RIESZ REPRESENTATION OF SUBHARMONIC FUNCTIONS ON COMPACT SETS T.L. PERKINS Abstract. The goal is to establish a Riesz Representation Theorem for subharmonic functions on compact sets.

1. Prerequisites First we establish a few classical results and definitions, c.f. [Conw91]. Let Ω be a domain in C. Recall that a function is φ : Ω → [−∞, ∞) is subharmonic if φ is upper semicontinuous and for every closed disk ¯ r) contained in Ω, we have the inequality B(a; Z 1 φ(a + reiθ )dθ. φ(a) ≤ 2π A function ψ : Ω → (−∞, ∞] is superharmonic if −ψ is subharmonic. We sa that a function φ : Ω → [−∞, ∞) satisfies the Maximum Principle if for every compact set K contained in Ω, φ ≤ h on K whenever h is a continuous function on K that is harmonic on int(K) and satisfies φ ≤ h on ∂K. It is known that an upper semicontinuous φ on Ω is subharmonic if and only if φ satisfies the Maximum Principle. If φ : Ω → R is a C 2 function such that ∆φ ≥ 0 on Ω, then φ is subharmonic. Then for u : ∂Ω → R is any bounded functions, we define the following set Pˆ = {φ : φ is subharmonic on Ω and lim sup φ(z) ≤ u(a) for every a in ∂Ω}. z→a

Also, we can define functions uˆ on Ω by uˆ(z) = sup{φ(z) : φ ∈ Pˆ (u, Ω)}. If u ∈ CR (∂Ω) and there is a solution h of the Dirichlet Problem with boundary values u, the uˆ = h. Furthermore, if a ∈ Ω, then the map u → uˆ(a) is a positive linear functional of norm 1. Date: November 23, 2008. 1991 Mathematics Subject Classification. Primary: 32F05; secondary: 32E25, 32E20. Key words and phrases. Subharmonic functions, potential theory. 1

For any bounded open set Ω and any point a in Ω, the unique probability measure ωa supported on ∂Ω and satisfying Z uˆ(a) = u dωa ∂Ω

for every u in CR (∂Ω) is called harmonic measure for Ω at a. In what follows we will let K ⊂ C be a connected compact set. Definition 1.1. A function is said to be subharmonic on K if it can be approximated uniformly by continuous subharmonic functions on neighborhoods of K. The space of functions subharmonic on K shall be denoted SH(K). A function u is said to be superharmonic on K if −u ∈ SH(K). A function is said to be harmonic on K if it can be approximated uniformly by functions harmonic in a neighborhood of K. The space of functions which are harmonic on K shall be denoted h(K). It follows that a function is harmonic on K if and only if it is both subharmonic on K and superharmonic on K, see Corollary 5.2. 2. On the Potential boundary in Rn In this section we will show that the potential boundary is infact just the topological boundary in Rn . A point x ∈ K is a (subharmonic) peak point of K if there exists a subharmonic function u on K so that ||u|| = 1, ||u(x)|| = 1, and ||u(y)|| < 1 for all y ∈ K − x. Let OK denote the set of peak points of K. The potential boundary, BK , is the closure of OK . Let Br (x) be the ball of radius r centered at x and Sr (x) = ∂Br (x). Let σ denote the surface measure on Sr (x). Let ωn denote the surface area of the unit n-sphere. Theorem 2.1. Let K ⊂ Rn be a compact set. The set of peak points is dense in the boundary of K. Proof. Consider x0 ∈ ∂K. Suppose we have an arbitrary ball centered at x0 . Then it contains a point y0 which does not belong to K. Suppose r0 = ||y0 − x0 ||. From now on we will call B = Br0 (x0 ). Let H be the hyperplane tangent to B at y0 . It is given by the equation H = {x ∈ Rn : (x, y0 − x0 ) = r02 }. Let us find the maximal t < r such that the hyperplane Ht = {x ∈ Rn : (x, y0 − x0 ) = tr0 } contains some point x1 ∈ K ∩ B. Since x0 ∈ K the number t ≥ 0. There are two possibilities: firstly, x1 ∈ B or, secondly, x1 ∈ ∂B. In the first case for every sufficiently small r > 0 all points y of the 2

sphere Sr (x1 ) for which (y, y0 − x0 ) > tr0 lie in the complement E of K. Hence σ(Sr (x1 ) ∩ E) > σ(Sr (x1 ))/2 and 1 σ(Sr (x1 ) ∩ E) ≥ . lim inf r→0 σ(Sr (x1 )) 2 In the second case, we take a small neighborhood V of y1 in ∂B, lying in the set {y ∈ ∂B : (y, y0 − x0 ) > tr0 } and note that due to convexity all points of the intervals connecting x1 with y ∈ V , except x1 , lie in the set {y ∈ B : (y, y0 − x0 ) > tr0 } and, consequently, in E. Since the rays x1 + sy, s > 0, y ∈ V , form a cone of positive aperture with vertex at x1 we see that there is a constant c > 0 such that σ(Sr (x1 ) ∩ E) > cωn rn when r > 0 is sufficiently small. Hence σ(Sr (x1 ) ∩ E) lim inf ≥ c > 0. r→0 σ(Sr (x1 )) Recall, [Helm69, Corollary 10.5, p. 211], that if E is thin at a point x then σ(Sr (x) ∩ E) = 0. lim inf r→0 σ(Sr (x)) Thus the E is non-thin at x1 and by of Poletsky [Pole97, Theorem 3.3] x1 is a peak point of K. Therefore in any arbitrary neighborhood of x we can find a peak point. Thus the set of peak points are dense in the topological boundary.  Theorem 2.2 ( [Pole96, Lemma 4.1] ). The set of peak points, OK is the set of subharmonic peak points on K. Corollary 2.3. The potential boundary equals the topological boundary for any compact set K ⊂ Rn . Proof. To see this we recall [Pole96, Lemma 4.1] that the set OK is also the set of (pluri)subharmonic peak points on K. Since the potential boundary, BK , is the closure of OK , which is dense in the topological boundary by the theorem, we have that the potential boundary must be the topological boundary, i.e. BK = ∂K.  3. Convergence Results Let K ⊂ C be a connected compact set in the plane. T We can find a decreasing sequence of domains Uj such that K = Uj . Then for 2 any u ∈ C ( Uj ) we have the Riesz (Green’s) representation [GiTr01, pages 17–19] Z Z u(z) = u(ζ) dωj (z, ζ) + Gj (z, ζ) ∆u(ζ), Uj

∂Uj

3

where ωj is the harmonic measure on the boundary of Uj and Gj is the Green function on Uj . One of the goals of this paper is to study what happens when j → ∞. In this paper we take our Green functions to be negative and subharmonic. 4. The Left Half Theorem 4.1. T If {Uj } a decreasing sequence of domains such that K ⊂ Uj , K = Uj , then ωj (z, ·) converges weak-∗. Definition 4.2. As the ωj are weak-∗ converging, we define the weak-∗ limit of ωj (z, ·) to be ωK (z, ·). For the proof we need the following lemma. Lemma 4.3. Let u ∈ C02 (C). Then u can be decomposed into a subharmonic function plus a superharmonic function. Proof. Since u is in C02 (C), we know that ∆u is continuous on K. As a continuous function on a compact set is uniformly bounded, we have that ∆u is bounded. Then there exists C > 0 such that ∆u < C. Define the subharmonic function usub = C|z|2 . Then define usup = u − usub , which has the property ∆usup = ∆(u − usub ) = ∆u − C < 0, and is therefore superharmonic. Therefore u has the desired decomposition, u = usub + usup .  For the moment we would like to only consider the left integral from above, Z u(ζ) dωj (z, ζ). ∂Uj

Proof of Theorem 4.1. Since u is a continuous function defined on K. We may continuously extend u to all of C. For convenience this extension shall also be called u. The solution of the Dirichlet problem on Uj with boundary value u will be denoted uj . Then Z uj (z) = u(ζ) dωj (z, ζ). ∂Uj

Assume first that the extension u is subharmonic, then uj ≥ u on Uj . Since uj+i = u on ∂Uj+1 and uj ≥ u = uj+1 on ∂Uj+1 , the maximum principal for harmonic functions implies that uj ≥ uj+1 on Uj+1 . Thus 4

{uj } is a decreasing sequence on K, which converges to a function say v. Since uj ≥ u implies that v ≥ u. If we repeat the process for a superharmonic u we get an increasing sequence {uj } of superharmonic functions, which converges to a function. Now for any u ∈ C02 (C) we may decompose it 4.3 into a subharmonic plus a superharmonic as say usub + usup . Therefore we see that there is convergence of Z u(ζ) dωj (z, ζ), ∂Uj

for this class of functions. In other words that the measures ωj (z, ·) are weak-∗ converging for functions in C02 (C). However as C02 (C) is dense in C(C) we may conclude that the measures are weak-∗ converging on all of C(C), or just weak-∗ converging.  Theorem 4.4. If {φj } is a monotone sequence of continuous functions on a compact set K ⊂ C which is converging pointwise to a continuous function φ, then φj converges uniformly to φ. Proof. Fix ε > 0. Let Vj = {z ∈ K : |φj (z) − φ(z)| < ε}, which are open as φj and φ are continuous. Since the φj are monotone, we have that Vj ⊂ Vj+1 . Since the φj are converging pointwise, for every z ∈ K there existsSa J > 0 so that for all j > J z ∈ Vj . Therefore we have that K = Vj . So we have an open covering of K which is compact, and there must be a finite subcovering. As the Vj are increasing there must be a largest one, V0 , in the subcovering. Then V0 = K. Thus for any ε > 0 there is a J > 0 so that for all j > J, |φj (z) − φ(z)| < ε for all z ∈ K, so that the φj are converging uniformly to φ.  5. Harmonic Functions Recall that we’ve said a function is harmonic on K if it can be uniformly approximated by functions harmonic in a neighborhood of K. Theorem 5.1. A continuous function h on K is harmonic on K if and only if it has the representation Z (1) h(z) = h(ζ) dωK (z, ζ) ∂K

for all z in K. Proof. Suppose h ∈ h(K). Then we have a sequence of functions hl each harmonic in a neighborhood of K and converging uniformly to h on K. Since hl is harmonic in a neighborhood of K there exists a 5

N > 0 such that for all j > N , hl will be harmonic on Uj . Then as hl is harmonic on Uj we have Z hl (z) = hl (ζ) dωj (z, ζ). ∂Uj

However the measures ωj (z, ·) are weak-∗ converging to ωK (z, ·), so that Z Z hl (z) = lim hl (ζ) dωj (z, ζ) = hl (ζ) dωK (z, ζ). j→∞ ∂Uj

∂K

Since the hl converge uniformly to h, we have the desired representation (1) Z Z h(z) = lim hl (z) = lim hl (ζ) dωK (z, ζ) = lim hl (ζ) dωK (z, ζ) l→∞

l→∞ ∂K

l→∞

∂K

Z =

h(ζ) dωK (z, ζ). ∂K

Now conversely, suppose that h has the representation (1). We ex˜ and then define the harmonic tend h continuously to all of U1 as h, functions Z ˜ h(ζ) dωj (z, ζ). hj (z) = ∂Uj

Notice that hj converges pointwise to h, Z Z ˜ ˜ h(ζ) dωj (z, ζ) = h(ζ) dωK (z, ζ) lim hj (z) = lim j→∞

j→∞ ∂Uj

∂K

Z =

h(ζ) dωK (z, ζ) = h(z). ∂K

Suppose that h is subharmonic and continuous on U1 . Then the hj form a decreasing sequence which converges pointwise to h, so that by Theorem 4.4 the sequence converges uniformly. Now similarly if h is superharmonic and continuous on U1 . Then the hj form an increasing sequence which converges pointwise to h, so that by Theorem 4.4 the sequence converges uniformly. Now for h ∈ C02 (C), by Lemma 4.3, we may decompose h as h = hsub + hsup where hsub is subharmonic and hsup is superharmonic. Then we have that hj = hsub + hsup where hsub j j j sup sub converges uniformly to h and hj converges uniformly to hsup , so that hj converges uniformly to h. 6

˜ our original extension. We have hj converging pointwise Now for h, ˜ Fix ε > 0. By density we can find f ∈ C 2 (U1 ) so that |fj −hj | ≤ ε to h. which shows that the fj are converging uniformly to fˆ. Hence |fj −f | ≤ ε for all j > j0 . Therefore |hj − hk | = |(hj − f ) − (hk − f )| ≤ |hj − f | + |hk − f | = |hj − fj + fj − f | + |hk − fk + fk − f | ≤ |hj − fj | + |fj − f | + |hk − fk | + |fk − f | ≤ 4ε, for all j, k > j0 .



Corollary 5.2. A function is harmonic on K if and only if it is subharmonic on K and superharmonic on K. Proof. Suppose h is in h(K). Then h can be uniformly approximated by functions harmonic in a neighborhood of K. Since a harmonic function is subharmonic and superharmonic, we have that h can be uniformly approximated by functions subharmonic in a neighborhood of K and h can be uniformly approximated by functions superharmonic in a neighborhood of K. Suppose h is in SH(K) and −h is in SH(K). Let {fl } be a sequence of functions subharmonic in a neighborhood of K which converges uniformly to h, and let {gl } be a sequence of functions subharmonic in a neighborhood of K which converges uniformly to −h. Since fl is subharmonic in a neighborhood of K, there exists a J > 0 such that fl is subharmonic on Uj for all j > J. Then Z fl (z) ≤ fl (ζ) dωj (z, ζ), ∂Uj

and the ωj (z, ·) are weak-∗ converging so that Z fl (z) ≤ fl (ζ) dωK (z, ζ). ∂K

Since the fl converge uniformly to h we have that Z h(z) ≤ h(ζ) dωK (z, ζ). ∂K

7

Similarly since gl is subharmonic in a neighborhood of K, there exists a J > 0 such that gl is subharmonic on Uj for all j > J. Then Z gl (z) ≤ gl (ζ) dωj (z, ζ), ∂Uj

and the ωj (z, ·) are weak-∗ converging so that Z gl (z) ≤ gl (ζ) dωK (z, ζ). ∂K

Since the gl converge uniformly to −h we have that Z −h(z) ≤ −h(ζ) dωK (z, ζ), ∂K

or equivalently that Z h(z) ≥

h(ζ) dωK (z, ζ). ∂K

Therefore

Z h(z) =

h(ζ) dωK (z, ζ). ∂K

Thus by Theorem 5.1 we have that h is harmonic on K.



Definition 5.3. A point z0 ∈ K is a harmonic peak point, if there exists h ∈ h(K) such that h(z0 ) = 1, ||h|| = 1 and |h(y)| < 1 for y ∈ K, y 6= z0 . The function h is said to peak at z0 . We will denote the set of harmonic peak points on K as Oh(K) . The following theorem is a generalization of Bishop’s “ 41 − 34 criterion.” The theorem and it’s proof come directly from Gamelin’s Uniform Algebras Chapter II Theorem 11.1 [Game84]. Theorem 5.4. Suppose there are constants 0 < c < 1 and M ≥ 1 with the following property: For every neighborhood U of z0 , there exists f ∈ h(K) such that f (z0 ) = 1, ||f ||K ≤ M , and |f (y)| ≤ c for y ∈ K \ U . Then z0 is a harmonic peak point of K. Proof. Choose 0 < s < 1 sufficiently close to 1 so that M − 1 + s(c − M ) < 0. Choose a sequence {n }∞ n=0 decreasing to 0 so that n (1 − sn ) + sn (M − 1 + s(c − M )) < 0, 8

n ≥ 1.

Let {Fn }∞ n=0 be a sequence of closed subsets of K such that ∞ [

Fn = K \ {z0 }.

n=0

We choose a sequence of functions {hn }∞ n=0 by induction, as follows. Take h0 = 1. Supposing h0 , . . . , hn have been chosen, let Wn = {y : max |hj (y)| ≥ 1 + n }. 1≤j≤n

Take hn+1 ∈ h(K) such that hn+1 (z0 ) = 1, ||hn+1 ||K ≤ M, and ||hn+1 ||Wn ∪Fn ≤ c. Define ∞ X h = (1 − s) sj hj . j=0

The h ∈ h(K), while h(z0 ) = 1. S Suppose y 6= z0 does not belong to ∞ n=0 Wn . Then |hj (y)| ≤ 1 for 0 ≤ j < ∞. Also, y ∈ Fj for some j, so |hj (y)| < 1 for at least one index j. Consequently S∞|h(y)| < 1. Now suppose y ∈ n=0 Wn . Since the Wn are an increasing sequence of closed sets, there is an index m such that y ∈ Wm while y 6∈ Wm−1 . Then we have the estimates |hj (y)| ≤ 1 + m−1 ,

1≤j ≤m−1

|hm (y)| ≤ M |hj (y)| ≤ c,

j > m.

Consequently, ( |h(y)| ≤ (1 − s) (1 + m−1 )

m−1 X j=0 m

m

sj + M s m + c

∞ X

) sj

j=m+1

= 1 + m−1 (1 − s ) + s [M − 1 + s(c − M )] < 1. Hence |h(y)| < 1 for all y ∈ K such that y 6= z0 , and h peaks at z0 .  Lemma 5.5. If z0 is a harmonic peak point of K, then for any constants 0 < c < 1 and M ≥ 1 and for every neighborhood U of z0 , there exists f ∈ h(K) such that f (z0 ) = 1, ||f ||K ≤ M , and |f (y)| ≤ c for y ∈ K \ U. Proof. Suppose z0 is a harmonic peak point of K. Then there exist a peaking function f ∈ h(K) which peaks at z0 . Fix the constants 0 < c < 1 and M ≥ 1 and a neighborhood U of z0 . Then for a large constant k, f˜ = k(f − 1) + 1, will satisfy f˜(z0 ) = 1, |f˜| ≤ M , and |f˜(y)| ≤ c for y ∈ K \ U and f˜ ∈ h(K).  9

Putting the two together we have the following theorem. Theorem 5.6. The point z ∈ K is a harmonic peak point of K if and only if for any constants 0 < c < 1 and M ≥ 1 and for every neighborhood U of z0 , there exists f ∈ h(K) such that f (z0 ) = 1, ||f ||K ≤ M , and |f (y)| ≤ c for y ∈ K \ U . Furthermore Gamelin shows us that this implies that the set of peak points is a Gδ . Corollary 5.7. The set of harmonic peak points of a compact set K in the plane is a Gδ -set. Proof. Let Un be the set of all x ∈ K for which there exists a function f ∈ h(K) satisfying ||f || = 1, |f (x)| > 12 , and |f (y)| < 41 when |x−y| ≥ 1 . Then each Un is open. n If x is a peak point, then there is a function f ∈ h(K) peaking at x. As in the previous lemma, we can find a large constant k so that f˜ = k(f − 1) + 1 will peak at x will be in h(K) and satisfy ||f˜|| = 1, |f˜(x)| > 12 , and |f˜(y)| < 14 when |x − y| ≥ n1 . So x ∈ Un for all n. Conversely, if x ∈ Un for all n, then appropriate multiples of the T U functions will satisfy the hypothesis of 5.4. Consequently ∞ n=1 n coincides with the set of peak points of h(K).  The next very useful theorem follows almost immediately from this result. Again the theorem and proof is taken (mostly) from Gamelin’s Uniform Algebras; this time it is Chapter II Theorem 11.3 [Game84] Theorem 5.8. Let K ⊂ C be a compact set. Then z0 is a harmonic peak point of K if and only if ωK (z0 , ·) = δz0 . Proof. Suppose h ∈ h(K) peaks at z0 . Then Z 1 = h(z0 ) = h(ζ) dωK (z0 , ζ). ∂K

So we have that

Z (h(ζ) − 1) dωK (z0 , ζ).

0= ∂K

By the definition of a peak function h(ζ) < 1, or rather that h(ζ) − 1 < 0, on ∂K − {z0 }. We put that together with the fact that ωK (z0 , ·) is a probability measure on ∂K to see that ωK (z0 , ∂K − {z0 }) = 0, otherwise Z (h(ζ) − 1) dωK (z0 , ζ) < 0, ∂K

10

a contradiction. However the statement ωK (z0 , ∂K − {z0 }) = 0, provides us with ωK (z0 , ·) = δz0 . Now suppose that ωK (z0 , ·) = δz0 . Let V be a neighborhood of z0 . Pick v ∈ C(K) so that v ≤ 0, v(z0 ) = 0, and v(y) < −2 for y ∈ X \ V . Then the Hahn-Banach theorem provides sup{h(z0 ) : h ∈ h(K), h ≤ v} = 0. In particular, there is an g ∈ h(K) so that g ≤ 0, g(z0 ) > −1, and g(y) < −2 for y ∈ K \ V . The f = eg−g(z0 ) ∈ h(K) satisfies f (z0 ) = 1, ||f ||K ≤ |e−g(z0 ) | ≤ e, and |f (y)| ≤ 1/e for y ∈ K \ V . Consequently the conditions of Theorem 5.4 are satisfied, with c = 1/e and M = e. This shows that z0 is a peak point of h(K).  6. Boundaries 6.1. The Choquet Boundary. The main ideas and results of this section are from the paper of Bishop and de Leeuw [BiDL59]. If B is a linear subspace of C(K) containing the constant functions, then by the Hahn-Banach theorem any L in B ∗ extends to a continuous linear functional of C(K) and thus by the Riesz Representation theorem there exists some signed Baire measure µ on K so that Z (2) L(f ) = f dµ all f in B. The Choquet Boundary of B is denoted by Cho(B) and consists of all points x in K having the following property: There is a unique positive Baire measure µ that represents in the sense of (2) for the linear functional Lx defined by (3)

Lx (f ) = f (x)

for all f in B.

This unique µ will of course be the unit point mass at x. We define the set Ext(B) to be the extreme points of S = {L : L ∈ B ∗ L(1) = ||L|| = 1}. Shortly we will show (6.2) a main result of Bishop and de Leeuw [BiDL59, Lemma 4.3] which states that Ext(B) = Cho(B). From this fact and the Krein-Milman theorem it follows that any L in B ∗ has a representation of the form (2) with µ a measure concentrated on the closure of Cho(B); this closure of Cho(B) is often called the ˇ Silov Boundary. Let M+ (K) be the class of all non-negative Baire measure on K. In the immediate future we shall need the following basic lemma of Bishop and de Leeuw. 11

Lemma 6.1. Let B be a subspace of C(K). Then each L in B ∗ has a representation of the form Z L(f ) = f dµ all f in B; where µ is a Baire measure which is of the form µ1 − µ2

µ1 , µ2 in M+ (K).

Furthermore the µ can be chosen to be in M+ (K) if and only if L(1) = ||L||. Proof. By the Hahn-Banach theorem, L can be extended to all of C(K) with the preservation of norm; i.e. there is a linear function L0 on C(K) with L0 (f ) = L(f ) all f in B, 0 and ||L || = ||L||. By the Riesz representation theorem, there is a signed Baire measure µ on K so that Z 0 L (f ) = f dµ all f in C(K). Each signed Baire measure on K is of the form µ1 − µ2 . If L(1) = ||L||, then ||µ|| = ||L0 || = ||L|| = L(1) = µ(K), so µ must be non-negative and thus in M+ (K). Conversely, if µ is in M+ (K), then the L defined by the integral representation satisfies L(1) = ||L||.  For any x ∈ K we define Mx+ (B) to be the subset of M+ consisting of all µ with Z f dµ = f (x), all f in B. Note that Mx+ (B) always contains the point mass at x and is nonempty. As B contains the constant functions, then for any µ in Mx+ (B) we have that µ(K) = 1. For any subset X of K, define iB (X) = {y : f (y) = f (x) for some x ∈ X for all f ∈ B}. If µ is any Baire measure in M+ , we denote by µ ˆ its regular Borel extension. This is the unique regular Borel measure on K that agrees with µ on the Baire set of K. It is defined by µ ˆ(X) = inf µ(U ) where U runs over all open Baire sets that contain X. 12

The Choquet boundary of B is defined to consist of those points x in K which are such that any µ in Observe that the Choquet boundary is the set of points x ∈ K such that Mx+ (B) satisfies µ ˆ(iB (x)) = 1. If B contains sufficiently many functions to distinguish a point x from all other points of K, iB (x) = {x}, so that x will be in the Choquet boundary of B if and only if the unit point mass at x is the only µ in Mx+ (B). Theorem 6.2. Let K be a compact set, and B a linear subspace of C(K). Let L0 be a linear functional in B ∗ . Then the following are equivalent: i. L0 is an extreme points of S, ii. There is a point x in Cho(B) so that L0 (f ) = f (x), for all f in B. Proof. To show that (ii) implies (i), suppose that L0 is not an extreme point of S. Then we can find L1 and L2 in S with L0 6= L1 such that L0 = 21 (L1 + L2 ). By Lemma 6.1 there are measures µ1 and µ2 in M+ (K) with µ1 (K) = µ2 (K) = 1 and Z L0 (f ) = f dµi , all f in B and i = 1, 2. Since L0 6= L1 , µ1 (iB (x)) < 1. Let µ = 21 (µ1 + µ2 ). Then µ is in Mx+ (B) and µ(iB (x)) < 1, so x is not in Cho(K). To show that (i) implies (ii), suppose that L0 is an extreme point of S. By Lemma 6.1 there is a measure µ in M+ (K) so that Z L0 (f ) = f dµ, all f in B. Let X1 be any Baire subset of K with 0 < µ(X1 ) < 1, and X2 = K−X1 . Then if the linear functionals L1 and L2 in S are defined by Z 1 Li (f ) = f dµ, all f in B, µ(Xi ) Xi

we have a representation of L0 as a convex combination L0 = µ(X1 )L1 + µ(X2 )L2 of points in S. Since L0 is an extreme point S, this must be a trivial representation, so Z Z f dµ = µ(X) f dµ X

for all f in B and all Baire X in K. Thus each f in B is constant almost everywhere with respect to µ, and if x is chosen to be in the 13

carrier C(µ), that is, the smallest closed subset of K whose complement has µ ˆ measure 0, that constant value must be f (x). This shows that L0 (f ) = f (x) for all f in B and that C(µ) ⊂ iB (x). Since µ could have been chosen to be any measure in Mx+ (B) shows that x is in Cho(B).  Now we move on to the more specific linear subspace of harmonic functions on K, that is the case when B = h(K). Theorem 6.3. The Choquet boundary is the set of harmonic peak points of K. 6.2. The Minimal Boundary. In this next part we will see the connection between the Choquet boundary and that of Bishop’s Minimal Boundary [Bish59] for the space of harmonic functions on K. We should note that for a uniform algebra the Choquet boundary and the minimal boundary are the same. However in general for just a linear subspace this need not be so; see Section VII - Examples of [BiDL59]. Definition 6.4. If µ is a nonnegative regular Borel measure on the compact Hausdorff space X and S is a Borel subset of X, we say that µ is supported by S if µ(X \ S) = 0. Since our favorite measure ωK (z, ·) is a probability measure, the property ωK (z, K \ S) = 0 is equivalent to ωK (z, S) = 1. Definition 6.5. If B is a subspace of C(K), a subset X of K will be called a boundary for B if for each f in B there is some x in X with |f (y)| = ||f ||. Definition 6.6. The smallest Gδ -set M such that ωK (z, M ) = 1 for all z ∈ K will be called the minimal boundary of K. We as an immediate consequence of the Krein-Milman theorem, we have that the Choquet boundary is of h(K) is a boundary of K. The following is Theorem 6.1 of [BiDL59] Theorem 6.7. The Choquet boundary of h(K) is a boundary of K. Proof. Consider the subset S of h(K)∗ defined as S = {L : L ∈ h(K)∗ L(1) = ||L|| = 1}. Then S is convex and weak-∗ compact. Choose any function h in h(K). Let L0 be a point of S with |L0 (h)| = max{|L(h)| : L ∈ S} 14

and S0 = {L : L ∈ S, L(h) = L0 (h)}. By the Krein-Milman theorem, the compact convex set S0 has an extreme point. This extreme point must also be a point of S. By Lemma 6.2, such an extreme point will be of the form Ly , Ly (f ) = f (y),

all f in h(K),

for some y in Cho(K). Since for each x in K, the Lx defined by Lx (f ) = f (x),

all f in h(K),

is in S, it follows from the choice of Ly that ||h|| = max{|h(x)| : x ∈ K} = max{|Lx (f )| : x ∈ K} = |Ly (h)| = |h(y)| Since y is a point of Cho(K), and h is an arbitrary function in h(K), Cho(K) is a boundary for h(K).  Theorem 6.8. The set of harmonic peak points of K is the minimal boundary of K. Proof. If z0 ∈ Oh(K) then by Theorem 5.8 ωK (z0 , ·) = δz0 . Therefore z0 ∈ M , and so Oh(K) ⊂ M . From the previous lemma we know that Cho(K) is a boundary. Since Cho(K) is the set of harmonic peak points, we have that Oh(K) is the minimal boundary of K.  7. Consequences to the Fine Topology The fine topology of classical potential theory is the coarsest topology on Rn which makes every superharmonic function on Rn continuous. If a set is closed in this topology, then we say it is fine closed. The following example shows that the set of peak points is not always fine closed. Corollary 7.1. Let K ⊂ Rn compact. If OK 6= ∂K, then the set OK is not fine closed. Proof. Showing the set of peak points is fine closed is equivalent to finding a subharmonic function which is one on OK and strictly less than one elsewhere. As we have shown the set of peak points is dense in the boundary. So we may take a sequence {xn } in OK converging 15

to y in K − OK . Now suppose there exists a subharmonic function u on K with u = 1 on OK . Then by upper semicontinuity 1 = lim sup u(xn ) ≤ u(y). xn →y

Then u(y) = 1, which is a contradiction. Thus the set of peak points of K is not fine closed.  8. The Right Half Now we will discuss the convergence properties of Z Gj (z, ζ) ∆u(ζ) Uj

as Uj decreases to K. In particular I’d like a way to define the Laplacian for subharmonic functions on compact sets. As the Uj are decreasing, the associated Green functions Gj (z, ζ) increase, and are bounded above by zero. Hence they converge to a function which we’ll call GK (z, ζ). This GK is what we will refer to as the Green function on the compact set K. This function is, generally not upper but lower semicontinuous and symmetric, i.e. GK (z, ζ) = GK (ζ, z). Dominated convergence provides us with Z Z Z Gj (z, ζ) ∆u(ζ). GK ∆u(ζ) = lim χUj Gj (z, ζ) ∆u(ζ) = lim j→∞

K

j→∞

Uj

C

We say that z ∼ w for points z, w ∈ K if GK (z, w) > 0. We use this equivalence to define the sets Q(z) = {w ∈ K : z ∼ w}. Theorem 8.1. The set Q(z) is a Gδ . Theorem 8.2. If z ∈ Q(z) then ωK (z, Q(z)) = 1. 9. Some Thoughts One goal of ours has been to find an appropriate formulation of the Laplacian to arbitrary compact sets. We have seen that GUj (·, ζ)∆uj is weak-∗ converging. So then if we have a sequence of functions {uj } where uj ∈ C 2 (Uj ) and uj is converging uniformly to a function u defined on K, then can we define ∆u as the weak-∗ limit of ∆uj ? It seems that the answer is no. More to be added. 16

10. On Jensen Measures Definition 10.1. Let Ω be an open subset of C and let x ∈ Ω. A Jensen Measure for x (with respect to Ω) is a Borel probability measure µ, supported on a compact subset of Ω, such that every subharmonic function u on Ω satisfies Z u(x) ≤ u dµ. For a compact set K ⊂ C, the Jensen measures for x(∈ K) on K are simply the Jensen measures for x which are supported by K. We shall denote the set of Jensen measures for x on K by Jx (K), and the set of Jensen measures on K is given by J (K) =

[

Jx (K).

x∈K

The set of Jensen measures at a point in K is convex, and weak-∗ compact, as is the set of Jensen measures on K. The harmonic measure is the basic example of a Jensen measure. For convenience from now on we shall adopt the notation that Z µ(φ) = φdµ for all measures µ and functions φ. Recall that the Gelfand transform, simply the evaluation homorˆ a closed subset of C ∗ (K) by the phism, allows us to embed K, as K, ∗ association of x ∈ K with xˆ ∈ C (K) by xˆ(f ) = f (x), for all f in C(K). We should also recall from classical functional analysis( [Swar92]) that the weak topology of an infinite dimensional normed linear space is not metrizable the closed unit ball can be metrizable in the weak topology as we see in the next theorem. In the following theorems X will be a normed linear space and X 0 the dual of X, and for x ∈ X and x0 ∈ X 0 we will denote < x0 , x >= x0 (x). Theorem 10.2 (Chapter 16, Theorem 11, [Swar92]). The closed unit ball S 0 of X 0 is metrizable in the weak-∗ topology if and only if X is separable. 17

Proof. Assume that X is separable and let {xk } be a countable dense subset of X. For x0 , y 0 ∈ S 0 set ∞ X | < x0 − y 0 , xk > | 0 0 . d(x , y ) = 2k (1 + | < x0 − y 0 , xk > |) k=1 Then d is a metric on S 0 which induces a topology on S 0 which is weaker than the weak-∗ topology. but S 0 is a compact Hausdorff space under the weak-∗ topology so the metric topology is exactly the weak-∗ topology on S 0 . Conversely, assume that the weak-∗ topology of S 0 is metrizable. Then there exists a sequence {Un } of weak-∗ neighborhoods on 0 in X 0 such that ∞ \ Un = {0}. n=1

We may assume that Un = {x0 ∈ X 0 : | < x0 , x > | < εn ∀x ∈ An }, where εn > 0 and An is a finite subset of X. Put ∞ [ A= An , n=1

and X1 = (A). It suffices to show that X1 = X. Suppose that x0 ∈ X 0 is such that < x0 , X1 >= 0. Then x0 ∈ Un for all n so that x0 = 0 and X1 = X.  We would like to define a new metric on K in the following way: ( sup{ inf{d(µ, η) : η ∈ Jy (K)} : µ ∈ Jx (K)} dJ (x, y) = max sup{ inf{d(µ, η) : η ∈ Jx (K)} : µ ∈ Jy (K)}, where d is as in the previous theorem. Then the dJ metric induces a topology on K which we call the J -topology. Theorem 10.3. The J -topology is finer than the usual topology on K. Proof. Suppose that {xj } is a sequence in K that is Cauchy in the J -topology. Then for any ε > 0 there exists a J > 0 such that for all j, j 0 > J we have that dJ (xj , xj 0 ) < ε. As δxj ∈ Jxj and δxj0 ∈ Jxj0 , we have that d(δxj , δxj0 ) < ε which implies that d(xj , xj 0 ) < δε where δε is going to zero with ε.  Recall that a point z of K is a peak point if and only if Jz (K) = {δz }. Theorem 10.4. The set of peak points of K is closed in the J -topology. 18

Proof. Suppose we have a sequence of peak points xj J -converging to the point x. As xj is a peak point, the set of Jensen measures at xj contains only delta function at xj , i.e. Jxj (K) = {δxj }. So we are just looking at a sequence of measures in C ∗ (K), as opposed to a sequence of sets. That C ∗ (K) is Hausdorff implies that this sequence has (at most) one limit point. Since Jx (K) contains δx , it must therefore equal δx . Thus x is a peak point.  With our new topology K need not be compact anymore so we take KJ to be the J -completion of K. We say that {xj } is fundamental if the sequence is Cauchy in the J -topology. Let {xj } ∼ {yj } if dJ (xj , yj ) → 0. Then formally the J -completion of K is given by KJ = { {xj } − fundamental }/ ∼ . Theorem 10.5. The J -completion of K, KJ , is compact. Proof. Let Uα be an open covering of KJ . The fibers of Uα , Uˆα are an open covering of J (K) in C ∗ (K). As J (K) is compact, we can find a finite subcovering from Uˆα which we renumber to be {Uˆ1 , . . . , Uˆn }. Then {U1 , . . . , Un } form a covering of J (K).  Now we define the set Jensen measures for x in KJ . Definition 10.6. For xj → x in KJ , we define Jx (KJ ) = {lim µj : µj ∈ Jxj (K)} where the limit is the weak-∗ limit on measures. Theorem 10.7. The set of Jensen measures on KJ at x is compact. Proof. By Banach-Alaoglu the unit ball in the space of measures on K is weak-∗ compact, and metrizable 10.2 with d. We have that the space Jx (KJ ) is d closed and a subset of the unit ball. Therefore Jx (KJ ) is compact.  Theorem 10.8. The set of Jensen measures on KJ at x is convex. Proof. Let xj → x in Kj and µ, µ0 ∈ Jx (KJ ). Then we can find ∗ ∗ µj , µ0j ∈ Jxj (K) so that µj → − µ and µ0j → − µ0 . Then for any a ∈ [0, 1], b = 1 − a, we have that aµj + bµ0j ∈ Jxj (K) by the convexity of Jxj (K). Then the (weak-∗) lim(aµj + bµ0j ) = a lim µj + b lim µ0j = aµ + bµ0 . Therefore aµ + bµ0 ∈ Jx (KJ ).  Theorem 10.9. The multifunction x → Jx (KJ ) is continuous on KJ . Proof.

 19

11. Quasi-Subharmonic Functions 11.1. Quasi-Subharmonic Functions on K. Definition 11.1. Let K be a compact set in the plane. Then a function u : K → [−∞, ∞) is quasi-subharmonic if (a) u is locally bounded above, and for each y ∈ R the set {x ∈ K : u(x) < t} is analytic, (b) for eachR x ∈ K and each Jensen measure µ for x, we have u(x) ≤ u dµ. Definition 11.2. We define the upper-semicontinuous regularization of u to be u∗ (x) = inf (sup u(ζ)), N

ζ∈N

where the infimum is being taken over all J -neighborhoods N of x in KJ . Note that equivalently we may define u∗ (x) = max{u(x), lim sup u(ζ)} ζ→x ∗

and also that u ≤ u . Furthermore we say that u satisfies the subaveraging inequality if Z u(z0 ) ≤ udµ for all µ ∈ Jz0 (K). Theorem 11.3. If φ is bounded above and satisfies the sub-averaging inequality, then φ∗ is subharmonic. Proof. By construction φ∗ is J -upper-semicontinuous. As such we may ∗ find a sequence of continuous R functions R φj∗ converging down to φ . Then by monotone convergence φj dµ → φ dµ. Now we show that φ∗ satisfies the sub-averaging inequality Z ∗ φ (z0 ) ≤ φ∗ dµ for all µ ∈ Jz0 (K). Since φ∗ (z0 ) = lim supw→z0 φ(w), then we can find a sequence wk (it is possible that wk = z0 ) such that φ(wk ) converges to φ∗ (z0 ), and there exists corresponding Jensen measures µk of wk weak-∗ converging to µ. Then Z Z Z ∗ φ(wk ) ≤ φdµk ≤ φ dµk ≤ φj dµk . 20

As the φj are continuous, Z lim

Z φj dµk =

k→∞

φj dµ.

Furthermore lim φ(wk ) = φ∗ (z0 ).

k→∞

Therefore

Z

∗

φ (z0 ) ≤

φj dµ

and by monotone convergence Z

∗

φ (z0 ) ≤ lim

j→∞

Z φj dµ =

φ∗ dµ. 

11.2. Quasi-Subharmonic Functions on KJ . Definition 11.4. Let K be a compact set in the plane. Then a Suslin function u : KJ → [−∞, ∞) is quasi-subharmonic if (a) u is locally bounded above, (b) for eachR x ∈ KJ and each Jensen measure µ for x, we have u(x) ≤ u dµ. We say that u is subharmonic on KJ if u is upper-semicontinuous. Furthermore we let i denote the embedding i : K → KJ , and for a function f on KJ we will use the notation i(f ) = f ◦ i. Theorem 11.5. For a (quasi-) subharmonic u on KJ , i(u) is (quasi−) subharmonic on K. Proof.

 12. Capacity

12.1. Example: We would like to find an adequate definition of capacity in this setting. We find the following example illustrative. Consider the function ( 1 : x∈K \N φ(x) = 0 : x∈N where N is no-where dense. Then φ∗ ≡ 1. Then for x ∈ N there is a sequence {yn } ⊂ K \ N such that yn → x. Then lim inf ( sup µ(N ) ) = 0 yn →x

µ∈Jyn

21

R R as if N had positive µ measure then φdµ < 1 but 1 = φ(yn ) ≤ φdµ a contradiction. Therefore we define two functions Cx (N ) = lim inf ( sup µ(N ) ) y→x

µ∈Jy

and a local version Cxloc (N ) = lim inf ( sup{µ(N ) : µ ∈ Jy , supp(µ) ⊂ B(x, r)} ) y→x

In our example we see that Cx (N ) ≡ 0. However, we have a few problems with taking Cx (N ) as our definition of capacity. Namely we have to account for the fact that N may not be measurable in our ideal application. To account for this we will use the outer-measure of µ. Definition 12.1. The outer measure of µ is defined as µ0 (E) = inf{µ(F ) : F measurable, F ⊃ E}. Then we take Cx0 (N ) = sup{µ0 (E) : µ ∈ Jx } to be the outer capacity. Lemma 12.2. If E1 ⊂ E2 then Cx0 (E1 ) ≤ Cx0 (E2 ). Proof. Since E1 ⊂ E2 , if we have any F such that E2 ⊂ F then E1 ⊂ F . So that inf{µ(F ) : E1 ⊂ F } ≤ inf{µ(F ) : E2 ⊂ F } for any measure µ and therefore µ0 (E1 ) ≤ µ0 (E2 ). Then the sup{µ0 (E1 ) : µ ∈ Jx } ≤ sup{µ0 (E2 ) : µ ∈ Jx }, and we have that Cx0 (E1 ) ≤ Cx0 (E2 ).



Lemma 12.3. If Kj is aTsequence of compact sets decreasing to the compact set K, i.e. K = Kj , then Cx0 (K) = lim Cx0 (Kj ). j→∞

Proof. By the previous lemma 12.2, we have that Cx0 (Kj ) is a decreasing sequence (bounded below by zero), which, of course, must converge. In fact, the sequence is bounded below by Cx0 (K). So we have that Cx0 (K) ≤ lim Cx0 (Kj ). j→∞

Let a = lim Cx0 (Kj ). j→∞

Then we can find a sequence of measures µj ∈ Jx such that µj (Kj ) ≥ a−2−j . Then measures must have a weak-∗ limit in Jx that we will call 22

∗

µ, i.e. µj → − µ. Also we can find continuous functions φl converging to χK , the characteristic function on K. Then µ(φl ) = lim µj (φl ) ≥ a j→∞

and so we have that µ(K) = µ(χK ) = lim µ(φl ) ≥ a. l→∞

Thus Cx0 (K) ≥ µ(K) ≥ lim Cx0 (Kj ). j→∞

 Lemma 12.4. If Fj is an increasing sequence of sets, and F = then Cx0 (F ) = lim Cx0 (Fj ).

S

Fj ,

j→∞

Proof. Again we apply the earlier lemma (12.2) and we have that Cx0 (Fj ) is a increasing sequence (bounded above by Cx0 (F )), which, of course, must converge. So we have that lim Cx0 (Fj ) ≤ Cx0 (F ).

j→∞

All that remains to be shown is that the sequence converges to Cx0 (F ) ≤ lim Cx0 (Fj ). j→∞

For any δ > 0, we can find a Jensen measure at x so that µ0 (F ) ≥ Cx0 (F ) − δ. Then there is a measurable set F ∗ such that µ0 (F ) = µ(F ∗ ) and F ⊂ F ∗ . Similarly for each Fj we can find measurable Fj∗ such that µ0 (Fj ) = µ(Fj∗ ) and Fj ⊂ Fj∗ . However the sequence may not be increasing, but this can be fixed by redefining \ Fk∗ . Fj∗ ≡ k ≥ j

Then the new have the properties that µ0 (Fj ) = µ(Fj∗ ), Fj ⊂ Fj∗ , ∗ . Then and Fj∗ ⊂ Fj+1 [  µ0 (F ) = µ(F ∗ ) ≤ µ Fj∗ = lim µ(Fj∗ ). Fj∗

j→∞

Also we see that

Cx0 (Fj )

0

≥ µ (Fj ) =

µ(Fj∗ ).

Therefore

lim Cx0 (Fj ) ≥ lim µ(Fj∗ ) ≥ µ(F ∗ ) ≥ Cx0 (F ) − δ

j→∞

j→∞

for any δ > 0. Thus lim Cx0 (Fj ) ≥ Cx0 (F ).

j→∞

23

 Theorem 12.5 (Choquet’s Capacitability Theorem). Let C()˙ be a function, defined for all E ⊂ Rn , taking values on the extended real line, and satisfying: (a) C(∅) = 0; (b) E1 ⊂ E2 ⇒ C(E1 ) ≤ C(E2 ); (c) If Ki is a decreasing sequence of compact sets, then \  C Ki = lim C(Ki ); i→∞

(d) If Ei is a decreasing sequence of arbitrary sets, then [  C Ei = lim C(Ei ). i→∞

Then all Suslin sets, and in particular all Borel sets, are capacitable for C. Corollary 12.6. The function Cx0 (·) is a capacity. Definition 12.7. A set E is polar if there is a Gδ set F ⊃ E such that Cx0 (F ) = 0 for all x ∈ K E. This next classical result will be used to prove the following theorem which is a generalization of the works if Cole and Ransford [CoRa97]. Theorem 12.8 (Edwards’ Theorem). Let φ : M → (−∞, ∞] be a lower semicontinuous function. Then, for each x ∈ M  Z φ dµ : µ ∈ Jx sup{v(x) : v subharmonic, v ≤ φ} = inf Theorem 12.9. If u is quasi-subharmonic, then N = {u∗ 6= u} is polar. Proof. Assume that Cx0 (N ) 6= 0 for some x ∈ / N . As Cx0 is a capacity, Cx0 (N ) = sup{Cx0 (E) : E ⊂ N , E compact}. Therefore there exists a compact set E ⊂ N so that Cx0 (E) > 0. We define the following two functions: pE (z) = sup{v(z) : v continuous, subharmonic onK, andv ≤ −χE } qE (z) = sup{w(z) : e quasi − subharmonic onK, andw ≤ −χE } The following facts lead to an immediate contradiction. (a) There exists an z ∈ E such that p∗E (z) ≤ −1. (b) For all x ∈ K we have qE∗ (x) > −1. (c) For all x ∈ K we have p∗E (x) = qE∗ (x).  24

Proof of (a).



Proof of (b). We can find a t so that u ≤ t < u∗ on E. Pick s > max(t, sup u), and set u−s . w= s−t Then w is quasi-subharmonic on K, and w ≤ −χE on K, so qE ≥ w on K. It follows that qE∗ ≥ w∗ on K. Now u∗ − s t−s w∗ = > = −1 onE, s−t s−t so qE∗ > −1 on K as desired.  Proof of (c). We shall in fact prove the stronger result that pE = qE on K, using the duality theorem of Edwards. Applying Edwards’ theorem with φ = −χE we deduce that for x ∈ K,  Z Z pE (x) = −χE dµ : µ ∈ Jx Now let w be a quasi-subharmonic function on K with w ≤ −χE on K. Then, for each x in K and each Jensen measure µ for x, we have Z Z w(x) ≤ w dµ ≤ −χE dµ. Taking the infimum over all such µ, we deduce that w ≤ pE on B. Taking the supremum over all such w, we conclude that qE ≤ pE on K. As the reverse inequality is obvious, the proof is complete.  13. The Dirichlet Problem on KJ Let u : OK → R be a J -continuous function. We say that there is a solution to the Dirichlet problem for u, e.g. the Dirichlet problem is solvable for u, if there exists a function h : KJ → R that is J continuous and harmonic (on K) with h = u on OK . Following the classical methods, we define two sets: Pˆ = {φ : φ is subharmonic on K and lim sup φ(z) ≤ u(a) for every a in OK } z→a

Pˇ = {ψ : ψ is superharmonic on K and lim sup ψ(z) ≥ u(a) for every a in OK }. z→a

Then we have uˆ = sup{φ : φ ∈ Pˆ } and uˇ = inf{ψ : ψ ∈ Pˇ }. By construction we see that uˆ is subharmonic and that uˇ is superharmonic, and also that uˆ ≤ uˇ. Theorem 13.1. For u : OK → R J -continuous. The Dirichlet problem is solvable for u if and only if uˆ = uˇ. 25

Proof. Suppose there exists a solution h to the Dirichlet problem for u. Then h is subharmonic and J -continuous with the lim sup h(ζ) ≤ u(a) for a ∈ OK , and so h ≤ uˆ. Similarly h is superharmonic and J continuous with the lim inf h(ζ) ≥ u(a) for a ∈ OK , and so uˇ ≤ h. Therefore we have that h ≤ uˆ ≤ uˇ ≤ h. Thus uˆ = uˇ. Suppose uˆ = uˇ. Define h = uˆ Then h is J -continuous, harmonic and limζ→a h(ζ) = f (a) for all a ∈ OK .  Observe that since ωK (x, ·) is a Jensen measure, we have that Z uˆ(x) ≤ u(ζ) dωK (x, ζ) ≤ uˇ(x). OK

14. Smaller to Larger Consider x ∈ KJ . Then x = lim xj for some xj in K. Then for u function on K, we can extend u as u∗ to KJ by u∗ (x) = inf ( sup u(ζ)) N

ζ∈N ∩K

where the N are J -neighborhoods of x. Lemma 14.1. If u is subharmonic on K, then u∗ is subharmonic on KJ . Proof. That u∗ is upper semi-continuous follows from construction. That u∗ satisfies the subaveraging inequality follows from previous arguments.  15. Larger to Smaller Lemma 15.1. If u is subharmonic on KJ , then u|K satisfies the subaveraging inequality. Since the J -topology is finer than the usual topology, it does not follow that u|K will remain upper-semicontinous. Theorem 15.2. If u is quasi-subharmonic on KJ , then u|K is quasisubharmonic. 16. Other The following definitions and theorem follow from the work of Nihat Gogus [Gogu05]. Definition 16.1. Given a function φ which is locally bounded above, we define Iφ(z) =

inf

µ(φ),

Sφ(z) =

µ∈Jz (K)

sup µ(φ). µ∈Jz (K)

26

Theorem 16.2 ( [Gogu05, Theorem 3.2] ). The function Iφ (respectively Sφ) is upper semicontinuous at a point z0 ∈ D for all φ ∈ C(D) if the set E is lower semicontinuous (respectively upper semicontinuous) at z0 . If the fibers Ez are convex for all z ∈ D (respectively if the fiber Ez0 is convex), then the converse is also true. References [Bish59] Bishop, E., A Minimal Boundary for Function Algebras, Pacific J. Math., 9 (1959), 629–642 [BiDL59] Bishop, E., de Leeuw, K. The representations of linear functionals by measures on sets of extreme points, Ann. Inst. Fourier, Grenoble, 9 (1959), 305–331 [CoRa97] Cole, B. J., Subharmonicity without Upper Semicontinuity, Journal of Functional Analysis, Volume 147, (1997), 420-442 [Conw91] Conway, J. B., The Theory of Subnormal Operators, AMS Mathematical Surveys and Monographs, 1991 [Game84] Gamelin, T. W., Uniform Algebras, AMS Chelsea Publishing Company, 1984 [GiTr01] Gilbarg, D., Trudinger, N.S., Elliptic Partial Differential Equations of Second Order, Springer, New York, 2001 [Gogu05] Gogus, N. G., Continuity of plurisubharmonic envelopes, Annales Polonici Mathematici, 86.3, (2005), 197–217 [Helm69] Helms, L. L., Introduction to Potential Theory, Wiley-Interscience, 1969 [Phel01] Phelps, R. R., Lectures on Choquet’s Theorem, Second Edition, Springer, 2001 [Pole96] Poletsky, E. A., Analytic geometry on compacta in Cn , Mathematische Zeitschrift, Volume 222, (1996), 407–424 [Pole97] Poletsky, E. A., Approximation by Harmonic Functions, Transactions of the AMS, Volume 349, Number 11, (1997), 4415–4427 [Rans95] Ransford, T., Potential Theory in the Complex Plane, Cambridge University Press, Cambridge, 1995. [Swar92] Swartz, C., An Introduction to Functional Analysis, Marcel Dekker, Inc., 1992 Department of Mathematics, 215 Carnegie Building, Syracuse University, Syracuse, NY 13244-1150

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