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Write $g_2=60G_4$ and $g_3=140G_6$, with the lattice sums taken over nonzero lattice elements. Reindexing the normally convergent series shows that the Weierstrass elliptic function $\wp$ is an even function, and consequently ${\wp }^{\prime }$ is an odd function. Expanding $(z-w{)}^{-2}$ near zero and cancelling the odd lattice sums gives the Laurent coefficients of the Weierstrass elliptic functionThus the elliptic function ${\wp }^{\prime 2}-4{\wp }^3+60G_4\wp$ has no pole at zero and its constant term is $-140G_6$. By periodicity it has no poles anywhere. An entire elliptic function is bounded on a fundamental parallelogram and hence everywhere, so Liouville theorem makes it constant. This proves the Weierstrass elliptic differential equation

We first verify that the cubic is smooth, rather than presupposing it when using its group law. There are three nonzero two-torsion points of a complex torus $h_j\in (\frac{1}{2}\Lambda )/\Lambda$. Oddness and periodicity imply ${\wp }^{\prime }(h_j)=0$. Since ${\wp }^{\prime }$ has a triple pole and no other poles on the torus, value multiplicity of an elliptic function says that these three distinct zeros exhaust its zeros and are simple. Put $e_j=\wp (h_j)$. Each $\wp -e_j$ has a double zero at $h_j$, since ${\wp }^{\prime }(h_j)=0$ and the zero of ${\wp }^{\prime }$ is simple. If $e_i=e_j$ with $i\ne j$, this same function would have at least four zeros counted with multiplicity, although its only pole is double. Therefore the half-period values of the Weierstrass elliptic function are distinct. They are roots of $4x^3-g_{2}x-g_3$, so all its roots are distinct and the affine curve is a smooth algebraic curve. Its projective closure isIt has the single point $O=(0:1:0)$ at infinity. The derivative with respect to $Z$ of the defining homogeneous polynomial is nonzero at $O$, proving smoothness there too.

For any finite value $a$, value multiplicity of an elliptic function gives exactly two solutions of $\wp (z)=a$, counting multiplicities. Evenness supplies the pair $z,-z$. At a nonzero half-period these coincide and the zero is double; elsewhere they are distinct. An even elliptic function $f$ therefore descends to a meromorphic function of $x=\wp (z)$ on the Riemann sphere. To justify descent at a half-period, put $z=h+u$: both $f(h+u)$ and $\wp (h+u)$ are even in $u$, and $\wp (h+u)-\wp (h)$ is $u^2$ times a nonvanishing even analytic function. The local Laurent series of $f$ is consequently meromorphic in the coordinate $x-e$. At zero the same argument uses $1/x=z^2+O(z^6)$. Every meromorphic function on the Riemann sphere is a rational function, proving that even elliptic functions are rational in the Weierstrass function.

For an arbitrary elliptic function $f$, its even part $(f(z)+f(-z))/2$ is $A(\wp )$. Its odd part divided by ${\wp }^{\prime }$ is an even meromorphic function, hence is $B(\wp )$ with $B$ rational. We have therefore proved the elliptic function-field decompositionUniqueness follows by taking even and odd parts. In particular the field of elliptic functions is $\mathbb{C}(\wp ,{\wp }^{\prime })$.

Define $\Phi (0)=O$ and $\Phi (z)=(\wp (z),{\wp }^{\prime }(z))$ elsewhere on $\mathbb{C}/\Lambda$. Periodicity makes the map well defined. Near zero, the projective chart $Y\ne 0$ has coordinates $X/Y=\wp /{\wp }^{\prime }=-z/2+O(z^5)$ and $Z/Y=1/{\wp }^{\prime }=-z^3/2+O(z^7)$, so the extension is holomorphic. The fiber description above proves injectivity: the two candidates $z,-z$ are distinguished by ${\wp }^{\prime }$ unless they coincide at a half-period. It also proves surjectivity, since every $x$ occurs and the two derivatives are the two allowed $y$ values, with $y=0$ at a branch value.

Finally, a nonvertical line $y=ax+b$ pulls back to the elliptic function ${\wp }^{\prime }-a\wp -b$, whose only pole is triple at zero. Its three zeros $z_1,z_2,z_3$, counted with intersection multiplicity, satisfy $z_1+z_2+z_3=0$ modulo $\Lambda$ by the zero-pole sum of an elliptic function. On the cubic, reflection $y\mapsto -y$ corresponds to $z\mapsto -z$. The chord-and-tangent group law therefore gives $\Phi (z_1)+\Phi (z_2)=\Phi (-z_3)=\Phi (z_1+z_2)$. Vertical lines give inverse pairs; tangencies are included through repeated zeros, and the identity cases follow from the extension at zero. This completes the Weierstrass uniformization by half-period values: $\Phi$ is a group isomorphism.

For an entire function, the Cauchy derivative formula on a positively oriented circle of radius $R$ about $z_0$ isThe resulting Cauchy estimate is $|f^{(n)}(z_0)|\le n!M_R/R^n$, where $M_R$ is the maximum of $|f|$ on that circle.

The Liouville theorem says that a bounded entire function is constant. Indeed, if $|f|\le M$ on the plane, the Cauchy estimate with $n=1$ gives $|f^{\prime }(z_0)|\le M/R$. Sending $R$ to infinity gives $f^{\prime }(z_0)=0$ at every point, hence $f$ is constant.

For the growth assumption, apply the Cauchy estimate on circles centred at zero. For every integer $n>k$,All sufficiently high Taylor series vanish. Since the Taylor series of an entire function converges everywhere, $f$ is a polynomial, of degree at most $\lfloor k\rfloor$ when $k\ge 0$. If a negative finite exponent is interpreted for nonzero $z$, the same estimate forces every coefficient to vanish, giving $f=0$.

The complex Liouville theorem states that a bounded entire function is constant. Indeed, if $|f|\le M$, the Cauchy estimate on any disc of radius $R$ centered at $z$ gives $|f^{\prime }(z)|\le M/R$. Letting $R\to \infty$ gives $f^{\prime }(z)=0$ everywhere.

The analogous conclusion for bounded real analytic functions on $\mathbb{R}$ is false. For example, $\sin x$ is bounded, nonconstant, and real analytic on the whole real line. Boundedness only on that line does not bound its holomorphic extension on the complex plane.

Invertibility is stable under sufficiently small perturbations, and inversion is continuous. Let $1$ be the identity of the complex unital Banach algebra $A$, with its submultiplicative algebra norm. For $\Vert h\Vert <1$, completeness makes the Neumann seriesconverge in $A$. Multiplying either side of the partial sums by $1-h$ gives $1-h^{N+1}$, which tends to $1$, proving that the sum is a two-sided inverse.

Fix $a\in G(A)$, the group of invertible elements of a Banach algebra. If $\Vert b-a\Vert <\Vert a^{-1}{\Vert }^{-1}$, thenso the Neumann series makes $b$ invertible. This proves that $G(A)$ is open. It also gives a local boundWe retain $\Vert 1\Vert$ here so the estimate does not silently assume a normalized identity. The inverse identitythen impliesThis proves continuity of inversion in a Banach algebra.

For a unital Banach algebra, the spectrum of an element isFor a nonunital Banach algebra, use its unitization of an algebra $A^{+}=A\oplus \mathbb{C}$ with multiplication and normIt is a unital Banach algebra, with identity $(0,1)$, and define ${\sigma }_A(x)={\sigma }_{A^{+}}((x,0))$. In this nonunital convention, $0$ belongs to the spectrum: the scalar coordinate of $(x,0)$ is zero, so it cannot be invertible in $A^{+}$.

The spectrum is nonempty and compact. It suffices to work in a nonzero unital Banach algebra, since unitization reduces the other case to this one. The group of invertible elements of a Banach algebra is open, so the complement of the spectrum of an element is open. For $|\lambda |>\Vert x\Vert$, the Neumann series givesThus ${\sigma }_A(x)$ is closed and lies in $\{\lambda :|\lambda |\le \Vert x\Vert \}$, hence is compact.

For nonemptiness of the Banach-algebra spectrum, suppose that $R$ were defined on all of $\mathbb{C}$. At any ${\lambda }_0$, factoring $\lambda 1-x=({\lambda }_{0}1-x)(1+(\lambda -{\lambda }_0)R({\lambda }_0))$ gives a locally convergent power seriesFor every bounded linear functional $\ell \in A^{\ast }$, the scalar function $\ell (R(\lambda ))$ is therefore entire. The bound at infinity makes it bounded outside a disk, while continuity makes it bounded on the disk. By the Liouville theorem, it is constant; since it tends to zero at infinity, it is identically zero. The version of the Hahn-Banach theorem used here says that bounded linear functionals separate points of a normed vector space: if $z\ne 0$, there is $\ell$ with $\ell (z)\ne 0$. Consequently $R(\lambda )=0$ for every $\lambda$, contradicting $(\lambda 1-x)R(\lambda )=1$. This proves the assertion.

Every nonzero complex unital normed division algebra is algebraically and topologically isomorphic to $\mathbb{C}$. This is the Gelfand-Mazur theorem; completeness is not needed in its statement. Let $D$ be a normed division algebra, and take its completion of a normed space using the algebra norm to obtain a unital Banach algebra $\hat{D}$. Submultiplicativity extends multiplication continuously to the completion, and the original identity remains its identity.

For $x\in D$, the spectrum of an element of $x$ in $\hat{D}$ contains some $\lambda$. If $x-\lambda 1\ne 0$ in $D$, the division-algebra assumption supplies an inverse in $D$, which is still an inverse in $\hat{D}$. This contradicts $\lambda \in {\sigma }_{\hat{D}}(x)$. Therefore $x=\lambda 1$. The map $\lambda \mapsto \lambda 1$ is a bijective complex algebra homomorphism, andIt and its inverse are continuous; when $\Vert 1\Vert =1$, it is an isometry.

A complete algebra norm on a function algebra dominates the supremum norm, even before continuity of point evaluations is known. Let $A$ be the given algebra of functions. Form the same artificial unitization $A^{+}$ even if $A$ already has an identity. For each $t\in K$, the algebraic mapis a unital multiplicative complex linear functional. No continuity has been assumed. Since ${\chi }_t(f-f(t)1)=0$, the element $f-f(t)1$ cannot be invertible: applying ${\chi }_t$ to an inverse equation would give $0=1$. Hence $f(t)\in {\sigma }_{A^{+}}(f)$, and the spectrum bound already proved yieldsTaking the supremum proves the supremum bound for a complete function-algebra norm:In particular every $f$ is bounded, and every point evaluation is a bounded linear functional of norm at most $1$. This is an instance of automatic continuity of characters: a character of an algebra on a Banach algebra is bounded because its value at any element belongs to that element's spectrum.

Let ${\pi }_i:{\mathbb{C}}^{g_i}\to X_i$ be the quotient covering maps. Since the source universal cover is simply connected, the lifting criterion for a covering space gives a lift $F:{\mathbb{C}}^{g_1}\to {\mathbb{C}}^{g_2}$ of $f\circ {\pi }_1$. The lift is a holomorphic map because ${\pi }_2$ is locally a biholomorphism.

For every $\lambda \in {\Lambda }_1$, the difference $F(z+\lambda )-F(z)$ belongs to the discrete Euclidean lattice ${\Lambda }_2$. It is a continuous function of $z$ on a connected space, hence is constant. Differentiating shows that each coefficient of $DF$ is ${\Lambda }_1$-periodic. Such a coefficient is bounded on a compact fundamental parallelepiped, and periodicity bounds it on all of ${\mathbb{C}}^{g_1}$. The several-variable Liouville theorem, obtained by applying the one-variable theorem on coordinate lines, makes every coefficient constant.

Consequently $DF=A$ for a constant complex-linear map $A$ and $F(z)=Az+x$. The period differences become $A\lambda \in {\Lambda }_2$. Thus the affine lift of a holomorphic map between complex tori givesEvery holomorphic map between complex tori is a homomorphism followed by a translation. The translation vector $x=F(0)$ is determined modulo ${\Lambda }_2$, while $A$ is the derivative of any lift and is unaffected by that ambiguity.

Write the period lattice as $\Lambda =\mathbb{Z}{\omega }_1+\mathbb{Z}{\omega }_2$, with ${\omega }_1,{\omega }_2$ real-linearly independent and $\mathrm{Im}({\omega }_2/{\omega }_1)>0$. A meromorphic function $f$ is an elliptic function for this lattice if $f(z+\lambda )=f(z)$ for every $\lambda \in \Lambda$. If it has no poles, it is an entire function bounded on the closure of a fundamental parallelogram. Periodicity makes it bounded on all of $\mathbb{C}$, so the Liouville theorem proves that an elliptic function without poles is constant.

Here are the two contour identities needed for counting its zeros and their sum. Choose a positively oriented fundamental parallelogram $P$ whose boundary avoids all zeros and poles, and put $h=f^{\prime }/f$. The argument principle givesbecause $h$ is periodic and opposite edges cancel. Here $N,P_f$ count zeros and poles with multiplicity. For the weighted integral, let $J_i$ be the integral of $h$ along the edge from a vertex $c$ to $c+{\omega }_i$. Translating the opposite edges givesThe endpoint values of $f$ agree along either edge. A continuous logarithm along the edge therefore changes by $2\pi i{n}_i$, with $n_i\in \mathbb{Z}$, and $J_i=2\pi i{n}_i$. On the other hand, the residue theorem evaluates the left side as the sum of the zero positions minus the sum of the pole positions, both with multiplicity. Thus the elliptic divisor-sum identity isUnder the stated pole hypothesis there is just one pole class, represented by zero, of order $m$. ConsequentlyChanging representatives of the zero classes changes their weighted sum by an element of $\Lambda$, so the congruence is well defined.

The Weierstrass elliptic function isOn a bounded set of $z$, the summand is $O(|\omega {|}^{-3})$ for large $|\omega |$. The lattice sum of $|\omega {|}^{-3}$ converges in real dimension two, proving Normal convergence of the Weierstrass elliptic-function series on compact sets away from $\Lambda$. Hence $\wp$ is a holomorphic function there, and at zero its principal part is $z^{-2}$, so it has a double pole. Replacing $\omega$ by $-\omega$ shows that $\wp$ is even. Differentiating the normally convergent series giveswhose absolute convergence permits reindexing by any $\lambda \in \Lambda$. Thus ${\wp }^{\prime }$ is periodic and $\wp (z+\lambda )-\wp (z)$ is constant. Substituting $-z-\lambda$ for $z$ and using evenness changes that constant to its negative, so it is zero. This proves that $\wp$ is an elliptic function, with double poles precisely at the lattice points.

The preceding zero count applied to $\wp -a$ says that it has exactly two zeros modulo $\Lambda$, counted with multiplicity. Evenness pairs any zero $z$ with $-z$. If these are distinct classes they must both be simple. If they coincide, $2z\in \Lambda$, and locally $\wp (z+t)=\wp (z-t)$: the zero has even order, which the total count forces to be two. In particular such a zero cannot be a lattice point, where $\wp$ has a pole. Every finite fibre of the Weierstrass function consists of two opposite simple points or one double nonzero half-period point.

To obtain the rational representation, first consider an even elliptic function $g$. The preceding description proves that $\wp :\mathbb{C}/\Lambda \to {\mathbb{P}}^1$ has exactly the fibres $\{z,-z\}$, so $g$ defines a function $A$ of $w=\wp (z)$. Away from the branch values a local inverse of $\wp$ makes $A$ meromorphic. At a half-period $z_0$, the local expansion is $w-w_0=c{t}^2+O(t^4)$, with $c\ne 0$ because the zero has order exactly two. The Laurent series of $g(z_0+t)$ contains only even powers, and $t^2$ is a holomorphic local coordinate as a function of $w-w_0$. Thus $A$ is meromorphic at that value too. At zero the same argument uses $1/w=t^2+O(t^6)$, proving meromorphy at infinity. A meromorphic function on the Riemann sphere is a rational function: subtract its finitely many principal parts and use compactness to make the remainder constant. This proves that even elliptic functions are rational in the Weierstrass function.

For an arbitrary elliptic function, split $f=f_{+}+f_{-}$, where $f_{\pm }(z)=(f(z)\pm f(-z))/2$. The even part is $A(\wp )$. Since ${\wp }^{\prime }$ is odd and not identically zero, $f_{-}/{\wp }^{\prime }$ is an even meromorphic elliptic function, including at the zeros of ${\wp }^{\prime }$ where the quotient may have poles. It is therefore $B(\wp )$ for a rational function $B$. The elliptic function-field decomposition isThe representation is unique: its even and odd parts determine $A$ and $B$, and $\wp$ takes every value on the Riemann sphere.

No such conformal bijection exists. A prescribed point is a polar point for planar Brownian motion: a planar Brownian motion started away from zero hits zero with probability zero. For example, inside an annulus the probability of reaching radius $\epsilon$ before radius $R$ isby the harmonic function $\log |w|$ and the optional stopping theorem. For fixed $R>|z^{\prime }|$ this tends to zero as $\epsilon \downarrow 0$. Any finite-time visit to zero would occur before leaving some sufficiently large disc; the countable union over integer $R$ still has probability zero.

The exit time from $\mathbb{C}\setminus \{0\}$ is therefore infinite almost surely, contradicting part (b) if this were the image of a bounded domain. Equivalently, a hypothetical inverse $\psi :\mathbb{C}\setminus \{0\}\to D$ would be bounded and holomorphic. The Riemann removable singularity theorem extends it over zero, and the Liouville theorem makes it constant, contradicting bijectivity.

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.