4.5.7 The Martin boundaryB6BAE13977A4

Proof. Compact-support functions separate interior points and recover their original neighborhoods by positive bumps, so the evaluation map is an embedding. Each bump's nonzero set in the closure lies in the image of its compact support; thus the interior image is open and its complement is the boundary. All compact-support coordinates vanish on the boundary, so the remaining coordinates separate its points. For any other such compactification the coordinate extensions define a continuous injection into the same product, with exactly this closure as image by density. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism, proving uniqueness fixing $\Omega $. $\square$

Proof. For $C\Subset W\Subset \Omega $ containing $a$, Harnack comparison in the pole variable, or sweeping onto a compact neighborhood of $C$, bounds $G(y,x)/G(y,a)$ uniformly for $x\in C$, $y\notin W$. For the sweeping proof, both pole functions equal their compact reductions off that neighborhood: they are bounded harmonic there and have the same boundary values outside a polar exception, so compactified uniqueness applies. Their boundary values on its compact interior boundary have uniform positive lower and finite upper bounds, giving the ratio bound. A general compact $C$ is enlarged to include $a$. Since $b=G(\cdot ,a)$ off a compact set, these bounds also hold for $k$. Normal convergence and the derivative estimates for bounded harmonic families show that every boundary kernel is harmonic, and that convergence in the compactification is locally uniform in $x$. This proves joint continuity at boundary points. Interior joint continuity is that of $G$, and $K_ y(a)=1$ on the boundary follows from the same identity off the cutoff compact. Positivity follows from the harmonic minimum principle.

Interior kernels are distinguished by their poles. A boundary kernel has no interior pole. For two boundary points, proportionality and the value one at $a$ imply equality of every defining coordinate, hence equality of the points. More explicitly, $b=G\lambda _0$ with compactly supported $\lambda _0$; therefore $\int K(y,x)d\lambda _0(x)=1$, first for interior $y$, then on the boundary by joint continuity near that compact support. This supplies the normalization and the ratio representation of defining coordinates used in Helms 12.4 as well. The countable coordinates in the displayed formula separate points: density and extended continuity in $x$ recover every kernel. They define a continuous injection of compact $M$ into a countable cube, hence a homeomorphism onto its image. The uniformly convergent sum metric on that cube induces the given topology. No sequential compactness is assumed before this argument. $\square$

Proof. Exhaust $F$ by compact $F_ j$. The compact reductions are potentials $G\nu _ j=K(b\nu _ j)$, with supports in $F_ j$. At a point $x$ where $u(x)$ is finite the strictly positive minimum of $K(\cdot ,x)$ on $M$ bounds their masses. Choose a weak limit $\mu $. For a sphere compactly inside $\Omega $, its average of $K(y,\cdot )$ is a finite continuous function of $y\in M$: away from the sphere use joint bounded continuity; on it use the proved continuity of spherical Green averages after cancellation of the integrable fundamental singularity. Thus weak convergence and monotone convergence of reductions identify all spherical averages of $K\mu $ and $\widehat R_ Fu$. Fatou first bounds $K\mu \le u$, so it is defined. Let the sphere radius tend to zero to get equality everywhere. For harmonic $u=h$, take $F=\Omega $ and compact neighborhoods $F_ j$ exhausting it. The reduction equals $h$ on their interiors and its measure is supported on their boundaries; these escape every interior compact, so the limit measure lies on $\Delta $.

The representing compact measure is unique by the interior Riesz identity. To prove continuity of $T_ C$, uniform continuity of $K$ on $\Delta \times C$ shows that $K_ y$ and $K_ z$ differ there by at most $\varepsilon $ when $y,z$ are close. Fix a positive harmonic kernel bounded below on $C$; its multiple $v$ majorizes one there. Order and subadditivity of reductions give $|K\gamma _{y,C}-K\gamma _{z,C}|\le \varepsilon \widehat R_ Cv$. Testing the signed bounded-density representation of $q\in C_ c^2$ above therefore bounds $|T_ Cq(y)-T_ Cq(z)|$ by a constant times $\varepsilon $. The masses are uniformly bounded since $\int K(t,a)d\gamma _{y,C}(t)=\widehat R_ CK_ y(a)\le 1$ and $\inf _{t\in C}K(t,a)>0$. Approximate restrictions of arbitrary $f\in C(M)$ uniformly on $C$ by smooth compactly supported functions; this proves continuity of $T_ Cf$. When the cutoff compact lies in $\operatorname {int}C$, the measure is on $\partial C$, where $K(t,a)=1$, and the reduction at $a$ equals $K_ y(a)=1$. Hence its mass is one, proving the last assertion. $\square$

Proof. Off a common polar set of irregular boundary points, closed sweeping is the identity on $F$ and the harmonic-measure integral off $F$, with zero on the original boundary. These kernels are Borel: for continuous compact tests their integrals are continuous in the pole; a countable monotone class gives indicators and all nonnegative Borel tests. Since the Euclidean second factor is second countable, its product Borel sigma algebra is generated by rectangles. Thus the reduced family is jointly Borel off that common polar set. Extend it by zero there and recover its actual canonical values by limits of ball averages; these averages are jointly Borel, so the recovered family is jointly Borel everywhere. The integral is lsc by Fatou and super-mean-valued by Tonelli, and is finite at $x_0$ because reduction is at most $u_ t$. Outside the common polar set, Tonelli in the harmonic-measure formula identifies it with the reduction of the integral. Canonical ball averages give equality everywhere. Only a sigma-finite effective parameter set is needed: the sets $\{ u_ t(x_0)>1/m\} $ have finite measure, and a nonnegative superharmonic function zero at $x_0$ vanishes identically. For potentials, lower harmonically along a regular exhaustion. Each minorant decreases to zero; its integral is dominated at $x_0$ by the integrable $u_ t(x_0)$. Dominated convergence gives zero greatest harmonic minorant of the integral. Interior uniqueness proves the measure identity. $\square$

Proof. We supply the uniqueness mechanism rather than assuming a simplex theorem. If a nonzero $K\mu $ generates an extreme ray of positive superharmonic functions, every restriction of $\mu $ to a neighborhood of a support point has potential proportional to $K\mu $. Normalize these restrictions to represent $K\mu $. At a finite interior test point away from the support point, positivity bounds their masses. Shrinking the neighborhoods makes their kernels converge uniformly at each other test point, so their limit is $c\delta _ y$ and $K\mu =cK_ y$. Nonproportionality of distinct kernels forces the support to be that single point. For harmonic $K\mu $ its support is on $\Delta $, so the point is in $\Delta _1$.

Choose nested compact neighborhoods $C_ j$ exhausting $\Omega $ and containing the cutoff compact in their interiors. Put $T_ j=T_{C_ j}$. For $y\in \Delta _1$, every weak cluster of the probability measures $\gamma _{y,C_ j}$ represents $K_ y$: their supports escape interior compacts, where the kernel singularities lie, and their potentials equal $K_ y$ eventually at each interior test point. Extremality and the preceding support assertion force this cluster to be $\delta _ y$. Thus $\gamma _{y,C_ j}\to \delta _ y$. Conversely, if this convergence holds along an exhaustion and $K_ y=K\mu +K\eta $, additivity and interior uniqueness give $\gamma _{y,C_ j}=\mu _{C_ j}+\eta _{C_ j}$. Weak clusters of these positive measures sum to $\delta _ y$ and represent the respective harmonic functions. Each is proportional to $\delta _ y$, proving extremality. This convergence criterion also makes $\Delta _1$ Borel: require convergence of $T_ jf_ l(y)$ to $f_ l(y)$ for a countable dense family, expressed by countable intersections and unions of the open error sets. These arguments also show that for any measure $\nu $ carried by $\Delta _1$, $\nu _{C_ j}\to \nu $ by dominated integration of $T_ jf\to f$. Consequently two such measures representing the same function agree.

For existence start with any boundary representing measure $\mu $ and take a subsequence $\mu _{C_ j}\to \nu $, still supported on $\Delta $ in the limit and representing $K\mu $. For this subsequence let $\mathcal L$ be the continuous functions $f$ on $M$ for which $T_ jf\to f$ in $\nu $-measure. It is a linear space containing constants. It is a lattice: if $g_1,g_2\in \mathcal L$, positivity gives $T_ j(g_1\wedge g_2)\le T_ jg_1\wedge T_ jg_2$. The right side tends in measure and integral to $g_1\wedge g_2$. The integral of the left side is $\int (g_1\wedge g_2)d\nu _{C_ j}= \int (g_1\wedge g_2)d\mu _{C_ j}$, tending to the same value by the chosen weak convergence. Their nonnegative difference therefore tends to zero in $L^1$, proving the lattice assertion. The space contains $C_ c(\Omega )$, whose restrictions and $T_ j$ values on $\Delta $ are eventually zero. It also contains every continuous function agreeing with $K(\cdot ,x)$ outside an interior compact: for large $j$, $T_ jf=K(\cdot ,x)=f$ on $\Delta $. These functions separate boundary points, while compact bumps separate the other points. Lattice Stone–Weierstrass makes $\mathcal L$ dense in $C(M)$. The unit contraction bound for $T_ j$ makes it uniformly closed, hence all of $C(M)$. For a countable dense family choose a common subsequence converging $\nu $-almost everywhere. At almost every $y$, $\gamma _{y,C_ j}\to \delta _ y$ along this subsequence, which is still an exhaustion; the converse criterion makes $y\in \Delta _1$. This proves existence and the mass identity follows by evaluation at $a$.

For the topology choose a countable dense family $f_ l$ in $C(M)$. The criterion just proved identifies $\Delta _1$ with

This formula requires simultaneous approximation of finitely many tests. Each inner finite intersection is relatively open; a diagonal subsequence proves its converse. Since $\Delta $ is closed in compact metric $M$, it is itself $G_\delta $, proving the assertion in $M$. $\square$

Proof. Compactness of $M$ and the interior strong minimum principle prove the boundary minimum principle, so the upper/lower Perron constructions and harmonic alternatives apply. If $y\in \Delta _1$ and $U$ is a neighborhood of $y$ in $M$, then $p=\widehat R_{\Omega \setminus U}K_ y$ is a potential. Indeed its greatest harmonic minorant, being below the minimal $K_ y$, is $tK_ y$, $0\le t\le 1$. The case $t=1$ would represent the extreme $K_ y$ on $M\setminus U$, impossible by the support assertion. Write $p=w+tK_ y$ with $w$ a potential. Idempotence and additivity of reduction give $w+tK_ y=\widehat R_{\Omega \setminus U}w+tw+t^2K_ y$. Uniqueness of Riesz decomposition yields $t=t^2$, hence $t=0$.

Extend $f$ continuously to $M$ and first assume $0\le f\le 1$. Partition $\Delta _1$ into the bins $E_ i=\{ (i-1/2)/j\le f<(i+1/2)/j\} $, $0\le i\le j$, and put $h_ i=K(\chi |_{E_ i})$, so $\sum h_ i=1$. Let $F_ i=\{ x\in \Omega :|f(x)-i/j|\ge 1/j\} $. The preceding local-potential assertion and the integral-sweeping lemma make $p_ i=\widehat R_{F_ i}h_ i$ a potential. On $F_ i$, $p_ i=h_ i$ quasi-everywhere; off $F_ i$, $(i-1)/j<f<(i+1)/j$. Therefore

Initially the inequalities hold quasi-everywhere; canonical ball averages and continuity of $f$ give them everywhere. The left and right sides are bounded lower and upper competitors respectively. Taking harmonic minorants removes their potential terms: every harmonic minorant of a potential is at most zero. Thus the lower/upper solutions are between $\sum _{i\ge 1}(i-1)h_ i/j$ and $\sum _{i\ge 0}(i+1)h_ i/j$, a gap at most $2/j$. The same bounds enclose $K(f\chi )$. Letting $j\to \infty $ proves the formula, and scaling and adding constants handle arbitrary continuous data.

For the ball, expand $G(r\theta ,x)$ in $R-r$, uniformly for $\theta $ near a specified boundary point. The reference normal derivative is nonzero and bounded away from zero there; the quotient therefore tends along every approach to the quotient of normal derivatives. The explicit ball kernel gives precisely the normalized Poisson kernel displayed above. It separates sphere points, so uniqueness of function compactification identifies $M$ with the closed ball. We also verify the uniqueness needed in this example. For continuous sphere data $g$, the symmetry $P_ B(t\zeta ,y)=P_ B(ty,\zeta )$ and the ball approximate-identity estimate give

uniformly in $y$ as $t\uparrow 1$. Uniformity follows from uniform continuity of $g$, kernel mass one and a uniformly vanishing distant tail. Finite signed Fubini therefore recovers every continuous integral of a boundary measure from its interior Poisson function; equal functions have equal boundary measures. Now if $0\le h\le K_ y$, represent both $h$ and $K_ y-h$ by positive ball boundary measures. Their sum is the point mass representing $K_ y$, by this uniqueness, so each is supported on $\{ y\} $. Thus the sphere kernels are minimal and $\Delta _1=\Delta $. This argument supplies uniform control even for tangential approaches. $\square$