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Write $dm=dt/(2\pi )$ in angular coordinates on the unit circle. The centered Hardy-Littlewood maximal function iswhere $I_r(\omega )$ is the centered circular arc of angular radius $r$, with $r=\pi$ giving the whole circle. The Hardy-Littlewood maximal inequality in this one-dimensional setting is the weak-type estimateThe precise constant is not important for differentiation, but we prove this one explicitly.

Let $E=\{Mf>\alpha \}$ and choose a compact subset $F\subset E$. For each point of $F$, choose a centered arc whose average of $|f|$ exceeds $\alpha$. A finite subfamily covers $F$. Order these arcs by decreasing length and retain an arc precisely when it is disjoint from the previously retained arcs. Every rejected arc meets a retained arc of at least its length and lies in the latter's triple, interpreted on the circle and truncated to the whole circle if necessary. If the retained disjoint arcs are $I_1,\dots ,I_N$, thenThis is the interval version of the Wiener covering lemma. The superlevel set is measurable: the average at each fixed radius is continuous in the center, by translation continuity in $L^1$, so $E$ is open. Inner regularity of Lebesgue measure now passes the compact-set bound to $E$, proving the Hardy-Littlewood maximal inequality. The same finite-interval argument gives the line version $|\{Mf>\alpha \}|\le 3\Vert f{\Vert }_1/\alpha$ for integrable functions on the real line.

To prove differentiation, first recall that continuous functions are dense in $L^1(m)$ on the circle. Approximate an integrable function by simple functions; for each of their measurable sets, inner and outer regularity give a compact subset and an open superset with arbitrarily small measure difference. A continuous cutoff between these sets approximates the indicator in $L^1$. Combining finitely many cutoffs proves the claimed density.

For a representative of $f$ finite almost everywhere, define its local absolute oscillation byIf $h$ is continuous and $u=f-h$, the triangle inequality and continuity of $h$ giveThus the Hardy-Littlewood maximal inequality and the elementary integral bound for a superlevel set yield, for every $\epsilon >0$,Choose continuous $h$ with $\Vert f-h{\Vert }_1$ arbitrarily small. It follows that $\Delta f=0$ outside a null set, by using positive rational $\epsilon$. In particular,for almost every $\omega$. This proves the requested Lebesgue differentiation theorem, including the stronger absolute-oscillation formulation, rather than assuming pointwise convergence of the approximating continuous functions.

For failure at a specified point, use coordinates $-\pi <t\le \pi$ and put $a_n=2^{-2^n}$. Let $f$ be one on the annuli $a_{2j+1}<|t|\le a_{2j}$ for $j\ge 0$, and zero elsewhere, including at zero. This bounded measurable function is integrable. Since $a_{n+1}/a_n\to 0$, the average $A_n$ over the centered interval $[-a_n,a_n]$ satisfiesIndeed, the outermost annulus occupies all but the fraction $a_{n+1}/a_n$ of that interval, and has value one or zero according to parity. The even-radius subsequence tends to one and the odd-radius subsequence to zero, so no centered-average limit exists at $\omega =1$. The normalization of $m$ cancels from these averages. This is how oscillating annuli obstruct differentiation at a point without contradicting almost-everywhere differentiation.

For $|\zeta |=1$ and $|z|<1$, the Poisson kernel on the circle satisfiesFor fixed $\zeta$, it is the real part of a holomorphic function of $z$, hence a harmonic function. On each compact subdisk, its derivatives are bounded uniformly in $\zeta$ because the denominator stays away from zero. Differentiation under the integral therefore proves that $P\varphi$ is harmonic, with the word interpreted componentwise when $\varphi$ is complex-valued.

The geometric seriesconverges uniformly in $\zeta$ for a fixed compact subdisk. With normalized Lebesgue measure, integration of its real part gives kernel mass one. Fix a boundary point $\omega$ and choose $\delta >0$ so that $|\varphi (\zeta )-\varphi (\omega )|<\epsilon$ when $|\zeta -\omega |<\delta$. ThenIf $z\to \omega$ from inside the unit disc, the remaining denominator is at least $(\delta /2{)}^2$ for sufficiently close $z$, while $1-|z{|}^2\to 0$. Hence $P\varphi (z)\to \varphi (\omega )$ along every interior approach, not just radial ones.

Orient the unit circle counterclockwise in the Cauchy transform. Its kernel and every $w$-derivative are uniformly bounded for $w$ in any compact subset of $\mathbb{C}\setminus \mathbb{T}$. Thus differentiation under the integral givesand proves that $\Phi$ is a holomorphic function on both components of its domain. Since $d\zeta =2\pi i\zeta dm(\zeta )$, for $0<r<1$ and $|\omega |=1$ the difference of its two kernels isConsequently the paired radial jump of a continuous Cauchy transform obeys the exact identityand the established Poisson integral boundary convergence givesThis proves the paired difference without assuming that the two individual boundary values exist for every continuous datum.

The printed direction $r\downarrow 1$ has the opposite sign. It is present in the original PDF, not merely in the converted TeX. With that direction $r>1$, so substituting $s=1/r<1$ in the exact identity givesThe simplest counterexample is $\varphi \equiv 1$: the Cauchy integral formula gives $\Phi (w)=1$ inside and $\Phi (w)=0$ outside, so the printed difference is constantly $-1$. The requested positive jump is corrected by taking $r\uparrow 1$, or by reversing the order of the difference when $r\downarrow 1$.

The holomorphic convex hull is usually defined analytically byWith that convention the requested equivalence is the definition. Its substantive planar content is the geometric descriptionOne can instead define the hull by the right side and prove the analytic equivalence; we do so after proving Runge theorem.

The prescribed-pole form of Runge theorem is as follows. If $K\subset \mathbb{C}$ is compact and $A\subset \mathbb{C}\setminus K$ meets each bounded complementary component, every holomorphic function on a neighborhood of $K$ is uniformly approximable on $K$ by rational functions with finite poles in $A$. Polynomial terms are allowed, accounting for the unbounded component. In particular, if every complementary component meets the complement of a domain $\Omega$, the approximants can be chosen as holomorphic functions on $\Omega$. Connected complement permits polynomial approximants.

For a proof, surround $K$ by the interior of a finite union of small grid squares whose closure lies in the neighborhood where $f$ is a holomorphic function. Orient their boundaries positively and cancel shared edges, obtaining finitely many piecewise linear boundary curves $\Gamma$ a positive distance from $K$. The Cauchy integral formula, applied to the union, givesUniform continuity of this integrand for $z\in K$ makes its Riemann sums converge uniformly on $K$. Each Riemann sum is a finite linear combination of reciprocals $(z-b{)}^{-1}$ with $b\in \Gamma \subset \mathbb{C}\setminus K$.

It remains to move these poles into the allowed components. Let $\mathcal{A}$ be the uniform closure on $K$ of rational functions with finite poles in $A$ and polynomial terms. This is a closed algebra. DefineFor $b\in E$ and $|c-b|<\mathrm{dist}(b,K)$, the uniformly convergent geometric serieslies in $\mathcal{A}$, so $E$ is relatively open. If $b_n\in E$ tends to $b\notin K$, the corresponding reciprocals converge uniformly on $K$, so $E$ is also relatively closed. Thus containing one point of a complementary component means containing the entire component. Each bounded component has its prescribed point in $A\subset E$. The unbounded component meets $E$ because, for $|b|>{\sup }_K|z|$,is uniformly approximated by polynomials. Therefore $E=\mathbb{C}\setminus K$. Approximate the finitely many reciprocal terms in a Riemann sum closely enough to control their combined error. This completes the proof of Runge theorem without assuming its conclusion during the pole-moving argument.

Now let $H$ denote the geometric filled set displayed above. For $w\in K$ the required inequality is automatic. If $w$ lies in a filled component $U$, then $f$ is a holomorphic function near its compact closure and $\partial U\subset K$. The maximum modulus principle gives $|f(w)|\le {\sup }_K|f|$. Hence $H\subseteq {\hat{K}}_{\Omega }$.

Conversely, suppose $w\in \Omega \setminus H$. Its component $V$ of $\mathbb{C}\setminus K$ meets $\mathbb{C}\setminus \Omega$, or is unbounded. To see the alternative, a bounded component contained in $\Omega$ has boundary in $K$, so its closure is compact in $\Omega$ and would have been filled. Apply the same open-and-closed reciprocal argument to the algebra generated by polynomials and reciprocals with poles outside $\Omega$. The component $V$ has a starting reciprocal either at a point outside $\Omega$ or near infinity. Thus there is $h\in \mathcal{O}(\Omega )$ approximating $(z-w{)}^{-1}$ as closely as desired on $K$. Let $M={\sup }_K|z-w|>0$ and choose the error less than $1/(2M)$. ThenThis holomorphic function separates $w$ from the analytic hull. Thus $w\notin {\hat{K}}_{\Omega }$, proving the equivalence with the geometric definition as well. For empty $K$, the filled hull is empty and the constant function one excludes every point when the supremum of the nonnegative modulus over the empty set is taken as zero; the nonempty case is the one used above.

Write $Y={\mathbb{P}}^1\setminus \{0,1,\infty \}$. One concrete uniformization uses the image in ${\mathrm{PSL}}_2(\mathbb{R})$ of the level-two principal congruence subgroupIt acts on the complex upper half-plane by Möbius transformations. Integer matrix entries make the group discrete, and the action is properly discontinuous. There are no nonidentity elliptic stabilizers: an elliptic element would have integer even trace with absolute value less than two, hence trace zero. But then $d=-a$ gives $bc=-(a^2+1)\equiv 2(\mathrm{mod}4)$, contradicting the evenness of $b,c$. Thus the quotient is an ordinary Riemann surface, with no orbifold points.

A fundamental region is the ideal quadrilateralIts vertical sides are identified by $T(\tau )=\tau +2$, and its semicircular sides by $U(\tau )=\tau /(1-2\tau )$. The ideal vertices are $-1,0,1,\infty$; $-1$ and $1$ represent one puncture, while zero and infinity give two others. Filling these three punctures gives genus zero: the identified quadrilateral has three vertices, two paired edges and one face, so its compactification has Euler characteristic $3-2+1=2$. Its complex compactification is a sphere, and a Möbius transformation puts the punctures at $0,1,\infty$. HenceThis is the thrice-punctured sphere as a modular quotient. Equivalently, the uniformization theorem applied to its universal covering map produces the upper-half-plane model: the sphere cannot cover this noncompact surface, and an entire covering from $\mathbb{C}$ would contradict Little Picard theorem by omitting both zero and one.

The curvature-minus-one metric $d{s}_{\mathbb{H}}=|d\tau |/\mathrm{Im}\tau$ is invariant under every real Möbius transformation preserving $\mathbb{H}$. Indeed, for determinant-one coefficients,Thus it descends to the hyperbolic metric on $Y$. This defines its distance by the infimum of path lengths; a choice of local lift does not affect lengths.

For the Schwarz-Pick lemma, first prove the Schwarz lemma. If $h:\mathbb{D}\to \mathbb{D}$ is a holomorphic function and $h(0)=0$, the function $h(z)/z$ extends as a holomorphic function across zero. On $|z|=r$ its modulus is at most $1/r$. The maximum modulus principle and $r\uparrow 1$ yield $|h(z)|\le |z|$ and $|h^{\prime }(0)|\le 1$. Equality at a nonzero point or in the derivative makes $h(z)/z$ a constant of modulus one, again by the maximum modulus principle, so $h$ is a rotation.

For a general holomorphic function $F:\mathbb{D}\to \mathbb{D}$, conjugate by the automorphisms of the unit disk $A_a(z)=(z-a)/(1-\overline{a}z)$, putting $a$ and $F(a)$ at zero. Applying the proved Schwarz lemma givesThe disk hyperbolic metric is $2|dz|/(1-|z{|}^2)$, and its distance is twice the inverse hyperbolic tangent of the displayed pseudohyperbolic distance. Both forms of Schwarz-Pick lemma therefore express metric contraction. Equality in the derivative at one point makes $F$ a disk automorphism by the equality case already proved.

For $X=\mathbb{D}\setminus \{0\}$ and a holomorphic function $f:X\to Y$, compose with a universal covering map of $X$ and lift to a universal covering map of $Y$. The simply connected source of the lift is a disk or half-plane. Schwarz-Pick lemma there bounds the lifted metric derivative, and the covering maps are local isometries, soThe second inequality follows by applying the length inequality to every joining path and taking the infimum. The hyperbolic metric on the punctured disk is its own intrinsic metric,obtained from the covering $z=e^{i\tau }$. It is not merely the ordinary disk metric restricted to the punctured disk.

Yes, equality can occur for distinct points. The cyclic subgroup $⟨T⟩$ lies in $\Gamma (2)$. Its quotient is conformally the punctured disk, using $z=e^{\pi i\tau }$, because this identifies precisely $\tau \sim \tau +2$. The subgroup inclusion induces a covering map that is a holomorphic functionIt is a local isometry. Choose a sufficiently small geodesically convex ball around any target point that is evenly covered. Two distinct sufficiently close points in one lift of that ball are joined by the lift of its minimizing target geodesic. The lift has the same length, giving $d_X\le d_Y$ for this pair, while contraction gives the reverse inequality. Thus equality holds.

For context, if the lifted map is not an automorphism, the derivative inequality is strict at every point; integrating along a minimizing geodesic in the complete punctured-disk metric gives a strict distance inequality for every distinct pair. Equality forces the covering-map case, but even a covering map need not preserve all distances globally because shorter paths can pass through other sheets. This is the distinction captured by equality in hyperbolic Schwarz-Pick contraction.

The zero-prescription claim requires a closed discrete subset, meaning no accumulation in $\Omega$. If “discrete” means only that each listed point is isolated, it is false: $\{1/n:n\ge 1\}\subset \mathbb{C}$ is a discrete subset, but any holomorphic function vanishing there is identically zero by the identity theorem for holomorphic functions, and then has additional zeros. In the usual zero-set interpretation, discreteness includes local finiteness. We prove the claim with this necessary interpretation.

First choose an increasing compact exhaustion $K_j$ of $\Omega$ by relatively filled sets: every component of $\hat{\mathbb{C}}\setminus K_j$ meets $\hat{\mathbb{C}}\setminus \Omega$. Such an exhaustion is obtained by exhausting with finite unions of small closed squares lying in $\Omega$, then adding all complementary components relatively compact in $\Omega$. The filled sets remain compact in $\Omega$: a complementary component containing a point within distance $\mathrm{dist}(K_j,\mathbb{C}\setminus \Omega )$ of the complement could join that complement through a ball disjoint from $K_j$, and hence would not be filled. Their boundedness follows by using a disk containing the initial compact. Nesting can be ensured by enlarging each stage first. This is also the geometric description of the holomorphic convex hull proved below.

A finite prescribed set is realized by a finite product of linear factors. For an infinite closed discrete subset, enumerate the distinct points as $a_n$. Every compact contains only finitely many of them, so there are indices $j(n)\to \infty$ with $a_n\notin K_{j(n)}$. Finitely many initial factors can be handled without an approximation requirement. For the remaining factors, let $U_n$ be the complementary component containing $a_n$. It also contains a point $c_n\notin \Omega$ or infinity. Join $a_n$ to $c_n$ by an arc in that component, or to infinity by a ray-like arc. DefineEach $p_n$ is a holomorphic function on $\Omega$, has exactly one simple zero, at $a_n$, and no other zeros there.

The arc is disjoint from $K_{j(n)}$, and $p_n$ has a holomorphic logarithm $L_n$ near that compact. To justify the logarithm, any closed curve avoiding the arc has equal winding numbers about its finite endpoints, so the integral of $p_n^{\prime }/p_n$ around it is zero. For a ray to infinity the winding number about $a_n$ is zero. A primitive of $p_n^{\prime }/p_n$, adjusted by a constant on each component, supplies the logarithm.

The Runge theorem proved in Question 4 approximates $L_n$ uniformly on $K_{j(n)}$ by $h_n\in \mathcal{O}(\Omega )$: all needed finite poles can be chosen outside $\Omega$ because every complementary component meets its complement. Choose the error small enough thatEach $F_n$ still has exactly the simple zero $a_n$. On every fixed compact, all sufficiently late factors satisfy the summable bound above. Hence their infinite product converges locally uniformly. It has a nonzero tail because, once $|F_n-1|<1/2$, its logarithmic tails satisfy $|\log F_n|\le 2|F_n-1|$ and converge absolutely. Thus multiplying by the finitely many earlier factors gives a holomorphic function with precisely the prescribed zeros, and no additional zeros. Repeating a factor to any prescribed finite positive multiplicity gives the corresponding multiplicity version. This is prescribed zeros in a plane domain via Runge approximation, not an invocation of the existence theorem being asked for.

For the specified infinite product, put $a_k=q^{-2k-1}$. On any compact annulus $0<r\le |z|\le R$,The logarithmic-tail estimate therefore proves locally uniform convergence on ${\mathbb{C}}^{\ast }$ and nonvanishing wherever no individual factor vanishes. At a zero of one factor, the product is zero and the remaining product converges to a nonzero value. ConsequentlyAll these zeros are simple and distinct: their moduli distinguish the two strings and their indices. At $z=0$ the reciprocal factors are undefined, so the given expression has no convergence value there. The infinite sequence of zeros tending to zero also rules out continuation across that point as a holomorphic function.

Set $P(z)={\prod }_{k\ge 1}(1-q^{-2k}z)(1-q^{-2k}{z}^{-1})$. Absolute local convergence allows reindexing:Since $1-z=-z(1-z^{-1})$, $\sigma (qz)=-z\sigma (q^{-1}z)$ for every $z\ne 0$. No division by a vanishing factor is involved, so the identity also holds at all the zeros.

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.