Riesz schauder spectrum of a compact operator

For $\lambda \mathrm{\ne }0$, $\lambda \mathrm{\notin }\sigma (K)$ if and only if $\ker (\lambda I-K)=\{0\}$: the operator $\lambda I-K$ is bijective exactly when it is injective, by the injective-iff-surjective part of [A1] applied to $A_{\lambda }$.

For $\lambda \mathrm{\ne }0$ the generalized eigenspace satisfies $G_{\lambda }(K)={\bigcup }_{n\ge 1}\ker (A_{\lambda }^n)=\ker (A_{\lambda }^m)$ for an $m$ given by [A2], hence $G_{\lambda }(K)$ is finite dimensional whenever it is nonzero, and it is nonzero exactly when $\lambda$ is an eigenvalue.

The resolvent set $\rho (K)$ is open: if $\lambda \in \rho (K)$ and $B:=\lambda I-K$ with inverse $B^{-1}$, then for every scalar $\mu$ with $\mathrm{\mid }\mu -\lambda \mathrm{\mid }\mathrm{\parallel }{B}^{-1}\mathrm{\parallel }<1$ the operator $\mu I-K=B+(\mu -\lambda )I$ is invertible with bounded inverse by [A3], so such $\mu$ lie in $\rho (K)$.

$\sigma (K)\subseteq \{\lambda :\mathrm{\mid }\lambda \mathrm{\mid }\le \mathrm{\parallel }K\mathrm{\parallel }\}$: if $\mathrm{\mid }\lambda \mathrm{\mid }>\mathrm{\parallel }K\mathrm{\parallel }$ then $\mathrm{\parallel }K\mathrm{/}\lambda \mathrm{\parallel }<1$ and $\lambda I-K=\lambda (I-K\mathrm{/}\lambda )$ is invertible with bounded inverse by [A3], so $\lambda \in \rho (K)$.

If $X$ is infinite dimensional then $0\in \sigma (K)$: if $0\in \rho (K)$ then $-K=0I-K$ has a bounded inverse $S$, and $I=(-K)S=K(-S)$ is compact by [A5]; then the closed unit ball of $X$, the image of itself under $I$, is compact, so $X$ admits an ordered basis of finite length by [A5], a contradiction.

If $\lambda \in \sigma (K)$ and $\lambda \mathrm{\ne }0$, then with $N:=G_{\lambda }(K)=\ker (A_{\lambda }^m)$ and $Y:=\mathrm{ran}{A}_{\lambda }^m$ from [A2] one has $X=N\oplus Y$, $A_{\lambda }^m$ vanishes on $N$, and $\lambda I-K$ restricted to $Y$ is invertible with bounded inverse; for every scalar $\mu \mathrm{\ne }\lambda$ the operator $\mu I-K$ is invertible on $N$, because on $N$ it equals $(\mu -\lambda )(I+(\lambda \mathrm{/}(\mu -\lambda ))A_{\lambda })$ and $A_{\lambda }{\mathrm{\mid }}_N$ is nilpotent.

Claim 1: if $\lambda \in \sigma (K)$ and $\lambda \mathrm{\ne }0$, then $\lambda$ is an eigenvalue with finite-dimensional generalized eigenspace: by [step 1.1] the kernel $\ker (\lambda I-K)$ is nonzero, and by [step 1.2] the generalized eigenspace is finite dimensional.

For real $\epsilon >0$ the set $S_{\epsilon }:=\{\lambda \in \sigma (K):\mathrm{\mid }\lambda \mathrm{\mid }\ge \epsilon \}$ is compact in $\mathbb{C}$: it is bounded by [step 1.4] and closed because $\rho (K)$ is open by [step 1.3], so under $\mathbb{C}={\mathbb{R}}^2$ it is a closed and bounded subset of ${\mathbb{R}}^2$, hence compact by [A4].

If $\lambda \in \sigma (K)$ and $\lambda \mathrm{\ne }0$, there is a real $\delta >0$ with $\mu \in \rho (K)$ for every $\mu$ with $0<\mathrm{\mid }\mu -\lambda \mathrm{\mid }<\delta$: choose $\delta >0$ with $\delta \mathrm{\parallel }((\lambda I-K){\mathrm{\mid }}_Y{)}^{-1}\mathrm{\parallel }<1$ for the $Y$ of [step 2.1]; then for such $\mu$ the restriction of $\mu I-K$ to $Y$ is invertible by [A3] and its restriction to $N$ is invertible by [step 2.1], and invertibility on both summands of $X=N\oplus Y$ gives invertibility on $X$.

Claim 2: $S_{\epsilon }$ is finite. Every point of $S_{\epsilon }$ lies in $\sigma (K)$ and is nonzero, so by [step 3.1] each $\lambda \in S_{\epsilon }$ has a ball $B(\lambda ,{\delta }_{\lambda })$ meeting $\sigma (K)$ only in $\lambda$; the sets $S_{\epsilon }\cap B(\lambda ,{\delta }_{\lambda })$, $\lambda \in S_{\epsilon }$, form an open cover of the compact set $S_{\epsilon }$ by [step 2.3], and each member contains only the single point $\lambda$, so a finite subcover exhibits $S_{\epsilon }$ as a finite set.

Claim 3 is [step 1.5], and claims 1, 2, 3 are respectively [step 2.2], [step 4.1] and [step 1.5]; the statement is proved.