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For a meromorphic function near $0<|z|\le 1$, and a target $a$ off its image of the unit circle, let $u_a=-\log k(f,a)$ for the chordal metric, $M_a(s)$ its normalized angular mean, and $I(a)=M_a^{\prime }(1)$. The weighted spherical area $T(r)=(4\pi {)}^{-1}{\int }_{r<|z|<1}\log (|z|/r)(f_{\mathrm{round}}^{\#}{)}^{2}dA$ satisfies the displayed formula, where $N$ counts $a$-points with multiplicity and weight $\log (|z|/r)$ and $\overline{m}=M_a(r)-M_a(1)$. Indeed $\Delta u_a=(f_{\mathrm{round}}^{\#}{)}^2/2-2\pi \sum {\mathrm{deg}}_z(f){\delta }_z$, and Green's second identity with the radial logarithmic weight proves the formula. The angular mean must include $1/(2\pi )$.

The printed proximity integral is missing a factor $1/(2\pi )$. Denote that literal integral by $m_{\mathrm{print}}$ and write $\overline{m}=m_{\mathrm{print}}/(2\pi )$. The correct Annular spherical Jensen formula, with the printed definitions of $N$ and $T$, isThe following derivation proves this identity and also isolates the normalization issue.

For the chordal metric put $u_a(z)=-\log k(f(z),a)$ and let $M_a(s)=(2\pi {)}^{-1}{\int }_0^{2\pi }{u}_a(s{e}^{i\theta })d\theta$. In a finite target coordinate,For $a=\infty$ use $u_{\infty }=\frac{1}{2}\log (1+|f{|}^2)$. Differentiating away from the preimages of $a$ gives half the squared spherical derivative of a meromorphic function. At an $a$-point of local degree $d$, $u_a=-d\log |z-z_0|+O(1)$; the second distributional derivatives give the Laplacian of $\log |z-z_0|$ is $2\pi {\delta }_{z_0}$. Thus, including poles by changing spherical coordinates,Because $a$ is off the image of $|z|=1$, $u_a$ is smooth near that circle and $I(a)=M_a^{\prime }(1)$ is finite and independent of $r$.

Apply Green's second identity to the harmonic weight $w(z)=\log (|z|/r)$ on $r<|z|<1$. On the outer boundary $w=\log (1/r)$ and ${\partial }_{n}w=1$; on the inner boundary $w=0$ and ${\partial }_{n}w=-1/r$. HenceThe distributional formula makes the left side $2\pi T(r)-2\pi N(r;a)$, and $M_a(r)-M_a(1)=\overline{m}(r;a)$, proving the corrected Annular spherical Jensen formula. Initially take an inner circle avoiding $a$-points. Logarithmic singularities have continuous angular means, and a point on the inner circle has zero logarithmic counting weight, so continuity extends the identity to every $0<r\le 1$.

There is no way to absorb the missing factor into a different constant $I(a)$. For the holomorphic function $f(z)=z$ and $a=\infty$, direct integration givesThe value of $(T-m_{\mathrm{print}})/\log (1/r)$ tends to $1/2$ as $r\downarrow 0$ and to $\pi$ as $r\uparrow 1$. Therefore the literal printed identity has no constant $I(\infty )$, whereas the corrected identity has $I(\infty )=1/2$.

For the singularity criterion, define the nondecreasing normalized spherical areaTonelli's theorem applied to the logarithmic weight gives $T(e^{-l})={\int }_0^{l}s(t)dt$. Averages of a nondecreasing function converge to its supremum: for fixed $t_0<l$, $(1-t_0/l)s(t_0)\le T(e^{-l})/l\le s(l)$. Thus the limit, not just the lower limit, exists in $[0,\infty ]$ and equalsIf $f$ has a removable singularity or pole at zero, it extends as a holomorphic map to the Riemann sphere; the pulled-back spherical density is smooth and bounded on the closed disk, hence its area is finite. Conversely an essential singularity gives, by the meromorphic version of the Great Picard theorem, infinitely many preimages of every spherical target with at most two exceptions in each punctured neighborhood. The area formula givesThe formula follows by applying change of variables on inverse-function neighborhoods away from the discrete critical points, then summing with multiplicity; the countable critical values have area zero. This proves the finite spherical area criterion for an isolated singularity and the requested equivalence between a finite lower limit and a removable singularity or pole.

With determinant-one normalization, the necessary and sufficient condition for sphere rotations as special-unitary Möbius transformations isEquivalently, the matrix $M=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ belongs to the special unitary group $SU(2)$. The two determinant-one representatives $M$ and $-M$ induce the same Möbius transformation, and both satisfy this condition.

Use north-pole stereographic projection in the direction from the plane to the sphere:Its chord-distance identity is $|\phi (z)-\phi (w)|=2\chi (z,w)$, where $\chi$ is the chordal metric. Direct subtraction, using $ad-bc=1$, givesThis is first calculated away from poles and then extended continuously to the Riemann sphere. A sphere rotation preserves chord distances, so for distinct finite nonpole pointsUsing three distinct points shows all these positive ratios equal one. Hence $H(z)=1+|z{|}^2$ everywhere. Comparing its coefficients givesThus $M$ is a unitary matrix. Comparing $M^{-1}=M^{\ast }$ with $M^{-1}=\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$ proves the boxed relations.

Conversely, these relations give $H(z)=1+|z{|}^2$, so the induced sphere map $T=\phi g{\phi }^{-1}$ preserves all Euclidean chord distances. Such a bijection of the unit sphere is the restriction of an orthogonal matrix: let $v_j=T(e_j)$; preservation of distances makes these an orthonormal basis, and $T(p)\cdot v_j=p\cdot e_j$ forces $T(p)={\sum }_j{p}_j{v}_j$. Finally the Möbius transformation is holomorphic with positive real Jacobian at every ordinary coordinate point. Conjugating by stereographic projection preserves orientation, so this orthogonal matrix has determinant $+1$. It is therefore a genuine rotation in three dimensions, proving sufficiency rather than just spherical isometry.

For the circle assertion, a nondegenerate generalized circle has an equationFor $A\ne 0$ this is an ordinary circle, and for $A=0$ it is a line with infinity included. On the sphere write coordinates $(p_1,p_2,p_3)$, with $z=(p_1+i{p}_2)/(1-p_3)$ and $|z{|}^2=(1+p_3)/(1-p_3)$. Multiplication by $1-p_3$ gives the planeIf $n=(2\mathrm{Re}B,-2\mathrm{Im}B,A-C)$ and $h=-(A+C)$, then $|n{|}^2-h^2=4(|B{|}^2-AC)>0$. Thus the plane intersects the unit sphere in a proper circle. It contains the north pole exactly when $A=0$, accounting for the point at infinity on a line.

Conversely, a circle on the sphere is the intersection with a plane $n\cdot p=h$, $|h|<|n|$. Set $A=(n_3-h)/2$, $C=(-n_3-h)/2$ and $B=(n_1-i{n}_2)/2$. The preceding equation is recovered and has $|B{|}^2-AC=(|n{|}^2-h^2)/4>0$. Its inverse stereographic image is an ordinary circle or a line with infinity attached. This proves both directions of the circle-plane relation under stereographic projection.

Use the extended-value convention for a normal family, which is essential here: every sequence in the holomorphic family has a subsequence converging uniformly on compact subsets either to a finite holomorphic function or to infinity. Convergence to infinity means ${\inf }_K|f_n|\to \infty$ for every compact $K\subset D$. Equivalently one uses local uniform convergence in the chordal metric. For holomorphic functions a nonconstant meromorphic limit cannot acquire a pole: near a supposed pole, the reciprocals of sufficiently large-index functions are holomorphic and nowhere zero, and converge to a holomorphic function with a zero. Hurwitz's theorem would force that reciprocal limit to vanish identically, giving the constant infinite limit instead.

Restriction proves that global normality implies normality near each point. For the converse, choose a countable cover by connected discs $U_j$ on which the family is normal, with smaller discs covering $D$ and having compact closure inside the corresponding $U_j$. Such a cover exists from the countable rational-disc basis and local normality. Starting with any sequence, extract a subsequence converging on $U_1$, then a further subsequence converging on $U_2$, and so on. The diagonal sequence converges on each $U_j$. On an overlap the limits agree; a finite holomorphic limit and infinity cannot both occur there. The overlap graph of this cover is connected, since otherwise its components would partition the connected domain into disjoint open subsets. Therefore all local limits are of the same type and patch to one holomorphic function or to infinity throughout $D$. Every compact set has a finite cover by the smaller discs, so convergence on their compact closures gives uniform convergence on that compact set. This proves that normality is a local property.

For the counterexample take the unit disc and the family $f_n(z)=n(1+z^2/2)$. Since $|1+z^2/2|\ge 1-|z{|}^2/2>1/2$, $f_n$ tends uniformly to infinity. Any sequence from this family either has a constant subsequence or has indices tending to infinity, so the family is normal. Its derivatives are $nz$. No subsequence of these derivatives can have a finite locally uniform limit, because their values at any fixed nonzero point grow without bound; nor can they tend locally uniformly to infinity, since every derivative vanishes at zero. Thus a holomorphic normal family need not have a normal derivative family.

Now impose the point bound at $z_0$. For any sequence of derivatives choose corresponding functions from the original family. Normality supplies a locally uniformly convergent subsequence, and the point bound excludes the infinite limit. Let its finite holomorphic limit be $f$. The Cauchy integral formula gives convergence of derivatives uniformly on compact subsets. Explicitly, for compact $K\subset D$ choose $\rho >0$ so its closed $\rho$-neighborhood lies in a compact $L\subset D$; thenEvery sequence of derivatives therefore has a locally uniform subsequential limit. This pointwise anchoring of a holomorphic normal family proves the derivative family is normal under the stated bound.

Let $R$ be a nonconstant rational map of the Riemann sphere, and write $R^n$ for its $n$-fold iterate, with $R^0$ the identity. The appropriate notion of convergence is uniform convergence on compact subsets in the chordal metric, which treats poles as ordinary values at infinity. A family of meromorphic maps is a normal family if every sequence in it has a subsequence converging locally uniformly in that metric; the limit is meromorphic or constant, possibly the constant infinity.

The Fatou set $F(R)$ consists of points with a neighborhood on which the iterate family is normal. The complementary Julia set is $J(R)=\hat{\mathbb{C}}\setminus F(R)$. Thus the sphere is the disjoint union of an open stability region $F(R)$ and a closed instability region $J(R)$. These definitions concern all subsequences of iterates, rather than assuming that the iterates themselves converge on every stable region.

Equicontinuity makes this distinction precise. On a compact subset $K$ of a neighborhood, the family is equicontinuous if, for every $ϵ>0$, some $\delta >0$ makes $\chi (R^n(z),R^n(w))<ϵ$ whenever $z,w\in K$ and $\chi (z,w)<\delta$, uniformly in $n$. The Arzelà-Ascoli theorem says that equicontinuity of continuous maps from a compact domain to a compact metric target gives relative compactness for uniform convergence. Apply it to a compact exhaustion and take a diagonal subsequence to obtain local normality. Locally uniform spherical limits of meromorphic maps are again meromorphic or constant, by using finite target charts and the locally uniform limit theorem for holomorphic functions. Conversely, if equicontinuity failed, there would be maps $f_j$ and points $z_j,w_j\to z$ with $\chi (f_j(z_j),f_j(w_j))$ bounded below by a positive number. A subsequence converging uniformly on a compact neighborhood contradicts this. Hence local normality and local equicontinuity are equivalent here.

Both sets are completely invariant:For the inverse direction, if the family is normal near $R(z)$, composition with $R$ makes $R^{n+1}$ normal near $z$, and adding the identity does not affect normality. For the forward direction, suppose the iterates are equicontinuous near $z$. A nonconstant holomorphic sphere map is open. Thus each sufficiently small neighborhood of $z$ maps onto a neighborhood of $R(z)$; every nearby target point $w$ has a preimage $x$ as close to $z$ as desired. Thenis uniformly small. Applying this at each point of the image neighborhood proves equicontinuity there. This argument works at rational critical points and requires no single-valued inverse branch. Surjectivity of $R$ supplies the asserted image equalities. Also $F(R^k)=F(R)$ and $J(R^k)=J(R)$ for $k\ge 1$: one direction follows by taking a subfamily, and the other by splitting $R^n$ into the finitely many residue classes $R^r\circ (R^k{)}^m$, $0\le r<k$.

For degree $d\ge 2$, the Julia set is nonempty. Otherwise the family would be equicontinuous on the whole compact sphere, giving a subsequence $R^{n_j}$ converging uniformly to a continuous sphere map $G$. Maps sufficiently close in chordal distance are homotopic, using the unique short spherical geodesic between their values. They therefore have equal topological degrees. This is impossible because $\mathrm{deg}{R}^{n_j}=d^{n_j}\to \infty$ whereas $G$ has one fixed degree.

It cannot be a nonempty finite set either. Complete invariance would make $R$ a permutation of that finite set, with just one sphere preimage for each of its points. That preimage would have local degree of a holomorphic map $d$. Every holomorphic cycle in this set would consequently be superattracting for its return map, hence lie in $F(R)$, a contradiction. In particular $J(R)$ contains at least three points.

The Montel theorem in the form used here says that a family of meromorphic maps on a domain, each omitting the same three distinct points of the sphere, is normal for spherical convergence. It also shows that $J(R)$ is perfect. If $a$ were isolated in $J(R)$, choose a disc $U$ with $U\cap J(R)=\{a\}$ and four distinct points of $J(R)$. For each $n$, complete invariance makes $R^n(U)\cap J(R)$ have at most the single point $R^n(a)$. Partition the iterates into finitely many families according to which of the four points, if any, that single value equals. Each family omits at least three fixed points, so each is normal by Montel; their finite union is normal. This would put $a$ in $F(R)$, a contradiction. Therefore the degree-at-least-two Julia set is compact, infinite and has no isolated points.

The same omitted-value argument explains its global reach. If an open $U$ meets $J(R)$, then ${\bigcup }_{n\ge 0}{R}^n(U)$ can omit at most two sphere points, since three omissions would imply normality on $U$. If $J(R)$ had an interior and were not the whole sphere, these image sets would remain in $J(R)$ and their complement would contain the nonempty open set $F(R)$, impossible. Thus a proper Julia set has empty interior and equals the boundary of $F(R)$. If the Julia set is the whole sphere, the Fatou set is empty instead.

An attracting periodic point and its holomorphic cycle has a contracting return map in a small disc, so its basin is in the Fatou set. Repelling periodic points belong to the Julia set: return iterates fix the point but have derivatives equal to unbounded powers of their periodic-orbit multiplier, contradicting derivative bounds for a normal subsequence near that fixed value. A standard further global property is that repelling periodic points are dense in the Julia set. Fatou components can also support convergence towards a parabolic cycle on the boundary, or rotation behavior in a Siegel disc or a Herman ring; normality need not mean attraction. For example, $R(z)=z^d$ has $J(R)=\{|z|=1\}$, with the interior converging to zero and the exterior to infinity. On the unit circle, arbitrarily close radii have incompatible limiting behavior, preventing normality there.

Degree one has a different picture. A rational map of degree one is a Möbius transformation. Up to Möbius conjugacy, an elliptic map is $z\mapsto az$ with $|a|=1$; its iterates are rotations and are equicontinuous on the sphere, so $F$ is the whole sphere and $J$ is empty (including the identity). For $z\mapsto az$ with $|a|>1$, all compact subsets away from zero converge to infinity, while $z_n=a^{-n}\to 0$ and $R^n(z_n)=1$ prevent equicontinuity at zero. Thus $J=\{0\}$; for $|a|<1$ the reciprocal coordinate gives $J=\{\infty \}$. A nonidentity parabolic map is conjugate to $z\mapsto z+1$. Its iterates converge locally uniformly to infinity on the finite plane, but $z_n=-n\to \infty$ has $R^n(z_n)=0$, so $J=\{\infty \}$. Holomorphic conjugacy carries these sets to the corresponding fixed points of the original map. These finite Julia sets explain why the degree restriction in many global Julia-set properties is essential.

For $f_c(z)=z^2+c$, the Mandelbrot set is the set of parameters for which the finite critical orbit $0,f_c(0),f_c^2(0),\dots$ remains bounded. It is a compact connected parameter set. Its main cardioid consists of parameters with an attracting fixed point, with a period-two bulb attached on the left, smaller bulbs attached recursively and fine filamentary structure. It is symmetric about the real axis and meets that axis in $[-2,1/4]$. The quadratic filled Julia set is connected exactly when $c$ belongs to this parameter set; outside it, the rational critical point escapes and the Julia set is a Cantor set. Thus it organizes the contrasting phase-space dynamics of the quadratic family, rather than itself being a Julia set in the dynamical $z$-plane.

The affine map $h(z)=1/2-z$, its own inverse, givesThis value is unique even among Möbius conjugacies. The finite fixed point of $P$ is zero and its periodic-orbit multiplier is one. Holomorphic conjugacy preserves that periodic-orbit multiplier. Infinity has periodic-orbit multiplier zero for a quadratic polynomial, so a corresponding multiplier-one fixed point of $w^2+c$ must be finite; it obeys $2w=1$ and $w^2+c=w$, forcing $c=1/4$. To locate it explicitly, a quadratic fixed point of periodic-orbit multiplier $\lambda$ gives $c=\lambda /2-{\lambda }^2/4$. The boundary parametrization $|\lambda |=1$ has its cusp at $\lambda =1$, so $c=1/4$ is the right-hand cusp on the Mandelbrot boundary. It is in the set: the critical orbit starts at zero, stays in $[0,1/2]$ and increases to the fixed point $1/2$. For real $c>1/4$, the increment $f_c(x)-x=(x-1/2{)}^2+c-1/4$ is strictly positive; a bounded critical orbit would have a limit solving $x^2+c=x$, which has no real solution. These nearby real parameters escape, proving that the cusp belongs to the boundary of the full Mandelbrot set as well.

The fixed points of $P$ on the sphere are zero and infinity. There is no neighborhood of zero on which all iterates converge to zero, even pointwise: a negative real initial point satisfies $P(x)=x-x^2<x<0$, so its iterates decrease without a finite limit and escape to infinity. Every neighborhood of zero contains such points. At infinity there is a uniformly attracting neighborhood in the chordal metric. In the reciprocal coordinate $w=1/z$,For $|w|\le 1/4$, its modulus is at most $|w|/3$. Thus the reciprocal iterates tend uniformly to zero there, proving uniform convergence to infinity on the corresponding sphere neighborhood.

For the given disc, $|z-1/2|<1/2$ is equivalent to $|z{|}^2<\Re z$, and hence to $\Re (1/z)>1$. Under $w=1/z$ the iteration becomesIf $\Re w>1$, then $\Re (1/(w-1))>0$, so $\Re T(w)>\Re w+1$. Consequently the half-plane is forward invariant andThis proves uniform convergence to zero on the disc and therefore $\Delta \subset F(P)$. It is the explicit parabolic disk for z minus z squared.

Let $U=F_{\Delta }$. The iterate family is normal on the connected open set $U$. Every subsequential limit along $n_j\to \infty$ equals zero on $\Delta$; by the identity theorem it is zero throughout $U$. If the full sequence failed to converge uniformly on some compact subset, a further normal subsequence would contradict that conclusion. Hence $P^n\to 0$ locally uniformly throughout $U$. Complete invariance of the Fatou set puts $P(U)$ in a Fatou component. Since $P(\Delta )\subset \Delta$, that component is $U$, so $P(U)\subseteq U$.

We prove the reverse-image assertion, rather than assuming that a basin has only one component. Each connected component $V$ of $P^{-1}(U)$ maps properly to $U$: for compact $K\subset U$, the sphere preimage $P^{-1}(K)$ is compact, and its intersection with $V$ is closed in it because $V$ is a relatively closed component of $P^{-1}(U)$. Its image is open by holomorphic openness. To prove it is closed in $U$, lift a convergent sequence of image points to preimages in $V$. Their images eventually lie in a compact neighborhood inside $U$; properness gives a convergent preimage subsequence with limit in $V$ and the desired limiting image. Thus the image is also closed, and equals the connected target $U$. Moreover $U$ itself is one such component: it lies in $P^{-1}(U)$ by forward invariance, and any connected inverse-image component containing it lies in $F(P)$ and hence in $U$ by maximality.

Now $1/4\in \Delta$ and its entire preimage is $\{1/2\}$, counted twice. Every component $V$ of $P^{-1}(U)$ must contain a preimage of $1/4$, so every one must be $U$. ThereforeThis is the Fatou component containing an entire fiber is completely invariant argument. The convergence on this component is locally uniform: it cannot be uniform on all of $U$, because $P^n(U)=U$ for every $n$ and $U$ contains nonzero points.

Finally suppose zero belonged to the Fatou set. A normal subsequence on a disc around zero would have limit zero, since that disc meets $\Delta$ in an open set where the entire sequence tends to zero. The identity theorem forces this limit to be zero on the whole disc. Uniform convergence in a smaller finite coordinate disc would then imply derivative convergence at zero by the Cauchy integral formula. Butfor every $n$, whereas the derivative of the zero limit is zero. This contradiction proves

Use the normalized chordal metricUnder stereographic projection it is half the Euclidean chord distance on the unit sphere, which proves the metric axioms and the agreement with the usual sphere topology. If $\rho$ is spherical geodesic distance, then $\rho =2\arcsin \chi$, so $2\chi \le \rho \le \pi \chi$.

A rational map is holomorphic, hence smooth, as a map of compact spheres. At a pole use $1/R$ as target coordinate, and at infinity use $1/z$ as source coordinate; neither is a singularity of the sphere map. Its differential in the round metric has a finite maximum norm $L$. Integrating its norm along a minimizing spherical geodesic gives $\rho (Rz,Rw)\le L\rho (z,w)$. The preceding comparisons implyThis proves that the rational map is Lipschitz in the chordal metric. For a constant map take constant zero; for a nonconstant map the bound is finite by compactness. In finite coordinates the differential norm is $R^{\#}(z)=|R^{\prime }(z)|(1+|z{|}^2)/(1+|R(z){|}^2)$, with its continuous sphere extension.

Now suppose $R,S$ are commuting maps. We first prove $S(F(R))\subseteq F(R)$. The family $\{R^n\}$ is locally equicontinuous on $F(R)$, by the normality equivalence in question 1. Commutation and the Lipschitz estimate give, for all $n$,For $x_0\in F(R)$ choose a source ball within that open set, small enough to make the right side less than any prescribed $ϵ$ uniformly in $n$. Holomorphic openness makes its image contain a ball about $S(x_0)$. Every point $w$ of that target ball has some preimage $x$ in the chosen source ball, so the same inequality proves equicontinuity of the iterates of $R$ at $S(x_0)$. Repeating at each point of the image gives local normality on that image. Thus the stated inclusion holds even where $S$ has a rational critical point.

Carefully state the permitted Montel theorem: a family of meromorphic maps on an open domain, all omitting the same three distinct values of the Riemann sphere, is normal for locally uniform spherical convergence. Since $\mathrm{deg}R\ge 2$, question 1 shows that $J(R)$ contains at least three distinct points. The forward inclusion just proved means every iterate of $S$ maps $F(R)$ into itself and therefore omits those three points there. Montel gives $F(R)\subseteq F(S)$. Exchanging the roles, using $\mathrm{deg}S\ge 2$, gives the opposite inclusion. ConsequentlyThis proves that commuting rational maps of degree at least two have the same Julia set without assuming either is an iterate of the other.

For commuting maps of degree one with different Julia sets, takeTheir two compositions are the identity, and the Julia sets follow from the expanding and contracting Möbius cases in question 1.

For equal Julia sets without commutation, takeBoth affine maps contract toward their finite fixed points and have a repelling periodic point at infinity. But $R\circ S=z/4+1/2$, while $S\circ R=z/4+1$, so they do not commute.

Finally the distinct Chebyshev polynomialshave degrees two and three and neither is a pure power of $z$. Direct expansion gives both compositions equal to $32z^6-48z^4+18z^2-1$. Equivalently $T_m(T_n(z))=T_{mn}(z)$, first seen at $z=\cos \theta$ and then as a polynomial identity. This provides the requested higher-degree pair.