284 CHAPTER 11. INTEGRATION ON MANIFOLDS

Definition 11.8.5 For U a bounded open set and g ∈C (∂U), define

wg (x)≡ sup{

u(x) : u ∈ Sg}

where Sg consists of those functions u which are subharmonic with u(y) ≤ g(y) for ally ∈ ∂U and u(y)≥min{g(y) : y ∈ ∂U} ≡ m.

Note that Sg ̸= /0 because u(x) ≡ m is in Sg. Also all functions in Sg have valuesbetween m and max{g(y) : y ∈ ∂U}. The fundamental result is the following amazingresult.

Proposition 11.8.6 Let U be a bounded open set and let g ∈ C (∂U). Then wg ∈ Sgand in addition to this, wg is harmonic.

Proof: Let B(x0,2r)⊆U and let {xk}∞

k=1 denote a countable dense subset of B(x0,r).Let {u1k} denote a sequence of functions of Sg with the property that limk→∞ u1k (xl) =wg (xl) . By Lemma 11.8.4, it can be assumed each ulk is a harmonic function in B(x0,2r)and continuous on B(x0,r) since otherwise, you could use the process of replacing u withux0,2r. Now define wk = (max(u1k, · · · ,ukk))x0,2r . Then each wk ∈ Sg, each wk is harmonicin B(x0,2r), and for each xl , limk→∞ wk (xl) = wg (xl) .

From the representation theorem for harmonic functions, 11.17, if x ∈ B(x0,r)

wk (x) =1

ωn−12r

∫∂B(x0,2r)

wk (y)r2−|x−x0|2

|y−x|ndσ (y) (11.18)

and so there exists a constant C which is independent of k such that for all i = 1,2, · · · ,nand x ∈ B(x0,r),

∣∣∣ ∂wk(x)∂xi

∣∣∣ ≤ C. Indeed, you could differentiate under the integral signwith respect to the xi. The wk are all bounded functions thanks to the maximum princi-ple Proposition 11.7.9. Therefore, this set of functions, {wk} is equicontinuous on B(x0,r)as well as being uniformly bounded, thanks to the mean value inequality, and so by theAscoli Arzela theorem, Theorem 3.10.5, it has a subsequence which converges uniformlyon B(x0,r) to a continuous function I will denote by w which has the property that for allk, w(xk) = wg (xk). Also since each wk is harmonic,

wk (x) =1

ωn−1r

∫∂B(x0,r)

wk (y)r2−|x−x0|2

|y−x|ndσ (y) (11.19)

Passing to the limit in 11.19 using the uniform convergence, it follows

w(x) =1

ωn−1r

∫∂B(x0,r)

w(y)r2−|x−x0|2

|y−x|ndσ (y) (11.20)

which shows that w is also harmonic. I have shown that w = wg on a dense set. Also, itfollows by definition of wg that w(x) ≤ wg (x) for all x ∈ B(x0,r). It remains to verifythese two functions are in fact equal. By Lemma 11.8.1 wg is lower semicontinuous on U .Let x ∈ B(x0,r) and pick xkl → x where

{xkl

}is a subsequence of the dense set, {xk}.

Thenwg (x)≥ w(x) = lim inf

l→∞w(xkl

)= lim inf

l→∞wg(xkl

)≥ wg (x) .

This proves w = wg and since w is harmonic, so is wg. ■It remains to consider whether the boundary values are obtained. This requires an

additional assumption on the set U.

284 CHAPTER 11. INTEGRATION ON MANIFOLDSDefinition 11.8.5 For U a bounded open set and g € C (OU), defineWe (x) = sup {u(a) :u € Se}where Sg consists of those functions u which are subharmonic with u(y) < g(y) for ally € OU and u(y) > min{g(y):y € OU} =m.Note that S, 4 @ because u(x) =m is in S,. Also all functions in S, have valuesbetween m and max{g(y):y € JU}. The fundamental result is the following amazingresult.Proposition 11.8.6 Let U be a bounded open set and let g € C(QU). Then we € Sgand in addition to this, wg is harmonic.Proof: Let B(ao,2r) C U and let {a,};_, denote a countable dense subset of B (20,7).Let {uj} denote a sequence of functions of S, with the property that limy_,.. U1, (#7) =We (a). By Lemma 11.8.4, it can be assumed each uj; is a harmonic function in B (a9, 2r)and continuous on B(a,r) since otherwise, you could use the process of replacing u withUay,2r- Now define wz = (max (ujx, °° Uk) )e92r- Then each wz € Sz, each w; is harmonicin B(ao,2r), and for each a7, limy—soo Wg (@7) = We (7).From the representation theorem for harmonic functions, 11.17, if x € B(ao,7r)1 ?? —|e%—a|*wz (a) = w, —_——_,—_do 11.18§ ©) = 5 igas 1g") yar ow) (11.18)and so there exists a constant C which is independent of k such that for all i= 1,2,--- ,nduplr)Xjwith respect to the x;. The wz are all bounded functions thanks to the maximum princi-ple Proposition 11.7.9. Therefore, this set of functions, {w,} is equicontinuous on B (a,r)as well as being uniformly bounded, thanks to the mean value inequality, and so by theAscoli Arzela theorem, Theorem 3.10.5, it has a subsequence which converges uniformlyon B(ao,r) to a continuous function I will denote by w which has the property that for allk, w(ax) = We (ax). Also since each wx is harmonic,and @ € B(a0,r),|<C. Indeed, you could differentiate under the integral sign1 r- |a — xo”Wz (x) = w ——,_do (11.19)12) = Doe, 9) yar 40)Passing to the limit in 11.19 using the uniform convergence, it follows1 r — |x —a|"w(a@) = w(y) —__——_,, —do 11.20©) = 5 cay Oy nar 00) (11.20)which shows that w is also harmonic. I have shown that w = wz on a dense set. Also, itfollows by definition of we that w(a) < we (x) for all 2 € B(a,r). It remains to verifythese two functions are in fact equal. By Lemma 11.8.1 wg is lower semicontinuous on U.Let x € B(ao,r) and pick a, > @ where {a,, } is a subsequence of the dense set, {x,}.ThenWe (x) > w(x) = lim inf w (xx,) = lim inf we (ax,) > We (a).This proves w = wg and since w is harmonic, so is wg.It remains to consider whether the boundary values are obtained. This requires anadditional assumption on the set U.