# 7.2.1: Conclusions from the Representation Formula

- Page ID
- 2185

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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.