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Cover the product by $U_0={\mathbb{C}}_z\times {\mathbb{C}}_w$ and $U_{\infty }={\mathbb{C}}_{\zeta }\times {\mathbb{C}}_w$, with $\zeta =1/z$ on their intersection ${\mathbb{C}}^{\ast }\times \mathbb{C}$. The permitted vanishing and the Dolbeault theorem make this an acyclic cover for both $\mathcal{O}$ and ${\Omega }^2$. The acyclic cover theorem lets its two-term Čech complex compute the cohomology. In particular, all groups with $q\ge 2$ vanish.

For $p=0$, a global holomorphic function is constant on each compact complex projective line fibre, by the maximum modulus principle. Its remaining dependence on $w$ is entire. Hence $H_{\bar{\partial }}^{0,0}=\mathcal{O}(\mathbb{C})$. For the first Čech group, a function on the intersection has a Laurent seriesEach coefficient is entire in $w$ by its Cauchy integral formula. The nonnegative powers extend to $U_0$, and the negative powers extend to $U_{\infty }$ in coordinate $\zeta$. These two series converge locally uniformly with the parameter $w$, by the usual Laurent estimates on compact parameter sets. Thus every intersection function is a Čech coboundary and $H_{\bar{\partial }}^{0,1}=0$.

For $p=2$, write a two-form on the intersection as $F(z,w)dz\wedge dw$. The other chart hasForms extending from $U_0$ have coefficient powers $j\ge 0$; those extending from $U_{\infty }$ have powers $j\le -2$. A globally defined two-form would need to have both types of expansion, so it is zero. In the first Čech quotient, precisely the $z^{-1}$ term remains, and its coefficient is an arbitrary entire function of $w$. ConsequentlyThe second nonzero group is represented in Čech cohomology by $a(w)z^{-1}dz\wedge dw$. This explicit residue description is the Dolbeault cohomology of the projective line times the affine line; it also identifies that group naturally with the holomorphic one-forms on the affine factor.

The printed assertion for every cohomological degree is false. The valid conclusion supplied by the stated Hartogs extension theorem is the degree-zero isomorphism. Indeed, around each removed point choose a small coordinate ball on which the holomorphic vector bundle $E$ is trivial. A holomorphic section on the punctured ball has finitely many holomorphic coefficient functions, each of which extends by Hartogs extension theorem when the complex dimension is at least two. Uniqueness of holomorphic extension makes these local extensions agree with the original section on overlaps, givingThis also identifies the direct image sheaf $j_{\ast }\mathcal{O}(E{|}_{X_0})$ with $\mathcal{O}(E)$ for the inclusion $j:X_0\hookrightarrow X$. It does not identify higher sheaf cohomology.

Here is an explicit higher-degree obstruction. Take $X={\mathbb{P}}^2$, remove $[0:0:1]$, and take the trivial line bundle. Cover $X_0$ by $U=\{x\ne 0\}$ and $V=\{y\ne 0\}$. On $U$ use coordinates $t=y/x$, $s=z/x$; on $V$ use $u=x/y=1/t$, $v=z/y=s/t$. Both charts are ${\mathbb{C}}^2$, and the overlap is ${\mathbb{C}}^{\ast }\times \mathbb{C}$. For every integer $m\ge 2$, the holomorphic functiondefines a Čech cocycle. It cannot be a Čech coboundary $b(1/t,s/t)-a(t,s)$ with $a,b$ entire on their charts. Taking the coefficient of $s^m$, the term from $a$ has only nonnegative powers of $t$, whereas the term from $b$ has the form $t^{-m}{b}_m(1/t)$ and only powers at most $-m$. Neither can provide the coefficient $t^{-1}$. Uniqueness of Laurent series proves the contradiction. The same argument applied to each fibre degree proves that all the classes $[g_m]$, $m\ge 2$, are linearly independent.

These classes remain nonzero in Čech cohomology of $X_0$, rather than merely of this cover. The low-degree Mayer-Vietoris sequence for sheaf cohomology injects the quotient of overlap sections by the two chart-section groups into $H^1(X_0,\mathcal{O})$. Concretely, a partition of unity gives smooth $f_U,f_V$ with $f_V-f_U=g$; their common Dolbeault operator defines a global $(0,1)$-form. If that form were $\bar{\partial }$-exact, subtracting its global smooth primitive would make $f_U,f_V$ holomorphic and split $g$. The nonsplitting just proved therefore gives the same obstruction in Dolbeault cohomology, and the Dolbeault theorem identifies it with sheaf cohomology. Thus $H^1(X_0,\mathcal{O})$ is infinite dimensional. In contrast, $H^1({\mathbb{P}}^2,\mathcal{O})$ is finite dimensional by compact Dolbeault Hodge decomposition for the Fubini-Study form. Hence the two degree-one groups cannot be isomorphic, disproving the printed all-degree request already in complex dimension two.

In complex dimension one, even the degree-zero conclusion fails. Take $X={\mathbb{P}}^1$, remove its point at infinity and use the trivial line bundle. Holomorphic functions on compact connected ${\mathbb{P}}^1$ are constant by the maximum modulus principle, whereas $X_0\cong \mathbb{C}$ has nonconstant entire functions such as $z$. Therefore the restriction map is not onto even in degree zero.

A weight-$k$ modular form for $\Gamma =S{L}_2(\mathbb{Z})$ is a holomorphic function $f$ on the complex upper half-plane satisfyingand holomorphic at a cusp. Since the full modular group has one cusp class, this last condition means that the period-one function has a convergent expansion $f(\tau )={\sum }_{n\ge 0}{a}_n{q}^n$, $q=e^{2\pi i\tau }$, near infinity. A cusp form has $a_0=0$. The element $-I$ makes a nonzero weight odd form impossible.

For weight zero, $f$ descends to a holomorphic function on the quotient of the complex upper half-plane by the modular group. At an elliptic fixed point, invariance under its finite stabilizer makes the local Taylor series a series in the quotient coordinate. At the cusp, the $q$ expansion extends it over $q=0$. The standard fundamental domain of the modular group, with its boundary identified and its cusp added, is a compact Riemann surface, the compactified modular curve $X(1)$. The maximum modulus principle now provesHolomorphy at the cusp is essential to this conclusion.

For positive even $k\ge 4$, define the unnormalized Eisenstein seriesThe real-linear map $(m,n)\mapsto m\tau +n$ is invertible, and on a compact subset of the complex upper half-plane its norm is uniformly comparable to $\sqrt{m^2+n^2}$. Thus the sum converges absolutely and locally uniformly for $k>2$, proving holomorphy. For $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$,The map $(m,n)\mapsto (ma+nc,mb+nd)$ is a bijection on the nonzero integer pairs, giving $G_k(\gamma \tau )=(c\tau +d{)}^k{G}_k(\tau )$.

For the expansion, the cotangent partial-fraction Fourier kernel follows by differentiating the partial-fraction expansion of $\pi \cot \pi z$ and its geometric-series expansion $-\pi i(1+2{\sum }_{r\ge 1}{e}^{2\pi irz})$, valid for $\mathrm{Im}z>0$. It gives, for even $k\ge 2$,The $m=0$ part of $G_k$ is $2\zeta (k)$, where $\zeta$ is the Riemann zeta function; positive and negative $m$ contribute equally. Apply the kernel at $z=m\tau$ for $m>0$, and regroup the absolutely convergent sum according to $N=mr$. With the divisor sum ${\sigma }_{k-1}(N)={\sum }_{r\mid N}{r}^{k-1}$, this provesThe expansion has no negative powers and converges for $|q|<1$, so it also proves holomorphy at a cusp and completes the proof of modularity.

The supplied zeta values yield the Fourier expansion of a normalized Eisenstein seriesBoth $E_4^3$ and $E_6^2$ are modular forms of weight twelve. Their constant terms cancel, so their difference divided by $1728$ is a cusp form. Integrality requires an argument: dividing an integral series by $1728$ alone would not suffice. Expanding directly gives the integrality identity for the modular discriminantFor every integer $d$, $5d^3+7d^5$ is divisible by twelve. Modulo three, $d^3\equiv d^5\equiv d$; modulo four, an even $d$ makes both powers divisible by four, while an odd $d$ has $d^3\equiv d^5\equiv d(\mathrm{mod}4)$. Thus each coefficient $(5{\sigma }_3(n)+7{\sigma }_5(n))/12$ is integral. The other three series visibly have integral coefficients, provingThis proves integrality without assuming a product expansion for the modular discriminant.

Write $g=g_K$ and $f=g^{-1}$. We use the real boundary bounds for a unit-disc H-hull from part (c), including their reflected version. SetStrict monotonicity and those bounds give $-2\le \alpha \le -1$ and $1\le \beta \le 2$. The inverse $f$ extends across $\mathbb{R}\setminus [\alpha ,\beta ]$ and maps these intervals onto $(-\infty ,-1)$ and $(1,\infty )$.

Consider the holomorphic function $F(w)=w-f(w)$ on $\mathbb{H}$. If $w$ approaches a real $u$ outside $[\alpha ,\beta ]$, its inverse tends to a real $x$ with $|x|>1$. Part (c) givesIf $u\in [\alpha ,\beta ]$, every cluster point of $f(w)$ as $w\to u$ lies in the closed unit disc. To justify this without a boundary regularity assumption, first note that $f(w)$ cannot tend to infinity for bounded $w$, since $g(z)=z+O(1/z)$ there. An interior cluster point in $H$ would be mapped to the real number $u$, impossible for $g:H\to \mathbb{H}$. A real cluster point $x$ with $|x|>1$ would, by the reflected extension and strict monotonicity, have $g(x)$ outside $[\alpha ,\beta ]$, also impossible. All remaining finite boundary points lie in $K$ or in $[-1,1]$, hence in the unit disc.

It follows for every such $u$ thatFinally $f(w)=w+O(1/w)$ at infinity, so $F(w)\to 0$ there. Apply the maximum modulus principle on large upper half-discs, using the boundary limsup just obtained, to conclude $|F(w)|\le 3$ everywhere in $\mathbb{H}$. Taking $w=g(z)$ proves the sharp displacement bound for a compact H-hull, including irregular hulls:

Fix $z_0\in \mathcal{U}$. We prove the Riemann mapping theorem in this case by maximizing a normalized derivative. Choose $a\notin \mathcal{U}$. Since $\mathcal{U}$ is a simply connected domain and $z-a$ never vanishes, it has a holomorphic square root $g$. This $g$ is injective, and $g(\mathcal{U})$ is disjoint from $-g(\mathcal{U})$: equality up to sign would first force the original points to coincide. Choose $b\in g(\mathcal{U})$ and $r>0$ with $B(b,r)\subset g(\mathcal{U})$. The ball $B(-b,r)$ is omitted, so $1/(g+b)$ is a bounded univalent function. Scaling and composing with an automorphism of the unit disk produces an injective holomorphic function $\varphi :\mathcal{U}\to \mathbb{D}$ with $\varphi (z_0)=0$ and ${\varphi }^{\prime }(z_0)>0$.

Let $\mathcal{F}$ be the family of all such normalized univalent functions. A Cauchy estimate in a small disk about $z_0$ bounds their derivatives, so $M={\sup }_{\varphi \in \mathcal{F}}{\varphi }^{\prime }(z_0)$ is finite and positive. Choose a maximizing sequence in this normal family. Montel theorem gives a subsequence converging uniformly on compact subsets to $\varphi$. Its derivative at $z_0$ is $M>0$. The maximum modulus principle puts its image in $\mathbb{D}$, and Hurwitz's theorem implies that a nonconstant limit of injective holomorphic functions is injective. Thus the maximum is attained.

If $a_1\in \mathbb{D}$ is omitted, then $a_1\ne 0$. Put $T_a(w)=(w-a)/(1-\overline{a}w)$. A holomorphic square root $v$ of $T_{a_1}\circ \varphi$ exists, is injective, and takes values in $\mathbb{D}$. Compose $v$ with $T_{v(z_0)}$ and a rotation to normalize it. Writing $s=|a_1|\in (0,1)$, the new derivative has magnitudea contradiction. Therefore $\varphi$ maps onto $\mathbb{D}$. The Cayley transform between the half-plane and disk now givesIts inverse is holomorphic by the holomorphic inverse function theorem, so this is a biholomorphism. The proper-subset hypothesis was used to choose $a\notin \mathcal{U}$; the whole complex plane cannot be mapped this way, by Liouville theorem.

The hypothesis says that $|f|$ attains a global, hence local, maximum at the interior point $a$. By the maximum modulus principle, a nonconstant holomorphic function cannot do this. Therefore $f$ is constant.

This is Schwarz lemma. Define $g(z)=f(z)/z$ for $z\ne 0$ and $g(0)=f^{\prime }(0)$. The removable singularity makes $g$ holomorphic. Applying the maximum modulus principle to $g$ on disks of radius $r<1$, then letting $r\uparrow 1$, gives $|g(z)|\le 1$. Hence

Evaluation is surjective as a sheaf morphism: near $p$, any prescribed value is supplied by a constant holomorphic function. We therefore have a short exact sequence of sheaveswhere ${\mathcal{I}}_p$ is the point ideal sheaf and ${\mathcal{O}}_X$ the sheaf of holomorphic functions. By part (c), the skyscraper sheaf is flasque, so $H^0(X,{\mathbb{C}}_p)=\mathbb{C}$ and $H^i(X,{\mathbb{C}}_p)=0$ for $i>0$.

On the connected compact elliptic curve, the maximum modulus principle makes every global holomorphic function constant. Consequently $H^0(X,{\mathcal{O}}_X)=\mathbb{C}$, and evaluation on global sections is an isomorphism. Also $H^1(X,{\mathcal{O}}_X)\cong \mathbb{C}$: by Serre duality for compact complex manifolds, its dual is $H^0(X,K_X)$, generated by the nowhere-zero form $dz$ descended from $\mathbb{C}$. Higher cohomology of ${\mathcal{O}}_X$ vanishes by the Dolbeault theorem in complex dimension one.

The long exact sequence in sheaf cohomology thus beginsIt follows that

Apply Serre duality for compact complex manifolds to $\mathcal{L}\otimes K_M$, where $K_M$ is the canonical bundle:We can show the group on the right vanishes directly from the embedding. Since ${\mathcal{L}}^{-1}=i^{\ast }\mathcal{O}(-1)$, the tautological bundle inclusion gives an inclusion of holomorphic vector bundlesA global holomorphic section of ${\mathcal{L}}^{-1}$ consequently determines $r+1$ global holomorphic functions on $M$. Compactness and the maximum modulus principle make these functions constant on every connected component. On such a component the section is therefore a constant vector $v$ lying in the tautological line $i(x)$ for every $x$.

If $v\ne 0$, all the points $i(x)$ equal $[v]$. This contradicts the fact that $i$ is an embedding of a component of positive dimension $n\ge 1$. Hence $v=0$ on every component, and $H^0(M,{\mathcal{L}}^{-1})=0$. Duality proves

For degree zero, compactness and the maximum modulus principle make holomorphic functions constant on each connected component. For degree two, a holomorphic differential form is $\bar{\partial }$-closed and its $\partial$ derivative has type $(3,0)$, which vanishes on a complex surface. Degrees above two have no nonzero holomorphic forms. It remains to treat a holomorphic differential form $\alpha$ of degree one.

Set $\beta =d\alpha =\partial \alpha$. Its coefficients are holomorphic, so $\beta$ is a holomorphic differential form of degree two, hence closed by the degree-two observation. Its conjugate is closed too. The Stokes theorem givesThe integrand is nonnegative in the complex orientation. In local coordinates, if $\beta =b(z)d{z}_1\wedge d{z}_2$ with $z_j=x_j+i{y}_j$, thenIt is strictly positive wherever $\beta$ is nonzero. Its zero integral therefore forces $\beta =0$ everywhere. Thus holomorphic forms on a compact complex surface are closed, even when the surface is not Kähler:

The Cauchy integral formula says that if $f$ is holomorphic on a neighbourhood of the closed disc $\overline{D(z_0;R)}$, then for $z\in D(z_0;R)$,In particular, at the centre,for every $0<r<R$.

If $f(z_0)=0$, the assumed bound gives $|f(z)|\le 0$, so $f=0$. Otherwise define $g=f/f(z_0)$. Then $g(z_0)=1$ and $|g|\le 1$. Applying the centre formula and taking real parts givesThe continuous nonnegative function $1-\mathrm{Re}g$ has integral zero, so it vanishes around the circle. Equality $\mathrm{Re}g=1$ together with $|g|\le 1$ forces $g=1$. This holds for every $r<R$, hence $f(z)=f(z_0)$ throughout the disc. This is the equality case underlying the maximum modulus principle.

The map $f$ is bounded by one near the puncture, so the Riemann removable singularity theorem extends it holomorphically to a map $\tilde{f}:\Delta \to \mathbb{C}$. The maximum modulus principle shows that its image still lies in $\Delta$: if $|\tilde{f}(0)|=1$, the extension would be constant, contradicting its values in $\Delta \setminus \{0\}$ unless that constant also had modulus less than one.

If $f$ is biholomorphic, its inverse $g:\Delta \setminus \{0\}\to \Delta \setminus \{0\}$ also extends holomorphically. Were $a=\tilde{f}(0)\ne 0$, continuity of $\tilde{g}\circ \tilde{f}$ and the identity $g(f(z))=z$ on the punctured disk would give $g(a)=0$. This is impossible because $a\in \Delta \setminus \{0\}$ and $g$ takes values in the punctured disk. Hence

If such an $f$ existed, boundedness would make its isolated singularity at zero removable. Let $F$ be the holomorphic extension to the disc. Continuity gives $1\le |F(0)|\le 2$. Equality at either endpoint contradicts the minimum-modulus principle applied to $F$ or the maximum modulus principle, since $F$ is nonconstant. Hence $F(0)\in A$. Surjectivity of $f$ supplies some nonzero $z$ with $f(z)=F(0)$, contradicting injectivity of the extension. Therefore

An $s$-stage Runge-Kutta method has stages and updateOn the Dahlquist test equation $y^{\prime }=\lambda y$, elimination of the stages gives the stability functionThe linear stability domain is the set where $|R(z)|\le 1$. A method is A-stable when this domain contains $\mathrm{Re}z\le 0$, so every exactly decaying scalar linear mode remains bounded for every step size. It is L-stable when it is A-stable and $R(z)\to 0$ as $|z|\to \infty$ in the left half-plane; this extra limit strongly damps unresolved stiff modes.

The rational function $R$ makes several useful conclusions immediate. No explicit Runge--Kutta method is A-stable because its stability function is a nonconstant polynomial and is therefore unbounded on the negative real axis. The implicit midpoint rule has $R(z)=(1+z/2)/(1-z/2)$ and is A-stable, but $R(z)\to -1$, so it is not L-stable. The Backward Euler method has $R(z)=(1-z{)}^{-1}$ and is L-stable. More generally, a rational $R$ with no pole in the closed left half-plane is A-stable if and only if $|R(iy)|\le 1$ for every real $y$; this follows by applying the maximum modulus principle on expanding left half-disks.

Scalar linear stability does not by itself control nonlinear perturbations. Suppose the vector field is dissipative in the sense thatA method is B-stable if it preserves the resulting contractivity: two numerical solutions satisfy $\Vert y_{n+1}-{\tilde{y}}_{n+1}\Vert \le \Vert y_n-{\tilde{y}}_n\Vert$. A practical sufficient condition is algebraic stability of a Runge-Kutta method: $b_i\ge 0$ andTo prove the implication, let $D_i=Y_i-{\tilde{Y}}_i$ and $F_i=f(Y_i)-f({\tilde{Y}}_i)$. Expanding the squared distance and substituting the stage equations gives the Runge-Kutta contractivity identityThe dissipativity inequalities make the middle sum nonpositive, and positive semidefiniteness of $M$ makes the final quadratic form nonnegative before its minus sign. The distance therefore cannot increase. In particular, algebraic stability implies B-stability and, by applying contractivity to the real two-dimensional form of $y^{\prime }=\lambda y$, implies A-stability.

Important collocation families illustrate these notions. Gauss methods are A-stable, symmetric, and have order $2s$, but they do not damp infinitely stiff modes. Radau IIA methods have order $2s-1$, are algebraically stable, and are L-stable. These properties explain why A-stability controls unrestricted linear decay, L-stability is useful for stiff transients, and algebraic or B-stability is the stronger tool for nonlinear dissipative equations.

Choose a local coordinate $z_i$ at each $p_i$ and map $f\in V$ to all coefficients of the negative-power terms in its Laurent expansions:The kernel consists of globally holomorphic functions on the compact Riemann surface $R$. By the maximum modulus principle, every such function is constant, so the kernel has dimension one. Rank--nullity givesFor $n=0$, this simply says $\dim V=1$.

Rouché's theorem states that if $g$ and $h$ are holomorphic functions on a neighbourhood of the closure of a bounded domain andon its positively oriented boundary, then $g$ and $g+h$ have the same number of zeros in the domain, counted with multiplicity.

The Open mapping theorem states that a nonconstant holomorphic function on a domain maps every open subset to an open subset. To prove it, fix $z_0$ and put $w_0=f(z_0)$. By the identity theorem, the zeros of $f-w_0$ are isolated. Choose $r>0$ so that the closed disc $\overline{D(z_0,r)}$ lies in the domain and $z_0$ is the only zero of $f-w_0$ in that disc. ThenWhenever $|w-w_0|<\delta$, the constant $w_0-w$ has modulus smaller than $f-w_0$ on the circle. The Rouche theorem therefore says thathas the same positive number of zeros in the disc as $f-w_0$. Thus every $w\in D(w_0,\delta )$ lies in the image of $f$, proving that the image is open.

If $|f|$ had a local maximum at $z_0$, take a small open disc $D$ on which $|f(z)|\le |f(z_0)|$. The restriction of $f$ to $D$ cannot be constant, since the identity theorem would then make $f$ constant on the whole domain. The complex open mapping theorem says that $f(D)$ is an open neighbourhood of $f(z_0)$, and such a neighbourhood contains points of modulus greater than $|f(z_0)|$, a contradiction. This proves the maximum modulus principle from the complex open mapping theorem.

Now let $\Omega$ be bounded. Its closure is compact, so the continuous function $|f|$ attains a maximum there. If $f$ is nonconstant, the maximum modulus principle excludes an interior maximum; if $f$ is constant, the claimed bound is immediate. Hencewhich is the maximum modulus principle on a bounded domain.

Finally let $\Omega =\{z:\mathrm{Re}z>1\}$ and suppose $|f|\le K$ throughout $\overline{\Omega }$. Fix $z_0\in \Omega$. The principal complex logarithm is holomorphic on the right half-plane, so for each positive integer $n$ the functionis holomorphic on $\Omega$ and continuous on its closure. Given $\epsilon >0$, choose $R>|z_0|$ so large thatApply the bounded-domain result to $g_n$ on $\Omega \cap D(0,R)$. On the vertical part of the boundary, $|z|\ge 1$ and hence $|g_n(z)|\le M$. On the circular part,It follows thatLetting $n\to \infty$ and then $\epsilon \to 0$ gives $|f(z_0)|\le M$. Since $z_0$ was arbitrary, this proves the bounded half-plane maximum principle.

The boundedness assumption is necessary. The functionis holomorphic on $\Omega$ and continuous on its closure, and $|f(1+iy)|=e$ on the boundary, but $|f(x)|=e^x$ is unbounded for real $x>1$. Thus the boundary estimate does not control an unbounded holomorphic function on this unbounded domain.

The local maximum modulus principle says that if $|f|$ has a local maximum at an interior point of a connected domain, then $f$ is constant. On a small circle about the maximum, the mean-value property andforce equality everywhere. Equality in the triangle inequality makes the boundary values identical; Cauchy's formula then makes $f$ constant locally, and the identity theorem makes it constant on the domain.

Write $f=u+iv$. The hypothesis gives $u-v\le 0$. Thereforewith equality at zero. The local maximum modulus principle makes the exponential constant. Differentiation then gives $f^{\prime }=0$, so $f$ is constant; since $f(0)=0$, it is identically zero.

The holomorphic functional calculus assigns to every function $f$ holomorphic on a neighbourhood of ${\sigma }_A(x)$ the elementwhere the oriented contour $\Gamma$ surrounds the spectrum inside that neighbourhood. The value is independent of the admissible contour, and $f\mapsto f(x)$ is a continuous unital algebra homomorphism sending the coordinate function $z$ to $x$.

Every $\phi \in {\Phi }_A$ commutes with the contour integral, so the Cauchy integral formula givesPart a applied to $f(x)$ now proves the spectral mapping theorem:

Let $\Omega$ be the unbounded component of $\mathbb{C}\setminus {\sigma }_A(x)$. On the spectrum, spectral mapping givesEvery remaining point of $\mathbb{C}\setminus \Omega$ lies in a bounded complementary component $D$. Its boundary is contained in ${\sigma }_A(x)$, and $f$ is holomorphic near $\overline{D}$. The maximum modulus principle therefore extends the same estimate from $\partial D$ to $D$. Hence

For a closed unital subalgebra $A\subseteq B$ containing $x$, invertibility in $A$ implies invertibility in $B$, so ${\sigma }_B(x)\subseteq {\sigma }_A(x)$. On a connected component of the resolvent set of $x$ in $B$, the set of $\lambda$ for which $(x-\lambda 1{)}^{-1}\in A$ is both open, by a local Neumann series, and closed, by closedness of $A$. It contains all sufficiently large $\lambda$, hence the entire unbounded component. Thus spectrum in a closed unital subalgebra says that ${\sigma }_A(x)$ is ${\sigma }_B(x)$ with some bounded complementary components filled in.

Now let $A$ be the Banach subalgebra generated by one element $x$ and put $K={\sigma }_A(x)$. If $D$ were a bounded component of $\mathbb{C}\setminus K$, choose $\lambda \in D$. Since $(x-\lambda 1{)}^{-1}\in A$, polynomials $p_n(x)$ converge to it. The polynomialssatisfy $q_n(\lambda )=0$ and $q_n(x)\to 1$. Applying the contractive Gelfand transform gives $q_n\to 1$ uniformly on $K$, hence on $\partial D$. But the maximum modulus principle applied to $1-q_n$ givesa contradiction. Therefore $\mathbb{C}\setminus K$ is connected.

The mapis continuous and surjective by part a. It is injective because characters agreeing on $x$ agree on every polynomial in $x$, hence by continuity on their norm closure $A$. The character space is compact by the Banach-Alaoglu theorem, while $K$ is Hausdorff, so this continuous bijection is a homeomorphism.

Under this identification, the Gelfand transform $\theta :A\to C(K)$ obeysby the calculation in part b. Since $f(x)\in A$, choose polynomials $p_n$ with $p_n(x)\to f(x)$. Contractivity of $\theta$ givesThus every function holomorphic near $K$ is uniformly approximable there by polynomials.