything except convergence of $$Tx_n$$ for a subsequence, which should follow from the compactness of $$T$$.
• Why $x_r \in X$ and not $x_r \in M$? why can't they be $0$? – user8469759 Mar 15 at 15:14
• You're correct, the $x_r$ are in $M$. – Gary Moon Mar 15 at 15:22