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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.

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.

If the canonical representative of a subharmonic distribution on a connected domain attains a finite global maximum at an interior point, it is constant. The ball mean inequality forces equality almost everywhere on a small ball around the maximum. The canonical representative then equals the maximum everywhere on that ball. The maximum set is open by this argument and closed by upper semicontinuity; connectedness makes it the whole domain.