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On a relatively compact coordinate disk $V$ in a Riemann surface, replace a continuous subharmonic function $u$ inside $V$ by the Poisson integral of its boundary values, leaving it unchanged outside. The resulting function $L_{V}u$ is continuous and subharmonic, equals $u$ off $V$, is harmonic on $V$, and satisfies $L_{V}u\ge u$. The maximum principle for subharmonic functions proves the inequality. The pasting assertion follows from harmonic comparison: a harmonic comparison function dominating the pasted function on the boundary of a test domain first dominates $u$, then dominates the disk lift on its part inside $V$.

A harmonic majorant of a real function $u$ is a real harmonic function $h$ with $h\ge u$ pointwise. A least harmonic majorant lies below every other harmonic majorant. A continuous subharmonic function need not have one: $u(z)=(1-|z{|}^2{)}^{-1}$ on the unit disk has increasing circular means diverging to infinity, whereas any harmonic majorant would have the same finite mean on every centered circle.

For continuous subharmonic function $u$ on the unit disk, let $H_r$ be the Poisson integral of its values on $|z|=r$, evaluated inside that disk. Harmonic comparison gives $H_s\ge H_r\ge u$ for $|z|<r<s<1$. The Harnack inequality for harmonic functions makes their limit either finite and harmonic everywhere or everywhere infinite. In the finite case every harmonic majorant dominates every $H_r$, so their limit is the least harmonic majorant. The supremum must restrict $r>|z|$; the Poisson expression evaluated outside its circle does not have this interpretation.

For a countable family of subharmonic functions, a locally integrable pointwise supremum is subharmonic in the distributional derivative sense. Finite maxima preserve subharmonicity: approximate the maximum by the smooth convex coordinatewise increasing function $(s+t+\sqrt{(s-t{)}^2+{\delta }^2})/2$, apply the chain rule after mollification, then pass to a local L1 norm convergence limit. The maxima $g_N=\max (f_1,\dots ,f_N)$ increase to $g$ and are bounded in absolute value by $|f_1|+|g|$ on compact sets, so the dominated convergence theorem completes the proof. The hypothesis cannot be omitted: $f_j=j$ has infinite supremum. To obtain a pointwise subharmonic function, take the canonical representative of a subharmonic distribution.

A continuous subharmonic function on the closure of a bounded domain attains its maximum on the boundary. If it is twice differentiable, the strict condition $\Delta u>0$ rules out an interior maximum immediately because the Hessian matrix there would be negative semidefinite.

Use the Nevanlinna proximity function $m(r,f)=(2\pi {)}^{-1}\int {\log }^{+}|f(r{e}^{i\theta })|d\theta$ and the Nevanlinna integrated counting function $N(r,f)$ for poles, counting multiplicity. If the pole order at zero is ${\nu }_0$, thenBoundedness of the Nevanlinna characteristic refers to $r\uparrow 1$, equivalently to a uniform bound on $r_0\le r<1$ for any fixed $r_0>0$. This avoids the harmless origin normalization when ${\nu }_0>0$.

The Nevanlinna first main theorem states, for every fixed finite value $a$ and every nonconstant meromorphic function on the disk,For $a=\infty$, this is the definition of $T$. The error is bounded independently of $r$. Translation changes $T$ by $O(1)$, while Jensen's formula gives $T(r,h)-T(r,1/h)=\log |c|$ when $h(z)=c{z}^{\nu }(1+O(z))$ near zero; these observations explain the first theorem.

The Nevanlinna second main theorem in the unit disc states that for $q\ge 3$ distinct values $a_j$ on the Riemann sphere,as $r\uparrow 1$ outside an exceptional set $E$ satisfying ${\int }_{E}dr/(1-r)<\infty$. The truncated Nevanlinna counting function $\overline{N}$ counts each distinct preimage once. The boundary-distance error cannot simply be suppressed: on the disk it need not be $o(T)$. The disk error and exceptional-set estimates follow from the logarithmic-derivative estimates in mathweb.tifr.res.in/Documents/Publications/Lectures/tifr17.pdf .

Suppose first that $T(r,f)$ is bounded. Since $m(r,f)\ge 0$, its poles satisfy the Blaschke condition, counting their orders: let $r\uparrow 1$ in their integrated count and use $1-|b|\le -\log |b|$. Let $B$ be their Blaschke product, including the origin factor if necessary. Then $g=Bf$ extends to a holomorphic function on the whole disk, and $|B|\le 1$ givesHere the last bound follows from bounded $T$ and the uniform lower bound $N(r,f)\ge {\nu }_0\log r_0$ on $r\ge r_0$.

The nonnegative subharmonic function ${\log }^{+}|g|$ has a finite harmonic majorant. Indeed its Poisson integrals $H_R$ on expanding disks dominate it by the maximum principle for subharmonic functions, increase with $R$, and have $H_R(0)\le C$. The Harnack inequality for harmonic functions bounds them on every compact subdisk, so their increasing limit is a finite harmonic function $H\ge {\log }^{+}|g|$. Since the disk is simply connected, $H$ has a harmonic conjugate, giving a holomorphic function $A$ with $\mathrm{Re}A=H$. ConsequentlyThis proves the constructive half of the quotient characterization of bounded characteristic.

Conversely, if $f=u/v$ with $u,v$ bounded holomorphic functions and $v\not\equiv 0$, each pole of $f$ is a zero of $v$ of at least the same order. The elementary inequality ${\log }^{+}|u/v|\le {\log }^{+}|u|+{\log }^{+}|1/v|$ and the Nevanlinna first main theorem giveAny cancellation at an origin zero contributes only a bounded multiple of $\log r$ on $r\ge r_0$, which is included in this $O(1)$. The zero function is immediate; constant nonzero $v$ is handled directly. Thus bounded characteristic is exactly a quotient of two bounded analytic functions.

An unbounded example with bounded characteristic isIt is a ratio of the bounded holomorphic functions $1$ and $1-z$. More explicitly, Jensen's formula gives the mean of $\log |1-r{e}^{i\theta }|$ as zero, so $m(r,1/(1-z))=m(r,1-z)\le \log 2$, although $g(r)\to \infty$.

Finally, set $Y=(1+|g{|}^2{)}^{p/2}\ge 1$. Pointwise ${\log }^{+}|g|\le p^{-1}\log Y$. Jensen's inequality for the concave logarithm givesThere is no pole term because $g$ is holomorphic. Hence the stated integral bound implies bounded characteristic. This argument in fact works for every $p>0$.

Use the absolute value around the logarithm in the original PDF. Set $v(z)={\log }^{+}|f(z)|=\max (0,\log |f(z)|)$. This is a nonnegative subharmonic function, including at zeros by its continuous extension. Its radial means are nondecreasing: Poisson integral comparison on a larger concentric circle bounds the smaller mean by the larger one. The hypothesis givesConsequently $T(r)=0$ for every $r$, and continuity of $v$ implies $v=0$ throughout the unit disc. Thus $|f|\le 1$ everywhere, which is stronger than the requested conclusion. In particular $f$ has bounded characteristic, because ${\sup }_{r}T(r)<\infty$; it belongs to the Nevanlinna class and to H-infinity space. The hypothesis also excludes the identically zero function, whose absolute logarithmic integral is infinite.

If $\varphi$ is identically zero there are no exceptions. Otherwise choose $x_0$ with $\varphi (x_0)>0$ and put $I_{+}=\{x\ge x_0:{\varphi }^{\prime }(x)>\varphi (x{)}^2\}$. Monotonicity implies $\varphi (x)>0$ thereafter. On this set, $1<{\varphi }^{\prime }/{\varphi }^2$, whenceAdding the finite initial interval $[0,x_0]$ provesThis derivative growth outside a finite-measure set controls exceptional radii in Nevanlinna estimates. For example let $T_s$ be the smooth spherical characteristic and $A(r)=r{T}_s^{\prime }(r)$ its increasing area-counting function. Apply the lemma to $T_s+1$ and $A+1$: outside the union of two finite-measure sets,Their logarithms are consequently $O(\log r+\log (T_s+1))$. Such derivative/area bounds in estimates from the Poisson-Jensen formula give the Nevanlinna logarithmic derivative lemma, $m(r,f^{\prime }/f)=O({\log }^{+}{T}_f+\log r)$ outside a finite-length set. The second main theorem uses that estimate as its error term. The elementary growth lemma controls that error and its exceptional set; it is not itself a substitute for the analytic estimates.

We use the truncated second main theorem stated in the preceding solution, together with the fact that a transcendental meromorphic function has $T_f(r)/\log r\to \infty$. Here is a justification of the latter fact. If there are infinitely many poles, fixing any arbitrarily large finite pole count gives $N(r;\infty )\ge A\log r-O_A(1)$. If there are only finitely many poles, multiply by a polynomial $p$ removing them to obtain a transcendental entire function $h=pf$. Its Taylor series has nonzero coefficients of arbitrarily high degrees $n$. Cauchy's coefficient estimate gives $\log M_h(r/2)\ge n\log r-O_n(1)$; the subharmonic function Poisson estimate gives $\log M_h(r/2)\le 3m(r,h)$. Since $T_h\le T_f+O(\log r)$, letting $n$ be arbitrarily large proves the claimed ratio limit. Thus the second-theorem error is $o(T_f)$ along large nonexceptional radii.

Let $N_{\mathrm{simple}}(r;a)$ count just the simple $a$-points. A multiple $a$-point of degree $m\ge 2$ contributes one to the truncated count and at least two to the full count; a simple point contributes one to each. Hence, with the same positive logarithmic weights for large radii,If each of the five values had only finitely many simple preimages, each simple counting function would be $O(\log r)$. Summing and using the first main theorem would give ${\sum }_{j=1}^5\overline{N}(r;a_j)\le (5/2)T_f+O(\log r)$. But the second theorem requires $3T_f\le \sum \overline{N}+o(T_f)$, a contradiction. This proves the stronger result five values force infinitely many simple preimages:In particular more than one simple zero is necessary, but more than one target with simple zeros is not necessary.

For the last distinction, take a nonsingular Weierstrass elliptic function scaled so that ${\wp }^{\prime 2}=4\wp (\wp -1)(\wp +1)$; it has branch values $-1,0,1,\infty$. These are the four totally ramified Weierstrass values: finite branch preimages have local degree two, and all poles are double. SetThe first four targets are the images of those branch values under the Möbius transformation. Their preimages remain double; at a lattice pole, for example, $f(z)\sim (z-z_{\ast }{)}^2$. The last target corresponds to $\wp =3$, where ${\wp }^{\prime 2}=96\ne 0$, so every such preimage is simple. The degree-two torus map attains that regular value, and periodicity produces infinitely many preimages in the plane. This nonconstant elliptic function is transcendental because it has infinitely many poles. Exactly one of the five selected finite targets therefore has simple preimages.

On a connected Riemann surface $R$, a Perron family of subharmonic functions is a nonempty family $\mathcal{P}$ of real continuous subharmonic functions closed under pairwise maxima and under harmonic lifting on every relatively compact coordinate disk. A harmonic lifting of a continuous subharmonic function replaces its values inside such a disk by the harmonic Poisson integral of its boundary values, leaving the function unchanged outside. Harmonic comparison shows that the lift dominates the original function. The lift is continuous and remains subharmonic by harmonic comparison across the circle.

Write $U={\sup }_{v\in \mathcal{P}}v$. Because the family is nonempty and its members are finite, $U$ cannot be minus infinity. Suppose first $U(p)<\infty$, and choose a relatively compact coordinate disk $V$ containing $p$. Choose $v_n\in \mathcal{P}$ approaching $U(p)$ at $p$. Lift $v_1$ on $V$ to get $h_1$, and recursively lift $\max (h_{n-1},v_n)$ to get $h_n$. These are members of the family, harmonic on $V$, and increasing. Also $h_n(p)\to U(p)$.

The nonnegative harmonic functions $h_n-h_1$ are bounded at $p$. The Harnack inequality for harmonic functions, on chains of compact subdisks, bounds them uniformly on every compact subset of $V$. The Poisson integral then bounds their derivatives on smaller subdisks. Compact convergence and the mean value property show that their increasing limit is finite and harmonic; denote the limit of $h_n$ by $h$.

Fix an arbitrary $v\in \mathcal{P}$, and let $k_n$ be the lift of $\max (v,h_n)$ on $V$. These functions increase, are harmonic on $V$, and satisfy $h_n\le k_n\le U$. They are bounded at $p$, so the same Harnack inequality for harmonic functions argument gives a harmonic limit $k$. At $p$, $k(p)=h(p)=U(p)$, whereas $k\ge h$ on $V$. The nonnegative harmonic difference has an interior zero, so the strong maximum principle for harmonic functions gives $k=h$. Since $v\le k_n$, we have $v\le h$. Taking the supremum over $v$ gives $U\le h$, and $h_n\le U$ gives the opposite inequality. Thus $U=h$ on $V$.

The finite locus of $U$ is open by this argument. It is also closed: if a point is a limit of finite points, choose a coordinate disk around it and a finite point in that disk; the preceding argument applied with that finite point shows that $U$ is finite on the entire disk. Connectedness now proves the dichotomyBoth occur. The family of all continuous subharmonic functions bounded above by zero is a Perron family of subharmonic functions, since the maximum principle for harmonic functions keeps every lift below zero; it contains zero, so its supremum is zero. The family of constant functions with positive integer values is also a Perron family of subharmonic functions, and its supremum is everywhere infinite.

For the final application, the radius must enclose the evaluation point. For $|z|<r<1$ setThis is the harmonic lifting of a continuous subharmonic function on the radius-$r$ disk. Hence $H_r\ge u$ there. If $|z|<r<s<1$, $H_s\ge u=H_r$ on the radius-$r$ boundary, so harmonic comparison gives $H_s\ge H_r$ throughout the smaller disk. The increasing limit is either everywhere infinite or finite and harmonic, by the preceding Harnack argument and propagation across overlapping disks. In the finite case it dominates $u$. Any harmonic majorant $h$ of $u$ dominates $H_r$ by boundary comparison and therefore dominates their limit. Thus the precise conclusion isprovided a finite harmonic majorant exists. This is the least harmonic majorant by expanding disk lifts.

There are two necessary qualifications to the printed formula. First, the supremum cannot include $r<|z|$. For example, if $u\equiv -1$ and $z\ne 0$, the kernel has normalized integral $1$ for $r>|z|$ and $-1$ for $r<|z|$. The displayed integral is consequently $-1$ in the first case and $+1$ in the second. The unrestricted supremum would be $+1$, although the least harmonic majorant is $-1$; at $r=|z|$ the kernel is undefined. Second, a finite harmonic majorant need not exist. The continuous subharmonic functionhas $H_r\equiv (1-r^2{)}^{-1}\to +\infty$. If $h\ge u$ were harmonic, its centered circular means would equal $h(0)$ and dominate $(1-r^2{)}^{-1}$ for every $r$, an impossibility. In this case the limit formula holds only in the extended sense, with value $+\infty$ and no finite least harmonic majorant.

For a $C^2$ function, let $M(r)=(4\pi {)}^{-1}{\int }_{S^2}u(y+r\omega )d\omega$ be its spherical mean. Differentiation and the divergence theorem giveIf $u$ is harmonic, this derivative is zero; since $M(0)=u(y)$ by continuity, every spherical mean equals the value at the centre. Integrating $4\pi r^{2}M(r)$ over $0<r<R$ gives the same statement for the volume mean on a ball. This proves the mean value property for harmonic functions.

If $u$ is subharmonic, $\Delta u\ge 0$, so $M^{\prime }(r)\ge 0$. ThusThese are the corresponding mean value inequalities. A subharmonic function has the maximum principle on a ball: apply the second-derivative test to $u+ϵ|x-y{|}^2$, whose Laplacian is strictly positive and hence cannot have an interior maximum, then let $ϵ\downarrow 0$.

For $w=|\varphi {|}^2$, the given equation says $\Delta \varphi =iV\varphi$. Since $V$ is real,The subharmonic maximum principle gives $w(x)\le {\max }_{\partial B}w$ throughout the ball. Taking square roots yieldsNo sign condition on the real potential $V$ is needed.

For a real locally integrable function $f$, the distributional definition of a subharmonic function is $\Delta f\ge 0$: for every nonnegative $\phi \in C_c^{\infty }(\Omega )$,A superharmonic function has the opposite inequality, equivalently $-f$ is subharmonic. A harmonic function has $\Delta f=0$ distributionally, equivalently equality for every test function. These definitions concern almost-everywhere classes; the given upper semicontinuous representative supplies the pointwise version for subharmonic functions.

For a real locally integrable function, the definitions are distributional:Thus the Laplacian is respectively a positive distribution, a negative distribution, or zero. A superharmonic function is the negative of a subharmonic function; being both is equivalent to being harmonic. For a $C^2$ function these conditions are $\Delta f\ge 0$, $\Delta f\le 0$, and $\Delta f=0$ pointwise.

An $L_{\mathrm{loc}}^1$ equivalence class does not itself specify point values. For the maximum principle use the canonical representative of a subharmonic distributionThese normalized integrals are ball averages. To justify this construction, first mollify $f$: its smooth local regularizations satisfy $\Delta f_{\epsilon }\ge 0$. For any smooth subharmonic function, the spherical mean $m_x(r)$ obeysThis follows by differentiating the spherical integral and applying the divergence theorem. Its ball mean also increases with radius: its derivative is $n/r$ times the spherical mean minus the ball mean. Local $L^1$ convergence of the regularizations passes the monotonicity of the ball means to $f$. The decreasing-radius limit therefore exists, possibly as minus infinity at an exceptional point, and equals $f$ almost everywhere by the Lebesgue differentiation theorem. It is upper semicontinuous as the local infimum of continuous ball averages and satisfiesFor a superharmonic function use the corresponding lower semicontinuous representative and reverse the inequality.

For a distributionally harmonic function, the smooth regularizations are harmonic and have constant spherical means. A radial mollifier ${\rho }_{\delta }$ of integral one therefore reproduces them by averaging these means. At fixed interior radius $\delta$, pass to the local $L^1$ limit to obtain $f=f\ast {\rho }_{\delta }$ almost everywhere on a smaller region. The right side is smooth and has zero distributional Laplacian, giving the smooth harmonic representative. This proves the relevant Weyl lemma. Changing a value at a single point can spoil a pointwise maximum principle, so the representative convention is essential.

Because $\overline{\Omega }$ is compact, $\varphi$ attains a maximum. If a maximizer lay in $\Omega$, its Hessian matrix would be negative semidefinite, so its trace would satisfy $\Delta \varphi \le 0$, contradicting $\Delta \varphi >0$. Hence every strictly subharmonic function under the stated hypotheses attains a maximum on $\partial \Omega$.

Statement (i) is true. For $ϵ>0$, the functionhas $\Delta {\varphi }_{ϵ}=4ϵ>0$, soLetting $ϵ\downarrow 0$ givesThis is the maximum principle for harmonic functions.

Let $B⋐\Omega$ and let $h$ be the harmonic function whose boundary values are $\max (u_1,u_2)$. The harmonic extensions $(u_i{)}_B$ satisfy $(u_i{)}_B\le h$ by the maximum principle for harmonic functions, while subharmonicity gives $u_i\le (u_i{)}_B$ in $B$. Hence $\max (u_1,u_2)\le h$, proving that the maximum of two subharmonic functions is subharmonic.

A nonempty collection of real continuous subharmonic functions on a connected Riemann surface is a Perron family if it is closed under pairwise maxima and harmonic lifting of a continuous subharmonic function on every relatively compact coordinate disk. Its pointwise supremum is either everywhere infinite or a harmonic function on a Riemann surface. At a point where the supremum is finite, take increasing disk lifts approaching that value. The Harnack inequality for harmonic functions gives a finite harmonic limit on the disk. Lifting their maxima with any other member gives a larger harmonic limit with the same value at the chosen point, hence the same limit by the strong maximum principle for harmonic functions. The supremum is therefore harmonic on that disk; connectedness propagates finiteness to the whole surface.

Potential theory studies harmonic functions, subharmonic functions, and the Laplace operator, including harmonic measure, regular boundary points, and the Dirichlet problem. The Brownian representation of the Dirichlet problem connects these analytic objects with exit probability distributions.

A locally integrable real function is superharmonic in the distributional sense if its Laplacian is nonpositive on every nonnegative test function. Equivalently its negative is a subharmonic function. The canonical lower semicontinuous representative satisfies the reverse ball mean inequality. A function both subharmonic and superharmonic has zero distributional Laplacian and a smooth harmonic function representative by the Weyl lemma.

For a holomorphic function on the unit disc, this boundary condition characterizes Blaschke products, including unimodular constants. The nondecreasing radial means of the nonnegative subharmonic function ${\log }^{+}|f|$ must vanish, so $|f|\le 1$. In the Blaschke factorization of a bounded holomorphic function $f=Bg$, both logarithmic means tend to zero. The mean value property for harmonic functions applied to $\log |g|$ gives $|g(0)|=1$, and the maximum modulus principle makes $g$ unimodular constant. The converse follows from the radial logarithmic mean of a Blaschke product. Omitting the outer absolute value changes the condition: $e^z$ has zero mean logarithmic modulus, but is not a Blaschke product.