Riesz spectral projection properties

Idempotence and commutation: $P^2={\chi }_E(T){\chi }_E(T)=({\chi }_E\cdot {\chi }_E)(T)={\chi }_E(T)=P$ by [L1] and [L2]; $PT={\chi }_E(T)\mathrm{i}\mathrm{d}(T)=({\chi }_E\cdot \mathrm{i}\mathrm{d})(T)=(\mathrm{i}\mathrm{d}\cdot {\chi }_E)(T)=\mathrm{i}\mathrm{d}(T){\chi }_E(T)=TP$ by [L1] and [L2].

The splitting: by [step 1.1] the operator $P$ is a bounded projection, so [L3] gives that $\mathrm{ran}(P)$ and $\ker (P)$ are closed with $X=\mathrm{ran}(P)\oplus \ker (P)$; since $T$ commutes with $P$, both summands are $T$-invariant.

Range spectrum, exclusion: assume $\mathrm{ran}(P)\mathrm{\ne }\{0\}$, as in claim 2, so its operator spectrum is defined. Let $\mu \mathrm{\notin }E$. Choose the neighbourhoods $U_1\supseteq E$, $U_0\supseteq \sigma (T)\setminus E$ of the definition so that $\mu \mathrm{\notin }{U}_1$ (possible since $\mu \mathrm{\notin }E$ and $E$ is compact). Then the germ $h:={\chi }_E\mathrm{/}(\mathrm{i}\mathrm{d}-\mu )$ is holomorphic near $\sigma (T)$: on $U_1$ it is $1\mathrm{/}(z-\mu )$ with $\mu \mathrm{\notin }{U}_1$, and on $U_0$ it is $0$. By [L1], $P=(T-\mu )h(T)=h(T)(T-\mu )$. The operator $h(T)$ commutes with $P$, so it preserves $\mathrm{ran}(P)$; on that nonzero summand its restriction is a two-sided inverse of $T-\mu$. Hence $\mu \mathrm{\notin }\sigma (T{\mathrm{\mid }}_{\mathrm{ran}(P)})$.

Kernel spectrum, exclusion: assume $\ker (P)\mathrm{\ne }\{0\}$, as in claim 3, so its operator spectrum is defined. Let $\mu \mathrm{\notin }\sigma (T)\setminus E$, that is, $\mu \in E$ or $\mu \mathrm{\notin }\sigma (T)$. Choose $U_0,U_1$ with $\mu \mathrm{\notin }{U}_0$ when $\mu \in E$, and define $q:=(1-{\chi }_E)\mathrm{/}(\mathrm{i}\mathrm{d}-\mu )$ on the complement part. Then $1-P=(1-{\chi }_E)(T)$, and the same computation with $1-{\chi }_E$ in place of ${\chi }_E$ shows that $T-\mu$ has the restriction of $q(T)$ as a two-sided inverse on $\ker (P)$. Hence $\mu \mathrm{\notin }\sigma (T{\mathrm{\mid }}_{\ker (P)})$.

Range spectrum, inclusion: continue under $\mathrm{ran}(P)\mathrm{\ne }\{0\}$. Let $\mu \in E$ and suppose that $T{\mathrm{\mid }}_{\mathrm{ran}(P)}-\mu$ were invertible on $\mathrm{ran}(P)$, with inverse $S$. Put $H:=\frac{1-{\chi }_E}{\mathrm{i}\mathrm{d}-\mu }(T)$, a bounded operator because the germ is holomorphic near $\sigma (T)$ (its numerator vanishes on $U_1\supseteq E$ and $\mu \mathrm{\notin }\sigma (T)\setminus E$), and put $V:=H+SP$ on $X=\mathrm{ran}(P)\oplus \ker (P)$. Then $(T-\mu )V=(1-P)+P=1$ and $V(T-\mu )=1$ by the same multiplicativity computation, so $T-\mu$ would be invertible on $X$, contradicting $\mu \in E\subseteq \sigma (T)$. Hence $\mu \in \sigma (T{\mathrm{\mid }}_{\mathrm{ran}(P)})$.

Kernel spectrum, inclusion: continue under $\ker (P)\mathrm{\ne }\{0\}$. Symmetrically, if $\mu \in \sigma (T)\setminus E$ and $T{\mathrm{\mid }}_{\ker (P)}-\mu$ were invertible with inverse $S$, then the germ $\frac{{\chi }_E}{\mathrm{i}\mathrm{d}-\mu }$, read as $\frac{1}{z-\mu }$ on a neighbourhood of $E$ avoiding $\mu$ and as $0$ on a neighbourhood of $\sigma (T)\setminus E$, is holomorphic near $\sigma (T)$. The operator $V:=\left(\frac{{\chi }_E}{\mathrm{i}\mathrm{d}-\mu }\right)(T)+S(1-P)$ would be a two-sided inverse of $T-\mu$ on $X$, contradicting $\mu \in \sigma (T)$. Hence $\mu \in \sigma (T{\mathrm{\mid }}_{\ker (P)})$.

Claims 2 and 3 follow from [step 2.2], [step 3.1] and [step 2.3], [step 3.2] respectively; claim 1 was proved in [step 1.1] and [step 2.1]; claim 4 is the convention recorded in the statement, applied to an empty spectral part, and no spectrum is claimed for the zero operator.