# 7.2.1: Conclusions from the Representation Formula

Similar to the theory of functions of one complex variable, we obtain here results for harmonic functions from the representation formula, in particular from (7.2.5), (7.2.6). We recall that a function \(u\) is called *harmonic* if \(u\in C^2(\Omega)\) and

\(\triangle u=0\) in \(\Omega\).

**Proposition 7.1.** *Assume \(u\) is harmonic in \(\Omega\). Then \(u\in C^\infty(\Omega)\).*}

*Proof.* Let \(\Omega_0\subset\subset\Omega\) be a domain such that \(y\in\Omega_0\). It follows from representation formulas (7.2.5), (7.2.6), where \(\Omega:=\Omega_0\), that \(D^lu(y)\) exist and are continuous for all \(l\) since one can change differentiation with integration in right hand sides of the representation formulas.

\(\Box\)

**Remark.** In fact, a function which is harmonic in \(\Omega\) is even real analytic in \(\Omega\), see an exercise.

**Proposition 7.2** (Mean value formula for harmonic functions). *Assume \(u\) is harmonic in \(\Omega\). Then for each \(B_\rho(x)\subset\subset\Omega\)
$$
u(x)= \frac{1}{\omega_n\rho^{n-1}}\int_{\partial B_\rho(x)}\ u(y)\ dS_y.
$$*

*Proof.*Consider the case \(n\ge3\). The assertion follows from (7.2.6) where \(\Omega:=B_\rho(x)\) since \(r=\rho\) and

\begin{eqnarray*}

\int_{\partial B_\rho(x)}\frac{1}{r^{n-2}}\frac{\partial u}{\partial n_y}\ dS_y&=&\frac{1}{\rho^{n-2}}\int_{\partial B_\rho(x)}\frac{\partial u}{\partial n_y}\ dS_y\\

&=&\frac{1}{\rho^{n-2}}\int_{B_\rho(x)}\ \triangle u\ dy\\

&=&0.

\end{eqnarray*}

\(\Box\)

We recall that a domain \(\Omega\in\mathbb{R}^n\) is called connected if \(\Omega\) is not the union of two nonempty open subsets \(\Omega_1\), \(\Omega_2\) such that \(\Omega_1\cap\Omega_2=\emptyset\). A domain in \(\mathbb{R}^n\) is connected if and only if its path connected.

**Proposition 7.3** (Maximum principle). *Assume \(u\) is harmonic in a connected domain and achieves its supremum or infimum in \(\Omega\). Then \(u\equiv const.\) in \(\Omega\).*

*Proof.* Consider the case of the supremum. Let \(x_0\in\Omega\) such that

$$

u(x_0)=\sup_\Omega u(x)=:M.

$$

Set

\(\Omega_1:=\{x\in\Omega:\ u(x)=M\}\) and \(\Omega_2:=\{x\in\Omega:\ u(x)<M\}\). The set \(\Omega_1\) is not empty since \(x_0\in\Omega_1\). The set \(\Omega_2\) is open since \(u\in C^2(\Omega)\). Consequently, \(\Omega_2\) is empty if we can show that \(\Omega_1\) is open. Let \(\overline{x}\in\Omega_1\), then there is a \(\rho_0>0\) such that \(\overline{B_{\rho_0}(\overline{x})}\subset\Omega\) and \(u(x)=M\) for all \(x\in B_{\rho_0}(\overline{x})\). If not, then there exists \(\rho>0\) and \(\widehat{x}\) such that

\(|\widehat{x}-\overline{x}|=\rho\), \(0<\rho<\rho_0\) and \(u(\widehat{x})<M\). From the mean value formula, see Proposition 7.2, it follows

$$

M=\frac{1}{\omega_n\rho^{n-1}}\int_{\partial B_\rho(\overline{x})}\ u(x)\ dS

<\frac{M}{\omega_n\rho^{n-1}}\int_{\partial B_\rho(\overline{x})}\ \ dS=M,

$$

which is a contradiction. Thus, the set \(\Omega_2\) is empty since \(\Omega_1\) is open.

\(\Box\)

**Corollary.** Assume \(\Omega\) is connected and bounded, and \(u\in C^2(\Omega)\cap C(\overline{\Omega})\) is harmonic in \(\Omega\). Then \(u\) achieves its minimum and its maximum on the boundary \(\partial\Omega\).

**Remark.** The previous corollary fails if \(\Omega\) is not bounded as simple counterexamples show.