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On a bounded regular domain, a bounded subset of $C^{k,\alpha }$ is relatively compact in $C^{l,\beta }$ whenever $k+\alpha >l+\beta$. This is an Arzelà-Ascoli theorem argument applied successively to the derivatives.

For a fixed bounded measurable uniformly elliptic operator in divergence form, solutions bounded by one have uniform local Caccioppoli inequalities and Hölder continuity estimates. A diagonal subsequence argument yields a local weak $W^{1,2}$ limit, while the Arzelà-Ascoli theorem gives uniform convergence on a fixed smaller closed Euclidean ball. The weak equation passes to the limit because the coefficients are fixed and bounded. A global supremum normalization does not force the interior limit to be nonzero.

On a fixed smooth bounded domain, zero-boundary solutions with forcing in a bounded subset of $C^{0,\mu }$ are uniformly bounded in $C^{2,\mu }$ by a maximum-principle barrier and the global Schauder estimate. The Arzelà-Ascoli theorem gives convergence in $C^2$. A closed Hölder-norm constraint passes to the limit by lower semicontinuity of a Hölder seminorm, so the constrained solution set is compact rather than merely relatively compact.

A subset of $L^1(\mathbb{R})$ is relatively norm compact exactly when it is bounded, its absolute-integral tails tend to zero uniformly, and ${\sup }_{f\in S}\Vert f(\cdot +h)-f{\Vert }_1\to 0$ as $h\to 0$. Necessity follows by taking a finite norm net and using the tail decay and translation continuity in Lp of its finitely many centers; the error contributes at most twice the net radius. For sufficiency, convolve with a triangular unit-integral kernel of width $\delta$. Translation continuity gives uniform approximation in $L^1$, while boundedness gives a common supremum bound and a common Lipschitz constant for the smoothed family. The original tail condition controls its tails outside a slightly enlarged interval. On that interval Arzelà-Ascoli theorem gives a finite uniform net. Extend its centers by zero outside the interval to obtain a finite $L^1$ net, and combine the smoothing and tail errors. Completeness of $L^1$ turns total boundedness into relative compactness.

If a thrice differentiable solution has both endpoint values and both endpoint first derivatives bounded in modulus by $N$, it obeys $|f|\le N$ and $|f^{\prime }|\le N$ throughout the interval. A positive interior maximum of $f$ would give $f^{\prime \prime }=f>0$, and a negative interior minimum would give $f^{\prime \prime }=f<0$, both impossible. For $w=f^{\prime }$, differentiation gives $w^{\prime \prime }=w+2w{w}^{\prime }$, so the same argument applies at an extremum of $w$. The resulting uniform bound and Lipschitz constant imply total boundedness of the solution family by the Arzelà-Ascoli theorem.

Use the direct method in the calculus of variations. The affine function $1-x$ is admissible with finite energy, and the energy is nonnegative since $2010$ is even. Thus $m=\inf I$ is finite. Choose a minimizing sequence $u_n$ with uniformly bounded energy. ThenThe fixed value $u_n(0)=1$ and the Cauchy-Schwarz inequality give $\Vert u_n{\Vert }_{\infty }\le 1+\Vert u_n^{\prime }{\Vert }_2$ and $|u_n(x)-u_n(y)|\le \Vert u_n^{\prime }{\Vert }_2|x-y{|}^{1/2}$. Hence the sequence is bounded in $W^{1,2}$ and uniformly bounded and equicontinuous.

By reflexivity and the Arzelà-Ascoli theorem, pass to a subsequence with $u_n\rightharpoonup u$ in $W^{1,2}$ and $u_n\to u$ uniformly. The two limits coincide; uniform convergence also preserves $u(0)=1$, $u(1)=0$. Since all functions stay in a common bounded interval, uniform convergence gives $u_n^{2010}\to u^{2010}$ uniformly, and therefore convergence of the potential integrals. The derivative term is lower semicontinuous for weak convergence, soAdmissibility gives $I(u)\ge m$. Thus

Work in the Banach space $C([0,1])$ with its supremum norm, and take the nonempty closed bounded convex set $M=\{u:100\le u(t)\le 121\text{ for all }t\}$. DefineFor $u\in M$ the absolute value is unnecessary, and $\boxed{101\le Tu(t)\le 101+12=113\le 121}$. Thus $T:M\to M$. Also, for $a,b\ge 100$,so $\Vert Tu-Tv{\Vert }_{\infty }\le \frac{1}{20}\Vert u-v{\Vert }_{\infty }$ and in particular $T$ is continuous.

The image is uniformly bounded, and $|Tu(t)-Tu(s)|\le 12|t-s|$, so it is equicontinuous. The Arzelà-Ascoli theorem says that a uniformly bounded equicontinuous family in $C([0,1])$ has relatively compact closure in the uniform norm. Hence $T$ is compact, and the Schauder fixed-point theorem supplies $\boxed{u=Tu}$, the required solution. In fact the displayed contraction estimate also lets the contraction mapping theorem give uniqueness within $M$ and convergence of the corresponding Picard iteration. The initial constant here is the printed integer $101$.

Since the kernel has integral one, The Tonelli theorem and the triangle inequality giveThis proves the stated convolution approximation estimate.

For the unlabelled compactness assertion let $M$ bound the norms of $S$, and fix $\eta >0$. Choose the kernel width $\delta$ smaller than the translation radius supplied for $\eta$, so that the same bound holds for all shifts of size at most this width. The preceding estimate gives $\Vert f\ast g-f{\Vert }_1<\eta$ uniformly. The smoothed family has uniform bound $M/\delta$ and uniform Lipschitz constant $M/{\delta }^2$. Choose $K$ with all original tails outside $[-K,K]$ below $\eta$, and put $L=K+\delta$. The Tonelli theorem givesOn $[-L,L]$ the smoothed family is relatively compact in the uniform norm by the Arzelà-Ascoli theorem. Take a finite uniform net with error below $\eta /(2L)$ and extend its centers by zero outside that interval. It is a finite $L^1$ net for the smoothed-and-truncated family with error $\eta$. Each original $f$ is within another $2\eta$ of its corresponding truncated smoothing. Thus $S$ is totally bounded in $L^1$. The displayed (a),(b) are the two hypotheses used together in this argument.

For the final family, Young convolution inequality gives $\Vert f_0\ast f{\Vert }_1\le \Vert f_0{\Vert }_1$. Its translations satisfywhich tends to zero uniformly over the family. Since $f$ vanishes outside $[-1,1]$,uniformly as well. Both compactness hypotheses apply, proving $S_0$ is totally bounded.

A subset is totally bounded if for every $\epsilon >0$ it has a finite cover by balls of radius $\epsilon$. If its closure is compact, the cover by all such balls has a finite subcover, proving total boundedness.

Conversely, cover $S$ by finitely many balls of radius $2^{-r}$ at stage $r$. Starting with all indices, at least one ball contains infinitely many of the currently indexed terms. Retain those indices as ${\sigma }_r$. These sets are infinite and nested, and two terms indexed by ${\sigma }_r$ have distance less than $2^{1-r}$. Choose increasing $n_r\in {\sigma }_r$; then $(x_{n_r})$ is Cauchy, because all its terms after stage $r$ lie in that same ball. If $M$ is complete it converges. To cover sequences in the closure as well, approximate their $n$th terms by points of $S$ at distance less than $1/n$ and apply this argument. The corresponding subsequence in the closure has the same limit. Sequential compactness therefore proves total boundedness implies relative compactness in a complete ambient space.

The Arzelà-Ascoli theorem says a subset of $C[0,1]$ is relatively compact exactly when it is uniformly bounded and equicontinuous. Pointwise boundedness together with equicontinuity is an equivalent version here. In the specified derivative-bounded family, the mean-value estimate gives $|f(t)-f(s)|\le |t-s|$, uniformly over the family, while $\Vert f{\Vert }_{\infty }\le 1$. Thus this family is relatively compact.

For $V$, the image of the unit ball has sup norm at most one and derivative $f$ of sup norm at most one; it lies in that family, so $V$ is compact.

For $W$, take $f_n=n{1}_{[0,1/n]}$, each of $L^1$ norm one. Then $W{f}_n(t)=\min (nt,1)$. Every subsequence converges pointwise to zero at $t=0$ and to one at every $t>0$. This limit is discontinuous, so no subsequence converges uniformly to a continuous function. Thus $W$ is not compact. The same integration formula has different compactness properties because the domain unit balls impose different controls.

A normal family of holomorphic functions on a domain is one from which every sequence has a subsequence converging locally uniformly to a holomorphic function, or locally uniformly to infinity in the extended convention. For a bounded family, only finite limits occur.

For functions mapping a plane domain $D$ into the unit disc, the Cauchy integral formula supplies uniform derivative bounds on each compact subset. More explicitly, if a closed disc of radius $r$ around $z$ lies in $D$, the Cauchy estimate gives $|f^{\prime }(z)|\le 1/r$. A slightly larger compact neighborhood of any given compact set therefore gives uniform boundedness and equicontinuity there. Apply the Arzelà-Ascoli theorem on a compact exhaustion and take a diagonal subsequence. Its limit is locally uniform; the Cauchy integral formula shows that the limit is holomorphic. This proves normality rather than merely invoking it.

Now let $|g|\le M$ on the complex upper half-plane. Every dilation $g_t(w)=g(tw)$ has the same bound, so the argument just given, after scaling the bound if necessary, proves that $\{g_t:t\ge 1\}$ is a normal family. For any sequence $t_j\to \infty$, every locally uniform subsequential limit $G$ satisfiesThe identity theorem for holomorphic functions forces $G\equiv \ell$. It follows that $g_t\to \ell$ locally uniformly as $t\to \infty$: otherwise a sequence violating convergence on one compact set would have a convergent subsequence with this same constant limit, a contradiction.

For $0<\epsilon <\pi /2$, the arc $K_{\epsilon }=\{e^{i\theta }:\epsilon \le \theta \le \pi -\epsilon \}$ is a compact subset of the complex upper half-plane. Write $z=|z|w$ with $w\in K_{\epsilon }$. Local uniform convergence of the dilations givesFor $\epsilon \ge \pi /2$ the specified sector is empty. This is the same compactness mechanism as the normal-family proof of angular boundary convergence.

To transfer the result to the unit disc, use the Cayley transform between the half-plane and diskIt sends $r\omega$ to $iy$ with $y=(1+r)/(1-r)$, so $g(iy)\to \ell$ as $y\to \infty$. Choose the curvature-minus-one normalization of the hyperbolic metric, for which the disc line element is $2|dz|/(1-|z{|}^2)$ and the half-plane line element is $|dw|/\mathrm{Im}w$. The Cayley transform between the half-plane and disk preserves these metrics. For $w\in \mathbb{H}$ and $y>0$,Thus ${\rho }_{\mathbb{D}}(z,r\omega )<k$ implies $\mathrm{Im}w/|w|>\mathrm{sech}k$. This is the hyperbolic tube around a radial segment estimate. It places $w$ in a fixed sector bounded away from the real axis. Moreover $z\to \omega$ implies $|C_{\omega }(z)|\to \infty$. Applying the sector result givesA different constant normalization of the hyperbolic metric only rescales $k$ and gives the same conclusion.

For a plane domain, the following are equivalent descriptions of a simply connected domain: every based loop contracts in the domain with its basepoint fixed; any two paths with the same endpoints are homotopic while those endpoints stay fixed; its complement in the Riemann sphere is connected; every holomorphic function has a single-valued primitive; and every nowhere-zero holomorphic function has a holomorphic logarithm. The complement must be taken in the extended plane, including infinity. An equivalent winding number formulation is that every closed curve has index zero about every point outside the domain.

Here is a direct equivalence proof for the first two topological formulations. If all paths with given endpoints are homotopic relative to those endpoints, apply this to a loop based at $a$ and the constant path at $a$, obtaining a contraction. Conversely, let ${\gamma }_0,{\gamma }_1$ join $a$ to $b$. The loop ${\gamma }_0\ast {\gamma }_1^{-1}$ contracts with basepoint fixed. Concatenating this contraction with ${\gamma }_1$ produces a homotopy from ${\gamma }_0\ast {\gamma }_1^{-1}\ast {\gamma }_1$ to ${\gamma }_1$, relative to $a,b$. The middle retracing excursion contracts by shortening its traversal in both directions, giving a homotopy from that first path to ${\gamma }_0$. Hence ${\gamma }_0$ and ${\gamma }_1$ are homotopic relative to their endpoints. Planar domains are path-connected, so no extra path-component condition is needed.

The analytic descriptions follow from path independence of the holomorphic integral. Local primitives exist on discs by integrating power series; a homotopy can be subdivided into pieces lying in such discs, and local path integrals cancel along its common edges. Thus integrals over homotopic paths agree, giving a global primitive. Applied to $h^{\prime }/h$ for a nowhere-zero function, a primitive $H$ makes $h{e}^{-H}$ constant and hence supplies a holomorphic logarithm. These observations are also the tools used in the mapping proof below.

The Riemann mapping theorem states that every nonempty simply connected proper domain $D⊊\mathbb{C}$ admits a conformal bijection onto the unit disc. Given $z_0\in D$, there is exactly one such map with $f(z_0)=0$ and $f^{\prime }(z_0)>0$. The exclusion of the whole plane is necessary: an entire map into the disc would be constant by Liouville's theorem.

An extremal proof has the following concrete steps. First construct an injective bounded holomorphic map. Choose $a\notin D$ and a holomorphic square root $h$ of $z-a$. It is injective, since equality of its values implies equality of their squares. Its image is disjoint from its negative: $h(z)=-h(w)$ would imply $z=w$ and then $h(z)=0$, impossible. If $b=h(z_0)$, choose $r>0$ with $B(b,r)\subset h(D)$; the disc $B(-b,r)$ is omitted. Therefore $1/(h+b)$ is injective and bounded. Scaling it and applying an automorphism of the unit disk gives a map into the disc taking $z_0$ to zero; a rotation makes its derivative positive.

Let $\mathcal{U}$ consist of all injective maps $D\to \mathbb{D}$ with this normalization. It is nonempty. The derivative at $z_0$ is bounded above by the Cauchy integral formula on a small disc compactly contained in $D$. Choose $f_n\in \mathcal{U}$ with derivatives tending to their positive supremum $L$. Boundedness, the Cauchy estimates and a diagonal Arzelà-Ascoli theorem argument give a subsequence converging locally uniformly to $f$; this is the bounded-family form of the Montel theorem. Then $f(z_0)=0$, $f^{\prime }(z_0)=L>0$, and the maximum modulus principle puts its image in the open disc. The limit is injective: if distinct $z,w$ had $f(z)=f(w)$, apply Hurwitz's theorem near $w$ to $f_n(\cdot )-f_n(z)$, which has no zero there; the limit would vanish identically, contradicting $f^{\prime }(z_0)>0$.

Finally suppose $a\in \mathbb{D}$ is omitted by $f$. It is nonzero since $f(z_0)=0$. Let $T_a(w)=(w-a)/(1-\overline{a}w)$ and choose a holomorphic square root $h$ of $T_a\circ f$. It is injective and takes values in the disc. Put $b=h(z_0)$, so $|b|=\sqrt{|a|}$, and normalize $T_b\circ h$ by a rotation. The new map's derivative magnitude at $z_0$ isThis square-root improvement of a normalized conformal map contradicts extremality. Thus $f$ is onto. Two normalized bijections differ by a disc automorphism fixing zero; Schwarz lemma applied to it and its inverse makes it a rotation, and positivity of the normalized derivatives forces that rotation to be the identity. This completes the requested proof outline.

In the supremum-norm metric, $S$ is totally bounded if for every $\epsilon >0$ finitely many norm balls of radius $\epsilon$ cover it. It is uniformly bounded if some $M$ satisfies $|f(x)|\le M$ for all $f\in S,x\in K$. It is equicontinuous if for each $x\in K$ and $\epsilon >0$ there is a neighbourhood $U$ of $x$ with $|f(y)-f(x)|<\epsilon$ for all $f\in S,y\in U$. This definition works for compact Hausdorff $K$ without assuming it is metrizable.

Suppose $S$ is totally bounded. A finite radius-one net makes $\Vert f{\Vert }_{\infty }$ uniformly bounded by one plus the largest norm of a net centre. For equicontinuity, choose a finite $\epsilon /3$ net. Continuity of each centre gives a common neighbourhood of $x$ on which all centre values vary by less than $\epsilon /3$. Approximating any $f$ by its centre and using the triangle inequality at $x$ and $y$ bounds its variation by $\epsilon$. Thus both conditions follow.

Conversely suppose those two conditions hold. For each $x$, choose a neighbourhood $U_x$ on which every function varies from its value at $x$ by less than $\epsilon /3$. Compactness provides a finite cover $U_{x_1},\dots ,U_{x_m}$. The vectors $(f(x_1),\dots ,f(x_m))$ form a bounded subset of ${\mathbb{C}}^m$, so partition their possible values into finitely many boxes of diameter less than $\epsilon /3$ in the maximum-coordinate norm. From each occupied box choose one function as a representative. If $f$ and its representative $h$ have sample values in the same box, and $y\in U_{x_j}$, thenThese finitely many representatives are a norm net, proving total boundedness is equivalent to uniform boundedness and equicontinuity. This supplies the finite-net part of the Arzelà-Ascoli theorem directly.

For a bounded set that is not totally bounded, take the standard unit vectors $S=\{e_n:n\ge 1\}$ in the Banach space ${\ell }^2$. Their norms are one and their pairwise distances are $\sqrt{2}$. A ball of radius less than $\sqrt{2}/2$ contains at most one of them, so finitely many such balls cannot cover $S$.

A normal family is a family such that every sequence has a subsequence converging locally uniformly in the spherical metric to a meromorphic limit, possibly the constant infinity. For a family uniformly bounded in modulus, the possible limits are finite holomorphic functions, and ordinary locally uniform convergence suffices. The limit need not take values in the open unit disc; a sequence of constants can approach its boundary.

Here is a direct compactness proof for disc-valued maps. For each compact subset $K\subset \Omega$, choose a slightly larger compact neighborhood contained in $\Omega$. The Cauchy estimate on a fixed-radius disk around each point bounds $|f^{\prime }|$ independently of $f$ because $|f|<1$. A finite covering of $K$ yields uniform boundedness and equicontinuity there. The Arzelà-Ascoli theorem, applied successively to a compact exhaustion and followed by a diagonal subsequence, gives a limit uniformly on every compact. The Cauchy integral formula passes to that limit on small circles, proving it is a holomorphic function. This proves the required normal family assertion rather than just invoking Montel theorem.

Every $g_n(z)=g(z/n)$ is a holomorphic function on the complex upper half-plane with values in the unit disc, so the same argument makes $(g_n)$ a normal family. If a subsequence converges locally uniformly to $h$, then for every $t>0$,The identity theorem for holomorphic functions gives $h\equiv \ell$. Since every convergent subsequence has this same limit, the whole sequence converges locally uniformly to $\ell$: otherwise a compact set and a subsequence with a fixed positive discrepancy would have a further locally uniformly convergent subsequence, a contradiction.

To obtain the nontangential limit, take any $z_j=x_j+i{y}_j\to 0$ with $|x_j|\le A{y}_j$ and $y_j>0$. Put $n_j=\lfloor 1/y_j\rfloor$. Then $n_j\to \infty$, $n_j{y}_j\to 1$, and the points ${\zeta }_j=n_j{z}_j$ lie, for large $j$, in a fixed compact rectangle inside the half-plane. ThereforeThis is the normal-family proof of angular boundary convergence; using integer rescalings requires this step because the points need not lie exactly on one fixed scaled ray.

For a tangential counterexample, useIf $z=x+iy$ with $y>0$, then $\mathrm{Re}(-i/z)=-y/(x^2+y^2)<0$, so $g$ is a holomorphic function and $|g|<1$. Also $g(iy)=e^{-1/y}\to 0$. ButTheir ratio $|\mathrm{Re}{z}_n|/\mathrm{Im}{z}_n=2\pi n^2$ tends to infinity. Thus the nontangential limit is zero while the unrestricted boundary limit does not exist.

Let $\mathbf{X}$ be a weak geometric p-rough path, $N=\lfloor p\rfloor$, with increments $X_{s,t}^I$ indexed by words $I=(i_1,\dots ,i_k)$ of length at most $N$. Choose a control function for rough paths $\omega$ with $|X_{s,t}^I|\le \omega (s,t{)}^{|I|/p}$. We use the standard sufficient vector-field hypothesis $V_i\in {\mathrm{Lip}}^{\gamma }$, $\gamma >p$, with globally bounded regularity norms; smooth bounded fields are a special case. Regard $V_i$ as the differential operator $V_i\cdot \nabla$, and put $F_I=V_{i_1}\cdots V_{i_k}\mathrm{id}$.

A pathlevel solution of a rough differential equation is a continuous state path $y:[0,T]\to {\mathbb{R}}^e$, with prescribed $y_0$, satisfying a local expansionfor some $\theta >1$ on sufficiently small control intervals. One can take $\theta =\min (\gamma ,N+1)/p>1$ under the stated regularity. This definition specifies the state path, without requiring its full rough-path enhancement as additional data. For $2<p<3$, the displayed expansion readsRepeated indices are summed.

We use the following discrete form of the Davie lemma. For a finite ordered grid and a continuous superadditive control, suppose nonnegative errors $a_{s,t}$ vanish on adjacent grid points and, for every grid triple $s<u<t$, satisfyThere are $\delta >0$ and $C$, depending on $L,A,p,\theta$ but not on the grid, such that $a_{s,t}\le C\omega (s,t{)}^{\theta }$ whenever $\omega (s,t)\le \delta$. This is the discrete higher-order error estimate: splitting close to half the control size makes the two $\theta$-power errors contract; a crossing single grid interval contributes zero adjacent error.

Define the local Euler mapFor small $\omega (s,t)$ it has Lipschitz constant at most $1+L\omega (s,t{)}^{1/p}$. Taylor expansion, the Chen identity and the shuffle identity for iterated integrals give the composition estimateHere is the cancellation underlying the estimate. Assign degree $k$ to a level-$k$ increment. Expanding each $F_I(E_{s,u}(z))$ and grouping terms through degree $N$ gives the differential-operator coefficients $F_{IJ}(z)$ multiplying $X_{s,u}^I{X}_{u,t}^J$. Products of coefficients from the first increment are combined by the shuffle identity, which is the iterated Leibniz rule. Chen's identity says that the sum of these products is $X_{s,t}^{IJ}$. Thus every term of degree at most $N$ agrees with the corresponding term in $E_{s,t}$. The remaining Taylor terms have control order at least $\theta$; bounded ${\mathrm{Lip}}^{\gamma }$ norms make the estimate uniform in $z$. For smooth fields the remainder has order $(N+1)/p$.

On a partition $\pi$, construct the Euler sequence by $y_{t_{j+1}}^{\pi }=E_{t_j,t_{j+1}}(y_{t_j}^{\pi })$. For grid endpoints letIt vanishes on adjacent points. Insert $y_u^{\pi }=E_{s,u}(y_s^{\pi })+I_{s,u}^{\pi }$ into the next local map. The Lipschitz and composition estimates giveThe Davie lemma therefore yields a bound $|I_{s,t}^{\pi }|\le C\omega (s,t{)}^{\theta }$ independent of $\pi$ on small control intervals. Together with the Euler expansion this gives $|y_t^{\pi }-y_s^{\pi }|\le C^{\prime }\omega (s,t{)}^{1/p}$. A finite subdivision into small-control intervals gives a uniform global bound as well.

Interpolate each Euler sequence linearly in time, and take partitions with mesh tending to zero. Continuity of the control makes these paths uniformly equicontinuous: the grid estimates apply to the interval enlarged to its neighboring grid endpoints, and the enlargement tends to zero uniformly with the mesh. The paths are uniformly bounded. The Arzelà-Ascoli theorem supplies a uniformly convergent subsequence with continuous limit $y$.

For arbitrary $s<t$ in a small control interval, choose neighboring grid endpoints converging to $s,t$. Continuity of $\mathbf{X}$, the coefficient functions and the control lets us pass to the limit in the grid error estimate. We obtain the defining expansion with $|R_{s,t}|\le C\omega (s,t{)}^{\theta }$, and $y(0)=y_0$. This constructs a pathlevel RDE solution. No continuity theorem for the RDE solution map was used to obtain existence.

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.

The Arzelà-Ascoli theorem for $C([0,1])$ states that a family has compact closure in the uniform norm exactly when it is uniformly bounded and equicontinuous. To prove sufficiency, choose a countable dense set. From any sequence of functions, successive subsequence extraction at each dense point and a diagonal selection give convergence at all those points. Given $\epsilon >0$, equicontinuity gives a common $\delta$; cover the interval by finitely many $\delta$-balls centered at dense points. On a sufficiently late tail the functions differ by less than $\epsilon /3$ at those finitely many centres, and equicontinuity makes each of the other two differences less than $\epsilon /3$. Thus the selected subsequence is uniformly Cauchy. Completeness of $C([0,1])$ gives a continuous uniform limit. In a metric space this proves compactness of the closure. Conversely, a compact family has finite uniform $\epsilon$-nets. The finitely many net functions are bounded and uniformly continuous; approximation by them gives a common bound and equicontinuity for the family. This proves both directions.

For $f\in S_N$, a positive interior maximum would have $f^{\prime }=0$ and $f^{\prime \prime }\le 0$, contradicting $f^{\prime \prime }=f>0$. A negative interior minimum likewise contradicts $f^{\prime \prime }=f<0$. The endpoint bounds therefore give $|f|\le N$ everywhere.

Now put $w=f^{\prime }$. Differentiating the equation, using the stipulated third derivative, gives $w^{\prime \prime }=w+2w{w}^{\prime }$. At an interior extremum of $w$, $w^{\prime }=0$, so $w^{\prime \prime }=w$. The same maximum/minimum argument and the derivative endpoint bounds imply $|f^{\prime }|=|w|\le N$. Consequently $|f(x)-f(y)|\le N|x-y|$ by the mean value theorem. Thus $S_N$ is uniformly bounded and equicontinuous, and the theorem makes it relatively compact, hence totally bounded in $C([0,1])$. Its closure need not be asserted to remain thrice differentiable.

For a compact space $K$, the Arzelà-Ascoli theorem says that a family $\mathcal{F}\subset C(K)$ is relatively compact in the supremum norm if and only if it is pointwise bounded and equicontinuous. In the usual formulation $K$ is compact Hausdorff; the elementary argument below also works for any compact $K$ for real-valued continuous functions.

Pointwise bounded means ${\sup }_{f\in \mathcal{F}}|f(x)|<\infty$ for every $x\in K$. Equicontinuity means that for every $x\in K$ and $\epsilon >0$ there is a neighborhood $U$ of $x$ such that $|f(y)-f(x)|<\epsilon$ for every $y\in U$ and $f\in \mathcal{F}$. Relative compactness means compactness of the closure in $C(K)$, where $\Vert f{\Vert }_{\infty }={\sup }_K|f|$. This is a complete metric vector space, since uniform limits of continuous functions are continuous. Equivalently, every sequence in the family has a uniformly convergent subsequence, with limit in $C(K)$, though not necessarily in the original family.

Assume first that $\overline{\mathcal{F}}$ is compact in the supremum norm. A finite norm-net shows that the norms of its members are bounded, so pointwise boundedness follows. Given $x$ and $\epsilon >0$, choose a finite $\epsilon /3$-net $g_1,\dots ,g_N$ in the closure. Their continuity gives a common neighborhood $U$ of $x$ where $|g_j(y)-g_j(x)|<\epsilon /3$ for every $j$. For any $f\in \mathcal{F}$, choose $g_j$ with $\Vert f-g_j{\Vert }_{\infty }<\epsilon /3$; the triangle inequality then gives $|f(y)-f(x)|<\epsilon$. Thus the family is equicontinuous.

Conversely, suppose pointwise boundedness and equicontinuity. For a prescribed $\epsilon >0$, choose neighborhoods $U_x$ where every $f$ varies from $f(x)$ by less than $\epsilon /3$. Compactness supplies a finite subcover $U_{x_1},\dots ,U_{x_N}$. Pointwise boundedness makes the set of vectors $(f(x_1),\dots ,f(x_N))$ bounded in ${\mathbb{R}}^N$. Partition a bounding box into finitely many boxes of coordinate diameter less than $\epsilon /3$, and choose one function from each occupied box. If $f,g$ belong to the same box and $y\in U_{x_j}$, thenThe chosen functions are therefore a finite $\epsilon$-net in the supremum norm. Thus $\mathcal{F}$ is totally bounded. Its closure is complete and totally bounded, hence compact. To recall the last metric-space fact, successively selecting subsequences in balls of radii $2^{-j}$ produces a Cauchy diagonal subsequence from every sequence; completeness supplies its limit. This proves both directions of the Arzelà-Ascoli theorem.

As usual, the bounded set is assumed Lebesgue measurable and an $L^q$ exponent is positive; the conventional Banach-space assertion concerns $1\le q<2^{\ast }=2n/(n-2)$. No boundary regularity of $\Omega$ is needed. Choose a smooth compactly supported cutoff $\eta$ equal to one on a ball containing $\Omega$. Weak convergence and the Uniform boundedness principle give a common $H^1$ bound for $f_j$, hence also for $g_j=\eta f_j$. These functions have a common compact support.

We prove their local $L^2$ compactness directly. Translation and the fundamental theorem along line segments give, by density from smooth functions,For a compactly supported mollifier ${\rho }_{\epsilon }$, integration of this estimate givesAt fixed $\epsilon$, Hölder's inequality bounds $\Vert g_j\ast {\rho }_{\epsilon }{\Vert }_{\infty }$ and $\Vert \nabla (g_j\ast {\rho }_{\epsilon }){\Vert }_{\infty }$ uniformly by the $L^2$ norm of the corresponding smoothing kernels. Their supports lie in one fixed compact set, so the Arzelà-Ascoli theorem makes this smoothed family precompact in $L^2$. The uniform $O(\epsilon )$ approximation makes the original family totally bounded in $L^2$: choose a finite net for the smoothed family and add the approximation error. Completeness gives relative compactness.

Every strongly convergent subsequence of $g_j$ has limit $\eta f$, since multiplication by $\eta$ preserves weak $L^2$ convergence. Consequently the whole sequence converges strongly: otherwise a subsequence with distance at least a fixed positive number would have a further convergent subsequence and contradict that unique limit. In particular $\Vert {\chi }_{\Omega }(f_j-f){\Vert }_2\to 0$.

The critical Sobolev inequality supplies a uniform bound in $L^{2^{\ast }}$. For clarity, this estimate follows from part 2(a), not from compactness at the critical exponent. For a compactly supported smooth $v$, integrating by parts against the vector field $(x-y)/|x-y{|}^n$ givesIts distributional divergence is the sphere-area multiple of a point mass; applying the Hardy–Littlewood–Sobolev inequality with $\alpha =1,p=2$ yields $\Vert v{\Vert }_{2^{\ast }}\le C\Vert \nabla v{\Vert }_2$. Density extends this to $H^1({\mathbb{R}}^n)$.

For $2<q<2^{\ast }$ choose $0<\theta <1$ with $1/q=\theta /2+(1-\theta )/2^{\ast }$. Hölder's inequality givesFor $1\le q<2$, use $\Vert v{\Vert }_{L^q(\Omega )}\le |\Omega {|}^{1/q-1/2}\Vert v{\Vert }_{L^2(\Omega )}$. The same integral estimate also works for $0<q<1$ if $L^q$ is understood with its usual quasi-norm. Thus ${\chi }_{\Omega }{f}_j\to {\chi }_{\Omega }f$ strongly for every positive $q<2^{\ast }$. We have never asserted that multiplying by the possibly nonsmooth indicator preserves $H^1$.

The Riemann mapping theorem asserts that a nonempty simply connected domain $\Omega ⊊\mathbb{C}$ is biholomorphic to the unit disk $\mathbb{D}$. More precisely, for each $z_0\in \Omega$ there is a unique biholomorphism $f:\Omega \to \mathbb{D}$ satisfying $f(z_0)=0$ and $f^{\prime }(z_0)>0$. We prove existence through an extremal derivative, deriving the compactness and injectivity steps rather than assuming special lemmas for the theorem.

First construct a bounded univalent function on $\Omega$. Choose $a\notin \Omega$ and a holomorphic logarithm of $z-a$, which exists by the permitted logarithm result on the translated simply connected domain. PutThen $u(z{)}^2=z-a$, so $u$ is univalent and never zero. Its image $G=u(\Omega )$ is open, and $G\cap (-G)=\varnothing$: equality $u(x)=-u(y)$ would imply $x=y$ after squaring and then $u(x)=0$.

Fix $w_0\in G$ and choose $r>0$ with $B(w_0,r)\subset G$. It follows that $B(-w_0,r)\subset -G$ is disjoint from $G$, so $|u(z)+w_0|\ge r$ throughout $\Omega$. Henceis a univalent function with $|h|\le 1/2$. For $b\in \mathbb{D}$, defineThe identityand the inverse $T_b^{-1}(w)=(w+b)/(1+\overline{b}w)$ show directly that $T_b$ is an automorphism of the unit disk. Apply $T_{h(z_0)}$ to $h$ and then a rotation to make its nonzero derivative at $z_0$ positive. The family $\mathcal{A}$ of univalent maps $\Omega \to \mathbb{D}$ with value zero and positive derivative at $z_0$ is therefore nonempty.

LetIt is positive. Choose $\rho >0$ with $\overline{B(z_0,\rho )}\subset \Omega$. The Cauchy estimate gives $v^{\prime }(z_0)\le 1/\rho$ for every $v\in \mathcal{A}$, so $M$ is finite. Take a sequence $v_j\in \mathcal{A}$ with $v_j^{\prime }(z_0)\to M$.

We now justify a locally uniform convergence subsequence. On any compact subset of $\Omega$, the bound $|v_j|\le 1$ and the Cauchy estimate on slightly larger interior disks give a uniform derivative bound. They therefore give equicontinuity. Apply the Arzelà-Ascoli theorem on a compact exhaustion of $\Omega$, then take a diagonal subsequence for locally uniform convergence. Its limit $v$ is holomorphic: pass to the limit in the Cauchy integral formula on every small interior circle. The same formula for derivatives givesThus $v$ is nonconstant. Initially $|v|\le 1$, and the maximum modulus principle improves this to $|v|<1$ everywhere.

For completeness, the needed locally uniform limit of univalent functions argument is short. If $x\ne y$ and $v(x)=v(y)$, the nonconstant holomorphic function $v-v(y)$ has an isolated zero at $x$. Choose a small closed disk around $x$ excluding $y$, with no zero on its boundary. Uniform convergence there gives $v_j-v_j(y)\to v-v(y)$. The Rouche theorem then makes $v_j-v_j(y)$ have a zero inside that disk for all sufficiently large $j$. Its only possible zero, by injectivity of $v_j$, is $y$, outside the disk. This contradiction proves injectivity of $v$. Therefore $v\in \mathcal{A}$ and attains the supremum $M$.

It remains to prove surjectivity. Suppose $c\in \mathbb{D}$ is omitted by $v$; since $v(z_0)=0$, necessarily $c\ne 0$. The domain $T_c(v(\Omega ))$ is simply connected, because both constituent maps are homeomorphisms onto their images, and it excludes zero. The permitted holomorphic logarithm gives a holomorphic square rootThis maps $\Omega$ into $\mathbb{D}$ and is univalent, since equality of its values implies equality after squaring and hence equality of the original arguments. Write $\gamma =g(z_0)$. Then ${\gamma }^2=T_c(0)=-c$ and $|\gamma |=\sqrt{|c|}$. Normalize again with $V=T_{\gamma }\circ g$ and a rotation. Its derivative magnitude isThe strict inequality follows from $(1-\sqrt{|c|}{)}^2>0$. This square-root improvement of a normalized conformal map contradicts extremality. Thus $v(\Omega )=\mathbb{D}$. An injective holomorphic map has nonzero derivative and a holomorphic local inverse; these inverses agree, so the resulting bijection is biholomorphic. We have proved existence.