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Let ${\rho }_t$ be uniform measure on the circle of radius $e^{-t}$ around the origin in the unit disc. The process $X_t=(h,{\rho }_t)$ is the circle-average process. The Domain Markov property of the Gaussian free field and the mean value property for harmonic functions imply that its increments are independent and stationary.

A locally uniformly bounded sequence of harmonic functions with a pointwise limit converges uniformly on compact subsets, and its limit is harmonic. A radial mollifier of fixed local radius reproduces each harmonic function by the mean value property for harmonic functions. Differentiating that convolution gives uniform interior gradient bounds, hence equicontinuity. A finite covering of a compact set converts pointwise convergence to uniform convergence. Passing to the limit in the radial convolution identity proves smoothness; passing to the limit against the Laplacian of each test function proves harmonicity.

For a $C^2$ harmonic function $u$ and Brownian motion stopped before leaving its domain, Itô formula gives $du(B_t)=\nabla u(B_t)\cdot d{B}_t+\frac{1}{2}\Delta u(B_t)dt$. The drift vanishes. On a compact subdomain, the stopped process is a bounded martingale. Optional stopping theorem at an almost-surely finite exit time therefore gives the boundary mean formula. At the centre of a disk, rotational invariance of planar Brownian motion makes the exit law uniform on the boundary circle. This proves the mean value property for harmonic functions probabilistically.

If $u\ge 0$ is a harmonic function on a neighbourhood of $\overline{B_{4r}(y)}$, thenFor $x,z\in B_r(y)$, use $B_r(x)\subset B_{3r}(z)\subset B_{4r}(y)$ and the mean value property for harmonic functions. Nonnegativity lets us enlarge the integral, and the ratio of volumes is $3^n$. This compares values without requiring derivative bounds.

For a nonnegative harmonic function on the unit disc, the displayed inequalities are sharp. Apply the Poisson integral on the unit disk on radius $R>|z|$, bound its positive kernel between $(R-|z|)/(R+|z|)$ and its reciprocal, and use the mean value property for harmonic functions. Letting $R\uparrow 1$ proves the assertion. Positive Poisson kernels with one boundary pole give equality in the appropriate radial directions.

Divide $f$ by its finitely many interior zero factors, with each factor repeated according to its multiplicity of a zero. The resulting holomorphic function $g$ has no zeros of a function in the closed disk, so $\log |g|$ is a harmonic function and its circle average is $\log |g(0)|$ by the mean value property for harmonic functions. For each removed factor $z-a$, with $0<|a|<R$, the circle average of $\log |R{e}^{i\theta }-a|$ is $\log R$: factor out $R{e}^{i\theta }$ and apply the mean value property for harmonic functions to $\log |1-(a/R)w|$ on a neighbourhood of the closed unit disk. Reassembling the factors gives $\log (R/|a|)$ for each zero, proving Jensen's formula.

For a bounded domain whose boundary points are all regular boundary points, continuous Dirichlet boundary data $f$ have a unique solutionHere ${\tau }_D$ is the Brownian exit time. The Strong Markov property gives the mean value property for harmonic functions; a barrier for the Dirichlet problem gives continuity at the boundary. This is an existence theorem as well as the Brownian representation of the Dirichlet problem for an already known solution.

For a fixed finite target $a$, the displayed identity compares growth with proximity to and occurrences of $a$. If $g(z)=c{z}^{\nu }(1+O(z))$ near zero, where $\nu$ is its signed zero order, the exact reciprocal identity is $T(r,g)-T(r,1/g)=\log |c|$. It follows by factoring out zeros and poles on a disc, applying the mean value property for harmonic functions to the remaining nonvanishing factor, and using ${\log }^{+}u-{\log }^{+}(1/u)=\log u$. Translation changes $T$ by a bounded amount, proving the general assertion.

A harmonic function on ${\mathbb{R}}^2\setminus \{0\}$ bounded on one side is constant. Shift and change sign to make it nonnegative. Its circular mean has the form $a\log r+b$; positivity for every $r>0$ forces $a=0$. For $|x|=r$, the mean value property for harmonic functions bounds its value by the mean integral over $r/2<|z|<3r/2$, giving at most $8b$. A removable singularity for a bounded harmonic function fills the origin, and the harmonic Liouville theorem finishes the proof. The dimensional distinction matters: $|x{|}^{2-n}$ is a positive nonconstant harmonic function on punctured ${\mathbb{R}}^n$ for $n\ge 3$.

The Harnack inequality on the unit disk states that a nonnegative harmonic function satisfiesOne proof applies the Poisson integral on the unit disk on a circle of radius $R>|z|$: the Poisson kernel on the circle lies between $(R-|z|)/(R+|z|)$ and its reciprocal. The mean value property for harmonic functions identifies the mean boundary value with $u(0)$; let $R\uparrow 1$.

A harmonic conjugate on the simply connected unit disc makes $H=u+iv$ holomorphic. The Carathéodory derivative bound for the right half-plane gives $|H^{\prime }(0)|\le 2u(0)$. The Cauchy-Riemann equations give $|H^{\prime }|=|\nabla u|$, so $|\nabla u(0)|\le 2u(0)$. For a fixed $z$, use the disk automorphismThe chain rule and conformality multiply the gradient length by $|{\psi }^{\prime }(0)|$. Applying the estimate at zero to $u\circ \psi$ proves the sharp gradient bound for positive harmonic functions:Sharpness follows from $u(\zeta )=\mathrm{Re}[(1+\zeta )/(1-\zeta )]$, which has equality at zero, and then composing with ${\psi }^{-1}$. These extremizers are positive multiples of the Poisson kernel on the circle with its boundary pole fixed at one point.

Write a Blaschke product as $B(z)=\lambda z^m{\prod }_n{b}_{a_n}(z)$, where $a_n\ne 0$, $|\lambda |=1$, and the zeros are counted with multiplicity. With normalized angular measure $dm=d\theta /(2\pi )$, Jensen's formula for each Blaschke factor givesIndeed, the mean of $\log |a-r{e}^{i\theta }|$ is $\log \max (r,|a|)$, while $\log |1-\overline{a}r{e}^{i\theta }|$ has mean zero by the mean value property for harmonic functions. Each factor has modulus at most one. Applying the monotone convergence theorem to the negative logarithms of the finite products yields the radial logarithmic mean of a Blaschke product:The Blaschke condition implies ${\sum }_n-\log |a_n|<\infty$: only finitely many zeros have modulus below $1/2$, and $-\log t\le 2(1-t)$ for $1/2\le t<1$. Every summand tends to zero and is bounded in absolute value by $-\log |a_n|$. The dominated convergence theorem for this series, together with $m\log r\to 0$, therefore givesThis also covers finite products and constant unimodular products. Zeros on a circle create integrable logarithmic singularities and do not change the conclusion.

A harmonic function on $\Omega$ is a twice continuously differentiable function satisfying $\Delta u=0$ there. For complex-valued functions this means that both its real part and its imaginary part are harmonic functions. The mean value property for harmonic functions characterizes harmonicity: a continuous function is harmonic exactly when its value at each point equals its average over every sphere, or equivalently every ball, whose closed ball lies in $\Omega$. In the forward direction, the derivative of the spherical average vanishes by the divergence theorem. Conversely, radial smoothing leaves a continuous function with the mean value property unchanged locally; it is therefore smooth. Expanding its small-ball average gives $u(x)+r^2\Delta u(x)/(2(d+3))+o(r^2)$ in dimension $d+1$, and hence $\Delta u=0$.

Write $M=\Vert u{\Vert }_{h^1}$ and $v_{d+1}$ for the volume of the unit ball in ${\mathbb{R}}^{d+1}$. We first obtain a pointwise estimate directly from the mean value property for harmonic functions, before using any boundary representation. For the ball of radius $r=t/2$ centered at $(x,t)$,Thus every upward shift of $u$ is a bounded harmonic function, with a continuous bounded boundary value.

The required map is the Poisson integralThe Poisson kernel for the upper half-space is positive, has integral one, and is harmonic as a function of $(x,t)$. Its normalization follows by polar coordinates and $r=\tan \theta$, which reduces the radial integral to ${\int }_0^{\pi /2}{\sin }^{d-1}\theta d\theta$; direct differentiation gives ${\Delta }_{x,t}{P}_t=0$. Differentiation under the integral on compact subsets proves that $P\mu$ is a harmonic function. Tonelli theorem givesHere the total variation norm of a measure is the norm on finite signed or complex Borel measures.

The Poisson kernel for the upper half-space is an approximate identity: for $\psi \in C_0({\mathbb{R}}^d)$, $P_t\ast \psi \to \psi$ uniformly. Indeed, uniform continuity controls translations smaller than a fixed radius, and the kernel mass outside that radius tends to zero. ConsequentlyThe dual characterization of the total variation norm of a measure, as the supremum over $\psi \in C_0$ of supremum norm at most one, now gives the reverse inequality $\Vert \mu {\Vert }_{\mathrm{TV}}\le \Vert P\mu {\Vert }_{h^1}$. Thus $P$ is a linear isometry and is injective.

For surjectivity, fix $s>0$ and put $u_s(x)=u(x,s)$. Bounded harmonic uniqueness in the upper half-space givesTo justify this uniqueness, subtract the Poisson integral of the bounded continuous boundary value from the upward-shifted harmonic function. Their difference is bounded and harmonic and has zero continuous boundary values. Its odd reflection across the boundary is an entire bounded harmonic function; the reflection is harmonic across the flat boundary by harmonic odd reflection across a hyperplane. The Liouville theorem for harmonic functions makes it constant, and oddness makes it zero. The initial pointwise estimate supplies the boundedness needed here. To see the reflection locally, take a ball centered on the hyperplane and extend the boundary data on its upper hemisphere oddly to its lower hemisphere. The harmonic solution on this ball is odd and zero on its equatorial disk. The maximum principle for harmonic functions identifies it with the given difference on the upper half-ball, proving that the reflected function is harmonic across the boundary.

Choose $s_j\downarrow 0$. The measures $u_{s_j}dx$ have total variation norm of a measure at most $M$. By the Banach-Alaoglu theorem, and separability of $C_0({\mathbb{R}}^d)$, a subsequence converges in the weak-star topology to a finite measure $\mu$ with $\Vert \mu {\Vert }_{\mathrm{TV}}\le M$. For fixed $(x,t)$ and $s_j<t$,The kernels in this display belong to $C_0$ and converge uniformly to $P_t(x-\cdot )$. The uniform measure norm bound therefore allows passage to the limit, giving $u(x,t)=P\mu (x,t)$. This proves the Poisson representation of harmonic h1 by finite measures, including onto-ness.

Finally $P_t(x)\le c_d{t}^{-d}$, so the Poisson representation of harmonic h1 by finite measures improves the preliminary pointwise constant:The last occurrence of $h^1({\mathbb{R}}^{d+1})$ in the printed question is understood as the harmonic Hardy space of the upper half-space defined earlier in that question.

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.

Write ${\log }^{+}x=\max (0,\log x)$. The Nevanlinna proximity function and Nevanlinna integrated counting function areThe pole sum counts multiplicity; it is finite for each $r$ because $f(0)$ is finite and nonzero and poles of a meromorphic function are isolated. Equivalently $N(r,f)={\int }_0^{r}n(t,f)dt/t$, where $n(t,f)$ counts poles in the disc. The Nevanlinna characteristic isFor a general nonzero meromorphic function $g$ with signed order $\nu$ at zero, use $g(z)=c{z}^{\nu }(1+O(z))$, $c\ne 0$. The general counting convention includes a pole at zero:where $n(0,g)=\max (0,-\nu )$. This convention will also cover $g=f-a$ when the target $a=f(0)$.

Here is a direct proof of the exact reciprocal identity underlying the Nevanlinna first main theorem. Choose a circle with no zero or pole of $g$ on it, and a slightly larger disc with the same interior zeros and poles. Divide $g$ by $z^{\nu }$ and its nonzero zero factors and multiply by its nonzero pole factors, obtaining a holomorphic, nowhere-zero function $H$ on that disc. For any factor $z-a$ with $|a|<r$,Indeed, factor out $r{e}^{i\theta }$; the remaining logarithm has real part equal to that of $-{\sum }_{k\ge 1}(a/r{)}^k{e}^{-ik\theta }/k$, whose angular mean is zero. Also, $\log |H|$ is harmonic, so the mean value property for harmonic functions equates its circle mean to $\log |H(0)|$. Putting back the zero and pole factors provesThe identity ${\log }^{+}x-{\log }^{+}(1/x)=\log x$ now yieldsZeros or poles on the circle cause only integrable logarithmic singularities; the same identities hold at those radii by limits. A boundary zero or pole contributes $\log (r/r)=0$ to its counting term.

For a finite target $a$, set $g=f-a$. Its finite poles and their orders equal those of $f$. The elementary estimatesgive $|T(r,f-a)-T(r,f)|\le \log (1+|a|)$. Combining this with the reciprocal identity proves the first main theorem:The error is bounded independently of $r$, with an explicit bound $\log (1+|a|)+|\log |c_a||$, where $c_a$ is the first nonzero local coefficient of $f-a$ at zero. The count on the right records the occurrences of the value $a$ with multiplicity. The target infinity is the defining identity $T=m+N$. If $f$ is a constant equal to $a$, the reciprocal expression is undefined; the value-distribution assertion assumes $f-a\not\equiv 0$, as it automatically does for nonconstant $f$. The exact reciprocal identity remains valid for every nonzero constant $f$.

Taking $g=f$, whose signed order is zero and whose local coefficient is $f(0)$, rearranges that exact identity into Jensen's formula:Thus the zero and pole terms, their signs and the initial-value constant follow from the same proved identity, rather than from a theorem citation without proof.

Finally let $f$ be a nonzero bounded holomorphic function on the unit disc, with $|f|\le M$. If its order at zero is $m$, write $f(z)=z^{m}h(z)$, $h(0)\ne 0$. The factorization argument above works on every compact subdisc and gives, for its nonzero zeros,For $1/2\le r<1$ this is a uniform finite upper bound. As $r\uparrow 1$, monotone convergence givesSince $1-u\le -\log u$ for $0<u<1$, and there are only $m$ zeros at zero, we obtainThis is precisely the Blaschke condition, so the zeros, counted with multiplicity, form a Blaschke sequence.

The strong maximum principle for harmonic functions states that a $C^2$ harmonic function on a connected domain which attains its supremum at an interior point is constant. Applying it to $-u$ gives the corresponding assertion for an attained infimum.

Let $M=u(x_0)={\sup }_{\Omega }u$. The ball mean value property for harmonic functions gives, for every sufficiently small ball about $x_0$,The integrand is continuous and nonnegative. If it were positive at any point, it would be bounded below by a positive number on a smaller ball, contradicting the zero integral. Therefore $u=M$ on the whole ball. The set $A=\{x\in \Omega :u(x)=M\}$ is relatively closed by continuity and relatively open by this same argument at each of its points. It is nonempty. Connected space forces $A=\Omega$, proving the principle. If the open set is disconnected, constancy is asserted on the relevant connected component.

The mean value property for harmonic functions states that if $u$ is harmonic on a neighborhood of a closed ball, then its average on the ball or its boundary sphere equals its value at the center. To prove it in three dimensions, define $M(r)=(4\pi {)}^{-1}{\int }_{S^2}u(x+r\omega )d\omega$. Differentiation and the divergence theorem giveSince $M(r)\to u(x)$ as $r\to 0$, the spherical average equals $u(x)$. Integrating $4\pi r^{2}M(r)$ from zero to the ball radius gives the volume version,For balls merely contained in the open harmonicity domain, apply the result first to smaller radii and pass to the radius of interest by continuity.

A local form of the Harnack inequality for harmonic functions is: if $u\ge 0$ is harmonic on a neighbourhood of $\overline{B_{4r}(y)}$, thenFor $x,z\in B_r(y)$, the triangle inequality gives $B_r(x)\subset B_{3r}(z)\subset B_{4r}(y)$. The ball mean value property for harmonic functions and nonnegativity therefore implyTaking the supremum over $x$ and infimum over $z$ proves the inequality, including the case where the infimum is zero. No strict positivity assumption was used.

On a connected domain $\Omega$, the equivalent compact-set statement is that for each nonempty compact $K⋐\Omega$ there exists $C(K,\Omega ,n)$ withHere is the passage from balls to this statement. Cover $K$ by finitely many small balls $B_{r_j}(y_j)$ whose quadrupled closed balls lie in $\Omega$. Join their centres to a fixed centre by paths in $\Omega$. Each of the finitely many paths has compact image and positive distance from the boundary, so subdividing it gives a finite chain of overlapping small balls whose quadrupled closures also lie in $\Omega$. The local inequality compares any two points in each ball; using a point in each successive overlap propagates comparisons along the chain. The number of comparisons is bounded uniformly over this finite collection of chains and covering balls. Multiplying the factors $3^n$ gives one finite constant for all pairs $x,z\in K$, proving the compact-set inequality. In particular a nonnegative harmonic function vanishing at one interior point vanishes throughout a connected domain.

This is Weyl lemma, with the representative condition needed for pointwise smoothness. Choose a smooth radial nonnegative mollifier ${\rho }_{\epsilon }$ of mass one. On interior domains, $f_{\epsilon }=f\ast {\rho }_{\epsilon }$ is smooth and harmonic because $\Delta f_{\epsilon }=(\Delta f)\ast {\rho }_{\epsilon }=0$.

Fix a smaller interior neighborhood and a radial mollifier ${\rho }_r$ whose support stays inside the domain there. The mean value property for harmonic functions implies $f_{\epsilon }=f_{\epsilon }\ast {\rho }_r$ when $\epsilon$ is sufficiently small: a radial average is a weighted average of spherical means, all equal to the center value.

As $\epsilon \downarrow 0$, $f_{\epsilon }\to f$ in $L_{\mathrm{loc}}^1$, while $f_{\epsilon }\ast {\rho }_r\to f\ast {\rho }_r$ locally uniformly. The latter follows by bounding the difference by $\Vert {\rho }_r{\Vert }_{\infty }$ times the local $L^1$ error on a slightly larger compact set. Thus $f=f\ast {\rho }_r$ almost everywhere locally. The right side is smooth, and its Laplacian is zero, providing a smooth harmonic representative. These representatives agree on overlapping neighborhoods since continuous functions agreeing almost everywhere agree everywhere.

The given uniqueness of the canonical upper semicontinuous representative identifies this smooth function with $\tilde{f}$. Since the question assumes $f=\tilde{f}$, it follows that $\boxed{f\in C^{\infty }(\Omega )}$ pointwise. An arbitrary null-set modification of a harmonic distribution would not have that conclusion.