Pome524A2588C8CERecall from section 2.5 that $B(x,r)$ is the open ball with center $x$ and radius $r>0$; its closure $\overline{B(x,r)}$ is the closed ball. We write $S^{d-1}=\{ y\in \mathbb R^ d:|y|=1\} $ for the unit sphere. The Laplacian is $\Delta =\sum _{j=1}^ d\partial _{x_ j}^2$. A real function $h\in C^2(\Omega )$ is harmonic if $\Delta h=0$, and a complex function is harmonic if its real and imaginary parts are harmonic. In dimension one the harmonic functions are the functions that are affine on each component interval of $\Omega $, because $h''=0$ on an interval forces $h'$ to be constant there.
We define surface measure on the unit sphere through the Lebesgue measure of cones: for a Borel set $A\subset S^{d-1}$ put
The cone over $A$ is the inverse image of $A$ under the continuous map $y\mapsto y/|y|$ on $B(0,1)\setminus \{ 0\} $, so it is a Borel set; moreover, the cone over a countable union is the union of the cones, and cones over disjoint sets are disjoint. Hence $S$ is a finite Borel measure on $S^{d-1}$, the surface measure. It is invariant under every orthogonal transformation $Q$ of $\mathbb R^ d$: the cone over $QA$ is the image under $Q$ of the cone over $A$, and Lebesgue measure is invariant under orthogonal transformations. By linearity and monotone convergence (proposition 2.2.3), it follows that $\int _{S^{d-1}}g\circ Q\, \mathrm dS=\int _{S^{d-1}}g\, \mathrm dS$ for every nonnegative or integrable Borel function $g$.
We write $s_{d-1}=S(S^{d-1})$ for the area of the unit sphere, so that the unit ball has volume $s_{d-1}/d$. For instance, $s_0=2$, because $S$ is counting measure on $S^0=\{ \pm 1\} $, and $s_1=2\pi $, because the unit disc has area $\pi $. On a sphere $\partial B(x,r)$ we use the measure obtained by translation and dilation, again denoted $\mathrm dS$:
Next we express Lebesgue measure in polar coordinates. Let $0\leq a<b$ and let $A\subset S^{d-1}$ be a Borel set. By dilation, the cone $\{ t\omega :0<t<b,\ \omega \in A\} $ has Lebesgue measure $b^ dS(A)/d$. The set $\{ t\omega :0<t\leq a,\ \omega \in A\} $ is the intersection of the decreasing cones $\{ t\omega :0<t<a+1/n,\ \omega \in A\} $, so its measure is $\lim _ n(a+1/n)^ dS(A)/d=a^ dS(A)/d$. Subtracting and translating, we find that for every $x\in \mathbb R^ d$ the annular sector $\{ x+t\omega :a<t<b,\ \omega \in A\} $ has Lebesgue measure
Now fix $x\in \mathbb R^ d$ and $R>0$, let $\lambda $ be the measure $t^{d-1}\, \mathrm dt$ on the Borel sets of $(0,R)$, and let $\Phi (t,\omega )=x+t\omega $. The map $\Phi $ is a homeomorphism from $(0,R)\times S^{d-1}$ onto $Z=B(x,R)\setminus \{ x\} $, with inverse $y\mapsto (|y-x|,(y-x)/|y-x|)$, and the annular sectors with center $x$ contained in $Z$ are the images under $\Phi $ of the rectangles $(a,b)\times A$ with $b\leq R$. These sectors generate the Borel $\sigma $-algebra of $Z$. Indeed, the sets $(a,b)\times (S^{d-1}\cap B(\omega ,q))$, where $0<a<b<R$ are rational, $\omega $ runs through a countable dense subset of $S^{d-1}$ and $q$ through the positive rationals, form a countable base of the topology of $(0,R)\times S^{d-1}$. Hence every open subset of $Z$ is a countable union of their images, which are sectors.
The sets $E\subset Z$ for which $\Phi ^{-1}(E)$ belongs to the product $\sigma $-algebra form a $\sigma $-algebra containing the sectors, and hence all Borel subsets of $Z$. Consequently $\mu (E)=(\lambda \times S)(\Phi ^{-1}(E))$ defines a finite Borel measure $\mu $ on $Z$, and $\mu $ agrees with Lebesgue measure on the sectors. The sectors are closed under finite intersections, and $Z$ itself is a sector, so the two finite measures agree on every Borel subset of $Z$ by part (2) of lemma 2.2.6. Hence $\int _ Zf\, \mathrm dy=\int f\circ \Phi \, \mathrm d(\lambda \times S)$ for indicator functions of Borel sets, by linearity for nonnegative simple Borel functions, and by monotone convergence (proposition 2.2.3) for all nonnegative Borel functions $f$. Since the point $x$ is a Lebesgue null set, Tonelli's theorem (proposition 2.2.7) yields the formula of integration in polar coordinates
AAD3B6ABC9B1
3.1.2
for every nonnegative Borel function $f$ on $B(x,R)$. Applying it to the positive and negative parts of $\operatorname{Re}f$ and $\operatorname{Im}f$ and using Fubini's theorem, we obtain (3.1.2) for every Borel function $f$ which is integrable on $B(x,R)$. Letting $R\to \infty $ and using monotone convergence, we see that (3.1.2) also holds with $R=\infty $, where $B(x,\infty )=\mathbb R^ d$. In particular, every sphere $\partial B(x,r)$ is a Lebesgue null set, since its indicator function vanishes at $x+t\omega $ unless $t=r$.
Next we prove the divergence theorem on balls. Let $B=B(c,R)$, let $\nu (y)=(y-c)/R$ be the outward unit normal on $\partial B$, and let $F=(F_1,\ldots ,F_ d)$ be a $C^1$ vector field on an open neighborhood $U$ of $\overline B$, with divergence $\operatorname {div}F=\sum _{j=1}^ d\partial _ jF_ j$. Then
4CF38DCFDD7E
3.1.3
It suffices to prove, for each $j$, that $\int _ B\partial _ jF_ j\, \mathrm dy=\int _{\partial B}F_ j\nu _ j\, \mathrm dS$. To do so, we approximate $\mathbf1_ B$ by smooth radial cutoff functions $\chi _\varepsilon $ and integrate $\partial _ j(F_ j\chi _\varepsilon )$ over $\mathbb R^ d$. Let $0<\varepsilon <R$, and let $\psi _\varepsilon $ be a smooth nonnegative function on $\mathbb R$ with support in $[R-\varepsilon ,R]$ and integral one, for instance a translate and dilate of the mollifier $\rho $ of section 2.5 in dimension one. The function $\theta _\varepsilon (t)=\int _ t^\infty \psi _\varepsilon (s)\, \mathrm ds$ is smooth with values in $[0,1]$, equals $1$ for $t\leq R-\varepsilon $ and $0$ for $t\geq R$, and satisfies $\theta _\varepsilon '=-\psi _\varepsilon $. Hence $\chi _\varepsilon (y)=\theta _\varepsilon (|y-c|)$ is smooth on $\mathbb R^ d$, because $|y-c|$ is smooth for $y\neq c$ and $\chi _\varepsilon =1$ on $B(c,R-\varepsilon )$. Moreover, $\chi _\varepsilon $ vanishes outside $\overline B$, and
The function $F_ j\chi _\varepsilon $, extended by zero outside $U$, is a $C^1$ function with compact support on $\mathbb R^ d$, because it is $C^1$ on $U$ and vanishes on the open set $\mathbb R^ d\setminus \overline B$. Integrating first in the variable $y_ j$ (Fubini's theorem) and using the fundamental theorem of calculus, we get $\int _{\mathbb R^ d}\partial _ j(F_ j\chi _\varepsilon )\, \mathrm dy=0$. Since $\partial _ j\chi _\varepsilon $ vanishes outside $B$, it follows, by (3.1.2) with center $c$, that
where
Since $F_ j$ is uniformly continuous on $\overline B$, the function $g$ is continuous on $(0,R]$, and as $\psi _\varepsilon \geq 0$ has integral one and support in $[R-\varepsilon ,R]$, we have $|\int _0^ R\psi _\varepsilon g\, \mathrm dt-g(R)|\leq \sup _{[R-\varepsilon ,R]}|g-g(R)|$. Hence the right side tends to $g(R)$ as $\varepsilon \to 0$. On the left, $\chi _\varepsilon \to \mathbf1_ B$ pointwise and $|\chi _\varepsilon \partial _ jF_ j|\leq |\partial _ jF_ j|\mathbf1_{\overline B}$, so by dominated convergence the left side tends to $\int _ B\partial _ jF_ j\, \mathrm dy$. Finally, $\nu (c+R\omega )=\omega $, so the definition of surface measure on $\partial B$ gives $g(R)=\int _{\partial B}F_ j\nu _ j\, \mathrm dS$. Summing over $j$ proves (3.1.3).
We also need the corresponding identity on an annulus. Let $0<r<R$, and let $F$ be a $C^1$ vector field on an open neighborhood $U$ of the closed annulus $\{ r\leq |y-c|\leq R\} $. Choose a smooth function $\eta $ with compact support in $U$ which equals one near the annulus (lemma 2.5.1(4)). Then $\eta F$, extended by zero outside $U$, is a $C^1$ vector field on $\mathbb R^ d$, and it equals $F$ near the closed annulus. We apply (3.1.3) to $\eta F$ on $B(c,R)$ and on $B(c,r)$ and subtract. On the left we obtain the integral of $\operatorname {div}(\eta F)$ over $\{ r\leq |y-c|<R\} $, which is the integral of $\operatorname {div}F$ over the open annulus, because the sphere $\partial B(c,r)$ is a Lebesgue null set. Hence
DEF1A85BF9E8
3.1.4
where $\nu (y)=(y-c)/|y-c|$ on both spheres.
For a continuous function $u$ on $\Omega $ and a closed ball $\overline{B(x,r)}\subset \Omega $, we write
for the spherical mean of $u$ over the sphere $\partial B(x,r)$. We say that $u$ has the sphere mean-value property if $u(x)=M_ u(x,r)$ for every closed ball $\overline{B(x,r)}\subset \Omega $, and the ball mean-value property if $u(x)=|B(x,r)|^{-1}\int _{B(x,r)}u\, \mathrm dy$ for every such ball.
Proposition 3.1.5.620D5E4DD6EC Let $u\in C^2(\Omega )$ and $\overline{B(x,R)}\subset \Omega $. Then, for $0<r<R$,
2B78C72EAD6C
3.1.6
In particular, harmonic functions have both the sphere and the ball mean-value properties.
Proof. Fix $0<r<R$. For $\omega \in S^{d-1}$ and small $h\neq 0$, the fundamental theorem of calculus gives
Since $\nabla u$ is uniformly continuous on the compact set $\overline{B(x,R)}$, these difference quotients converge to $\nabla u(x+r\omega )\cdot \omega $ uniformly in $\omega $ as $h\to 0$. Integrating over the finite measure space $(S^{d-1},S)$ and using the definition of surface measure on $\partial B(x,r)$, where $\nu (x+r\omega )=\omega $, we obtain
Applying (3.1.3) on $B(x,r)$ to $F=\nabla u$, which is $C^1$ on $\Omega $ with divergence $\Delta u$, gives (3.1.6).
Now let $u$ be harmonic. Then $M_ u(x,\cdot )$ has zero derivative, and hence is constant, on $(0,R)$. Since $|M_ u(x,r)-u(x)|\leq \max _{|y-x|=r}|u(y)-u(x)|$, which tends to zero as $r\downarrow 0$ because $u$ is continuous at $x$, the constant is $u(x)$. If $\overline{B(x,r)}\subset \Omega $, then the compact set $\overline{B(x,r)}$ has positive distance from the closed set $\mathbb R^ d\setminus \Omega $, so there is $R>r$ with $\overline{B(x,R)}\subset \Omega $, and therefore $u(x)=M_ u(x,r)$; this is the sphere mean-value property. Since $M_ u(x,t)=u(x)$ for $0<t\leq r$, (3.1.2) gives $\int _{B(x,r)}u\, \mathrm dy=s_{d-1}r^ du(x)/d=|B(x,r)|\, u(x)$, which is the ball mean-value property. $\square$
Theorem 3.1.7.A60BA3B2DBFC A continuous function on $\Omega $ is harmonic if and only if it has the sphere mean-value property. Every harmonic function is smooth. If $h$ is harmonic on $\Omega $ and $\overline{B(x,r)}\subset \Omega $, then for every multi-index $\alpha $,
D63569C18649
3.1.8
where the constant $C_{d,\alpha }$ depends only on $d$ and $\alpha $. Moreover, every harmonic function $h$ on $\Omega $ satisfies $h*\rho _\varepsilon =h$ on $\Omega _\varepsilon $ for every $\varepsilon >0$, where $\rho _\varepsilon $ and $\Omega _\varepsilon $ are as in section 2.5.
Proof. As in section 2.5, we write $u_\varepsilon =u*\rho _\varepsilon $. Let $u$ be continuous with the sphere mean-value property, and let $\overline{B(x,\varepsilon )}\subset \Omega $. The compact set $\overline{B(x,\varepsilon )}$ has positive distance from $\mathbb R^ d\setminus \Omega $, so $x\in \Omega _\varepsilon $, and $M_ u(x,r)=u(x)$ for $0<r\leq \varepsilon $. Since $\rho $ is radial, we may write $\rho _\varepsilon (y)=\phi _\varepsilon (|y|)$; as $\rho _\varepsilon $ is even and vanishes outside $B(0,\varepsilon )$, (3.1.2) gives
EB7C9EF52470
3.1.9
Every point of $\Omega _\varepsilon $ is the center of such a closed ball, so $u=u_\varepsilon $ on the open set $\Omega _\varepsilon $, and $u_\varepsilon $ is smooth there (section 2.5). As $\Omega $ is the union of the sets $\Omega _\varepsilon $, the function $u$ is smooth.
Next we show that $\Delta u=0$. Fix $x\in \Omega $. Taylor's formula gives
uniformly in $\omega \in S^{d-1}$. We average over the sphere. The reflections $\omega \mapsto -\omega $ and $\omega _ j\mapsto -\omega _ j$ preserve $S$, so $\int \omega _ j\, \mathrm dS=0$ and $\int \omega _ j\omega _ k\, \mathrm dS=0$ for $j\neq k$. Permutations of the coordinates also preserve $S$, so the integrals $\int \omega _ j^2\, \mathrm dS$ are all equal, and their sum is $s_{d-1}$ because $\sum _ j\omega _ j^2=1$. Hence
Since $M_ u(x,r)=u(x)$ for all small $r$, it follows that $\Delta u(x)=0$. Thus $u$ is harmonic. Conversely, harmonic functions have the sphere mean-value property by proposition 3.1.5; in particular they are smooth, and they satisfy $h=h_\varepsilon =h*\rho _\varepsilon $ on $\Omega _\varepsilon $ by the first part of the proof.
Now let $h$ be harmonic and $\overline{B(x,r)}\subset \Omega $. Then $x$ lies in the open set $\Omega _ r$, and $h=h_ r$ on $\Omega _ r$ by the first part of the proof. By the differentiation formula $\partial ^\alpha h_ r=h*\partial ^\alpha \rho _ r$ of section 2.5, and since $\partial ^\alpha \rho _ r$ vanishes outside $B(0,r)$, we get
Since $\partial ^\alpha \rho _ r(z)=r^{-d-|\alpha |}(\partial ^\alpha \rho )(z/r)$, this proves (3.1.8) with $C_{d,\alpha }=\sup |\partial ^\alpha \rho |$, where $\rho $ is the mollifier on $\mathbb R^ d$ fixed in section 2.5. $\square$
Proposition 3.1.10.B304785CC98D Let $h$ be a real harmonic function on an open set $D\subset \mathbb R^ d$.
If $D$ is connected and $h$ attains its maximum or its minimum over $D$ at a point of $D$, then $h$ is constant.
If $D$ is bounded and $\varlimsup _{D\ni x\to \zeta }h(x)\leq 0$ for every $\zeta \in \partial D$, then $h\leq 0$ on $D$.
Consequently, a harmonic function on a bounded open set which is continuous on its closure is determined by its boundary values.
Proof. For the first assertion, suppose that $h$ attains its maximum $m$ at a point of $D$, and let $E=\{ x\in D:h(x)=m\} $. This set is nonempty and closed in $D$. If $x\in E$ and $\overline{B(x,r)}\subset D$, the ball mean-value property gives $\int _{B(x,r)}(m-h)\, \mathrm dy=0$. The integrand is continuous and nonnegative, so it vanishes on $B(x,r)$. Thus $E$ is open, and $E=D$ by connectedness. For the minimum, apply this to $-h$.
For the second assertion, let $\varepsilon >0$, and suppose that $h>\varepsilon $ somewhere in $D$; we derive a contradiction. Each $\zeta \in \partial D$ has a ball $B(\zeta ,r_\zeta )$ such that $h<\varepsilon $ on $D\cap B(\zeta ,r_\zeta )$. Since $D$ is bounded, $\partial D$ is compact, and finitely many of these balls cover it; let $W$ be their union. Then $K=\overline D\setminus W=D\setminus W$ is compact. Since $h<\varepsilon $ on $D\setminus K=D\cap W$, the set $K$ is nonempty. The maximum of $h$ over $K$ is attained at some point $x_0$, and $h(x_0)>\varepsilon $; hence $h(x_0)$ is also the maximum of $h$ over $D$. Let $D_0$ be the component of $D$ containing $x_0$; it is open, because every point of $D$ lies in a ball contained in $D$, and balls are connected. By the first assertion, $h=h(x_0)$ on $D_0$.
Since $\mathbb R^ d$ is connected and $D_0$ is a nonempty, bounded, open subset, $D_0$ is not closed, so it has a boundary point $\zeta $. This point does not lie in $D$: otherwise the component of $D$ containing $\zeta $ would be an open neighborhood of $\zeta $, so it would meet $D_0$ and hence equal $D_0$, which does not contain its boundary point $\zeta $. Thus $\zeta \in \partial D$ and $\varlimsup _{D\ni x\to \zeta }h(x)\geq h(x_0)>0$, contradicting the hypothesis. Hence $h\leq \varepsilon $ on $D$, and letting $\varepsilon \downarrow 0$ gives $h\leq 0$.
Finally, if two harmonic functions on a bounded open set are continuous on its closure and agree on its boundary, apply the second assertion to the real and imaginary parts of their difference and of its negative. $\square$
Theorem 3.1.11 (Weyl's lemma).C9127FBA9AE8 Let $T\in \mathcal D'(\Omega )$ satisfy $\Delta T=0$. Then there is a unique smooth harmonic function $h$ on $\Omega $ such that $\langle T,\chi \rangle =\int _\Omega h\chi \, \mathrm dx$ for every $\chi \in C_ c^\infty (\Omega )$. In particular, a locally integrable function whose distributional Laplacian vanishes agrees almost everywhere with a harmonic function.
Proof. Recall from section 2.5 that $(T*\phi )(y)=\langle T,\phi (y-\cdot )\rangle $ for a test function $\phi $ on $\mathbb R^ d$ supported in $\overline{B(0,s)}$ and $y\in \Omega _ s$, and that $T_\varepsilon =T*\rho _\varepsilon $. By part (1) of proposition 2.5.3, $T*\phi $ is smooth on $\Omega _ s$ and $\partial ^\alpha (T*\phi )=T*\partial ^\alpha \phi $; moreover $\Delta T_\varepsilon =(\Delta T)_\varepsilon =0$, so $T_\varepsilon $ is a smooth harmonic function on $\Omega _\varepsilon $. We show that $T_\varepsilon $ converges to the required $h$, locally uniformly with all derivatives, by writing $T_\varepsilon =T*(\rho _\varepsilon *\rho _ r)$ near each point for a fixed $r>0$.
Fix $x\in \Omega $ and $r>0$ with $\overline{B(x,3r)}\subset \Omega $, and let $0<\varepsilon <r$. Since $\operatorname {dist}(x,\mathbb R^ d\setminus \Omega )>3r$, every point of $B(x,2r)$ has distance greater than $r$ from $\mathbb R^ d\setminus \Omega $, and every point of $B(x,r)$ has distance greater than $2r$; thus $B(x,2r)\subset \Omega _\varepsilon $ and $B(x,r)\subset \Omega _{2r}$. Let $y\in B(x,r)$. The closed ball $\overline{B(y,r)}$ lies in $B(x,2r)$, where $T_\varepsilon $ is harmonic, so the last assertion of theorem 3.1.7, applied to $T_\varepsilon $ on $B(x,2r)$, gives $T_\varepsilon (y)=\int T_\varepsilon (z)\rho _ r(y-z)\, \mathrm dz$. The test function $\chi =\rho _ r(y-\cdot )$ is supported in $\overline{B(y,r)}\subset \Omega _\varepsilon $, so (2.5.4) gives $T_\varepsilon (y)=\langle T,\chi *\rho _\varepsilon \rangle $. Substituting $z=w+v$ and using that $\rho _\varepsilon $ is even, we find
The function $\rho _\varepsilon *\rho _ r$ is a test function supported in $\overline{B(0,\varepsilon +r)}\subset \overline{B(0,2r)}$, so we have shown that
Since $\rho _\varepsilon *\rho _ r=\rho _ r*\rho _\varepsilon $, part (3) of lemma 2.5.1, applied on $\mathbb R^ d$ to $\rho _ r$, shows that $\partial ^\alpha (\rho _\varepsilon *\rho _ r)\to \partial ^\alpha \rho _ r$ uniformly for every multi-index $\alpha $ as $\varepsilon \to 0$. Let $C$ and $m$ be the constants of definition 2.4.1 for $T$ and the compact set $\overline{B(x,3r)}$, which contains the supports of $(\rho _\varepsilon *\rho _ r)(y-\cdot )$ and $\rho _ r(y-\cdot )$ for $y\in B(x,r)$. For $y\in B(x,r)$ and a multi-index $\alpha $, the difference $\partial ^\alpha T_\varepsilon (y)-\partial ^\alpha (T*\rho _ r)(y)$ is the pairing of $T$ with $w\mapsto \psi (y-w)$, where $\psi =\partial ^\alpha (\rho _\varepsilon *\rho _ r-\rho _ r)$, and the derivatives of order $\beta $ of this test function are $\pm (\partial ^\beta \psi )(y-w)$. Hence
which tends to zero. Thus $T_\varepsilon \to T*\rho _ r$ on $B(x,r)$, uniformly with all derivatives.
Every point of $\Omega $ is the center of such a ball. Therefore $h(y)=\lim _{\varepsilon \to 0}T_\varepsilon (y)$ exists for every $y\in \Omega $, and on each ball $B(x,r)$ as above it equals $T*\rho _ r$ and is the limit of $T_\varepsilon $ uniformly with all derivatives. Thus $h$ is smooth and $\Delta h=\lim _{\varepsilon \to 0}\Delta T_\varepsilon =0$. Every compact subset of $\Omega $ is covered by finitely many of these balls, so $T_\varepsilon \to h$ uniformly on compact sets, and part (2) of proposition 2.5.3 gives $\int h\chi \, \mathrm dx=\lim _{\varepsilon \to 0}\int T_\varepsilon \chi \, \mathrm dx=\langle T,\chi \rangle $ for every test function $\chi $. If two continuous functions define the same distribution, they agree almost everywhere by part (4) of proposition 2.5.3, and hence everywhere, since nonempty open sets have positive measure. This proves uniqueness. Finally, if $T$ is given by a locally integrable function $u$, then $u$ and $h$ define the same distribution, so $u=h$ almost everywhere by the same proposition. $\square$
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