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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.

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.

The local Harnack inequality for harmonic functions states that if $u\ge 0$ is harmonic on a neighbourhood of $\overline{B_{4r}(a)}$, thenMore generally, for any compact $K$ in a connected open domain $\Omega$, there is $C=C(K,\Omega ,n)$ such that ${\sup }_{K}u\le C{\inf }_{K}u$ for every nonnegative harmonic $u$ on $\Omega$. Positivity is essential; the assertion is not for arbitrary sign-changing functions.

First prove the mean value property for harmonic functions. For a smooth harmonic function the spherical mean has derivativeso it equals $u(x)$; integration in the radius gives the ball mean property. The preceding Weyl lemma supplies smoothness if harmonicity was initially distributional.

For $x,z\in B_r(a)$ the inclusion $B_r(x)\subset B_{3r}(z)\subset B_{4r}(a)$ and nonnegativity giveTaking the supremum over $x$ and the infimum over $z$ proves the local estimate directly. To obtain the compact-set form, connect centres of a finite ball cover of $K$ to a fixed interior point by paths in $\Omega$. The compact union of those paths stays a positive distance from the complement of $\Omega$. Subdivide the paths into finitely many steps small enough that the local estimate applies at consecutive points. Multiply the finitely many comparison constants, first from one point of $K$ to the reference point and then to any other point of $K$. This gives a constant independent of $u$. In particular a nonnegative harmonic function is either everywhere positive or identically zero on each connected component.

The interior elliptic regularity assertion for the Laplace equation is that a harmonic function is smooth, in fact real analytic, throughout $U$. No boundary regularity of its unspecified boundary values is implied.

First let $u\in C^2(U)$ and $\Delta u=0$. On a ball compactly contained in $U$, differentiating its spherical average and applying the divergence theorem expresses that derivative as a constant factor times ${\int }_{B_r}\Delta u$, which is zero. The spherical average tends to $u$ at the center as $r\to 0$. This proves the mean value property for harmonic functions.

Choose a radially symmetric smooth mollifier ${\eta }_r$, supported in $B_r$, with integral one. By integrating the spherical mean value property,whenever $B_r(x)⋐U$. For fixed $r$ the right side is a smooth convolution, since all derivatives can be placed on ${\eta }_r$. Thus $u$ is smooth. The same argument proves the Weyl lemma for a distributionally harmonic $u\in L_{\mathrm{loc}}^1$: first mollify $u$, apply the fixed-radius identity, and let the mollification radius tend to zero in distributions to obtain the same smooth representative.

To prove real analytic regularity, differentiating the fixed-radius convolution gives the interior derivative estimate for a harmonic functionfor any harmonic $v$. All derivatives of $u$ are harmonic. On nested balls between $B_R(a)$ and $B_{R/2}(a)$, apply this estimate $k$ times, decreasing the radius by $R/(2k)$ each time. For $|\alpha |=k$,The inequality $k^k\le e^{k}k!$ follows by integrating $\log s$ below the sum defining $\log (k!)$. Apply the one-dimensional Taylor theorem along each segment, expanding directional derivatives by the multinomial formula. The remainder is bounded by $M(A{\sum }_j|h_j|{)}^k$, so it tends to zero for sufficiently small $h$. This gives a locally convergent multivariate Taylor series. A harmonic function is real analytic in the interior.

The usual inhomogeneous elliptic regularity statement also follows: if $\Delta u=f$ and $f$ is smooth, take a cutoff $\chi$ equal to one near a given point and set $w=\Phi \ast (\chi f)$, where $\Phi$ is a fundamental solution of the Laplace equation with $\Delta \Phi =\delta$. Moving every derivative to the compactly supported smooth function $\chi f$ shows $w$ is smooth. Locally $u-w$ is harmonic, so $u$ is smooth there too.

On a relatively compact ball, mollification commutes with the Laplacian, so $u_{\epsilon }=u\ast {\eta }_{\epsilon }$ is a smooth harmonic function. Repeated interior derivative estimates, together with the Caccioppoli inequality, bound every derivative of $u_{\epsilon }$ on a smaller ball by the local $L^2$ norm of $u$, uniformly as $\epsilon \downarrow 0$. The Arzelà-Ascoli theorem and a diagonal argument give a smooth local limit, while mollification gives $u_{\epsilon }\to u$ in $L_{\mathrm{loc}}^2$. Thus the limit equals $u$ almost everywhere. After choosing this smooth representative, $u$ is harmonic pointwise. This is the Weyl lemma for an $H^1$ weak solution.

For $u\in L_{\mathrm{loc}}^2(\Omega )$, the correct formulation is the distributional identityThus $\Delta u=0$ as a distribution. The Weyl lemma applies already to locally integrable distributions, so $u$ agrees almost everywhere with a smooth harmonic function.

For an open subdomain $U\subset D$, identify $H_0^1(U)$ with the closed subspace of $H_0^1(D)$ obtained by zero extension. Its orthogonal complement isIf $f\in {\mathcal{H}}_U$, then testing against $C_0^{\infty }(U)$ and integrating by parts gives $\Delta f=0$ in $U$ in the distributional derivative sense. The Weyl lemma therefore gives a representative harmonic on $U$.

The orthogonal decomposition by a closed subspace givesProject the isonormal process defining $h$ onto these two orthogonal subspaces. The projections are jointly Gaussian and uncorrelated, hence independent. The first projection is a zero-boundary Gaussian free field $h_U$ on $U$; the second is a random distribution $h^{\mathrm{har}}$ that is harmonic on $U$. Thuswith independent summands. This proves the Domain Markov property of the Gaussian free field.

Suppose $u_1$ and $u_2$ are weak solutions with the same trace, and put $w=u_1-u_2\in H_0^1(\Omega )$. The weak formulation permits $w$ itself as a test function, givingThus $w$ is almost everywhere constant, and its zero trace makes that constant zero. This proves uniqueness.

The weak identity also says that $\Delta u=0$ in the sense of distributions. The Weyl lemma therefore gives $u\in C^{\infty }(\Omega )$ and $\Delta u=0$ pointwise. The assumed continuity on $\overline{\Omega }$ retains the prescribed boundary values, so the weak solution is the unique classical solution in $C^{\infty }(\Omega )\cap C^0(\overline{\Omega })$.

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.

An element $h\in H_{\mathrm{loc}}^1(U)$ is weakly harmonic on an open set $U$ when ${\int }_U\nabla h\cdot \nabla \varphi =0$ for every test function $\varphi \in C_c^{\infty }(U)$. This is the statement $\Delta h=0$ in distributions. It requires no regularity of the domain boundary. The Weyl lemma, a form of elliptic regularity, identifies such an element with a smooth harmonic function.