The Mean-Value Property Characterizes Harmonic Functions

Derives Lemma A16.3. Fix \(\vect{x} \in U\) and choose \(r>0\) with \(\overline{B(\vect{x},r)} \subset U\). The set \(\overline{B(\vect{x},r)}\times Q\) is compact, so \(\pp F/\pp x^{i}\) is uniformly continuous on it: given \(\varepsilon>0\) there is \(\delta \in (0,r)\) such that

for every \(\vect{y} \in Q\), the bound being uniform in \(\vect{y}\) — which is the whole content of uniform continuity on the product and the reason the compactness of \(Q\) is needed.

Let \(\vect{e}_{i}\) be the \(i\)-th coordinate vector and let \(0 < \abs{h} < \delta\). For each fixed \(\vect{y}\) the mean value theorem applied to the function \(s \longmapsto F(\vect{x}+s\vect{e}_{i},\vect{y})\) on the interval between \(0\) and \(h\) supplies \(\theta = \theta(h,\vect{y}) \in (0,1)\) with

Both sides of Equation (A16.5) are continuous in \(\vect{y}\) — the left side manifestly, and hence the right side as well, whatever the measurability of \(\theta\) — so both are Riemann integrable over \(Q\). Subtracting Equation (A16.3) and using Equation (A16.4) with \(\abs{\theta h} < \delta\),

with \(\abs{Q}\) the volume of the box. Since \(\varepsilon\) was arbitrary the difference quotient converges, which is Equation (A16.3). Continuity of \(\pp_{i}g\) follows from the same estimate with \(\vect{x}'' = \vect{x}\) and \(\vect{x}'\) a nearby point, the integral of a quantity bounded by \(\varepsilon\) being bounded by \(\varepsilon\abs{Q}\).

Derives Lemma A16.5. Derivatives for \(t>0\). We claim that for every \(n \ge 0\) there is a polynomial \(p_{n}\) with

For \(n=0\) take \(p_{0} = 1\). Differentiating Equation (A16.8) and using \(\dd(1/t)/\dd t = -1/t^{2}\),

so Equation (A16.8) holds at \(n+1\) with \(p_{n+1}(s) = s^{2}\bigl(p_{n}(s) - p_{n}'(s)\bigr)\).

Decay at the origin. For \(s>0\) and any integer \(m \ge 0\) the exponential series gives \(\ee^{s} \ge s^{m+1}/(m+1)!\), hence \(s^{m}\ee^{-s} \le (m+1)!/s \longrightarrow 0\) as \(s \to +\infty\). Consequently, for any polynomial \(p\),

Derivatives at the origin. We show by induction that \(f^{(n)}\) exists everywhere, is continuous, and \(f^{(n)}(0)=0\). For \(n=0\) this is Equation (A16.10) with \(p=1\). Assume it at \(n\). For \(t<0\) the difference quotient of \(f^{(n)}\) at \(0\) vanishes identically; for \(t>0\) it is

which tends to \(0\) by Equation (A16.10) applied to the polynomial \(s\,p_{n}(s)\). Hence \(f^{(n+1)}(0)\) exists and is \(0\). It is continuous at \(0\) because \(f^{(n+1)}(t) = p_{n+1}(1/t)\ee^{-1/t} \to 0\) as \(t\to0^{+}\) by Equation (A16.10) again, and vanishes for \(t<0\). Positivity for \(t>0\) is immediate.

Derives Lemma A16.7. \(\abs{\vect{x}}^{2} = x^{2}+y^{2}+z^{2}\) is a polynomial, hence \(C^{\infty}\), and the composition of a \(C^{\infty}\) function of one variable with a \(C^{\infty}\) function of three is \(C^{\infty}\) by the chain rule (Proposition 11.104); this is where Lemma A16.5 is used, and it is the only place where smoothness that is not analyticity is needed. The argument \(1-\abs{\vect{x}}^{2}\) is positive exactly on the open unit ball and non-positive outside it, so \(\rho > 0\) there and \(\rho = 0\) elsewhere; in particular \(\rho\) vanishes on a whole neighbourhood of every point of \(\abs{\vect{x}} = 1\) together with all its derivatives. The normalizing integral in Equation (A16.12) is the integral of a continuous non-negative function that is positive on an open set, hence finite and strictly positive, so \(c_{0}\) is well defined. Radial dependence is manifest. Finally the substitution \(\vect{x} = \varepsilon\vect{z}\), whose Jacobian is \(\varepsilon^{3}\), gives \(\int\rho_{\varepsilon}(\vect{x})\dd^{3}x = \int\rho(\vect{z})\dd^{3}z = 1\), and rescales the support to \(\abs{\vect{x}} \le \varepsilon\).

Derives Proposition A16.9. Fix \(\vect{x}_{0} \in \Omega_{\varepsilon}\) and \(\sigma>0\) so small that \(\overline{B(\vect{x}_{0},\sigma+\varepsilon)} \subset \Omega\); let \(U = B(\vect{x}_{0},\sigma)\) and let \(Q\) be a closed box containing \(\overline{B(\vect{x}_{0},\sigma+\varepsilon)}\). Extend \(u\) from \(\overline{B(\vect{x}_{0},\sigma+\varepsilon)}\) to \(Q\) by any continuous function — the extension is irrelevant, because for \(\vect{x} \in U\) the factor \(\rho_{\varepsilon}(\vect{x}-\vect{y})\) already vanishes for \(\vect{y}\) outside \(\overline{B(\vect{x}_{0},\sigma+\varepsilon)}\), so

for every \(\vect{x} \in U\). In Equation (A16.15) the whole dependence on the parameter \(\vect{x}\) sits in the factor \(\rho_{\varepsilon}\), which by Lemma A16.7 is \(C^{\infty}\); hence \(F\) and all its \(\vect{x}\)-derivatives of every order,

are continuous on \(U\times Q\), being products of continuous functions. Lemma A16.3 and Remark A16.4 therefore apply and give \(u_{\varepsilon} \in C^{\infty}(U)\) with \(\pp^{\alpha}u_{\varepsilon}(\vect{x}) = \int_{Q}\bigl(\pp^{\alpha}\rho_{\varepsilon}\bigr) (\vect{x}-\vect{y})\,u(\vect{y})\,\dd^{3}y\). Since \(\vect{x}_{0} \in \Omega_{\varepsilon}\) was arbitrary, the conclusion holds on all of \(\Omega_{\varepsilon}\).

Derives Proposition A16.10. Write \(\rho_{\varepsilon}(\vect{z}) = \varepsilon^{-3}\tilde\rho\bigl(\abs{\vect{z}}/\varepsilon\bigr)\), where \(\tilde\rho(s) = c_{0}f(1-s^{2})\) is the radial profile supplied by Lemma A16.7. Decompose the first integral of Equation (A16.14) into spheres, \(\dd^{3}z = s^{2}\,\dd s\,\dd\Omega\) — the same decomposition already used in Lemma 14.58 to pass between a solid and a spherical integral, and legitimate here because the integrand is continuous on the compact ball:

The inner integral is unchanged under \(\vect{n} \longmapsto -\vect{n}\), which is a symmetry of the unit sphere and of its solid-angle element, so it equals \(\int_{S^{2}}u(\vect{x}+s\vect{n})\,\dd\Omega = 4\pi\,M_{u}(\vect{x},s)\). For \(\vect{x} \in \Omega_{\varepsilon}\) and \(0 < s \le \varepsilon\) the closed ball \(\overline{B(\vect{x},s)}\) lies in \(\Omega\), so the mean-value property Equation (A16.1) gives \(M_{u}(\vect{x},s) = u(\vect{x})\) — a constant, which comes out of the \(s\)-integral:

the middle step being Equation (A16.18) read backwards with \(u \equiv 1\), and the last being the normalization of Lemma A16.7.

Proof of Theorem A16.2. Derives Theorem A16.2. Smoothness. Let \(\vect{x}_{0} \in \Omega\). Since \(\Omega\) is open there is \(\varepsilon>0\) with \(\overline{B(\vect{x}_{0},2\varepsilon)} \subset \Omega\), and then \(B(\vect{x}_{0},\varepsilon) \subset \Omega_{\varepsilon}\). On \(\Omega_{\varepsilon}\) we have \(u = u_{\varepsilon}\) by Proposition A16.10, and \(u_{\varepsilon} \in C^{\infty}(\Omega_{\varepsilon})\) by Proposition A16.9. Hence \(u\) is \(C^{\infty}\) on a neighbourhood of \(\vect{x}_{0}\). Smoothness being a local property and \(\vect{x}_{0}\) arbitrary, \(u \in C^{\infty}(\Omega)\). Note that the smoothing parameter \(\varepsilon\) is allowed to depend on the point, as it must: no single \(\varepsilon\) works near \(\pp\Omega\).

Harmonicity. Now that \(u \in C^{2}(\Omega)\) is established, the argument of Theorem 14.69 applies without change, and we repeat it for completeness. Suppose \(\nabla^{2}u(\vect{x}_{0}) > 0\) for some \(\vect{x}_{0} \in \Omega\). By continuity of \(\nabla^{2}u\) there is \(r>0\) with \(\overline{B(\vect{x}_{0},r)} \subset \Omega\) and \(\nabla^{2}u > 0\) throughout that ball. Darboux's identity Equation (14.55) then gives, for \(0 < s \le r\),

so \(s \longmapsto M_{u}(\vect{x}_{0},s)\) is strictly increasing on \((0,r]\). But the mean-value property makes it constant, equal to \(u(\vect{x}_{0})\), on that whole interval — a contradiction. The case \(\nabla^{2}u(\vect{x}_{0}) < 0\) is the same argument with the inequalities reversed. Hence \(\nabla^{2}u \equiv 0\) on \(\Omega\), which is Definition 14.66.

The two classes coincide. A harmonic function is \(C^{2}\) by definition and has the mean-value property by the direct half of Theorem 14.69; conversely a continuous function with the mean-value property is harmonic by what has just been proved.

Derives Corollary A16.12. A harmonic \(u\) is continuous and has the mean-value property (Theorem 14.69); Theorem A16.2 then returns \(u \in C^{\infty}(\Omega)\).

Derives Corollary A16.13. \(u\) is continuous, being a locally uniform limit of continuous functions. Fix \(\vect{x}\) and \(r\) with \(\overline{B(\vect{x},r)}\subset\Omega\); the sphere \(\abs{\vect{y}-\vect{x}} = r\) is compact, so \(u_{n} \to u\) uniformly on it and the averages converge: \(M_{u}(\vect{x},r) = \lim_{n} M_{u_{n}}(\vect{x},r) = \lim_{n} u_{n}(\vect{x}) = u(\vect{x})\). Thus \(u\) is continuous with the mean-value property, and Theorem A16.2 applies. The corresponding statement with \(C^{2}\) convergence would be trivial; the point is that mere uniform convergence suffices, and it does so only because the characterization proved here does not mention derivatives at all.