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If the right-hand side and initial data are jointly holomorphic, the local solution is jointly holomorphic in its independent variable and parameters. On a smaller closed polydisc, bound $|A|$ by $M$ and $|{\partial }_{w}A|$ by $L$. Choose $r$ so that $rM$ stays inside the dependent-variable disc and $rL<1$. The integral contraction mapping has jointly holomorphic iterates converging uniformly, so locally uniform convergence of holomorphic functions proves the assertion. Shifting the dependent variable turns initial values into parameters.

A unital Banach star-algebra is a complex unital Banach algebra with a continuous conjugate-linear algebra involution satisfyingThe algebra norm is complete and submultiplicative, with the unit conventionally normalized to norm one. Many definitions additionally require the involution to be isometric, $\Vert a^{\ast }\Vert =\Vert a\Vert$; the function algebra here satisfies this stronger convention as well. A Banach star-algebra does not by definition require the extra C-star identity $\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$.

Here is the lattice proof of the Stone-Weierstrass theorem in its complex self-adjoint form. Let $B$ be the uniform closure of the given unital star-subalgebra in $C(\Omega )$, and let $B_{\mathbb{R}}$ be its real-valued part. Closure preserves multiplication and complex conjugation, so $B$ is still a unital star-algebra. Real and imaginary parts lie in $B_{\mathbb{R}}$, and separation by the complex algebra implies separation by its real-valued part.

For $h\in B_{\mathbb{R}}$, polynomial approximation of absolute value on a bounded interval gives $|h|\in B_{\mathbb{R}}$. For example, this polynomial approximation follows from the binomial series for the square root, whose coefficients give uniform convergence on the relevant closed interval. Thusalso lie in $B_{\mathbb{R}}$. It is therefore a lattice of functions.

To see how the lattice gives density, fix a real continuous target $f$ and a tolerance $ϵ>0$. For any two distinct points $x,y$, separation and constants provide $h_{xy}\in B_{\mathbb{R}}$ with $h_{xy}(x)=f(x)$ and $h_{xy}(y)=f(y)$, by an affine rescaling of a separating function. For $x=y$, use the constant $f(x)$.

Fix $x$. The sets where $h_{xy}>f-ϵ$ cover $\Omega$ as $y$ varies. A finite subcover and a pointwise maximum give $g_x\in B_{\mathbb{R}}$ with $g_x>f-ϵ$ everywhere and $g_x(x)=f(x)$. The neighborhoods where $g_x<f+ϵ$ now cover $\Omega$ as $x$ varies. A second finite subcover and a pointwise minimum give $h\in B_{\mathbb{R}}$ satisfyingThis is lattice approximation from two-point approximation. Using half the desired tolerance gives strict uniform error below the desired bound. Since $B_{\mathbb{R}}$ is closed, it contains every real continuous function. Splitting a complex function into real and imaginary parts proves $\boxed{B=C(\Omega )}$.

The annulus test. For the closed annulus, the family holomorphic in the interior is not dense in the full continuous-function algebra. A uniform limit on the annulus is locally uniform in its interior, so locally uniform convergence of holomorphic functions makes that limit holomorphic there. Thus the proposed family is closed in $C(\Omega )$. It is proper because $z\mapsto \overline{z}$ is continuous but not holomorphic on the annulus interior.

There is also an explicit continuous functional detecting the obstruction. Choose a small closed disk of radius $r$ contained in the annulus interior and let $L(h)$ be the integral of $h(z)dz$ around its positively oriented boundary. The Cauchy integral theorem gives $L(h)=0$ for every member of the holomorphic family, while parametrizing the circle gives $L(\overline{z})=2\pi i{r}^2\ne 0$. Since $|L(h)|\le 2\pi r\Vert h{\Vert }_{\infty }$, no uniform approximation of $\overline{z}$ by that family is possible.

Choose $0<R<R_1$, $0<S<S_1$ and $0<T<T_1$. On this smaller closed polydisc, the holomorphic function $A$ and its $w$-derivative are bounded; write $|A|\le M$ and $|{\partial }_{w}A|\le L$. Choose $0<r<R$ with $rM<S/2$ and $rL<1$, replacing a zero bound by any positive upper bound if necessary, and choose $0<\epsilon <T$.

Consider functions $u(z,\alpha )$ continuous on $\overline{D(0,r)}\times \overline{D(0,\epsilon )}$, jointly holomorphic in the interior, with $u(0,\alpha )=0$ and $\Vert u{\Vert }_{\infty }\le S/2$. This is a closed subset of a Banach space in the supremum norm. The integral operatoris well defined on the straight segment and is jointly holomorphic. Indeed, it equals $z{\int }_0^{1}A(tz,u(tz,\alpha ),\alpha )dt$, an integral of jointly holomorphic functions locally uniformly bounded in both variables. Convexity of the $w$-disc and the derivative bound giveThus the Banach fixed-point theorem gives a fixed point $f(z,\alpha )$. More explicitly, the successive integral iterates starting from zero converge uniformly, and locally uniform convergence of holomorphic functions makes their limit jointly holomorphic. Differentiating the integral equation gives the required complex ordinary differential equation and initial value. This proves local existence, uniqueness and joint holomorphic dependence on $(z,\alpha )$. Uniqueness among arbitrary local solutions follows first on a small disc where their values stay within $|w|<S$, by the same contraction mapping estimate, and then throughout their common connected domain by the identity theorem.

For holomorphic dependence of ordinary differential equations on parameters in the initial value, fix an admissible initial value $w_{\ast }$ and put $u=f-w_0$. The equation becomesIts right-hand side is jointly holomorphic in $(z,u,w_0)$ near $(0,0,w_{\ast })$. The preceding construction supplies a common disc and a jointly holomorphic solution for $w_0$ near $w_{\ast }$. Consequently $f=u+w_0$ is jointly holomorphic in $z$ and its initial value $w_0$.

For the unrestricted zero sequence, introduce the Weierstrass elementary factorsFor $|u|\le 1/2$, cancellation of the first $p$ terms of the logarithm power series gives a holomorphic logarithm of the factor:Define the canonical productFix any compact disc $|z|\le R$. Since $|z_n|\to \infty$, eventually $|z/z_n|\le 1/2$ uniformly on this disc. The tail logarithms are then bounded by $2(1/2{)}^{n+1}$, a summable bound independent of $z$. Their sum is holomorphic by locally uniform convergence of holomorphic functions, and its exponential is nowhere zero. Multiplication by the finite initial product therefore gives an entire function. Each factor has a simple zero exactly at its specified $z_n$, the exponential factor has none, and a tail beginning beyond any given compact disc is nonvanishing there. Thus $f$ has precisely the prescribed zeros and no others, each simple, and $f(0)=1$. This proves the required existence without an assumption on ${\sum }_n|z_n{|}^{-1}$.

For the final, unheaded product request, suppose now that ${\sum }_n|z_n{|}^{-1}<\infty$. On $|z|\le R$, all sufficiently late factors have $|z/z_n|\le 1/2$, so part (a) bounds their principal complex logarithms by $2R/|z_n|$. The logarithms converge absolutely and uniformly; infinite product convergence from logarithmic tails proves thatis entire and again has exactly the prescribed simple zeros. For its finite partial products,Passing to the limit provesIn particular $P$ is an entire function of exponential type. This argument uses a valid modulus estimate and does not require the false logarithm assertion in part (c).

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.

For the period lattice $\Lambda =⟨\lambda ,\mu ⟩$, define the Weierstrass elliptic function byTo prove Normal convergence of the Weierstrass elliptic-function series, let $K\subset \mathbb{C}\setminus \Lambda$ be compact and choose $R$ with $|z|\le R$ on $K$. For $|\omega |>2R$,The comparison series ${\sum }_{\omega \ne 0}|\omega {|}^{-3}$ converges by the stated lattice-sum criterion. The Weierstrass M-test gives uniform convergence on $K$; since $K$ was arbitrary, the series converges locally uniformly away from $\Lambda$. The locally uniform convergence of holomorphic functions makes its sum holomorphic there, and the displayed principal term gives a double pole at every lattice point.

The function $\wp$ is even and satisfiesThe three nonzero two-torsion points $z_1,z_2,z_3$ are distinct modulo $\Lambda$ and each equals its own negative modulo $\Lambda$. Therefore no two can have the same $\wp$-value, so $e_1,e_2,e_3$ are distinct. Equivalently, $\wp$ is a degree-two branched map and each half-lattice point already occupies its fiber with multiplicity two.

For a transitive example, take the equianharmonic latticeand define $\theta ([z])=[\omega z]$. Multiplication by $\omega$ preserves $\Lambda$, so $\theta :\mathbb{C}/\Lambda \to \mathbb{C}/\Lambda$ is a conformal equivalence. On the nonzero half-lattice points it acts bybecause ${\omega }^2=-1-\omega$. Thus it acts transitively.

For $\sigma =\Re s>1$,On every compact subset of $\Re s>1$, the terms are bounded by a convergent series $\sum n^{-1-\delta }$. The Weierstrass M-test gives locally uniform convergence, and the theorem on locally uniform convergence of holomorphic functions shows that the limit is an analytic function. Hence $D_f$ is analytic throughout $\Re s>1$.

Morera's theorem states that a continuous function on a domain is holomorphic if its contour integral around every triangle whose filled-in interior lies in the domain is zero.

The uniform limit $f$ is a continuous function, by the uniform limit theorem. For any such triangle $T$, the estimation lemma givesEach integral of $f_n$ vanishes by the Cauchy integral theorem. Hence the integral of $f$ vanishes too, and Morera's theorem proves $f$ is holomorphic on $D$. This is the basic locally uniform convergence of holomorphic functions argument; the given uniform convergence on all of $D$ is more than needed.

Let $R,\alpha >0$ and ${\mathcal{C}}_{\alpha }=\{(x,t)\in {\mathbb{C}}^2:|x|<R,|t|<\alpha (1-|x|/R)\}$. The displayed norm makes the holomorphic functions with finite norm a Banach space. Finiteness forces $u(x,0)=0$: for fixed $x$, the weight grows like $1/|t|$ near zero. A Cauchy sequence in this norm converges uniformly on every compact subset of ${\mathcal{C}}_{\alpha }$, including at $t=0$. Its limit is holomorphic by locally uniform convergence of holomorphic functions, and the pointwise norm bounds show convergence in the original norm. The shrinking domain compensates for loss of an $x$-derivative in a Cauchy-Kovalevskaya theorem argument.