3.3 Classical harmonic analysis4061AEE18413

Proof. Choose a closed ball $K\subset U$ whose interior contains $x$. All the functions in the hypotheses are bounded and uniformly continuous on $K\times C$. For sufficiently small $h\neq 0$, the fundamental theorem of calculus gives

This converges to $\partial _{x_ j}u(x,y)$ uniformly in $y\in C$. Integrating the uniform error against the finite measure $|\mu |$ proves the differentiation formula. The same uniform continuity, now applied to $\partial _{x_ j}u$, proves continuity of the derivative integral as $x$ varies inside $K$. Joint continuity of $u$ also makes $F$ finite and continuous. The ball $K$ can be chosen about every point of $U$, proving the assertion. $\square$

Proof. Fix $x_0\in \Omega _ r$ and choose $\varepsilon >0$ so that $\overline{B(x_0,r+\varepsilon )}\subset \Omega $. For $|x-x_0|<\varepsilon $,

The symmetric difference lies in $\{ y:r-|x-x_0|\leq |y-x_0|\leq r+|x-x_0|\} $, whose volume tends to zero. Absolute continuity of the integral of $|f|$ on the fixed compact ball therefore proves continuity.

Suppose now that $f$ is continuous. We prove the formula for a coordinate direction $e_ i$ by slicing the ball perpendicular to $e_ i$. For $w\in e_ i^\perp $, $|w|<r$, set $a(w)=\sqrt{r^2-|w|^2}$. Fubini and translation in the $i$th coordinate give, for small nonzero $h$,

The formula uses signed integrals and holds for either sign of $h$. On a fixed compact neighborhood of the closed ball, $f$ is uniformly continuous and bounded. Divide by $h$ and let $h\to 0$; dominated convergence on the $(d-1)$-dimensional disk yields

The graph area formula on the two hemispheres gives $\nu _ i\, dS_ r=dw$ on the upper graph and $\nu _ i\, dS_ r=-dw$ on the lower graph, where $\nu $ is the outward unit normal of $\partial B(0,r)$. The equator has surface measure zero. Consequently the last integral is $\int _{\partial B(0,r)} f(x+y)\nu _ i(y)\, dS_ r(y)$. Substitute $y=r\theta $: the surface element supplies $r^{d-1}$ and $\nu _ i(r\theta )=\theta _ i$. Dividing by $v_ dr^ d$ proves (3.3.5) for $\eta =e_ i$. The right side is continuous in $x$, by uniform continuity on compact sets; hence all coordinate derivatives are continuous and $A_ rf\in C^1$. Linearity in $\eta $ gives the stated directional formula.

If $f\in C^ k$, repeated differentiation under this compact sphere integral, justified by the same uniform convergence argument as in Lemma 3.3.2, shows that each $\partial _ i A_ rf$ has continuous derivatives of order at most $k$. Thus $A_ rf\in C^{k+1}$. The argument includes $k=0$ and $f=0$. $\square$

Proof. If $X$ is empty or $M=0$, every measure is zero and the assertion is immediate. First construct a countable dense subset of $C(X)$. For each positive integer $n$, choose a finite $1/n$-net $x_{n,1},\ldots ,x_{n,N_ n}$, and put

The denominator is everywhere positive. These continuous functions form a partition of unity, and $p_{n,i}(x)>0$ implies $\operatorname {dist}(x,x_{n,i})<2/n$. For $f\in C(X)$, choose rational $q_{n,i}$ within $1/n$ of $f(x_{n,i})$. Then $\sum _ iq_{n,i}p_{n,i}$ converges uniformly to $f$ by uniform continuity. All rational combinations for all $n$ form a countable dense family.

Enumerate that family as $(g_ l)$. For every $l$, the sequence $\int g_ l\, d\mu _ j$ is bounded by $M\| g_ l\| _\infty $. Successive subsequences and diagonal extraction yield a subsequence along which all these integrals converge. If $f\in C(X)$ and $g_ l$ is uniformly close to $f$, then

Hence the integrals of every $f$ are Cauchy. Their limit $L(f)$ is a real linear functional with $|L(f)|\leq M\| f\| _\infty $. Riesz representation gives the required $\mu $ with $\| \mu \| \leq M$. If the original measures are positive, $L(f)\geq 0$ for $f\geq 0$; the positivity part of Riesz representation makes $\mu $ positive. $\square$

Proof. Subtract $a\sigma $, whose Poisson integral is the constant $a$. Thus $F(s)=\mu (C_ s)=o(A(s))$, where $A(s)=\sigma (C_ s)$. In particular $\mu (\{ \zeta \} )=0$. Fix $0<\delta <2R$ so that $|F(s)|\le \varepsilon A(s)$ for $0<s\le \delta $. The ball Poisson kernel at $t\zeta $ is a function of $s=|y-\zeta |$:

Here $s_{d-1}$ is the area of the unit sphere. This function is decreasing and smooth for $s\ge 0$, $0<t<1$. Integration of $k_ t(s)=k_ t(\delta )+\int _ s^\delta (-k_ t'(v))\, dv$ against the pushforward of $\mu |_{C_\delta }$, using finite signed Fubini, gives

For each fixed $t$ all integrands are bounded and $\mu $ is finite, so this identity also holds when $F$ has jumps. Its absolute value is at most $\varepsilon \int _{C_\delta }k_ t\, d\sigma \le \varepsilon $, by positivity and normalization of the Poisson kernel. On $S\setminus C_\delta $, $\sup k_ t\to 0$, so the remaining integral tends to zero, bounded by $|\mu |(S)\sup _{s\ge \delta }k_ t(s)$. Let $\varepsilon \downarrow 0$. $\square$

Proof. If $A=\varnothing $, choose the empty sequence. Otherwise choose any $F_1$. After $F_1,\ldots ,F_ j$, stop if they cover $A$. If they do not, a point outside their closed union belongs to sufficiently small members $F$ with $N(F,\operatorname {diam}F)$ disjoint from that union. Let $e_ j>0$ be the supremum of their diameters, and choose such an $F_{j+1}$ with diameter $>2e_ j/3$. All chosen sets are disjoint. Fix $x\in A\setminus \bigcup _{i\le j}F_ i$ and choose a member $F$ containing $x$ whose diameter is so small that its diameter-neighborhood avoids the first $j$ sets. It must meet a later chosen $F_ k$ in that neighborhood. Otherwise it remains eligible forever, so all subsequent chosen diameters exceed $2\operatorname {diam}F/3$; choosing a point in each yields an infinite uniformly separated subset of compact $X$, impossible. At the first such $k$, $\operatorname {diam}F_ k>2\operatorname {diam}F/3$. Some point of $F_ k$ is within $\operatorname {diam}F$ of $F$, hence $d(x,F_ k)<2\operatorname {diam}F<3\operatorname {diam}F_ k$. This proves the covering assertion. In the measure case the outer measure of the uncovered set is at most $C\sum _{i>j}m(F_ i)$ for every $j$; the tail tends to zero because the sets are disjoint and $m(X)<\infty $. Completeness makes every subset of the resulting null set measurable. $\square$

Proof. Small spherical caps satisfy uniform estimates $c r^{d-1}\le \sigma (K)\le C r^{d-1}$, $r=\operatorname {diam}K$, and $\sigma (N_ S(K,3r))\le C'\sigma (K)$. To verify them, cover $S$ by finitely many graph charts over tangent hyperplanes on which the graph map and its inverse have bounded derivatives and the surface Jacobian is bounded above and below by positive constants. A small cap centered at $z$ contains and is contained in chart balls of radii constant multiples of $r$; its $3r$-neighborhood is contained in a chart ball of radius a constant multiple of $r$. Euclidean volume scaling gives the estimates; compactness supplies uniform constants. These observations also show that any small cap containing $\zeta $ lies in a chart cube centered at $\zeta $ of side $Cr$, while its area is at least $cr^{d-1}$.

The Euclidean Lebesgue differentiation theorem for $L^1_{\rm loc}$ functions on cubes therefore gives the stated absolute-oscillation limit for $f$ in each chart: the numerator is bounded by a constant times the integral of the coordinate oscillation on that cube, since the Jacobian is bounded; the denominator is bounded below by its volume times a positive constant. Null sets in coordinates and on $S$ agree. A finite chart cover proves the assertion almost everywhere. The precise external interface here is the checked cube Lebesgue differentiation theorem lebesgue differentiation of the Foundations source, not a boundary Fatou theorem.

For completeness the singular-measure step follows from the preceding covering lemma rather than an unstated differentiation theorem. If $\nu $ is a finite positive measure and

the good caps $\nu (K)>\lambda \sigma (K)$ form a fine cover of $E_\lambda $. The lemma, applied to $m=\sigma $, gives disjoint good caps covering this set up to a surface null set. Thus its outer surface measure is at most $\sum \sigma (K)\le \nu (S)/\lambda $. This argument does not presume measurability of $E_\lambda $. If $\nu $ is singular, choose a compact subset $L$ of a surface-null carrier with $\nu (S\setminus L)<\varepsilon $, by regularity. Outside $L$, every sufficiently small cap sees only $\nu |_{S\setminus L}$. The preceding bound makes the set where the upper density exceeds $\lambda $ have outer area at most $\varepsilon /\lambda $, because $\sigma (L)=0$. Let $\varepsilon \downarrow 0$ and then take rational $\lambda >0$. Hence the upper density is zero almost everywhere. Apply this to $|\mu _ s|$ and combine with the $L^1$ result. The derivative equals measurable $f$ off a null set, so every version defined on the remaining null set is measurable for the completed surface measure. $\square$

Proof. We first prove that a positive finite measure $\nu $ with zero spherical cap density at $\zeta $ has nontangential Poisson limit zero there. Put $t=R-|x|$ and $s=|y-\zeta |$. Since $|x-y|\ge t$ and $s\le |x-y|+At\le (1+A)|x-y|$,

The numerator in the exact ball kernel is at most $2Rt$; the displayed bound follows with an explicit finite constant depending only on $A,d$ and the unit-sphere area. Given $\varepsilon >0$, choose $\delta >0$ so that $\nu (C_ s)\le \varepsilon \sigma (C_ s)\le C_ R\varepsilon s^{d-1}$ for $0<s\le \delta $. For $2t<\delta $, integrate first on $C_ t$, then on the annuli $2^{j-1}t<s\le 2^ jt\le \delta $. The contributions are bounded by $C\varepsilon $ and $C\varepsilon 2^{-j}$, respectively. Choose the last radius $r=2^ jt$ so that $\delta /2<r\le \delta $; outside $C_ r$ the integral is at most $C t\delta ^{-d}\nu (S)$, which tends to zero. Thus the limsup is at most $C\varepsilon $, and then zero.

By spherical measure differentiation, outside a null set the positive measure $|\mu _ s|$ has zero cap density and $|f-f(\zeta )|\sigma $ has zero cap density. Since $P_ B\sigma =1$,

which tends to zero by the first part. The exceptional set used only these density conditions, not $A$; hence it works for every finite aperture. $\square$

Proof. This is a short consequence of the checked local Riesz decomposition. On a sufficiently small ball define $u=h+\Phi (\cdot -a)$ off $a$ and $u(a)=+\infty $. The fundamental function tends to infinity, and $h\ge 0$; thus $u$ is lsc and is superharmonic, using locality off $a$ and the trivial supermean inequality at $a$. It is locally integrable by the checked local integrability theorem. Its positive Riesz measure $-\Delta u$ is supported at $a$, so it equals $b\delta _ a$ with $b\ge 0$. Since $-\Delta \Phi =c_ d\delta _ a$, where $c_2=2\pi $ and $c_ d=(d-2)s_{d-1}$ otherwise, Weyl's lemma gives $u=(b/c_ d)\Phi +v$ almost everywhere. Equality holds everywhere off $a$ by continuity. Hence $h=(b/c_ d-1)\Phi +v$ there, and its nonnegativity near $a$ forces $c=b/c_ d-1\ge 0$. This also proves removability when $h$ is bounded, or when a one-sided bounded harmonic function has a finite limit along one approach to $a$. $\square$

Proof. First, odd reflection across a hyperplane preserves harmonicity for a harmonic function continuous and zero at that plane. The reflected function is continuous; at a center on the plane its spherical means vanish by the reflection symmetry, and away from the plane sufficiently small spheres have the usual harmonic mean property. The checked local mean characterization therefore proves harmonicity across the plane.

Translate $a=0$. Add a sufficiently large positive multiple of $\log (R/|x|)$ if $d=2$, or of $|x|^{2-d}-R^{2-d}$ otherwise, so that the result is nonnegative on the inner sphere. The minimum principle makes it nonnegative throughout the annulus. These added functions already extend harmonically across the outer sphere, so it suffices to treat $h\ge 0$. Choose $b$ on the outer sphere and invert about a sphere centered at $b$. The outer sphere minus $b$ becomes a hyperplane, and the inner closed ball becomes a compact ball strictly in one of its half-spaces. The Kelvin transform is harmonic outside that compact ball in the half-space, continuous and zero at its planar boundary. Reflect it oddly. Applying the inverse Kelvin transform gives a harmonic extension on a neighborhood of the original closed annulus, except possibly at $b$. On its exterior side this extension is nonpositive, since the reflected function is nonpositive and the Kelvin multiplier is positive. On the interior side it tends to zero at $b$, so it is bounded above on a punctured neighborhood of $b$. Apply the preceding lemma to a constant minus this extension. Its singular coefficient must vanish because the extension has limit zero along the interior approach. Thus the singularity at $b$ is removable. Compactness of the outer sphere gives one $\varepsilon >0$ valid around the whole sphere. Finally subtract the added radial harmonic function. $\square$