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For a bounded operator on a complex Banach space, partition a disconnected spectrum into two nonempty compact pieces. The holomorphic function equal to one near the first piece and zero near the second defines a Riesz projection. The holomorphic spectral mapping theorem makes it neither zero nor identity. Its closed range is a proper nonzero invariant vector subspace. This need not be an orthogonal projection, and the argument does not require a normal operator.

Define the holomorphic function $e$ on the disconnected open set $U$ to be one on $V$ and zero on $W$. Since $e^2=e$, its holomorphic functional calculus value $p={\Theta }_x(e)$ is an idempotent. The spectral mapping theorem and the nonempty spectral intersection with $V$ show $1\in \sigma (p)$; thus $p\ne 0$. These are the Riesz projection values selecting the indicated spectral component.

Part (b) gives $p_n={\Theta }_{x_n}(e)\to p$. If $\sigma (x_n)$ avoided $V$, spectral mapping would give $\sigma (p_n)=\{0\}$. A nonzero idempotent always has $1$ in its spectrum: if $1-p_n$ were invertible, $p_n(1-p_n)=0$ would force $p_n=0$. Thus avoiding $V$ forces $p_n=0$. That is impossible eventually, since $p\ne 0$ and $p_n\to p$. Therefore $\sigma (x_n)\cap V\ne \varnothing$ eventually. This is persistence of a separated spectral component, not a claim that arbitrary individual spectral points vary continuously.

Give $\mathcal{O}(U)$ the compact-open topology, that is, uniform convergence on compact subsets. Put $R(z)=(z1-x{)}^{-1}$. Choose a finite oriented contour $\Gamma$ in $U\setminus \sigma (x)$ with winding number one on $\sigma (x)$ and zero outside $U$; boundaries of a finite union of sufficiently small rectangles around the compact spectrum provide such a contour. Define the holomorphic functional calculus byThe integral is a Bochner integral along the piecewise smooth contour. It is independent of the admissible contour by the Cauchy integral theorem, since the resolvent is holomorphic off the spectrum. In particular, restricting $f$ to a smaller open spectral neighborhood leaves its value unchanged.

Linearity is immediate, andproves continuity for the compact-open topology. For the constant and coordinate functions, move the corresponding resolvent integrals to a large circle. The norm-convergent expansion $R(z)={\sum }_{n\ge 0}{x}^n{z}^{-n-1}$ there gives ${\Theta }_x(1)=1$ and ${\Theta }_x(u)=x$.

To check multiplication, take two nested admissible contours, with $z$ on the outer one and $w$ on the inner one. The resolvent identity givesIn the double integral for ${\Theta }_x(f){\Theta }_x(g)$, the term with $R(w)$ integrates in $z$ to $f(w)R(w)g(w)$ by the Cauchy integral formula. The term with $R(z)$ integrates in $w$ to zero, because $z$ lies outside the inner region. Thus ${\Theta }_x(fg)={\Theta }_x(f){\Theta }_x(g)$. Nested regions can be chosen with closures inside $U$, so the argument also handles disconnected $U$ and contours with several boundary components.

For uniqueness, use the following form of the Runge theorem: rational functions whose poles lie outside an open set $U$ are dense in $\mathcal{O}(U)$ for uniform convergence on compact subsets; polynomials are included, with a pole at infinity allowed. This follows by a Runge exhaustion of $U$; no connectedness of $U$ is required. Any unital algebra homomorphism sending $u$ to $x$ must send $(u-a{)}^{-1}$ to $(x-a1{)}^{-1}$ for $a\notin U$. Consequently it agrees with the constructed map on every such rational function. Continuity and density give uniqueness. This is continuity and uniqueness of holomorphic functional calculus.

For the holomorphic spectral mapping theorem, first suppose $\mu \notin f(\sigma (x))$. The reciprocal $g=1/(f-\mu )$ is holomorphic on some smaller open neighborhood $V$ of $\sigma (x)$, even if it is not holomorphic on all of $U$. Restriction consistency and multiplicativity on $V$ show that ${\Theta }_x(g)$ is an inverse of ${\Theta }_x(f)-\mu 1$. Thus $\sigma ({\Theta }_x(f))\subseteq f(\sigma (x))$.

Conversely, if $\lambda \in \sigma (x)$, thenis holomorphic on $U$, with the singularity at $\lambda$ removed. HenceIn a commutative algebra, an invertible product has invertible factors. The displayed product cannot be invertible because $x-\lambda 1$ is not. Therefore

For the operator conclusion, work over complex scalars and let $\sigma (T)=K_1\cup K_2$ be a partition into disjoint nonempty compact sets. To apply the preceding commutative-algebra result precisely, take the closed unital subalgebra of $\mathcal{B}(X)$ generated by $T$ and all its resolvents $(zI-T{)}^{-1}$, $z\notin {\sigma }_{\mathcal{B}(X)}(T)$. The generators commute, so this is a resolvent-generated commutative algebra. Its spectrum of $T$ is exactly the operator spectrum: every resolvent outside the latter was included, and an inverse inside the subalgebra would also be an inverse in $\mathcal{B}(X)$.

Choose disjoint open neighborhoods $U_1,U_2$ of $K_1,K_2$, and let $f$ equal one on $U_1$ and zero on $U_2$. The corresponding Riesz projection $P=f(T)$ satisfiesThe last spectrum is computed in the chosen commutative algebra and already ensures $P\ne 0,I$. Its range $Y=PX=\ker (I-P)$ is closed, nonzero and proper. The commutation identity makes it invariant under $T$. This proves that a disconnected spectrum yields a nontrivial invariant subspace. Using all resolvents avoids inadvertently replacing the operator spectrum by the possibly larger spectrum in a polynomial-generated subalgebra.

Factor the perturbed operator on $D(A)$ asSincethe Neumann series makes $I+B{A}^{-1}$ invertible. Therefore $0\notin \mathrm{Sp}(A+B)$ andThe geometric-series bound gives

Now choose a bounded open neighborhood $U$ of the isolated spectral component $X$ such thatand $\overline{U}$ meets no other component of the spectrum. Compactness of $\partial U$ givesFor all sufficiently large $n$, $\Vert B_n\Vert M<1$. Applying the first part to $A-zI$ shows uniformly that $\partial U\subset \rho (A+B_n)$.

The corresponding Riesz projections areThe resolvent identity and the uniform Neumann bound imply $\Vert P_n-P\Vert \to 0$. The projection $P$ is nonzero because $U$ contains the nonempty spectral component $X$. Projections at distance less than one have isomorphic ranges, so $P_n\ne 0$ for large $n$. Therefore $A+B_n$ has spectrum inside $U$, and any such point $z$ satisfies $\mathrm{dist}(z,X)<ϵ$. Thusfor every sufficiently large $n$.