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Equip holomorphic functions on an open spectral neighborhood with the compact-open topology. A fixed admissible contour gives a bound on $\Vert {\Theta }_x(f)\Vert$ by a constant times the supremum of $|f|$ on the contour, proving continuity. The Runge theorem makes rational functions with poles outside the open set dense. A unital homomorphism sending the coordinate function to $x$ must send each inverse coordinate difference to the corresponding resolvent, so its rational values and then all its holomorphic values are forced.

Let $S$ be a closed square and let $f$ be holomorphic on a neighbourhood of a slightly larger square. The Cauchy integral formula on the larger boundary is uniformly approximated on $S$ by Riemann sums of resolvents whose poles lie outside $S$. Far-away resolvents have convergent geometric-series expansions in $z$, and resolvent propagation across a complementary component moves this polynomial approximability to every pole outside the square. Approximating the finitely many resolvents in each Riemann sum by polynomials proves polynomial approximation on $S$ without invoking the general Runge theorem.

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.

The prescribed-pole form of Runge theorem is this: let $K\subset \mathbb{C}$ be compact, and let $E\subset \hat{\mathbb{C}}\setminus K$ meet every connected component of the complement in the Riemann sphere. If $f$ is holomorphic on a neighbourhood of $K$, then for every $ϵ>0$ there is a rational function with all its poles in $E$ whose uniform distance from $f$ on $K$ is less than $ϵ$. A polynomial is regarded as having its only possible pole at infinity. The empty compact set is trivial, so assume $K\ne \varnothing$.

First suppose $\infty \in E$. Let $A\subset C(K)$ be the uniform rational approximation algebra with prescribed poles, the uniform closure of rational functions whose finite poles lie in $E$. It is a commutative unital Banach algebra, and contains $u(z)=z$. Setwhere the inverse in this definition is the pointwise continuous function on $K$. The set $S$ is relatively open: if the inverse at $\lambda$ belongs to $A$, the Neumann series gives inverses at nearby points. It is relatively closed: when ${\lambda }_n\to \lambda \notin K$, the corresponding scalar functions converge uniformly on $K$, and $A$ is closed.

Each bounded complementary component meets $E$ in a finite point $a$, and $(u-a1{)}^{-1}$ is an allowed rational function. The unbounded component meets $S$ because for $|\lambda |>\Vert u\Vert$ the geometric series for $(u-\lambda 1{)}^{-1}$ belongs to $A$. Being both open and closed in the complement, $S$ therefore contains every component, so it is the entire complement. Conversely, evaluation at any $z\in K$ prevents $u-z1$ from being invertible. HenceApply the holomorphic functional calculus in $A$ to $f$ near $K$. For every $z\in K$, the evaluation character of an algebra and the contour formula from (b) give $f(u)(z)=f(z)$. Thus $f{|}_K\in A$. By the definition of this uniform closure, the required rational approximations exist.

If $\infty \notin E$, choose a finite point $a\in E$ in the component containing infinity. The Möbius transformation $w=(z-a{)}^{-1}$ sends $K$ to a compact subset $K^{\prime }$ of the plane and sends $E$ to a set $E^{\prime }$ containing infinity. It preserves complementary components, so $E^{\prime }$ meets every component of the complement of $K^{\prime }$. The transformed function $F(w)=f(a+1/w)$ is holomorphic near $K^{\prime }$, since $0\notin K^{\prime }$. The case already proved approximates $F$ by rational functions with poles in $E^{\prime }$. Pulling them back gives rational approximations to $f$ on $K$ with poles only in $E$: a pole at infinity in the $w$-plane becomes a pole at $a$, and every other pole has its prescribed preimage. In particular the pullbacks do not acquire a pole at the original infinity, since $0\notin E^{\prime }$.

This proves the full prescribed-pole theorem, including the case where a pole at infinity is forbidden. If $\mathbb{C}\setminus K$ is connected, take $E=\{\infty \}$ to obtain the polynomial Runge theorem as a corollary.

Runge approximation theorem in its polynomial form says that if $K\subset \mathbb{C}$ is compact with connected complement and $f$ is holomorphic on a neighborhood of $K$, then polynomials approximate $f$ uniformly on $K$.

The polynomial form of Runge theorem says that if $K\subset \mathbb{C}$ is compact, $\mathbb{C}\setminus K$ is connected, and $f$ is holomorphic on a neighbourhood of $K$, then for every $\epsilon >0$ there is a polynomial $P$ such that

Statement (i) is false. Take the entire function $f(z)=e^{-z}$. It is bounded on the closed first quadrant $E$. If polynomials $P_n$ converged uniformly to $f$ on $E$, then each sufficiently late $P_n$ would be bounded on the positive real axis. A polynomial bounded on that ray must be constant. No sequence of constants converges uniformly to $e^{-x}$ on $[0,\infty )$, since its values at zero and near infinity differ by one.

Runge theorem says that if $K\subset \mathbb{C}$ is compact and $\mathbb{C}\setminus K$ is connected, every function holomorphic near $K$ is uniformly approximable on $K$ by polynomials.

Choose a compact exhaustion $K_j$ of the slit disc $\Delta \setminus L^{+}$, rounding the two sides of the slit so that $0\notin K_j$ and $\mathbb{C}\setminus K_j$ is connected. Runge applied to $1/z$ gives a polynomial $p_j$ with ${\sup }_{K_j}|p_j-1/z|<1/j$. Every compact subset of the slit disc lies in all sufficiently large $K_j$, proving locally uniform convergence.