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Use the Wirtinger derivatives ${\partial }_z=({\partial }_x-i{\partial }_y)/2$ and ${\partial }_{\bar{z}}=({\partial }_x+i{\partial }_y)/2$, and area measure $dA=dxdy=(i/2)dz\wedge d\bar{z}$. The supplied boundary-integral identity is the planar Generalized Stokes theorem; the usual Poincare lemma is a different local exactness result.

Let $f\in C^1(\overline{D})$ and $z\in D$. Remove a disk of radius $ϵ$ around $z$, and apply Generalized Stokes theorem to the one-form $f(\zeta )d\zeta /(\zeta -z)$ on the punctured domain. Away from the puncture,The outer boundary is counterclockwise and the small inner circle clockwise. Its counterclockwise integral tends to $2\pi if(z)$. Passing to the limit gives the Cauchy-Pompeiu formulaThe weak $1/|\zeta -z|$ singularity is locally integrable. If the boundary term vanishes on expanding $D$ to the whole plane, the formula becomes $f(z)={\pi }^{-1}{\int }_{\mathbb{C}}{f}_{\bar{\zeta }}(\zeta )/(z-\zeta )dA(\zeta )$. In particular it yields the distributional normalization ${\partial }_{\bar{z}}[1/(\pi z)]={\delta }_0$. For holomorphic functions the area term vanishes and one recovers the Cauchy integral formula.

Non-characteristic. The first-order principal symbol is ${\nu }_t-i{\nu }_x$, which equals one on $(0,1)$.

Set $z=x+it$. The equation is equivalent to ${\partial }_{\bar{z}}\varphi =({\varphi }_x+i{\varphi }_t)/2=0$. Because $\varphi$ is continuously differentiable, the Cauchy-Riemann equations imply that it is a holomorphic function of $z$. Its convergent complex Taylor series restricts to a real power series along $t=0$, so $g$ is a real analytic function.

For smooth-data instability of the Cauchy-Riemann Cauchy problem, chooseThese are entire functions and satisfy the equation. Their initial derivatives obey ${\sup }_x|{\partial }_x^j{g}_n|\le n^j{e}^{-\sqrt{n}}$, so every finite sum tends to zero. ButThus analytic solutions can exist while continuous dependence on initial data fails for this smooth-data topology.

The Cauchy estimate follows directly from the differentiated Cauchy integral formula:The circle has length $2\pi r$, giving the bound for every integer $j\ge 0$.

Put $w=\varphi -\psi$ and integrate on the straight segment $z=\tau t/|t|$. For fixed $|x|<sR$, a disc centred at $x$ with radius approaching $R(\sigma (\tau )-s)$ stays inside $|\xi |\le \sigma (\tau )R$. Applying the first-derivative Cauchy estimate there givesIntegrating its modulus and taking the supremum over $|x|<sR$ is exactly the asserted integral estimate. The source cancels from $T\varphi -T\psi$.

To obtain a contraction mapping, use the weighted holomorphic norm on a shrinking time domain. Its finite-norm space consists of holomorphic functions on ${\mathcal{C}}_{\alpha }$ with zero initial value; the apparent quotient at $t=0$ is interpreted by a limit. Completeness follows because convergence in this norm implies locally uniform convergence of holomorphic functions, including near $t=0$, and the limiting pointwise bounds give convergence in the norm.

Here is the explicit contraction estimate on a shrinking holomorphic domain. Write $N=\Vert w{\Vert }_{\alpha }$, $r=|t|$, $A=\alpha (1-s)$, and chooseThen $\sigma -s=(A-\tau )/(2\alpha )$ and $\alpha (1-\sigma )-\tau =(A-\tau )/2$. The whole auxiliary polydisc lies inside ${\mathcal{C}}_{\alpha }$, and its supremum of $|w|$ is at most $2N\tau /(A-\tau )$. ConsequentlyThus $\Vert T\varphi -T\psi {\Vert }_{\alpha }\le (4\alpha /R)\Vert \varphi -\psi {\Vert }_{\alpha }$. Choose $0<\alpha <\min (\eta ,R/4)$. The source is bounded on the closed polydisc, say by $M_f$, and $\Vert T0{\Vert }_{\alpha }\le \alpha M_f$. Therefore $T$ maps the Banach space into itself and is a strict contraction mapping. The Banach fixed-point theorem gives a holomorphic fixed point with $\varphi (x,0)=0$. Differentiating the integral identity yields ${\varphi }_t-i{\varphi }_x=f$, establishing this case of the Cauchy-Kovalevskaya theorem.

Finally, near each real initial point, a real analytic function $g$ has a holomorphic extension $G$. Apply the same construction to $\varphi =G(x)+v$ with $v(x,0)=0$ and source $i{G}^{\prime }(x)$, restricting the discs if necessary. This gives a local real analytic function of $(x,t)$ with the prescribed initial value. Equivalently the solution is $\boxed{\varphi (x,t)=G(x+it)}$ wherever the extension is defined.

The two opens are $U_1=\mathbb{C}\times {\mathbb{C}}^{\ast }$ and $U_2={\mathbb{C}}^{\ast }\times \mathbb{C}$; their intersection is $({\mathbb{C}}^{\ast }{)}^2$. There is just one degree-one term in the Čech cochain complex, and no degree-two term. The Čech coboundary sends $(g_1,g_2)$ to $g_2-g_1$, soThe denominator denotes the sum of the restricted function spaces.

To make the quotient explicit, expand a holomorphic function on the intersection in a normally convergent two-variable Laurent series,The terms with $m\ge 0$ extend to $U_1$. Of the remaining terms, those with $n\ge 0$ extend to $U_2$. Both subseries converge normally on their stated domains, by the coefficient bounds from the Cauchy integral formula. The unique remaining representative is the doubly negative partNo nonzero series of this form lies in the denominator, since a function on $U_1$ has no negative $z_1$ exponents and a function on $U_2$ has no negative $z_2$ exponents.

Writing $w_i=z_i^{-1}$, these representatives are exactly $w_1{w}_{2}F(w_1,w_2)$ with $F$ an entire holomorphic function on ${\mathbb{C}}^2$. To see that $F$ is entire, integrate for the coefficients on arbitrarily small product circles: $|a_{-i,-j}|\le M(r_1,r_2)r_1^i{r}_2^j$. Choosing $r_i{R}_i<1$ gives absolute convergence for $|w_i|\le R_i$, for every finite pair $R_i$. Conversely every such entire $F$ supplies a normally convergent representative on the intersection. Therefore the quotient consists of convergent doubly negative Laurent series, not merely finite Laurent polynomials:This gives the holomorphic first cohomology of punctured complex two-space; for instance $1/(z_1{z}_2)$ represents a nonzero class.

For the comparison with sheaf cohomology, $U_1,U_2$ and their intersection are Stein manifolds. They can be realized as closed complex submanifolds of affine complex spaces by adding equations $z_i{w}_i=1$ for their nonzero coordinates. Cartan theorem B makes their higher cohomology with coefficients in the sheaf of holomorphic functions vanish. Thus the cover is acyclic for ${\mathcal{O}}_X$, and the acyclic cover theorem givesThe individual cover members are Stein; their union has the nonzero cohomology just computed.

Fix $\mathrm{Im}\lambda >-c$. Truncate $C_1$ at $\pm R-ic$ and close it with the upper semicircle centred at $-ic$, oriented counterclockwise. For large $R$ the only pole of the integrand inside is $z=\lambda$. Uniformly on the arc, $|z-\lambda |$ is comparable to $R$ and $f(z)=o(R^{-1})$, so the arc integral is $o(R^{-1})$ and tends to zero. The line integral is absolutely convergent at infinity for the same decay reason. The Cauchy integral formula givesThe right-hand side is holomorphic on this open half-plane by the hypothesis, proving the required holomorphic function. The contour never passes through $\lambda$ because its imaginary part is strictly above the line.

Fix $R>|z|$. The Cauchy integral formula and the uniformly convergent geometric series on $|\zeta |=R$ giveDefiningthe remainder after degree $N$ has modulus at most $M_R(|z|/R{)}^{N+1}/(1-|z|/R)$, which tends to zero uniformly on each smaller disk. Differentiating the integral formula at zero identifies $c_j=f^{(j)}(0)/j!$, independently of $R$. Since $f$ is an entire function, $R$ can be chosen larger than any given $|z|$, proving the global Taylor series.

For the requested radius $r$, the same coefficient integral gives the Cauchy estimateTo characterize equality in the Cauchy coefficient estimate, parametrize its circle:If $M_r=0$, the identity theorem gives $f=0$. Otherwise equality in the modulus bound requires the continuous integrand to have both constant modulus $M_r$ and constant complex argument. To see this directly, rotate its average to the positive real axis: its real part is bounded above by $M_r$, and equality of its average with $M_r$ forces that real part to equal $M_r$ everywhere. Thus $f(r{e}^{i\theta })=c_n{r}^n{e}^{in\theta }$ on the circle. The identity theorem extends that equality to the plane. Conversely every such monomial attains the bound. Thereforeincluding the zero function and, when $n=0$, arbitrary constant functions.

Work over the complex scalar field. If $\lambda \in {\sigma }_A(x)$, the ideal generated by $x-\lambda 1$ is proper: otherwise commutativity would make $x-\lambda 1$ invertible. By Zorn lemma extend it to a maximal ideal $M$. This ideal is closed. Indeed, if its closure were all of $A$, it would contain an element of $M$ sufficiently close to $1$ to be invertible, since the invertible group is open; a proper ideal cannot contain such an element. Its closure is therefore proper, and maximality gives $\overline{M}=M$.

The quotient $A/M$ is a complex unital Banach algebra and a division algebra. The Gelfand-Mazur theorem identifies it with $\mathbb{C}$, giving a character of an algebra $\phi$ with $\phi (x)=\lambda$. Conversely, the argument in (a) shows that every $\phi (x)$ is in the spectrum. ThusHere, as usual, a unital Banach algebra has a nonzero identity; the zero algebra is excluded.

The holomorphic functional calculus assigns to every function holomorphic on an open neighbourhood $U$ of ${\sigma }_A(x)$ an element $f(x)\in A$. It is a unital algebra homomorphism, sends the coordinate function to $x$, agrees with evaluation of rational functions having no poles on the spectrum, depends only on the germ near the spectrum, and is continuous for uniform convergence on a fixed surrounding contour. Its formula isHere $\Gamma$ is a finite oriented contour cycle in $U\setminus {\sigma }_A(x)$, with winding number one on the spectrum and zero off $U$. A finite union of suitable polygonal boundary cycles suffices; no connectedness of $U$ is required. Such cycles exist by compactness of the spectrum, and the calculus is independent of the choice. The inverse is the resolvent of an element.

For the exponential, use a circle $|z|=R>\Vert x\Vert$. The uniformly convergent Neumann expansion $(z1-x{)}^{-1}={\sum }_{n\ge 0}{x}^n{z}^{-n-1}$ may be integrated term by term. The scalar Cauchy integral formula gives $\frac{1}{2\pi i}\int e^z{z}^{-n-1}dz=1/n!$, provingThis series converges absolutely in the Banach algebra norm.

A character of an algebra is bounded and takes the resolvent to $(z-\phi (x){)}^{-1}$. Applying it inside the contour integral and using the scalar Cauchy formula gives $\phi (f(x))=f(\phi (x))$. The character of an algebra description of the spectrum, now applied to $f(x)$, proves the spectral mapping theorem:In particular this spectrum is contained in $V$ when $V\supset f(U)$.

For the composition rule for holomorphic functional calculus, choose a relatively compact open neighbourhood $D$ of $f({\sigma }_A(x))$ whose closure is contained in $V$, and a surrounding contour cycle $\Delta$ in $V$ with winding number one on $D$. Choose a sufficiently small neighbourhood $\Omega$ of ${\sigma }_A(x)$ with $\overline{\Omega }\subset U$ and $f(\overline{\Omega })\subset D$, and a contour $\Gamma$ in $\Omega$ surrounding the spectrum. The neighbourhoods and cycles may be finite unions, obtained from finite covers of the relevant compact sets. For $\zeta \in \Delta$ the function $h_{\zeta }(z)=(\zeta -f(z){)}^{-1}$ is holomorphic near $\overline{\Omega }$. Locality and multiplicativity of the calculus giveInsert this in the formula for $g(f(x))$ and interchange the two vector-valued integrals. The scalar inner integral is $g(f(z))$, soThis proves equality as algebra elements. Equality of their character of an algebra values alone would be insufficient in a Banach algebra with nonzero radical.

Finally, $\Vert x\Vert <1$ places the spectrum in the open unit disk. On that disk $f(z)=\log (1-z)=-{\sum }_{n\ge 1}{z}^n/n$ is a holomorphic branch. Its calculus value is the norm-convergent series $y=-{\sum }_{n\ge 1}{x}^n/n$: one may integrate the uniformly convergent scalar series on a circle with $\Vert x\Vert <R<1$. Since $\exp (f(z))=1-z$, the composition rule gives the logarithm of an element near the identity formula

On an annulus $r<|z-z_0|<R$, the Laurent theorem represents an analytic function uniquely asFor any positively oriented simple closed contour $C$ in the annulus winding once around $z_0$, the coefficients in the Laurent series areThe formula follows from the Cauchy integral formula applied term by term, and it is independent of the chosen contour by the Cauchy integral theorem.

The Cauchy integral formula and its differentiated form areIf $f$ is entire and $|f|\le M$, the Cauchy estimate gives $|f^{\prime }(a)|\le M/\rho$. Letting $\rho \to \infty$ yields $f^{\prime }(a)=0$ for every $a$. Thus $f$ is constant, which is Liouville theorem.

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.

Letand choose $0<\eta <\delta$. Write $C_{\eta }$ for the positively oriented boundary of the square with $|x|,|y|\le 1+\eta$. The Cauchy integral formula gives, for every $z\in S$,Because $C_{\eta }$ and $S$ are compact and separated by a positive distance, Riemann sums for this contour integral converge uniformly in $z\in S$. Thus, for any $\epsilon >0$, there are points ${\zeta }_j\in C_{\eta }$ and constants $c_j$ such that

It remains to approximate each resolvent by polynomials on $S$. Let $\Lambda$ be the set of $\lambda \notin S$ for which $(z-\lambda {)}^{-1}$ is polynomially uniformly approximable on $S$. If $|\lambda |>{\sup }_{z\in S}|z|$, the geometric seriesshows that $\lambda \in \Lambda$. Moreover, the resolvent propagation across a complementary component argument shows that $\Lambda$ is both open and closed relative to $\mathbb{C}\setminus S$. The complement of a square is connected, and $\Lambda$ is nonempty, so $\Lambda =\mathbb{C}\setminus S$.

In particular, every ${\zeta }_j$ lies in $\Lambda$. Approximate the finitely many resolvents accurately enough that their weighted sum differs by less than $\epsilon /2$. Combining this with the Riemann-sum estimate produces a polynomial $p$ satisfyingThis proves the claimed elementary polynomial approximation on a square.

Fix $a\notin [\gamma ]$ and letFor $|z-a|<d$ and $\lambda \in [\gamma ]$, the geometric series givesThe series converges uniformly on every smaller closed disc, so it may be integrated term by term along the curve. Hencewhich is a power series about $a$.

If $f$ is holomorphic on a neighbourhood of the closed disc $\overline{D}(a,r)$, the Cauchy integral formula saysDifferentiation under the integral is valid uniformly on smaller discs and gives the Cauchy derivative formulaThis proves inductively that every holomorphic function has complex derivatives of every order. In particular,

Now suppose $f_n\to f$ locally uniformly on $U$. Given a compact set $K\subset U$, choose finitely many closed discs whose slightly larger concentric discs remain in $U$ and whose interiors cover $K$. Applying the derivative formula to $f_n-f$ on the larger boundary circles gives a uniform Cauchy estimatewhere $L\subset U$ is the compact union of those circles. The right-hand side tends to zero, proving that

Finally, choose open neighbourhoods $U_j$ of the closed discs $D_j$ so small thatlies in the given neighbourhood on which $f$ is holomorphic. Inside that neighbourhood choose a positively oriented piecewise smooth contour $\Gamma$ surrounding $D_1\cap D_2$. It may be chosen as the boundary of a slightly enlarged lens and split into arcsso that ${\Gamma }_1$ stays away from $D_1$ and ${\Gamma }_2$ stays away from $D_2$. DefineandBecause each arc avoids the corresponding disc, these formulas define holomorphic functions on possibly smaller neighbourhoods $U_1,U_2$. On their overlap, the Cauchy integral formula for the full contour gives

The holomorphic functional calculus assigns to $a$ in a unital complex Banach algebra and every function holomorphic near $\sigma (a)$ the elementwhere $\Gamma$ winds once around the spectrum. It is a unital algebra homomorphism, extends polynomial evaluation, is independent of the admissible contour, and obeys the spectral mapping theorem

Let $z_K\in \mathcal{R}(K)$ be the coordinate function. For a character $\phi$, put $\lambda =\phi (z_K)$. If $\lambda \notin K$, then $(z_K-\lambda {)}^{-1}$ is one of the admitted rational functions, contradicting the fact that $z_K-\lambda$ lies in the kernel of $\phi$. Thus $\lambda \in K$. For every rational function $r$ without poles on $K$,Continuity of characters and uniform density extend this identity to every member of $\mathcal{R}(K)$. Conversely evaluation at every $\lambda \in K$ is a character. Therefore the character space of R(K) is naturally $K$.

Runge approximation theorem says that if $K\subset \mathbb{C}$ is compact, $f$ is holomorphic on a neighbourhood of $K$, and one chooses one point in every bounded component of $\mathbb{C}\setminus K$, then $f$ can be approximated uniformly on $K$ by rational functions whose finite poles belong only to the chosen points.

To prove it, surround $K$ by finitely many small rectangles contained in the domain of $f$ and apply the Cauchy integral formula on their oriented boundaries:Riemann sums approximate this integral uniformly on $K$ by rational functions with poles on $\Gamma$. If a pole $b$ lies in a component containing the selected point $a$, choose a polygonal path from $b$ to $a$ inside that component and subdivide it finely. The resolvent identityallows the pole to be moved step by step along the path, with arbitrarily small uniform error on $K$. Moving every pole proves the theorem.

For an open set $U$, choose a compact exhaustion $K_1\subset K_2\subset \cdots$ such that every component of $\mathbb{C}\setminus K_n$ meets $\mathbb{C}\setminus U$. Runge's theorem approximates a holomorphic function on $K_n$ by a rational function with all poles outside $U$. A diagonal choice gives convergence uniformly on every compact subset, proving that such rational functions are dense in the space of holomorphic functions $\mathcal{O}(U)$ with its compact-open topology.

Finally, equality of the Gelfand transforms givesLet this common compact spectrum be $K$. The divided differenceis holomorphic near $K\times K$. The two-variable holomorphic functional calculus for the commuting pair $(x_1,x_2)$ gives an element $u=g(x_1,x_2)$ satisfyingFor every character,An element of a commutative unital Banach algebra is invertible exactly when its Gelfand transform has no zero. Thus $u$ is invertible, and $f(x_1)=f(x_2)$ implies $x_1=x_2$. This is injectivity through a holomorphic functional calculus with nonvanishing derivative.

Fix $z_0\notin \Gamma$ and put $\delta =\mathrm{dist}(z_0,\Gamma )>0$. If $|z-z_0|<\delta /2$, thenuniformly for $w\in \Gamma$. Multiplication by the bounded continuous function $f$ and integration therefore prove continuity of $F$ at $z_0$.

Take $\Gamma$ to be the positively oriented unit circle and $f\equiv 1$. By the Cauchy integral formula, $F(z)=1$ inside the circle and $F(z)=0$ outside it, so no continuous extension across $\Gamma$ exists.

For the final assertion, cover $K$ by finitely many sufficiently small closed axis-parallel squares whose slightly enlarged squares lie in $\Omega$, choosing the grid so that no boundary meets $K$. Apply the square Cauchy formula to $f$ on every selected square and add the results. Integrals over shared edges cancel with opposite orientations. The remaining finitely many oriented boundary polygons can be split into contours ${\Gamma }_j\subset \Omega \setminus K$, and their sum isfor every $z\in K$.

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

Find functions analytic on the two sides of a contour whose boundary values obey a prescribed jump, with suitable normalization at infinity and other distinguished points. An additive jump $J$ on a counterclockwise closed contour is solved by the Cauchy integral formula $F(k)=(2\pi i{)}^{-1}\int J(\zeta )/(\zeta -k)d\zeta$, subject to the normalization and compatibility conditions. The Sokhotski–Plemelj theorem verifies the jump.