Subharmonic function

A twice differentiable function is strictly subharmonic where $\Delta u>0$. It cannot have an interior local maximum because the Hessian at such a maximum is negative semidefinite.

For every holomorphic function and $p>0$, $|f{|}^p$ is subharmonic. Away from its zeros the displayed Laplacian calculation proves this. At the zeros, use the smooth approximations $(|f{|}^2+\epsilon {)}^{p/2}$, whose Laplacians are nonnegative, and let $\epsilon \downarrow 0$ in their mean inequalities. This permits Poisson integral comparison for fractional modulus powers even when a global analytic fractional power does not exist.

Write a holomorphic function as $f=u+iv$. The Cauchy-Riemann equations make $u,v$ harmonic functions and give $u_x=v_y$, $u_y=-v_x$. The product rule therefore gives $\Delta (u^2+v^2)=2(|\nabla u{|}^2+|\nabla v{|}^2)=4|f^{\prime }{|}^2$. Thus its squared modulus is subharmonic and is harmonic only where $f^{\prime }$ vanishes.

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.

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 $V$.

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 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 $u\in C^2(D)\cap C(\overline{D})$ on a bounded domain in ${\mathbb{R}}^n$ with $\Delta u\le 0$, the minimum lies on the boundary. Define $u_{\epsilon }=u-\epsilon |x{|}^2$, so $\Delta u_{\epsilon }\le -2n\epsilon <0$. An interior minimum would have positive semidefinite Hessian matrix and nonnegative trace, a contradiction. Thus $u_{\epsilon }$ attains its minimum on the boundary. Boundedness controls $|u-u_{\epsilon }|$ uniformly by a constant times $\epsilon$; letting $\epsilon \downarrow 0$ proves the weak minimum principle for the original superharmonic function. The perturbation handles the non-strict inequality, including constant functions.

For locally integrable real $u$ with nonnegative distributional derivative $\Delta u$, its ball means increase with radius. To see this, mollify: for a smooth subharmonic function the derivative of its spherical mean is the integral of $\Delta u$ over the ball divided by the sphere area. The ball means inherit monotonicity; local $L^1$ convergence passes this to $u$. Their decreasing-radius limit exists, may be minus infinity at exceptional points, and equals $u$ almost everywhere by the Lebesgue differentiation theorem. It is upper semicontinuous as an infimum of continuous ball means and satisfies the mean-value inequality. Arbitrary changes of a representative on a null set can destroy its pointwise maximum principle.

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.

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.

For a harmonic function $w$ smooth inside a bounded domain and $C^1$ on its closure, $\Delta |Dw{|}^2=2|D^{2}w{|}^2\ge 0$. The weak maximum principle for elliptic operators and continuity of the gradient imply ${\sup }_{\Omega }|Dw|={\sup }_{\partial \Omega }|Dw|$.