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Consider the singularity of the injective entire function $f$ at infinity. If removable, it is bounded outside a disk and therefore bounded globally; Liouville theorem makes it constant, a contradiction. If essential, the Casorati-Weierstrass theorem says its values outside every large disk are dense. Choose a small neighbourhood $U$ of a finite point; the open mapping theorem puts a nonempty open disk inside $f(U)$. A point outside a disk containing $U$ would map into that open disk, duplicating a value already attained in $U$. This contradicts injectivity. Thus infinity is a pole, so $f$ is a polynomial. If its degree were greater than one, its derivative would have a root; the local power-series form at that point has local degree at least two and cannot be injective. HenceFor a nonconstant holomorphic map of connected compact Riemann surfaces of degree $d$, the Riemann-Hurwitz formula iswhere $g_R,g_S$ are genera and $e_p$ is the local mapping degree, the multiplicity in the target fibre. There are finitely many points with $e_p>1$.

For a degree-two sphere map, the formula gives total ramification two. Since $e_p\le 2$, there are exactly two distinct ramification points, each of local degree two. Their images are distinct, since a fibre has total multiplicity two. Choose a Möbius transformation $T$ taking $0,\infty$ to these domain points and $S_0$ taking their images to $0,\infty$. The resulting rational map has its only zero at zero of order two and its only pole at infinity of order two, hence is $c{z}^2$ with $c\ne 0$. Replace $S_0$ by $S(z)=c^{-1}{S}_0(z)$, obtaining

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 inequality implies $u(u-1)\ge v^2\ge 0$, so $u\le 0$ or $u\ge 1$. Since $u$ is continuous and $\mathbb{C}$ is connected, its image lies wholly in one of these intervals. If $u\ge 1$, then $g$ has no zero and $|1/g|\le 1$; Liouville theorem makes $1/g$, hence $g$, constant. If $u\le 0$, then $|g-1|\ge 1$, so the same argument applied to $1/(g-1)$ proves that $g$ is constant.

The uniformization theorem states that every simply connected Riemann surface is biholomorphic to exactly one of the Riemann sphere $\hat{\mathbb{C}}$, the complex plane $\mathbb{C}$, or the unit disc $\mathbb{D}$. A surface is uniformized by one of these when that surface is its universal cover. The only surface uniformized by $\hat{\mathbb{C}}$ is the sphere itself, because every nonidentity Möbius transformation has a fixed point and therefore cannot act as a deck transformation. The Riemann surfaces uniformized by the complex plane are

Let $U\subseteq \mathbb{C}$ have a complement containing at least two points. Its universal cover cannot be the sphere because $U$ is noncompact. If it were the plane, the covering map $\mathbb{C}\to U\hookrightarrow \mathbb{C}$ would be a nonconstant entire function omitting two values, contrary to the Little Picard theorem. The plane domain with two omitted points is hyperbolic, so

Now put $S=R\setminus \{P_1,\dots ,P_n\}$. The spherical possibility occurs only for $(g,n)=(0,0)$. By the preceding list, the plane-uniformized possibilities arewhich give respectively $2g-2+n=-1,0,0$. Every remaining pair has $2g-2+n>0$ and must have disc universal cover. Thus the uniformization of a punctured compact Riemann surface gives

For the complex torus $X=\mathbb{C}/\Lambda$, any analytic map $f:\mathbb{C}\to X$ lifts through the universal covering map to an entire function $F:\mathbb{C}\to \mathbb{C}$. If $F$ is constant, so is $f$. Otherwise Little Picard says that $F$ omits at most one complex number. For every $x\in X$, its lifts form an infinite coset $z+\Lambda$, so at least one lift is attained by $F$. Hence the holomorphic map from the complex plane to a one-dimensional complex torus satisfies

Finally, $\mathbb{C}$ and $\mathbb{D}$ are homeomorphic byThey are not conformally equivalent: a biholomorphism $\mathbb{C}\to \mathbb{D}$ would be a nonconstant bounded entire function, contradicting the Liouville theorem. This proves the claimed example of complex plane and unit disc are homeomorphic but not conformally equivalent.

Let $\pi :\mathbb{C}\to \mathbb{C}/\Lambda$ be the universal covering map. Since the complex plane is a simply connected domain, the lifting criterion for a covering space gives a holomorphic lift between one-dimensional complex tori $F:\mathbb{C}\to \mathbb{C}$ such thatFor every $\lambda \in \Lambda$, the difference $F(z+\lambda )-F(z)$ belongs to the discrete period lattice $\Lambda$. It depends continuously on $z$, so it is constant. Differentiating shows that the derivative $F^{\prime }$ is $\Lambda$-periodic. It is bounded on the compact closure of a fundamental parallelogram, and periodicity makes it bounded on all of $\mathbb{C}$. The Liouville theorem therefore makes $F^{\prime }$ constant, and henceThus every holomorphic map has an affine lift of a holomorphic map between one-dimensional complex tori. If “map of complex tori” means an identity-preserving map, choose $F(0)=0$; then $b=0$, so $F$ is the required linear map. Without that convention the statement must say affine, since a nonzero translation in a group of the torus lifts to $z\mapsto z+b$.

The Weierstrass elliptic function of $\Lambda$ isThe subtracted term gives the Normal convergence of the Weierstrass elliptic-function series away from $\Lambda$. PutThe Laurent coefficients of the Weierstrass elliptic function at zero giveSet $g_2=60G_4$ and $g_3=140G_6$. Direct substitution shows that the principal part and constant term ofvanish at zero. Since $H$ is an elliptic function, translation gives the same cancellation at every point of the period lattice. Every apparent isolated singularity is therefore a removable singularity, so $H$ is an entire function. It is periodic and hence bounded on the translates of a compact fundamental parallelogram. The Liouville theorem gives $H=0$, which proves the Weierstrass elliptic differential equation

Finally suppose that $f$ is a biholomorphic group homomorphism. Its identity-preserving lift has the form $F(z)=\zeta z$ by the first part. Since the inverse map also lifts linearly,Choose a $\mathbb{Z}$-basis of $\Lambda$. Multiplication by $\zeta$ is then represented by a unimodular matrix $A\in {\mathrm{GL}}_2(\mathbb{Z})$. As a real-linear transformation of $\mathbb{C}$, it has determinant $|\zeta {|}^2>0$; hence $\det A=1$ and $|\zeta |=1$. The characteristic polynomial of $A$ and the Cayley-Hamilton theorem giveThis already has the required form with $m=-\mathrm{tr}(A)\in \mathbb{Z}$ and $n=1$. Moreover $2\mathrm{Re}\zeta =\mathrm{tr}(A)$, so $|\zeta |=1$ forces $\mathrm{tr}(A)\in \{-2,-1,0,1,2\}$. Thus $\zeta$ is a root of unity, of order $1$, $2$, $3$, $4$, or $6$, completing the description of an automorphism of a one-dimensional complex torus.

SetThis is an elliptic function. Write $a=g_2/20$ and $b=g_3/28$ in the expansion from part (i). At a lattice point, the coefficient of $z^{-2}$ in $F$ is $g_2-20a=0$, and its constant coefficient is $g_3-28b=0$; there are no other singular terms. Every apparent pole is therefore a removable singularity, and $F$ extends to an entire elliptic function with value zero at the lattice points.

An entire elliptic function is bounded on a fundamental parallelogram and hence everywhere. By Liouville theorem it is constant, and its value at zero is zero. Thus $F\equiv 0$, proving the Weierstrass elliptic differential equation

Liouville theorem states that every bounded entire function is constant.

Because $\mathbb{C}$ is simply connected, the harmonic function $u$ has a global harmonic conjugate $v$, sois entire. Positivity givesLiouville's theorem makes $e^{-F}$ constant. Differentiating this nonzero constant gives $F^{\prime }=0$, so $F$, and hence $u$, is constant.

Since $\mathbb{C}$ is simply connected, writefor an entire function $f$. If $u$ were bounded above, thenwould be bounded, so the Liouville theorem would make $e^f$ constant. Its derivative $f^{\prime }{e}^f$ would then vanish, whence $f$ and $u$ would be constant. Applying the same argument to $e^{-f}$ rules out a lower bound. Thus a nonconstant $u$ is unbounded both above and below.

Given any $c\in \mathbb{R}$, choose $z_{-},z_{+}\in \mathbb{C}$ withThe restriction of $u$ to the line segment from $z_{-}$ to $z_{+}$ is continuous, so the intermediate value theorem supplies a point on that segment where $u=c$. ThereforeThis proves the surjectivity of a nonconstant entire harmonic function.

Suppose there were no singular point in the Riemann sphere. Then every finite point would be ordinary, so $p$ and $q$ would be entire functions. Ordinary behavior at infinity would force $p(z)=2/z+O(z^{-2})$ and $q(z)=O(z^{-4})$. In particular $p(z)\to 0$, so Liouville theorem would give $p\equiv 0$, contradicting the required leading term $2/z$. Thus infinity cannot also be ordinary, and every such differential equation has at least one singular point in $\overline{\mathbb{C}}$.

For an element $x$ of a unital complex Banach algebra $A$, its spectrum of an element isIf $|\lambda |>\Vert x\Vert$, the Neumann seriesconverges, so the spectrum is bounded. If ${\lambda }_0$ is in the resolvent set, thenis invertible for $\lambda$ sufficiently close to ${\lambda }_0$, again by a Neumann series. Thus the resolvent is open and the spectrum is closed, hence compact.

If the spectrum were empty, the resolvent $R(\lambda )=(\lambda 1-x{)}^{-1}$ would be an entire Banach-space-valued function and would tend to zero at infinity. For every $f\in A^{\ast }$, the scalar entire function $f(R(\lambda ))$ would be bounded and therefore constant by Liouville theorem. It would be zero because of the limit at infinity. The Hahn-Banach theorem would then force $R(\lambda )=0$, contradicting $(\lambda 1-x)R(\lambda )=1$. Therefore ${\sigma }_A(x)$ is nonempty.

Now let $B\subseteq A$ be a closed unital subalgebra containing $x$. Invertibility in $B$ implies invertibility in $A$, soOn each connected component of $\mathbb{C}\setminus {\sigma }_A(x)$, either the resolvent belongs to $B$ everywhere or nowhere. Indeed, membership holds on a neighbourhood of any one such point by its local Neumann expansion, and the same argument makes the set of such points relatively closed by taking limits in the closed subalgebra $B$. The unbounded component belongs to the first case because the geometric resolvent series lies in $B$ for large $|\lambda |$. ConsequentlyThis is the spectrum in a closed unital subalgebra theorem.

For the bilateral shift $T$ on ${\ell }^1(\mathbb{Z})$, both $T$ and $T^{-1}$ are isometries. The geometric-series argument applied to $T$ for $|\lambda |>1$ and to $T^{-1}$ for $|\lambda |<1$ givesFor $|\lambda |=1$, normalize the vector given by $x_n={\lambda }^{-n}$ on $-N\le n\le N$ and zero elsewhere. Only its two boundary coordinates contribute to $(T-\lambda I)x$, so the norm of that image tends to zero. Thus $T-\lambda I$ is not bounded below and cannot be invertible. Hence

The algebra generated by $T$ is the norm closure of polynomials in nonnegative powers of $T$. It does not contain $T^{-1}$: every $p(T)e_0$ is supported in nonnegative coordinates, whereas $T^{-1}{e}_0=e_{-1}$. Therefore $0\in {\sigma }_{\mathcal{A}}(T)$. The preceding hole theorem then fills the unique bounded component of $\mathbb{C}\setminus \mathbb{T}$, giving

The uniformization theorem says that every simply connected Riemann surface is conformally equivalent to exactly one of the Riemann sphere, the complex plane, and the unit disc. A proper simply connected plane domain is noncompact, so it is not the sphere. It cannot be uniformized by the plane: after choosing a point outside the domain, a branch of a square root and then a Möbius transformation would produce a nonconstant bounded entire function, contradicting the Liouville theorem. It is therefore conformally equivalent to the disc, which is the Riemann mapping theorem.

The Liouville theorem states that every bounded entire function is constant.

The domains are not conformally equivalent. If $F:\mathbb{C}\to \mathbb{H}$ were a conformal equivalence, composing it with a Cayley transform from the upper half-plane $\mathbb{H}$ to the unit disc would produce a bounded nonconstant entire function. This contradicts the Liouville theorem, proving that the complex plane is not conformally equivalent to the upper half-plane.

The domains are not conformally equivalent. Suppose that $F:\mathbb{C}\setminus \{0\}\to \{w:0<|w|<1\}$ were a conformal equivalence. Since $F$ is bounded near zero, the Riemann removable singularity theorem extends it holomorphically across zero. The extension is a bounded entire function, so the Liouville theorem makes it constant, contradicting bijectivity. This is the punctured plane is not conformally equivalent to the punctured unit disc obstruction.

Let the photon phase-space distribution be a Planck distribution whose local temperature is $T(\eta )[1+\Theta (\eta ,\mathbf{x},\hat{\mathbf{p}})]$. In the absence of collisions, Liouville theorem says that the distribution is constant along a photon null geodesic. Linearizing $df/d\eta =0$ about the homogeneous distribution and using $d\mathbf{x}/d\eta =\hat{\mathbf{p}}$ givesA spatial Fourier transform sends $\hat{\mathbf{p}}\cdot \nabla$ to $i\hat{\mathbf{p}}\cdot \mathbf{k}$, soThis is the Free-streaming photon Boltzmann equation: the second term transports angular structure, while the last term is the gravitational redshift source.

The inverse is nonconstant; locally at zero,If the assumed meromorphic function $\mathcal{H}$ had no poles, it would be entire. It is bounded on the compact closure of a fundamental parallelogram, and its two periods then make it bounded throughout the complex plane. The Liouville theorem would force it to be constant, a contradiction. Therefore $\mathcal{H}$ has at least one pole, as stated by the general result that a nonconstant elliptic function has a pole.

The interior Schauder estimate iswhere $C$ depends only on $n$, $\alpha$, the ellipticity constant, and the stated coefficient bounds.

It is enough first to prove the estimate for the Hölder seminorm of $D^{2}u$. The Interpolation inequality in Holder spaces controls $Du$ and $D^{2}u$ by a small multiple of $[D^{2}u{]}_{\alpha }$ plus $\Vert u{\Vert }_0$; the small term is later absorbed. Freezing $a^{ij}$ at the centre of each ball and moving coefficient differences and lower-order terms to the right reduces the local estimate toOnce this is known, the Simon absorption lemma gives the desired estimate on $B_{1/2}$.

For completeness, prove the frozen-coefficient estimate by contradiction. If it failed, choose solutions $u_k$, points $x_k$, and scales ${\rho }_k\downarrow 0$ at which the scale-invariant Hölder quotient is almost maximal. Let $q_k$ be the quadratic Taylor polynomial of $u_k$ at $x_k$ and defineThen $v_k(0)=D{v}_k(0)=D^2{v}_k(0)=0$, the Hessians have uniformly bounded local $C^{0,\alpha }$ seminorms, and the normalization makes their oscillation nonzero on a fixed ball. The localized equation iswhere the coefficient matrices converge locally uniformly to one constant positive-definite matrix and the given normalized error $g_k$ tends locally uniformly to zero.

The Arzela-Ascoli theorem produces a locally $C^2$ convergent subsequence with limit $v$ satisfying a constant-coefficient elliptic equation on ${\mathbb{R}}^n$. A linear rotation and scaling turn it into a harmonic function. The normalization gives $D^{2}v(0)=0$ but a nonconstant Hessian, while maximality of the scaled quotient gives growth at most $C(1+|y{|}^{2+\alpha })$. Applying the Liouville theorem to derivatives shows that every second derivative is constant because $0<\alpha <1$. This contradicts the normalized Hessian oscillation. The frozen estimate follows, and interpolation plus absorption completes the proof.

At $\alpha =0$, bounded Hessians have no uniform modulus of continuity, so the Arzela-Ascoli theorem does not supply the compact $C^2$ limit used in the contradiction. At $\alpha =1$, the blow-up has cubic growth, and the Liouville theorem no longer forces its Hessian to be constant: nonzero harmonic cubic polynomials are possible. These are the two endpoint failures behind the restriction $0<\alpha <1$; ordinary Schauder theory replaces neither endpoint by a general $C^{2,\alpha }$ estimate of the displayed form.

For $x$ in a unital complex algebra $A$, the spectrum of an element isFor a nonunital algebra one uses its unitization.

Now let $A$ be a Banach algebra. The invertible group is open, so the resolvent set is open and the spectrum is closed. If $|\lambda |>\Vert x\Vert$, the Neumann seriesconverges, so ${\sigma }_A(x)$ is contained in the closed disc of radius $\Vert x\Vert$ and is therefore compact.

If the spectrum were empty, $R(\lambda )=(\lambda 1-x{)}^{-1}$ would be an entire $A$-valued function. For each $f\in A^{\ast }$, the scalar function $f(R(\lambda ))$ is bounded: it tends to zero at infinity by the Neumann series and is bounded on every compact disc. The Liouville theorem makes it identically zero. Since the Hahn-Banach theorem separates points, this would give $R(\lambda )=0$, contradicting its invertibility. Hence the spectrum is nonempty.