Let $n \in \mathbb{N}$ and $\Omega \subset \mathbb{R}^n$ a bounded domain. Let $\alpha(n)$ be the volume of $B(0,1)$.

Proposition 1: Let $u \in C^2(\Omega)$ and $x \in \Omega$. Let $r_0 \in (0, \infty)$ such that $ B(x,r_0 ) \subset \Omega$. Define $\varphi : [0, r_0) \to \mathbb{R}$ by \begin{equation} \varphi (r):= \begin{cases} u(x), \ \text{if} \ r=0, \\ \frac{1}{n \alpha(n)r^{n-1}} \int_{\partial B(x,r)} u dS, \ \text{else.} \end{cases} \end{equation} Then $\varphi$ has the following properties:

  1. $\varphi$ is continuous.
  2. $\varphi$ is differentiable on $(0,r_0)$ and $\forall r \in (0, r_0)$: \begin{equation} \varphi^\prime (r) = \frac{1}{n \alpha(n) r^{n-1}} \int_{B(x,r)} \Delta u dV. \end{equation} In particular $\varphi^\prime$ can be extended continuously to $0$ by $0$.
  3. (the continuous extension of) $\varphi^\prime$ is differentiable and \begin{equation} \varphi^{\prime \prime }(0) = \frac{1}{n} \Delta u (x). \end{equation}
  4. \begin{equation} \Delta u (x) = \lim_{(r_0, 0) \ni r \to 0} \frac{2n}{r^2} \bigg (\frac{1}{n \alpha(n) r^{n-1}}\int_{\partial B(x,r) } u(y)dy -u (x) \bigg). \end{equation}

Proof: By continuity of $u$, $\lim_{r \to 0} \varphi (r) = u(x) = \varphi(0)$ and so $\varphi$ is continuous at $0$ (using that $r^{n-1} n \alpha(n)$ is the surface area of $\partial B(x,r)$). Inserting spherical coordinates shows that $\forall r \in (0, r_0)$: \begin{equation} \varphi (r) = \frac{1}{n \alpha(n)} \int_{\partial B(0,1)} u(x+ry) dS(y). \end{equation} By the theorem of differentiation under the integral sign $\varphi$ is twice continously differentiable on $(0,r_0)$ and $\forall r \in (0, r_0)$: \begin{equation} \varphi^\prime (r) = \frac{1}{n \alpha(n)} \int_{\partial B(0,1)} u^\prime (x+ry) y dS(y). \end{equation} Inserting spherical coordinates and using the divergence theorem shows that $\forall r \in (0,r_0)$: \begin{equation} \varphi^\prime (r) = \frac{1}{n \alpha(n)r^{n-1}} \int_{\partial B(x,r)} u^\prime (z) \frac{z-x}{r} dS(y) = \frac{1}{n \alpha(n)r^{n-1}} \int_{B(x,r)} \Delta u d V. \end{equation} Let $r \in (0, r_0)$. Then by continuity of $\Delta u$ together with the fact that $\alpha(n) r^n$ is the volume of $B(x,r)$: \begin{equation} \varphi^\prime (r) = \underbrace{\frac{r}{n}}_{\to 0 , \ r \to 0} \underbrace{\frac{1}{\alpha(n) r^n}\int_{B(x,r)} \Delta u d V}_{ \to \Delta u (x), \ r \to 0}. \end{equation}

And so $\varphi^\prime$ can be extended continuously to $0$ by $0$. This also shows item 3. By Taylor’s theorem \begin{equation} \frac{\varphi^{\prime \prime}(0)}{2} = \lim_{(0, r_0) \ni r \to 0 } \frac{1}{r^2} \bigg (\varphi (r) -\varphi (0) - \varphi^\prime (0) r \bigg). \end{equation} Since $\varphi^\prime (0) =0$ item 4 follows from 3. $\square$

Proposition 2 (harmonic functions have the mean value property): Let $u \in C^2(\Omega)$ with $\Delta u =0$. Then for all $x \in \Omega$ and $r \in (0, \infty)$ with $\bar B(x,r) \subset \Omega$: \begin{equation} u (x) = \frac{1}{n \alpha(n)r^{n-1}} \int_{\partial B(x,r)} u dS. \end{equation} Proof: This is an immediate consequence of proposition 1. $\square$

Corollary: In the same situation: \begin{equation} u(x) = \frac{1}{ \alpha(n)r^{n}} \int_{B(x,r)} u dV. \end{equation}

Proof: Using spherical coordinates: \begin{equation} \frac{1}{ \alpha(n)r^{n}} \int_{B(x,r)} u dV = \frac{1}{ \alpha(n)r^{n}} \int_0^r \int_{\partial B(x,\rho)} u dS d\rho. \end{equation} and so by the mean value property \begin{equation} \frac{1}{ \alpha(n)r^{n}} \int_{B(x,r)} u dV = \frac{1}{ \alpha(n)r^{n}} u(x) n \alpha (n) \int_0^r \rho^{n-1} d \rho = u(x). \quad \square \end{equation}

Proposition 3 (mean value property implies $C^\infty$ and harmonicity): Let $u \in C(\Omega)$ such that for all $x \in \Omega$ there is $r_0 \in (0, \infty)$ such that for all $r \in (0, r_0)$: $\bar B(0,r) \subset \Omega$ and \begin{equation} \frac{1}{n \alpha(n) r^{n-1}} \int_{\partial B(x,r)} u dS = u(x). \end{equation} Then $u \in C^\infty (\Omega)$ and $\Delta u =0$.

Proof: Let $\rho : \mathbb{R} \to \mathbb{R}$ \begin{equation} \rho (x):= c \begin{cases} \exp \bigg ( \frac{1}{x^2 - 1} \bigg), \ \text{if} \ x \in B(0,1), \\ 0, \ \text{else}. \end{cases} \end{equation} With $c \in (0, \infty)$ chosen such that $\int_{\mathbb{R}^n} \rho(\|x\|) dV(x) = 1$.

Let $\eta : \mathbb{R}^n \to \mathbb{R}$, $\eta (x):= \rho (\|x\|)$. Then $\eta \geq 0, \eta \in C^\infty (\mathbb{R}^n)$, $\int_{\mathbb{R}^n} \eta dV= 1$ and the support of $\eta$ is contained in $\bar B(0,1)$. For $\varepsilon \in (0, \infty)$ define $\eta_\varepsilon : \mathbb{R}^n \to \mathbb{R}$ by \begin{equation} \eta_\varepsilon (x) := \varepsilon^{-n} \eta ( x/\varepsilon). \end{equation} Then for all $\varepsilon \in (0, \infty)$: $\eta_\varepsilon \in C^\infty(\mathbb{R}^n)$, $\int_{\mathbb{R}^n} \eta_\varepsilon dV =1$ and the support of $\eta_\varepsilon$ is contained in $\bar B(0,\varepsilon)$.

Let $x_0\in \Omega$ and $r_0 \in (0, \infty)$ such that $\bar B(x_0,r_0) \subset \Omega$ and $\forall r \in (0, r_0]$: \begin{equation} u(x_0) = \frac{1}{n \alpha(n) r^{n-1}} \int_{\partial B(x_0,r)} u dS. \end{equation} For $\varepsilon \in (0, \infty)$ let $u_\varepsilon : \mathbb{R}^n \to \mathbb{R}$, \begin{equation} u_\varepsilon (x) := \int_{\bar B(x_0,r_0)} \eta_\varepsilon(x-y) u(y) dV(y). \end{equation} For all $\varepsilon \in (0, \infty)$: $u_\varepsilon \in C^\infty (\mathbb{R}^n)$ by a corollary to the theorem of differentiation under the integral sign.

Let $\varepsilon := r_0/2$. Let $x \in B(x_0, \varepsilon)$. Then the support of $\eta_{\varepsilon}(x-\bullet)$ is contained in $\bar B(x,\varepsilon )\subset \bar B (x_0, r_0)$ by the triangle inequality. Therefore \begin{equation} u_\varepsilon (x) = \int_{B(x, \varepsilon)} \eta_\varepsilon(x-y) u(y) dV(y). \end{equation} Inserting the definition of $\eta_\varepsilon$ and using spherical coordinates: \begin{equation} u_\varepsilon(x) = \varepsilon^{-n} \int_0^\varepsilon \rho (r/\varepsilon) \int_{\partial B(x,r)}u d S dr. \end{equation} Therefore using the assumption on $u$, inserting the definition of $\alpha(n)$ and using spherical coordinates again \begin{equation} u_\varepsilon (x) = u(x) n \alpha(n) \varepsilon^{-n} \int_0^\varepsilon \rho ( r/\varepsilon) dr = u(x) \int_{\mathbb{R}^n} \eta_\varepsilon dV = u(x). \end{equation} This shows $u \in C^\infty (\Omega)$. Finally $\Delta u =0$ follows at once from item 4 in proposition 1. $\square$

Corollary: Harmonic functions are $C^\infty$ and any partial derivative (of any order) of a harmonic function is harmonic (because the partial derivatives commute with the laplacian). Furthermore the local uniform limit of a sequence of harmonic functions is harmonic (since the limit satisfies the mean value property).

Proposition 4 (strong maximum principle): Let $u \in C(\bar \Omega) \cap C^2(\Omega)$, $\Delta u=0$. Then:

  1. if $u$ attains its maximum in $\Omega$ then $u$ is constant,
  2. \begin{equation} \max_{x \in \bar \Omega} u(x) = \max_{x \in \partial \Omega} u(x). \end{equation}

Proof: $\bar \Omega$ and $\partial \Omega$ are compact, because $\Omega$ is bounded (together with the Heine-Borel theorem). Therefore $u$ attains its maximum. To 1.: Assume $u$ attains its maximum in $\Omega$. Let \begin{equation} S:= \{ x \in \Omega : u(x) = \max_{y \in \bar \Omega} u(y) \}. \end{equation} Then $S \neq \varnothing$. The idea is to show that $S$ is open and closed (relative to $\Omega$). $S$ is closed (in $\Omega)$ being the preimage of a closed set (a singleton) by a continuous map. Let $ x \in S$. Let $ r_0\in (0, \infty)$ such that $B(x,r_0) \subset \Omega$. Let $r \in (0, r_0)$. Then \begin{equation} 0 = u(x) - \frac{1}{n \alpha(n) r^{n-1}} \int_{\partial B(x,r)} u dS = \frac{1}{n \alpha(n) r^{n-1}} \int_{\partial B(x,r)} \underbrace{u(x) - u(y)}_{\geq 0} dS. \end{equation} Which implies that $u$ is constant on $\partial B(x,r)$. Since $r$ was arbitrary it follows that $u$ is constant on $B(x,r_0)$ and so $S$ is open. Therefore $S = \Omega$ and so $u$ is constant on $\Omega$ which in turn implies that $u$ is constant on $\bar \Omega$ by continuity. To 2.: Follows from 1., because 2. is true if $u$ is constant and if not, then $u$ attains its maximum on $\partial \Omega$ by item 1. $\square$

Proposition 5 (uniqueness of solution to the Dirichlet problem): Let $f \in C(\Omega)$ and $g \in C(\partial \Omega)$. Then there exists at most one $u \in C(\bar \Omega) \cap C^2(\Omega)$ solving the Dirichlet problem: \begin{equation} \begin{cases} - \Delta u = f \quad \text{on} \ \Omega, \\ u = g \quad \text{on} \ \partial \Omega. \end{cases} \end{equation}

Proof: Let $u_1, u_2 \in C(\bar \Omega) \cap C^2(\Omega)$ be solutions to the Dirichlet problem. Then $v := u_1 - u_2$ satisfies $\Delta v=0$ in $\Omega$ and $v = 0$ on $\partial \Omega$. The function $-v$ satisfies the same. Let $x \in \bar \Omega$. By the maximum principle applied to $v$: \begin{equation} v(x) \leq \max_{y \in \bar \Omega} v(y) = \max_{y \in \partial \Omega} v(y) = 0. \end{equation} By the maximum principle applied to $-v$: \begin{equation} -v (x) \leq \max_{y \in \bar \Omega} -v(y) = \max_{y \in \partial \Omega} -v(y) = 0. \end{equation} Therefore $v(x) \geq 0$. And so in total $0 \leq v(x) \leq 0$, which implies that $v(x) = 0$. $\square$

Proposition 6 (estimates for partial derivatives of harmonic functions): Let $u \in C^2(\Omega)$ with $\Delta u =0$. Let $j \in {1,\dots,n }$. Then for all $x \in \Omega$ and $r \in (0, \infty)$ such that $\bar B(x,r)\subset \Omega$: \begin{equation} |\partial_j u(x)| \leq \frac{n}{r} \sup_{y \in \partial B(x,r)} |u(y)|. \end{equation}

Proof: By the corollary to the mean value property ($\partial_j u$ is harmonic by another corollary above) and the divergence theorem: \begin{equation} \partial_j u(x) = \frac{1}{\alpha (n) r^{n}} \int_{B(x,r)} \partial_j u dV = \frac{1}{\alpha (n) r^{n}} \int_{\partial B(x,r)} u(y) \frac{y_j-x_j}{r} dS(y). \end{equation} and therefore \begin{equation} |\partial_j u(x) | \leq \frac{n}{r} \sup_{y \in \partial B(x,r)} |u(y)|. \quad \square \end{equation}

Proposition 7 (Liouvilles theorem): Let $u \in C^2(\mathbb{R}^n)$ with $\Delta u=0$ and $u$ bounded. Then $u$ is constant.

Proof: Let $j \in {1,\dots,n }$. Let $x \in \mathbb{R}^n$. Then by the preceding proposition: $\forall r \in (0, \infty)$: \begin{equation} |\partial_j u(x) |\leq \frac{n}{r} \sup_{y \in \mathbb{R}^n} |u(y)| \to 0 \ (r \to \infty). \end{equation} Therefore $u^\prime =0$ and so $u$ is constant. $\square$

Proposition 8 (representation formula for bounded solutions of poisson problem in $\mathbb{R}^n$): Let $n \geq 3$. Let $f \in C_c (\mathbb{R}^n)$ and $u \in C^2(\mathbb{R}^n)$ bounded with $-\Delta u = f$. Then there exists $c \in \mathbb{R}$ such that for all $x \in \mathbb{R}^n$: \begin{equation} u(x) = \int_{\mathbb{R}^n} \Phi(y) f(x-y) dV(y) +c, \end{equation} where $\Phi$ is the fundamental solution of poissons equation (see previous blog post).

Proof: Let $v : \mathbb{R}^n \to \mathbb{R}$, \begin{equation} v(x):= \int_{\mathbb{R}^n} \Phi(x-y) f(y) dV(y). \end{equation} Then $v \in C^2 (\mathbb{R}^n)$, $-\Delta v =f$ and $v$ is bounded by the previous blog post. Now $\Delta (u-v) =0$ and $u-v$ is bounded, so $u-v$ is constant by Liouvilles theorem. $\square$

Proposition 8 (estimates for the higher order partial derivatives of harmonic functions): Let $u \in C^2(\Omega)$ with $\Delta u =0$ and $\alpha \in \mathbb{N}^n$. Then for all $x \in \Omega$ and $r \in (0, \infty)$ such that $\bar B(x,r)\subset \Omega$. \begin{equation} |D^\alpha u(x)| \leq \bigg(\frac{n |\alpha|}{r}\bigg)^{|\alpha|} \sup_{y \in B(x,r)} |u(y)|. \end{equation}

Proof: By induction on the order of the multi-index. Order $=1$ is the preceding proposition. Let $k \in \mathbb{N}$, $k>0$ and assume the proposition holds for all multiindices of order $<k$. Let $\alpha \in \mathbb{N}^n$ with $|\alpha| =k$. Then $D^\alpha u $ is harmonic, because $u \in C^\infty$ and, because the partial derivatives commute. Furthermore there exists $j \in {1,\dots, n}$ and $\beta \in \mathbb{N}^n$ with $|\beta| = k-1$ such that $D^\alpha u = \partial_j D^\beta u$. Let $x \in \Omega$ and $r \in (0, \infty)$ such that $\bar B(x,r)\subset \Omega$. By proposition 6: \begin{equation} |D^\alpha u(x)| \leq \frac{n k}{r}\sup_{y \in \partial B(x,r/k)}| D^\beta u(x)| \end{equation} Let $y \in \partial B(x,r/k)$. Then $B(y,r(k-1)/k) \subset B(x,r)$, since if $z \in B(y,r(k-1)/k)$, then \begin{equation} \| z-x \| \leq \| z-y\| + \|y-x\| = r (k-1)/k + r/k < r. \end{equation} Therefore by induction hypothesis \begin{equation} |D^\beta u(y) | \leq \bigg(\frac{n (k-1)}{r (k-1)/k}\bigg)^{k-1} \sup_{z \in B(y,r(k-1)/k)} | u(z)|. \end{equation} and so in total \begin{equation} |D^\alpha u(x)| \leq \bigg ( \frac{k }{r}\bigg)^k \sup_{z \in B(x,r)} |u(z)|. \quad \square \end{equation}

Corollary: Let $(u_m)_{m \in \mathbb{N}}$ be a sequence of harmonic functions that converge locally uniformly to $u$ (which is known to be harmonic too). Then for any $\alpha \in \mathbb{N}^n$ the sequence $(D^\alpha u_n)_{m \in \mathbb{N}}$ converges locally uniformly to $D^\alpha u$.

Proposition 9 (harmonic functions are analytic): Let $u \in C^2(\Omega)$ with $\Delta u =0$. Then the Taylor series of $u$ is locally uniformly convergent to $u$.

Proof: Let $x_0 \in \Omega$. Let $r \in (0, \infty)$ such that $\bar B(x_0,r) \subset \Omega$. For $k \in \mathbb{N}$ let $R_k: B(x_0,r) \to \mathbb{R}$ \begin{equation} R_k (x) := u(x) - \sum_{\alpha \in \mathbb{N}^n , |\alpha| = k} \frac{D^\alpha u(x_0)}{\alpha!} (x-x_0)^\alpha. \end{equation} Then for all $k \in \mathbb{N}$ and $x \in B(x_0,r)$ there is $t \in [0,1]$ with \begin{equation} R_k(x) = \sum_{\alpha \in \mathbb{N}^n, |\alpha| = k+1} \frac{D^\alpha u(x_0 + t(x-x_0))}{\alpha !} (x-x_0)^\alpha \end{equation} by the Lagrange remainder formula. Therefore by the multinomial theorem for all $k \in \mathbb{N}$, $\rho \in (0, r)$ and $x \in B(x_0,\rho)$: \begin{equation} |R_k(x) |\leq \frac{\| x-x_0\|^{k+1}_1}{(k+1)!}\max_{\alpha \in \mathbb{N}^n : |\alpha| = k+1} \sup_{x \in B(x_0, \rho)} | D^\alpha u (x)|. \end{equation} By the estimates for the partial derivatives for all $k \in \mathbb{N}$ (together with the triangle inequality): \begin{equation} \max_{\alpha \in \mathbb{N}^n : |\alpha| = k} \sup_{ x \in \bar B(x_0,r/2) } | D^\alpha u (x)| \leq \bigg(\frac{n k}{r/2}\bigg)^{k} \underbrace{\sup_{y \in B(x_0,r)} |u(y)|}_{=:C}. \end{equation} Now let $\rho := r/(4 n^{3/2} e)$. Then for all $k \in \mathbb{N}$ and $x \in B(x_0, \rho)$: \begin{equation} |R_{k-1} (x)| \leq \frac{(\sqrt{n} \rho)^k}{k!} C\bigg(\frac{n k}{r/2}\bigg)^{k} \leq C \bigg(\frac{r}{4n e}\bigg)^k \bigg(\frac{2 n e}{r}\bigg)^{k} = C \frac{1}{2^k} \to 0 \ ( k \to \infty), \end{equation} where $ \|\cdot \|_1 \leq \sqrt{n} \| \cdot \|_2$ and $k^k /k! \leq e^k$ (by the exponential series) was used. $\square$

Proposition 10 (Harnack’s inequality): For every domain $V \subset \Omega$ with $\bar V \subset \Omega$ there exists $C \in (0, \infty)$ such that for all $u \in C^2(\Omega)$ with $u\geq 0$ and $\Delta u =0$: \begin{equation} \sup_{x \in V} u(x) \leq C \inf_{x \in V} u. \end{equation} Proof: Let $y \in \Omega$ and $r \in (0,\infty)$ such that $B(y,4r) \subset \Omega$. Let $x_1, x_2 \in B(y,r)$. Then using the mean value property, $u \geq 0$ and $B(x_1,r) \subset B(x_2,3r) \subset B(y,4r)$: \begin{equation} \begin{split} u(x_1) &= \frac{1}{\alpha(n) r^n} \int_{B(x_1,r)} u dV \leq \frac{1}{\alpha(n) r^n} \int_{B(x_2,3r)} u dV \\ &= \frac{3^n}{\alpha(n) (3r)^n} \int_{B(x_2,3r)} u dV = 3^n u(x_2). \end{split} \end{equation} Let $\rho \in (0, \infty)$ such that $\forall x \in \bar V$: $B(x,4 \rho) \subset \Omega$ (exists since $\bar V \subset \Omega$ is compact). Again by compactness we can find $x_1, \dots x_N \in \bar V$ such that $ \bar V \subset \bigcup_{j=1}^N B(x_j,\rho)$. Finally let $a,b \in V$ arbitrary. Let \begin{equation} \begin{split} S&:=\{ y \in V :\exists M \in \{1,\dots, N\} , \sigma: \{1, \dots, M\} \to \{1, \dots, N \} \ \text{injective}: \\ &a \in B(x_{\sigma(1)},\rho), \ y \in B(x_{\sigma(M)}, \rho)), \ \forall j \in \{1, \dots, M-1\}: \\ &B(x_{\sigma(j)},\rho) \cap B(x_{\sigma(j+1)},\rho) \neq \varnothing \}. \end{split} \end{equation} Then $S\neq \varnothing$, because $a \in S$ and $S$ is open since for any $y \in S$ the final ball in the chain intersected with $V$ is an open neighborhood of $y$ that is contained in $S$. Let $y \in V \setminus S$. Then there exists some $j \in \{1,\dots, N\}$ with $y \in B(x_j, \rho)$. Let $z \in V \cap B(x_j, \rho) \cap S$. If the chain of balls connecting $a$ to $z$ contains the ball $B(x_j, \rho)$ at any point we can truncate the chain to show that $y \in S$. If it does not contain $B(x_j, \rho)$ then we can define a new chain by adding $B(x_j,\rho)$ at the end. Therefore $S$ is closed relative to $V$ and so $S=V$ by connectedness.

Therefore \begin{equation} u(a) \leq 3^{n N} u(b). \end{equation}

Proposition 11 (Harnack’s convergence theorem): Let $(u_m)_{m \in \mathbb{N}}$ be a point-wise non-decreasing sequence of harmonic functions in $\Omega$. Let $u : \Omega \to (0, \infty]$ be the point-wise limit. Then either $u = \infty$ or $u$ is never infinite and $(u_m)_{m \in \mathbb{N}}$ is converging locally uniformly to $u$ (and therefore $u$ is harmonic).

Proof: Let $x_0 \in \Omega$ such that $u(x_0) <\infty$. Let $x \in \Omega$. Let $V \subset \Omega$ a region with $\bar V \subset \Omega$, $x_0,x \in V$. The existence of $V$ follows by taking a union of balls along a continuous curve connecting $x_0$ with $x$ (similar to the proof of Harnack’s inequality). Let $C \in (0, \infty)$ be the constant from Harnack’s inequality associated to $V$. Let $\varepsilon \in (0, \infty)$. Let $N \in \mathbb{N}$ such that for all $m,k \in \mathbb{N}$ with $m\geq k \geq N$: $ u_m(x_0)- u_k (x_0)< \varepsilon$ (possible since $\lim_{m \to \infty} u_m (x_0) \in \mathbb{R}$ and convergent sequences are Cauchy). Then by Harnack’s inequality for all $k,m\in \mathbb{N}$ with $m>k\geq N$ and $y \in V$: \begin{equation} |u_m (y) - u_k(y)| = (u_m - u_k)(y) \leq C (u_m (x_0) - u_k(x_0))< \varepsilon. \end{equation} Therefore $u$ is real valued in $\bar V$ and $(u_m)_{m \in \mathbb{N}}$ is a Cauchy sequence in $C(\bar V)$ and therefore uniformly convergent to $u$ in $\bar V$. $\square$