Spectral projection of an isolated eigenvalue agrees with the riesz projection

First $C_b(K)$ is Banach in the supremum norm. If $(f_m)$ is Cauchy, then $(f_m(\zeta ))$ converges for every $\zeta$; call the limit $f(\zeta )$. Fixing one sufficiently late index bounds $f$ uniformly, and letting the other index tend pointwise to its limit in the Cauchy estimate gives $\mathrm{\parallel }{f}_m-f{\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$. Thus $f$ is continuous by [A7], proving completeness. Now let $\delta =\mathrm{dist}(\gamma ,K)>0$, positive since the two compact sets are disjoint. For $z,w$ on the circle, $g_z\in C_b(K)$, $\mathrm{\parallel }{g}_z{\mathrm{\parallel }}_{\mathrm{\infty }}\le {\delta }^{-1}$ and $\mathrm{\parallel }{g}_z-g_w{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\mid }z-w\mathrm{\mid }{\delta }^{-2}$, by subtracting the reciprocals pointwise. Thus $t\mapsto g_{\gamma (t)}\gamma \mathrm{\prime }(t)$ is continuous, indeed uniformly continuous, into $C_b(K)$ on $[0,2\pi ]$. Step functions on successively finer equal subdivisions, with endpoint values as coefficients, approximate it uniformly, hence also in integral norm (error at most $2\pi$ times the uniform error); they show strong measurability and Bochner integrability by definition. Applying the bounded map ${\mathrm{\Phi }}_E:C_b(K)\to \mathcal{B}(H)$ also proves continuity and Bochner integrability of the resolvent contour integrand, since ${\mathrm{\Phi }}_E(g_z)=(zI-T{)}^{-1}$.

The scalar Cauchy kernel of the contour is the indicator of the enclosed disc: for every $\zeta \in \mathbb{C}\setminus \gamma ([0,2\pi ])$ one has $c(\zeta ):=\frac{1}{2\pi i}{\oint }_{\gamma }(z-\zeta {)}^{-1}dz=1$ if $\mathrm{\mid }\zeta -\lambda \mathrm{\mid }<r$ and $c(\zeta )=0$ if $\mathrm{\mid }\zeta -\lambda \mathrm{\mid }>r$, by the Cauchy integral formula applied to $f\equiv 1$ in the first case and Cauchy's theorem on the convex disc $D(\lambda ,\mathrm{\mid }\zeta -\lambda \mathrm{\mid })$ in the second.

By step 1.1 and Bochner commutation, $\frac{1}{2\pi i}{\oint }_{\gamma }(zI-T{)}^{-1}dz={\mathrm{\Phi }}_E(\frac{1}{2\pi i}{\int }_0^{2\pi }{g}_{\gamma (t)}\gamma \mathrm{\prime }(t)dt)$. For every $\zeta \in K$, evaluation $u\mapsto u(\zeta )$ is bounded linear on $C_b(K)$ with norm at most one; applying Bochner commutation once more identifies the function inside ${\mathrm{\Phi }}_E$ pointwise with $c{\mathrm{\mid }}_K$. All contour integrals here include the derivative of the parametrization.

Evaluation on the spectrum: the closed disc $\overline{D(\lambda ,r)}$ meets $\sigma (T)$ only at $\lambda$, so for $\zeta \in \sigma (T)$ the value $c(\zeta )$ is $1$ exactly at $\zeta =\lambda$ and $0$ otherwise; hence $c{\mathrm{\mid }}_{\sigma (T)}={\mathbf{1}}_{\{\lambda \}}$ and the contour integral equals ${\mathrm{\Phi }}_E({\mathbf{1}}_{\{\lambda \}})=E(\{\lambda \})$.

To check the defining Riesz cycle conditions, choose $R>r$ with $K\setminus \{\lambda \}$ disjoint from $\overline{D(\lambda ,R)}$: compactness gives such an $R$ if this complement is nonempty, and any $R>r$ works otherwise. Choose $r<R_1<R_2<R$, set $U_1=D(\lambda ,R_1)$ and $U_0=\{z:\mathrm{\mid }z-\lambda \mathrm{\mid }>R_2\}$, and define $\chi =1$ on $U_1$, $\chi =0$ on $U_0$. These are disjoint open neighborhoods of the respective spectral parts. The circle lies in $U_1\setminus K$, has index one at $\lambda$, zero at the other spectral points and zero outside $U_1\cup U_0$, by step 1.2. Thus it is a permitted cycle in the Riesz definition and $\chi =1$ on it. Consequently $P_{\{\lambda \}}=\frac{1}{2\pi i}{\oint }_{\gamma }(zI-T{)}^{-1}dz=E(\{\lambda \})$.

The isolated spectral point is an eigenvalue. Indeed ${\mathbf{1}}_{\{\lambda \}}$ is a nonzero continuous function on $K$ because the singleton is clopen there, so isometry of the continuous calculus gives $\mathrm{\parallel }{\mathbf{1}}_{\{\lambda \}}(T)\mathrm{\parallel }=1$. Agreement of the calculi and step 3.1 identify this operator with $E(\{\lambda \})$, which is therefore nonzero; since its range equals $\ker (T-\lambda I)$, that eigenspace is nonzero.

The spectral projection of the isolated eigenvalue $\lambda$ is therefore exactly the Riesz projection computed from the resolvent $(zI-T{)}^{-1}$ along a positively oriented circle separating $\lambda$ from the rest of the spectrum.