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This logarithmic area integral measures the accumulated pullback hyperbolic metric area of a holomorphic disc map. Its nonnegative kernel makes it increasing in $R$. Schwarz-Pick theorem bounds $H_g(z)$ by $2/(1-|z{|}^2)$, which controls its characteristic through a critical-point Jensen identity.

Use the curvature-minus-one normalization of the hyperbolic metric:The associated distance is $d(z,w)=2\mathrm{artanh}|(z-w)/(1-\overline{w}z)|$. A connected Riemann surface has its complete canonical Poincare metric on a Riemann surface precisely when its universal cover is conformally the unit disc. By the uniformization theorem, the other simply connected covering types are the complex plane and Riemann sphere. Disc automorphisms preserve $\lambda (z)|dz|$, as direct differentiation of $(z-a)/(1-\overline{a}z)$ shows; therefore the metric descends through the deck group. This is the uniformization notion of hyperbolicity, not the different potential-theoretic Green-function classification.

First prove the needed Schwarz lemma. If $F:\mathbb{D}\to \mathbb{D}$ is holomorphic and $F(0)=0$, then $F(z)/z$ extends holomorphically at zero. On $|z|=r<1$ its modulus is at most $1/r$. The maximum modulus principle and $r\uparrow 1$ give $|F(z)|\le |z|$ and $|F^{\prime }(0)|\le 1$.

Write ${\psi }_a(z)=(z-a)/(1-\overline{a}z)$. Conjugating $f$ by the source and target disc automorphisms, ${\psi }_{f(a)}\circ f\circ {\psi }_a^{-1}$ fixes zero. Applying the derivative bound givesIntegrate along any piecewise smooth path and then take the infimum of lengths: $d(f(z),f(w))\le d(z,w)$. This proves the Schwarz-Pick theorem. Contraction means nonexpansion, not a uniform Lipschitz constant strictly smaller than one; automorphisms preserve all distances.

If both $R$ and $S$ have disc universal covers, lift $g\circ {\pi }_R$ to $\tilde{g}:\mathbb{D}\to \mathbb{D}$ through ${\pi }_S$. The disc is simply connected, so the lift exists, and it is holomorphic because the covering maps are local biholomorphisms. Both covering maps are local hyperbolic isometries. The disc contraction therefore descends to $g^{\ast }d{s}_S\le d{s}_R$ and, by the path-length argument, $d_S(g(p),g(q))\le d_R(p,q)$. If only $S$ is known to have a hyperbolic metric, a nonconstant $g$ forces $R$ to have one too: otherwise lift from the plane or sphere universal cover of $R$ to the disc cover of $S$. A bounded entire function is constant by the Liouville theorem, and a holomorphic map from the compact sphere into the disc is constant by the maximum modulus principle. Thus a nonconstant map into a hyperbolic surface is distance-nonincreasing between the canonical hyperbolic metrics; a nonhyperbolic source admits only constant such maps.

For the prescribed point, Schwarz-Pick theorem gives $|{\psi }_{w_0}(f(z))|\le |{\psi }_{z_0}(z)|$. Thereforeis holomorphic after filling the removable singularity at $z_0$, and $|h|\le 1$. Inverting the target automorphism provesConversely every holomorphic $h$ with this bound gives an admissible disc map: $|{\psi }_{z_0}(z)h(z)|<1$, its inverse target automorphism remains in the disc, and the denominator cannot vanish.

Fix $z$ and put $r=|{\psi }_{z_0}(z)|<1$. The possible intermediate values fill $|v|\le r$, since constant choices $h\equiv c$, $|c|\le 1$, attain them all. Thus the one-point value region of a holomorphic disc map is $|{\psi }_{w_0}(w)|\le r$. Completing the square in $|w-w_0{|}^2\le r^2|1-{\overline{w}}_{0}w{|}^2$ gives its Euclidean centre and radius:This closed disc lies strictly inside the unit disc because $r<1$. For $z=z_0$ it degenerates to the single point $w_0$. Its boundary is attainable by unimodular constant $h$; requiring $f$ to map into the open disc does not remove that boundary.

There are two necessary qualifications to the printed formula. Critical points satisfy $g^{\prime }=0$, not $g^{\prime }=1$. At a point with $g^{\prime }=1$, the local degree is one and the printed summand is zero. In addition, $\log \Vert g^{\prime }(0)\Vert$ is finite only when $g^{\prime }(0)\ne 0$; constant maps and origin-critical maps need separate treatment.

These are genuine issues even for otherwise ordinary disc maps. The map $g(z)=(z-1/2{)}^2/4$ maps the unit disc into itself and has $g^{\prime }(0)\ne 0$. Its derivative never equals one in the disc, so the printed count is zero, but it has a ramification point at $1/2$ of multiplicity one. For $R>1/2$ the correct identity has the additional term $\log (2R)$, proving that the displayed identity with the literal printed count is false. Also $g(z)=z^2$ shows the undefined basepoint logarithm in an origin-critical example.

Prove the intended identity first for a nonconstant map with $g^{\prime }(0)\ne 0$. Let $H(z)=2|g^{\prime }(z)|/(1-|g(z){|}^2)$ be its hyperbolic derivative density and $u=\log H$. Away from derivative zeros, $\log |g^{\prime }|$ is harmonic, and direct differentiation givesAt a zero $a$ of $g^{\prime }$ of multiplicity $m_a$, factor $g^{\prime }=(z-a{)}^{m_a}h$ with $h(a)\ne 0$. Its logarithm contains $m_a\log |z-a|$, whose distributional Laplacian is $2\pi m_a{\delta }_a$. ThusThe local map degree is $m_a+1$, so this is the required excess-degree weighting.

For a smooth function, differentiating its circular mean and applying the divergence theorem gives ${\overline{u}}^{\prime }(r)=(2\pi r{)}^{-1}{\int }_{|z|<r}\Delta udA$. Integrating in $r$ gives the logarithmic Green-Jensen formula. Factoring the isolated logarithmic singularities extends it to the present $u$:Substitution yields the critical-point Jensen identity for the hyperbolic derivativewhere $N_{\mathrm{crit}}={\sum }_{|a|<R:g^{\prime }(a)=0}{m}_a\log (R/|a|)$. This is the requested formula with the critical-point typo corrected. Initially choose a circle containing no ramification point on its boundary; continuity and integrability of logarithmic boundary singularities extend the result to the remaining radii.

By Schwarz-Pick theorem, $H(z)\le 2/(1-|z{|}^2)$. ThereforeThe normalized hyperbolic distance is $\rho (R)=\log ((1+R)/(1-R))$, andConsequentlyOne can take $C_1=1$ and $C_2=\log (2/H(0))\ge 0$. Since the corrected count is nonnegative in the noncritical-basepoint case, this also proves the growth bound for $T_{\mathbb{D}}$ with the literal printed, identically zero count, though not its false identity.

For completeness, if $g^{\prime }$ vanishes to order $m\ge 1$ at zero, put $c={\lim }_{z\to 0}H(z)/|z{|}^m>0$ andApply the same formula to $\log H-m\log |z|$, which is regular at zero. It givesThe origin contribution is a regularized Jensen term, not the undefined $\log (R/0)$. If one instead omits the origin term, the upper bound gains $-m\log R$; this is bounded for $R\ge 1/2$, while the characteristic and off-origin count are bounded on any smaller compact disc. Thus a bound $C_1\rho +C_2$ still holds for that nonnegative-count convention after enlarging $C_2$. For a constant map $H=0$ and $T_{\mathbb{D}}=0$, so its characteristic growth bound is trivial; the logarithmic identity and a finite critical-divisor count are not defined. These qualifications resolve all maps allowed by the original opening sentence without asserting an undefined equality.

A compact H-hull is a bounded relatively closed set $K\subset \mathbb{H}$ whose complement $D=\mathbb{H}\setminus K$ is a simply connected domain. Equivalently one can retain its compact closure in $\overline{\mathbb{H}}$; real boundary points do not affect the mapping-out function. The Riemann mapping theorem and Schwarz reflection principle near infinity give a unique conformal map $g_K:D\to \mathbb{H}$ with hydrodynamic normalization at infinity. Its Laurent series isThe half-plane capacity is $\mathrm{hcap}(K)=a_K$. The missing constant term is part of the normalization, not a choice to be made again when comparing different hulls.

Here are the conformal map facts we use to justify positivity and comparison. The inverse $f_K:\mathbb{H}\to D\subset \mathbb{H}$ has expansion $f_K(w)=w-a_K/w+O(w^{-2})$. The upper-half-plane Schwarz-Pick theorem, applied to $w$ and $iY$ and then letting $Y\to \infty$, implies $\mathrm{Im}{f}_K(w)\ge \mathrm{Im}w$. Indeed the inequality between pseudohyperbolic distances givesand multiplying by $Y$ gives the height comparison in the limit. At $w=iY$, the inverse expansion reads $\mathrm{Im}{f}_K(iY)=Y+a_K/Y+O(Y^{-2})$, so $a_K\ge 0$.

For $r>0$, uniqueness of the mapping-out function givesConsequentlyTranslation by a real number $b$ similarly gives $g_{K+b}(z)=b+g_K(z-b)$ and leaves half-plane capacity unchanged.

If $K\subset A$ are compact H-hulls, their mapping-out functions compose as $g_A=g_L\circ g_K$, where $L$ is the filled image of $A\setminus K$. Comparing Laurent series yieldsThis proves monotonicity of half-plane capacity, rather than assuming it from inclusion alone. The filled half-disc $A=\{z\in \mathbb{H}:|z-b|\le R\}$ has mapping-out functionwhich takes its semicircular boundary to the real line and is a conformal bijection onto $\mathbb{H}$. Thus its half-plane capacity is $R^2$. Enclose $K$ in such a half-disc, compare capacities, and minimize over real centres and radii. This proves the sharp boundIf one defines the radius by an infimum, take $R$ strictly larger than that infimum and then let $R$ decrease; no attainment assumption is needed.

A connected Riemann surface whose universal cover is the unit disc inherits the complete curvature-minus-one metric displayed on its cover. Möbius transformations preserving the disc are isometries, so deck transformations preserve the metric and it descends to the surface. This uniformization notion must be distinguished from the potential-theoretic existence of a Green function. Holomorphic maps between such surfaces contract their metric by lifting to Schwarz-Pick theorem on the covers.

On the unit disc, the pseudohyperbolic distance isIt is invariant under disc Möbius transformations. For the curvature-minus-one hyperbolic metric with length element $2|dz|/(1-|z{|}^2)$, $\delta (z,w)=\tanh (\rho (z,w)/2)$ and $e^{-\rho (z,w)}=(1-\delta )/(1+\delta )$. The Schwarz-Pick theorem says that holomorphic functions from the disc into itself contract this distance. A Blaschke factor has modulus $\delta (z,a)$, so the distance also controls the size of a product with prescribed zeros.