Perturbation of the Laplace operator

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1 Perturbation of the Laplace operator WOLFHARD HANSEN Prague, November 2009 Contents 1 Introduction 1 2 Kato measures, potential kernels, and ( µ)-harmonic functions 3 3 Dominant functions and resolvents 7 4 ( µ)-regular sets and ( µ)-superharmonic functions 10 5 A general minimum principle and first consequences 14 6 Characterization of ( µ)-superharmonic functions and ( µ)- regular sets 17 7 ν-eigenvalues of µ 19 8 Harnack s inequalities 22 1 Introduction In this short course, we shall discuss perturbations of the Laplace operator = d 2 x 2 j=1 j on R d, d 1. We recall that a function h on an open set W is called harmonic, if h C 2 (W ) and h = 0 or equivalently if h C 2 (W ) and (1.1) h ϕ dλ d = 0 for every ϕ Cc (W ) W (where λ d denotes Lebesgue measure on R d and Cc (W ) the space of all real functions on W with compact support which are differentiable infinitely many times). In fact, it suffices to know that h C(W ) and h satisfies (1.1) to conclude that h is harmonic on W. 1

2 A relatively compact open set in R d is regular if, for every continuous real function ϕ on the boundary, there exists a unique extension to a continuous real function h on the closure, the solution to the Dirichlet problem for and ϕ, which is harmonic on. It is positive provided ϕ 0. For example, every open ball is regular and the solution to the corresponding Dirichlet problem is given by the Poisson kernel. We shall study solutions (and supersolutions) to equations (1.2) u = uµ, where µ is a signed Radon measure on R d belonging to a certain class (we shall assume that µ is a (local) Kato measure; for the definition see Section 2). If µ has a density, say, with respect to λ d, then L p loc(r d ), p < d/2, is a sufficient property; see Corollary 2.7). DEFINITION 1.1. A continuous real function h on an open set W in R d is called ( µ)-harmonic, if u uµ = 0 in the sense of distributions, that is, if h ϕ dλ d uϕ dµ = 0 for every ϕ Cc (W ). The set of all ( µ)-harmonic functions on W will be denoted by (W ). A relatively compact open set in R d will be called ( µ)-regular if, for every ϕ C( ), there exists a unique function h () such that h ϕ at, and h 0 provided ϕ 0. It will turn out every regular set is ( µ)-regular, if we have a positive perturbation, that is, if µ 0 (see Theorem 4.2). For negative perturbations, that is, if µ 0, this may fail. To get some feeling for these possibilities let us first look at the case d = 1 and µ = αλ 1, that is, at the equation (1.3) u = αu, where α R. Let = (a, b), < a < b <, and ϕ a real function on = {a, b}. a) If α = 0, then the solutions to (1.3) are the functions which are (locally) affinely linear and, of course, h := b x b a ϕ(a) + x a b a ϕ(b) is the solution to the (classical) Dirichlet problem. A slightly different approach (which will be useful when dealing with α 0) is the following. For x [a, b], let h 1 (x) := b x and h 2 (x) := x a. Then h 1, h 2 are positive affinely linear functions on [a, b], (1.4) h 1 (a) > 0, h 1 (b) = 0, h 2 (a) = 0, h 2 (b) > 0. 2

3 Hence (1.5) h := ϕ(a) h 1 (a) h 1 + ϕ(b) h 2 (b) h 2 is the solution to the Dirichlet problem. b) Let us now assume that α > 0 (positive perturbation) and let γ := α. Then the solutions to u = αu are linear combinations of e γx and e γx. So the functions h 1 := e γ(b x) e γ(x b) and h 2 := e γ(x a) e γ(a x) are positive solutions to u = αu. Since they satisfy (1.4), (1.5) yields a solution to the Dirichlet problem for u = αu, a solution which is positive provide ϕ 0 (and even strictly positive on (a, b), if at least one of the values of ϕ is stricly positive); the uniqueness is easily verified. So is ( αλ 1 )-regular. Let us note that now the functions h 1, h 2 are strictly convex, and hence the solution to the Dirichlet problem is strictly smaller than the solution in the case α = 0. c) Finally, let us assume that α < 0 (negative perturbation), and let γ := α. Then the solutions to u = αu are linear combinations of sin γx and cos γx and, up to multiples, the functions h 1 := sin γ(b x) and h 2 := sin γ(x a) are the only solutions which vanish at b and a, respectively. We observe that, for j = 1, 2, h j 0 on if and only if γ(b a) π. Therefore = (a, b) is ( αλ 1 )-regular if and only if γ(b a) < π, and then, again, the solution to the Dirichlet problem for the given function ϕ is obtained by (1.5). Now the solution is strictly concave, and hence strictly larger than the solution in the case α = 0. Moreover, we see that, assuming ϕ 0 and ϕ 0, the solution h is increasing to as γ π/(b a). If γ(b a) = π, then h 0 := h 1 = h 2 = sin γ(x a) is a stricty positive function in () C 0 () (where C 0 () := {f C(): f 0 at }). If γ(b a) = kπ, k N, k 2, then still sin γ(b a) () C 0 (), but sin γ(b a), as well as every solution to u = αu on, changes its sign. If γ(b a) > π and γ(b a) is not a multiple of π, a function h, defined by (1.5), still satisfies h(a) = ϕ(a), h(b) = ϕ(b), and h = αh, but h changes its sign (even if ϕ(a) and ϕ(b) are both positive). 2 Kato measures, potential kernels, and ( µ)- harmonic functions For x R d, d 1, and r > 0, let (x, r) denote the open ball with center x and radius r, that is, (x, r) := {y R d : y x < r}. 3

4 We define a (Green) function G: R d R d (, ] by 1 (1 x y ), d = 1, 2 1 G(x, y) := 2π ln 1 x y, d = 2, 1 x y 2 d, d 3. (d 2)ω d Let us observe that G > 0, if d 3, whereas, for d 2, G(x, y) > 0 if and only if x y < 1. Given a Radon measure ρ on R d with compact support, we may define a function Gρ: R d (, ] by (2.1) Gρ(x) := G(x, y) dρ(y), x R d, and note that Gρ is superharmonic on R d. It is not difficult to verify that G(, y) = ε y, y R d, (in the sense of distributions), and hence, by Fubini s theorem, (2.2) Gρ = ρ. In particular, Gρ is harmonic outside the support of ρ. DEFINITION 2.1. A signed Radon measure on R d is called a (local ) Kato measure, if, for every open ball in R d, (2.3) G(1 µ ) C(R d ). REMARKS 2.2. It is easily verified that, for every signed Radon measure µ on R d the following holds. 1. µ is a Kato measure if and only if µ + and µ are Kato measures. 2. If ( n ) is a sequence of open balls covering R d, then µ is a Kato measure provided (2.3) holds for every ball n, n N. 3. If µ is a Kato measure and f is a locally bounded orel measurable function on R d, then fµ is a Kato measure. For every regular set in R d, let H denote the harmonic kernel for, that is, H (x, ) is the harmonic measure µ x, if x, and H (x, ) = ε x, if x c. If, are regular sets such that, then (2.4) H H = H. DEFINITION 2.3. Let µ be a signed Radon measure on R d. For every regular set and every f b (), let (2.5) K µ f := G(fµ) H G(fµ) (where 2.1 is extended in an obvious way to signed measures). 4

5 LEMMA 2.4. Let A n, n N { }, be bounded orel sets such that A n A. Then G(1 An µ) G(1 A µ) locally uniformly as n. Proof. We may assume that µ 0. y bounded convergence, G(1 An µ) G(1 A µ) pointwise. If d 3, then the proof is immediately finished by Dini s lemma. If d 2, we may consider a ball such that A 1. Then, by Dini s lemma, K µ 1 A n K µ 1 A locally uniformly on. Moreover, H G(1 An µ) H G(1 A µ) locally uniformly on. Since K µ 1 A n + H G(1 An µ) = G(1 An µ), n N { }, the claim follows. The following statements are easily verified (the last property follows immediately from (2.4) and (2.5)). PROPOSITION 2.5. If µ is a Kato measure and is a regular set, then the following holds: 1. K µ± (+ b ()) S+ b (). 2. K µ ( b()) C 0 (). 3. If (f n ) is a bounded sequence in b () such that f n f pointwise, then K µ f n K µ f (even uniformly). 4. If is a regular set and, then (2.6) K µ = Kµ H K µ. If d 2, then Kato measures do not charge points. Indeed, for every x R d, 0 µ ({x})g(, x) G(1 (x,1) µ ) <, Knowing that G(x, x) = we see that µ({x}) = 0. PROPOSITION 2.6. For every signed Radon measure µ on R d, which does not charge points, the following statements are equivalent: (i) µ is a Kato measure. (ii) For every R > 0, lim ε 0 sup x R G(1 (x,ε) µ ) = 0. (iii) For every R > 0, lim ε 0 sup x R K µ (x,ε) 1 = 0. (iv) For every x R d, lim ε 0 K µ (x,ε) 1 = 0. Proof. We may assume without loss of generality that µ 0. (i) (ii): Let R > 0, A := (0, R), x A, and η (0, 1/2). Knowing that G > 0 on (x, η) (x, η), the functions G(1 (x,ε) µ), 0 < ε η, are continuous, and µ({x}) = 0, we obtain, by Dini s theorem, that there exists ε(x) (0, η) such that G(1 (x,ε(x)) µ) < η. We may choose points x 1,..., x m A such that A is covered by the balls (x j, ε(x j )), 1 j m. Now let ε := (1/2) min{ε(x 1 ),..., ε(x m )}. If x A, then (x, ε) (x j, ε(x j )) for some 1 j m, and hence 0 G(1 (x,ε) µ) G(1 (xj,ε(x j )µ) < η. (ii) (iii): Obvious, since 0 K µ (x,ε) 1 G(1 (x, ε)µ), if ε (0, 1/2). 5

6 (iii) (iv): Trivial. (iv) (i): Let be regular, x, η > 0, and ε (0, 1/2) such that := (x, ε) and K µ 1 η. Then p := G(1 µ) = G(1 \ µ) + G(1 µ) = G(1 \ µ) + H G(1 µ) + K µ 1, where the first two terms on the right side are harmonic on, and hence continuous on. So there exists δ > 0 such that p(y) p(x) < 3η provided x y < δ. COROLLARY 2.7. Let p > d/2 and L p loc (Rd ). Then λ d is a Kato measure. Proof. We may suppose that 0 and shall prove the result only for d 3 (leaving the cases d = 1 and d = 2 to the reader). Let 1/p + 1/q = 1. Then (2.7) 1 q = 1 1 p > 1 2 d = d 2 d. Let R > 0, x R, ε (0, 1/2), y (0, ε), and := (y, 2ε). Then (x, ε) (0, R + 1). Hence, by Hölder s inequality, G(1 (x,ε) λ d )(y) G(1 λ d 1 )(y) = y z 2 d (z) dλ d (z) (d 2)ω d ( 1/p I ε p (z) dλ (z)) d, (d 2)ω d (0,R+1) where the integral is finite and ( I ε := ) 1/q 2ε y z (2 d)q dλ d (z) = ω d t d 1 (d 2)q dt. 0 y (2.7), d (d 2)q > 0. Thus lim ε 0 I ε = 0. In the following let µ be a signed Radon measure on R d which is a Kato measure. If d = 1, we assume that µ does not charge points. PROPOSITION For every real function h on an open set W in R d, the following statements are equivalent: (i) h (W ). (ii) For every regular with W, (2.8) h b () and h + K µ h is harmonic on. (ii ) There exists a covering of W by regular sets such that W and (2.8) holds. 2. For every real function h on a regular set the following statements are equivalent: (i) h b (). (ii) h b () and h + K µ h H(). 6

7 Proof. Straightforward using K µ ( b()) C 0 () and (h + K µ h) = h hµ on. Consequence: If is regular and ϕ C( ), then h () and h tends to ϕ at if and only if (2.9) h + K h = H ϕ. So we are led to the following questions: Is I + K µ invertible on the space of all H ϕ, ϕ C( )? If yes, is (I + K µ ) 1 H ϕ 0 provided ϕ 0? efore starting to answer these questions let us note a convergence property of ( µ)-harmonic functions. COROLLARY 2.9. Let W be an open set and (h n ) an increasing sequence of ( µ)-harmonic functions such that h := sup h n is locally bounded. Then h is ( µ)-harmonic. Proof. Let be a ball, W. Then H h n = h n + K µ h n, for every n N. y dominated convergence, lim H h n = H h and lim n K µ h n = K µ h. Thus H h = h+k µ h. Since H h is harmonic on, we obtain that h is ( µ)-harmonic on. PROPOSITION Let be a regular set, s S + (), and suppose that µ 0. Let f b () such that s K µ f on {f > 0}. Then s Kµ f on. Proof. y assumption, (2.10) s + K µ f K µ f + on {f + > 0}. Then t := s + K µ f S + (). Let A be a compact set in {f > 0}, W := \ A, and g := K µ (1 Af). Clearly, g is harmonic on W. y (2.10), t g on A. Moreover, for every z, lim inf x z (t(x) g(x)) 0, since t 0 and g C 0 (). y the minimum principle, t g 0 on W. Hence t g on. This clearly implies that t K µ+ f on. 3 Dominant functions and resolvents In this section, we shall provide general results which in view of Proposition 2.10 will establish that the answers to the questions raised at the end of Section 2 are always YES, if µ is positive. Let L be a bounded kernel on a measurable space (E, E). DEFINITION 3.1. A function u E + is called L-dominant if, for every f E b, u Lf provided u Lf on {f > 0}. 7

8 REMARKS If u is L-dominant and α, β > 0, then αu is (βl)-dominant (trivial). 2. If u is a finite L-dominant function, then all functions αu + Lg, α 0, g E + b are L-dominant. Indeed, if f E b and αu + Lg Lf on {f > 0}, then, for every α > α, α u L(f g) on the set {f > 0} which contains the set {f g > 0}, and hence α u L(f g) on E. Letting α α we obtain that αu + Lg Lf. REMARK 3.3. Let us note the following identities which will be useful. If S, T are bounded operators on (E, E b ) such that I + S and I + T are invertible, then (3.1) (I + S) 1 (I + T ) 1 = (I + T ) 1 (T S)(I + S) 1 (to see this, it suffices to apply I + S from the right, and I + T from the left), in particular (taking T = 0), (3.2) (I + S) 1 = I (I + S) 1 S. The following lemma is very simple, but powerful. LEMMA 3.4. Let u be L-dominant and f, g E b such that f u and Lf = g+lg. Then g u. Proof. Since g = L(f g), we know that u L(f g) on the set {u g} which contains the set {f g > 0}. Thus u L(f g), that is, u g. PROPOSITION 3.5. Suppose that I+L is invertible and u is a bounded L-dominant function. Then 0 (I + L) 1 L L, the operator (I + L) 1 L is σ-continuous, and (3.3) 0 (I + L) 1 Lu u and 0 (I + L) 1 u u. Proof. Let v be any bounded L-dominant function and g := (I + L) 1 Lv so that g + Lg = Lv. y Remark 3.2.2, 0 is L-dominant (as well as every Lf, f E + b ). y Lemma 3.4, g v and g 0 (using v 0). So (3.4) 0 (I + L) 1 Lv v, 0 (I + L) 1 v v, where the second inequalities follow by (3.2). Thus (3.3) holds and, applying (3.4) to the functions v = Lf, f E + b, we see that 0 (I + L) 1 L L. Finally, let f n E + b, f n 0. Since 0 (I + L) 1 Lf n Lf n and Lf n 0, we obtain that lim n (I + L) 1 Lf n = 0. For the remainder of this section, let us assume that there exists a bounded L-dominant function u 1. PROPOSITION 3.6. I + L is invertible. Proof. Clearly, I + αl is invertible, if 0 < α < L 1 = L1 1. Let us fix a real c > u and suppose that, for some α > 0, the operator I + αl is invertible. y Proposition 3.5, (I + αl) 1 L α 1 (I + αl) 1 (αl)u α 1 u < c/α. 8

9 So, for every β [α, (1 + 1/c)α], and hence (β α)(i + αl) 1 L < (β α)c α 1, I + βl = (I + αl)(i + (β α)(i + αl) 1 L) is invertible. A straightforward induction shows that I + αl is invertible for every α > 0. For every α 0, let so that, by (3.2), L α := (I + αl) 1 L (3.5) (I + αl) 1 = I αl α. PROPOSITION 3.7. (L α ) α 0 is a resolvent, that is, for all α, β > 0, (3.6) L β L α = (α β)l α L β (resolvent equation). If v E + b is L-dominant, then αl α v v and α αl α v is increasing. Moreover, α L α is decreasing and L = L 0 = sup α>0 L α. Proof. y Proposition 3.2, the resolvent equation follows from (3.1), taking S = βl, T = αl, and then applying L from the right. Moreover, we see that, for every L-dominant v E + b, the mapping α (I + αl) 1 v is decreasing and positive, and therefore, by (3.5), the mapping α αl α v is increasing and bounded by v. The resolvent equation shows that α L α is decreasing and L = L α + αl α L, where lim α 0 αl α L = 0, since αl α L1 αl 2 1. Thus L = L 0 = sup α>0 L α. PROPOSITION 3.8. Let D := d/dα and f E b. Then, for all n N and α > 0, (3.7) D n L α f = ( 1) n n! L n+1 α f and D n (αl α )f = ( 1) n+1 n! L n α(i αl α )f. Proof. Let g E b. Then, for all i, j N {0} and α, β (0, ), by the resolvent equation (L β L α )L i βl j α = (β α)l i+1 β Lj+1 α g, where α g g L i+1 Lj+1 α u α (i+1) β (j+1) g u. L i+1 β Lj+1 Let n N. Recalling that L n β Ln α = (L β L α )(L n 1 β that lim β α L n β g = Ln αg and (3.8) D(L n αg) = nl n+1 α g. β +L n 2 β y induction, the first equality in (3.7) follows. Moreover, and, for every every n N, by (3.8), D(L n 1 D(αL α g) = L α g αl 2 αg = L α (I αl α )g, L α +... Lα n 1 ), we obtain α (I αl α )g) = D(Lα n 1 g αl n αg) = (n 1)L n αg L n αg + nαl n+1 α g = nl n α(i αl α )g. y induction, the proof of (3.7) is finished. 9

10 COROLLARY 3.9. For every bounded L-dominant v > 0, (I + L) 1 v v exp( Lv/v) > 0. Proof. Let us fix x E, ε (0, 1), and define ϕ(α) := ε + (I + αl) 1 v(x) = ε + v(x) αl α v(x), α > 0. Then ϕ > 0 and, by Proposition 3.8, ( 1) n ϕ (n) 0, for all n N. So, by ernstein s theorem, ψ := ln ϕ is convex. In particular, ψ(1) ψ(ε) + (1 ε)ψ (ε). 1 Since ψ = ϕ /ϕ, we obtain that ϕ(1) ϕ(ε) exp((1 ε)ϕ (ε)/ϕ(ε)). We have ϕ(1) = ε + (I + L) 1 v(x), ϕ(ε) = ε + (I + εl) 1 v(x), and ϕ (ε) = L ε (I εl ε )v(x) whence lim ε 0 ϕ (ε) = Lv(x). Thus the proof is finished letting ε 0. REMARK If we only want to know that (I + L) 1 v > 0 (and this will be sufficient for our applications of Corollary 3.9), we may proceed as follows. Let x E and ϕ(α) := (I + αl) 1 v(x), α 0. Then ϕ 0, ϕ(0) > 0, and ϕ is decreasing. Assuming that ϕ is not strictly positive, there hence exists α 0 > 0such that ϕ > 0 on [0, α 0 ) and ϕ = 0 on (α 0, ). Since ϕ can be represented by a Taylor series around α 0, we obtain a contradiction. 4 ( µ)-regular sets and ( µ)-superharmonic functions PROPOSITION 4.1. If µ 0, then, for every regular set, I + K µ and, for every s S + b (), 0 (I + Kµ ) 1 s s. is invertible Proof. y Proposition 2.10, every u S + b () is Kµ -dominant. Hence the result follows immediately from Proposition 3.5. THEOREM 4.2. If µ 0, then every regular is ( µ)-regular, the corresponding harmonic kernel is (4.1) = (I + K µ ) 1 H, and, for every ϕ C + ( ) with H ϕ > 0, H ϕ ϕ e Kµ H ϕ/h ϕ H ϕ. In particular, 1 1 e Kµ 1. Proof. If ϕ C + ( ), then s := H ϕ H + b () S+ b follow from Proposition 4.1 and Corollary 3.9. (). Hence the statements 1 For the moment, we do not want to worry about limits at 0. 10

11 To treat the general case let us note that, for every regular set, (4.2) I + K µ = (I + Kµ+ )(I (I + Kµ+ ) 1 K µ ), where L µ is a bounded kernel and L µ Kµ. := (I + Kµ+ ) 1 K µ LEMMA 4.3. Let L be a bounded kernel on a measurable space (E, E). Then the following statements are equivalent. (i) The function n=0 Ln 1 is bounded. (ii) The operator I L on (E b, ) is invertible and (I L) 1 = n=0 Ln. (iii) The operator I L on (E b, ) is invertible and its inverse is positive. (iv) There exists g E + b such that 1 + Lg g. (v) There exist c > 0 and γ (0, 1) such that, for every n N, L n 1 cγ n. (vi) The spectral radius ρ(l) := lim n L n 1/n is strictly less than 1. Proof. (i) (ii) (iii): Trivial. (iii) (iv): The function g := (I L) 1 1 E + b satisfies 1 + Lg = g. (iv) (v): Of course, c := g 1. Since c 1 g + Lg 1 + Lg g, and hence Lg (1 c 1 )g, we obtain that, for every n N, L n g (1 c 1 ) n g c(1 c 1 ) n. (v) (vi): Trivial, since L n = L n 1, n N, and lim n c 1/n = 1. (vi) (i): Let γ (ρ(l), 1). Then there exists m N such that L m 1 γ, and hence n=0 Ln 1 (1 γ) 1 m 1 j=0 Lj 1 is bounded. DEFINITION 4.4. A regular set is called µ-admissible, if I L µ is invertible and its inverse is positive. REMARK 4.5. In particular, every regular set satisfying K µ 1 < 1, is µ-admissible. So, by Proposition 2.6, there exists a base of µ-admissible balls. PROPOSITION 4.6. For every regular set the following properties are equivalent: 1. is µ-admissible. 2. I + K µ is invertible and (4.3) (I + K µ ) 1 = n=0 (L µ )n (I + K µ+ ) I + K µ is invertible and, for every s S+ b (), (I + Kµ ) 1 s 0. 11

12 Proof. (1) (2): Immediate consequence of Lemma 4.3 and (4.2). (2) (3): Recall that, by Proposition 4.1, (I + K µ+ ) 1 s 0, for every s S + b (). (3) (1): y (4.2), I L µ is invertible and (I L µ ) 1 = (I + K µ ) 1 (I + K µ+ ). Let s := 1 + K µ+ 1. Then s S+ b (), hence g := (I L µ ) 1 1 = (I + K µ ) 1 s 0. Thus is µ-admissible, by Proposition 4.3. y the considerations at the end of Section 2, we now obtain the following. COROLLARY 4.7. If is regular and µ-admissible, then is ( µ)-regular and = (I + K µ ) 1 H = (I L µ ) 1 + = (L µ )n (I + K µ+ ) 1 H. n=0 In particular, b ( ) into b (). + (whence REMARK 4.8. For all µ-admissible balls and x, (4.4) supp( (x, )) =. H, if µ 0) and Indeed, if ϕ C + ( ) and ϕ 0, then h := H ϕ > 0 on and hence maps (ϕ) + (ϕ) h exp( K µ+ h/h) > 0. PROPOSITION 4.9. Let ν be a Kato measure on R d such that µ ν, and let be a regular set. Then the following holds: 1. For every s S + b 2. L ν Lµ. (), (I + Kν+ ) 1 s (I + K µ+ ) 1 s. 3. If is µ-admissible, then is ν-admissible and, for every s S + b (), (I + K ν ) 1 s (I + K µ ) 1 s. In particular, H ν. Proof. Of course, µ + ν +, ν µ. For every s S + b (), by (3.1), (I + K µ+ ) 1 s (I + K ν+ ) 1 s = (I + K ν+ ) 1 K ν+ µ + (I + K µ+ ) 1 s 0. Moreover, for every f + b (), L ν f = (I + K ν+ ) 1 K ν f (I + K µ+ y Lemma 4.3 and Corollary 4.7, the proof is finished. ) 1 K ν f (I + K µ+ ) 1 K µ f = Lµ f. 12

13 The following converse generalizes what we know in the case, where ν is the measure µ +. COROLLARY Let ν be a Kato measure on R d such that µ ν. Let be a regular set which is ν-admissible, and let L be the positive operator (I+K ν ) 1 K ν µ. Then the following statements are equivalent: 1. is µ-admissible. 2. n=0 Ln 1 is bounded. 3. I + K µ is invertible and (I + Kµ ) 1 = n=0 Ln (I + K ν ) 1. Proof. Obviously, (4.5) I + K µ = (I + Kν )(I L), Let us suppose that is µ-admissible. Then the operator I L is invertible and (I L) 1 = (I + K µ ) 1 (I + K ν ). There exists a > 0 such that h := a 1 1 on. Obviously, s := h + K ν h = a + K ν µ h S + b (). Therefore g := (I L) 1 h = (I + K µ ) 1 s 0. Thus (2) holds by Lemma 4.3. The implications (2) (3) (1) follow immediately from (4.5), Lemma 4.3, and Proposition 4.6. DEFINITION Given an open set W in R d, a lower semicontinuous function s: W (, ], which is finite on a dense subset of W, is called ( µ)- superharmonic, if, for every µ-admissible regular set with W, s s. Let S µ (W ) denote the set of all ( µ)-superharmonic functions on W. The following is easily verified. PROPOSITION For every open set W the following holds. 1. S µ (W ) is a convex cone. 2. If s, t S µ (W ), then s t S µ (W ). 3. If (s n ) is an increasing sequence in S µ (W ), and s := sup s n < on a dense subset of W, then s S µ (W ). 4. S µ (W ) ( S µ (W )) = (W ). 5. If s S µ (W ) and W is an open set in W, then s W S µ ( W ). 6. If ν is a Kato measure such that µ ν, then (4.6) S µ + (W ) S ν + (W ) and S ν + (W ) S µ + (W ). 13

14 PROPOSITION Let be a regular set and f b (). 1. If f + K µ f S(), then f S µ (). 2. If f C() and f fµ 0, then f S µ (). Proof. Let be regular and µ-admissible,. Then, by (2.6), s := (I + K µ )(f H µ f) = (f + K µ f) H (f + K µ f). If f + K µ f S(), then f is lower semicontinuous and s S+ b (), hence f f = (I + K µ ) 1 s 0. This proves (1), and (2) is a consequence of (1). Later on, we shall be able to prove that the converse directions hold as well. PROPOSITION Let W be an open connected set and s S µ (W ), s 0, s 0. Then s > 0. Proof. Let A := {s = 0}. Then A is closed in W. Suppose that there is a point x W A. Let r > 0 such that := (x, r) is µ-admissible (which is true, if r is sufficiently small). Since s(x) s(x) = 0 and supp( (x, )) =, we see that A. So A contains a neighborhood of x, A is open. Since A W, we finally conclude that A =, that is, s > 0. 5 A general minimum principle and first consequences Let us first establish a general minimum principle (which is of independent interest). PROPOSITION 5.1. Let W be a relatively compact open set, s 0 C(W ) such that inf s 0 (W ) > 0, and s 1 : W (, ] such that s 1 is lower semicontinuous and lim inf x z s 1 (x) 0, for every z W. Assume that there exists a closed half-space H in R d such that 0 H and, for every x W, there exists a measure σ on W satisfying σ(w \ (x + H)) > 0 and Then s 1 0. σ(s j ) s j (x), j = 0, 1. Proof. Suppose that not s 1 0, and let u := s 1 /s 0. Then u: W (, ], u is lower semicontinuous, and lim inf x z u(x) 0, for every z W. Hence < α := inf u(w ) < 0 and A := {u = α} is a non-empty compact set in W. Shifting H we see that there exists x A with A x + H. 14

15 We choose a corresponding measure σ and define Then hence τ(1) = σ(s 0) s 0 (x) τ := s 0 s 0 (x) σ. 1 and τ(u) = 1 s 0 (x) σ(s 1) u(x), α = u(x) τ(u) ατ(1) α. Therefore τ(1) = 1 and τ is supported by A. However, τ(w \ (x + H)) > 0, by assumption. Since A x+h, we have a contradiction, and the proof is finished. Let W be an open set in R d. For technical reasons we shall now introduce a class of functions which is seemingly larger than S µ (W ), but will turn out to be equal to S µ (W ). DEFINITION 5.2. Let S µ (W ) denote the set of all lower semicontinuous function s: W (, ], which are finite on a dense subset of W, such that, for every x W, there exist µ-admissible balls n satisfying x n, n W, diam n < 1/n, and n s(x) s(x), n N. These functions satisfy the following minimum principle. PROPOSITION 5.3. Let W be relatively compact open such that there exists a function s 0 S µ (W ) C(W ) satisfying inf s 0 (W ) > 0, and let s S µ (W ) such that lim inf x z s(x) 0, for every z. Then s 0. Proof. Let H be any half-space in R d such that 0 H. Given x W, we may take σ := (x, ), where is any µ-admissible ball such that x, W, and s(x) s(x). Since supp(σ) =, we know that σ(w \ (x + H)) > 0. So the minimum principle follows from Proposition 5.1. The following result is preliminary. For a characterization of ( µ)-regular sets see Theorem 6.4 and Theorem 6.5. COROLLARY 5.4. Let be a relatively compact open set and let us consider the following properties. (i) is ( µ)-regular. (ii) There exists h () such that inf h() > 0. (iii) There exists s S µ () C() such that inf s() > 0. (iv) If s S µ () and lim inf x z s(x) 0, for every z, then s 0. Then (i) (ii) (iii) (iv). 15

16 Proof. (i) (ii). Let h := 1. Then h > 0 on, by Proposition Moreover, lim x z h(x) = 1, for every z. Hence inf h() > 0. (ii) (iii). Trivial. (iii) (iv). Proposition 5.3. PROPOSITION 5.5. Let W be an open set in R d. Then S µ (W ) = S µ (W ). Moreover, for every s S µ (W ) and every ( µ)-regular set with W, s s. Proof. Let v S µ (W ) and let be a ( µ)-regular set, W. We consider f C(W ), f v, and define s := v f. Then s S µ (), s is lower semicontinuous on W, and s = v f 0 on. Hence, by Corollary 5.4, s 0, f v. The proof is finished choosing f n C(W ) such that f n v. COROLLARY 5.6. The mapping W S µ (W ) is a sheaf: If W is the union of open sets W i, i I, and s: W R, then s S µ (W ) if and only if, for all i I, s Wi S µ (W i ). Proof. Clearly, the statement holds for W S µ (W ). So the result follows immediately from Proposition 5.5. COROLLARY 5.7. Let W be an open set, s S µ (W ), and let be ( µ)- regular, W. Then s S µ (W ). Proof. Let s := s, and f n C(W ), f n s. Then f n s s. Therefore s is lower semicontinuous. f n C(W ) and Let x W and let be a µ-admissible ball such that x and W. Then s(x) s(x) s(x), where s(x) = s(x), if x W \. If x and, then (x, ), and hence s(x) = s(x). Thus s S µ (W ), by Proposition 5.5. (x, ) = PROPOSITION 5.8. Let W be an open set, s S µ (W ) and t S µ (W ) such that t s and both s and t are locally bounded. 2 Then there exists a function h (W ) such that t h s. Proof. The proof is the same as in classical potential theory. Let := { m : m N} be a base of ( µ)-regular sets which are relatively compact in W and cover W. Let ( n ) be a sequence in containing every m, m N, infinitely many times. We define s n := n n s, n N. y Corollary 5.7, the sequence (s n ) is contained in S µ (W ) and, of course, it is decreasing. Moreover, it follows immediately by induction that s n t, n N. Let h := lim n s n. Then s h t. Given m N, we know that s n = m s n 1 for infinitely many n N, and hence m h = h. Since h is bounded on m, we obtain that h is ( µ)-harmonic on each m, m N. Thus h (W ). 2 It follows from the proof and Corollary 8.4 that the last assumption can be omitted. 16

17 6 Characterization of ( µ)-superharmonic functions and ( µ)-regular sets LEMMA 6.1. Let be a regular µ-admissible set and s S µ (), s 0. Then s is (I + K µ ) 1 K µ -dominant. Proof (See the proof of Proposition 2.10). Let L := (I + K µ ) 1 K µ such that s Lf on {f > 0}, that is, and f b() (6.1) s + Lf Lf + on {f + > 0}. Then (I + K µ )Lf = K µ f S + b (), and hence Lf S µ (), by Proposition So t := s + Lf S µ (). Let A be a compact set in {f > 0}, W := \ A, and g := L(1 A f). If is any regular set such that W, then (I + K µ )g = (I + Kµ )g H K µ g, and hence g is ( µ)-harmonic on. So g is ( µ)-harmonic on W. y (6.1), t g on A. Moreover, for every z, lim inf x z (t(x) g(x)) 0, since s 0 and g C 0 (). y Corollary 5.4 and Proposition 5.1, t g 0 on W. Hence t g on. This clearly implies that t Lf on. THEOREM 6.2. Let be a regular set and f b (). 1. f S µ () if and only if f + K µ f S(). 2. If f is continuous, then f S µ () if and only if f fµ 0. Proof. Let be a regular set such that and K µ 1 < 1/3. Then is µ-admissible and (I + K µ ) 1 < (1 (1/3)) 1 = 3/2. Hence S := (I + K µ ) 1 K µ+ and T := (I + K µ ) 1 K µ are bounded operators on b ( ) having a norm which is strictly smaller than 1/2. y Proposition 4.6, S 0 and T 0. So we may define and know that L < 1. Let us fix f S µ () and let L := (I + T ) 1 S v := (I + K µ )(f H µ f), w := f f. Then w S µ ( ), w 0. If g + b ( ), then, by Proposition 4.13, Sg S µ ( ), since (I + K µ )Sg = Kµ+ g is superharmonic on. If t is any positive function in S µ ( ), then t is T -dominant, by Lemma 6.1, and hence (I + T ) 1 t 0, 17

18 by Proposition 3.5. In particular, (I + T ) 1 w 0 and Lg 0, for every g + b ( ). We have I + K µ = (I (I + K µ ) 1 K µ ) 1 = (I S + T ) 1 = (I L) 1 (I + T ) 1 = L n (I + T ) 1. Thus v = (I + K µ )w 0. Since, by (2.6) (cf. also the proof of Proposition 4.13), n=0 v = (f + K µ f) H (f + K µ f), we see that f + K µ f S(). If, conversely, f + K µ f S(), then f S µ (), by Proposition Again, (2) is a consequence of (1). For the characterization of ( µ)-regular sets we shall need the following result which will also be useful in connection with a discussion of eigenvalues. PROPOSITION 6.3. Let be a regular set. Then K µ on b (). is a compact operator Proof. It clearly suffices to consider the case µ 0. Let ε > 0 and x. y Proposition 2.6, there exists an open ball A in, centered at x, such that K µ 1 A ε. There exists an open ball à A, centered at x, such that, for all harmonic functions g on A satisfying g K µ 1 and all y Ã, (6.2) g(y) g(x) < ε (see, for instance, [?, Lemma 1.5.6]). Now let f b (), f 1. Then K µ f = Kµ (1 \Af) + K µ (1 Af), where g := K µ (1 \Af) is harmonic on A, g K µ 1, and Kµ (1 Af) K µ 1 A ε. Hence, by (6.2), for all y Ã, K µ f(y) Kµ f(x) g(y) g(x) + 2ε < 3ε. Therefore the set {K µ f : f b(), f 1} is equicontinuous at x. Since K µ ( b()) is contained in C 0 (), the claim follows by Arzéla-Ascoli s theorem. THEOREM 6.4. For every regular set, the following statements are equivalent. (i) is ( µ)-regular. (ii) There exists h () such that inf h() > 0. (iii) There exists s S µ () C() such that inf s() > 0. (iv) If s S µ () and lim inf x z s(x) 0, for every z, then s 0. (v) is µ-admissible. 18

19 Proof. y Proposition 5.4, (i) (ii) (iii) (iv). (iv) (v): Let f b () and f + K µ f = 0 so that f H µ () C 0 (). Hence, by (iv), ±f 0, f = 0. So I + K µ is injective. y Proposition 6.3, Kµ is compact. Therefore I + K µ is bijective. Now let s S + b () and t := (I + Kµ ) 1 s. Then t S µ (), by Proposition Since K µ t C 0(), we know that, for every z, lim inf x z t(x) = lim inf x z s(x) 0. y (iv), t 0. Therefore is µ-admissible, by Proposition 4.6. (v) (i): Proposition 4.7. THEOREM 6.5. Let be a relatively compact open set in R d. Then is ( µ)- regular if and only if is regular and µ-admissible. Proof. y Proposition 4.7 and Theorem 6.4, it suffices to show that is regular if it is ( µ)-regular. So let us suppose that is ( µ)-regular, and let ϕ C( ), 0 ϕ 1. a) Let us first consider the case, where µ 0, and let t := ϕ, t 0 := (1 ϕ), s := 1 t 0. y Theorem 4.2 (applied to regular sets with ), t, t 0 are subharmonic on. Therefore s t is superharmonic on. Since both s and t tend to ϕ at, the (classical) minimum principle implies that s t 0, that is, t s. y Proposition 5.8, there exists a harmonic function g on such that t g s. Of course, g tends to ϕ at. So is regular. b) Let us now consider the general case. Let be a ball such that. If a > 0 is sufficiently large, then h 0 := a(+ 1) 1 on. Let u := ϕ, u 0 = (h 0 ϕ), v := h 0 u 0. y Corollary 4.7, u, u 0 are ( µ + )-superharmonic on. Therefore u v is ( µ + )-superharmonic on. Since both u and v tend to ϕ at, we obtain, by Proposition 5.3, that u v 0, that is, v u. y Proposition 5.8, there exists a ( µ + )-harmonic function h such that u h v. Of course, h tends to ϕ at. Hence is ( µ + )-regular. y (a), is regular. 7 ν-eigenvalues of µ Throughout this section let µ be a signed Kato measure, ν a positive Kato measure on R d, and a connected regular set. PROPOSITION 7.1. Suppose that ν charges every non-empty finely open 3 set in 4. Then there exists α > 0 such that is (µ + αν)-admissible. 3 The fine topology is the coarsest topology τ on R d such that every superharmonic function on R d is τ-continuous. 4 This holds, for example, if ν λ d, > 0 on. 19

20 Proof. y Lemma 2.4, there exists a compact set A in such that K µ 1 \A 1/3. Let s := K µ 1 A and define functions v α C + b (), α > 0, by v α := (I + K µ+ +αν ) 1 s. We know that the mapping α v α is decreasing (e. g., by Proposition 4.9). Let v := inf α>0 v α. Since (7.1) v α + K µ+ +αν v α = s, we obtain that w := sup α>0 K µ+ +αν v α S + b () and v + w = s. Therefore v is finely continuous. Moreover αk ν v αkν v α s, for every α > 0, and hence K ν v = 0. Thus v = 0. Since the functions v α are continuous and α v α is decreasing, we conclude that v α tends to zero locally uniformly on as α. So there exists α > 0 such that v α 1/3 on A. Then, by (7.1), K µ+ +αν v α + 1/3 s on A, and hence on, by Proposition Therefore v α 1/3, and hence (I + K µ+ +αν ) 1 K µ 1 Kµ 1 \A + v α 2/3, Thus is (µ + αν)-admissible, by Proposition nless stated otherwise we shall assume from now on that is µ+αν-admissible for some α R and that ν() > 0. DEFINITION 7.2. Let Γ µ, ν () denote the set of all α R such that is (µ+αν)- admissible and α 0 := α 0 (µ, ν, ) := inf Γ µ, ν (). For every α R, let E α µ, ν() denote the set of all h C 0 () such that that is h hµ = αhν, E α µ, ν() = αν () C 0 () = {h b (): h + K µ + αkν h = 0}. α is called a ν-eigenvalue of µ, if E α µ, ν() {0}, and then the functions in E α µ, ν() \ {0} are called ν-eigenfunctions of µ. Let E µ, ν () denote the set of ν-eigenvalues of µ. THEOREM 7.3. There exists α 0 R such that Γ µ, ν () = (α 0, ). Moreover, α 0 is the maximal ν-eigenvalue of µ. Proof. y assumption, Γ µ, ν (). Let α Γ µ, ν (). Then [α, ) Γ µ, ν () by Proposition 4.9. For every 0 < ε < (I + K µ + αkν ) 1 K ν 1, (I + K µ + (α ε)kµ ) 1 = [ε(i + K µ + αkν ) 1 ] n (I + K µ + αkν ) 1, n=0 and hence α ε Γ µ, ν (). This proves that Γ µ, ν () = (α 0, ), where α 0 <. 20

21 We show next that α 0 >. Let α Γ µ, ν (), L := (I + K µ + αν) 1 K ν, and let C be a compact set in such that ν(c) > 0. Then K ν 1 C > 0 on and hence the ( µ)-superharmonic function L1 C is strictly positive on. So there exists β (0, ) such that βl1 C 1 C. y induction, (βl) n 1 C 1 C for every n N, and hence n=0 (βl)n 1 = on C. Therefore, by Corollary 4.10 (with µ + (α β)ν, µ instead of µ, ν, respectively), α β Γ µ, ν (), α 0 α β. Finally, assume that the operator I + K µ + α 0K ν is injective. Then, by Proposition 6.3, I + K µ + α 0K ν is invertible and, for every s S+ b (), (I + K µ + α 0K ν ) 1 s = lim n (I + K µ + (α 0 + (1/n))K ν ) 1 s 0. So α 0 Γ µ, ν (), by Proposition 4.6, contradicting the fact that Γ µ, ν () = (α 0, ). This finishes the proof. COROLLARY 7.4. Let be a regular set. Then is µ-admissible if and only if I + K µ + αkν is injective for every α 0. THEOREM 7.5. The set E α µ, ν() is at most countable, it is contained in (, α 0 ] and has no finite accumulation point. For each α E α µ, ν(), the eigenspace E α µ, ν() is finite-dimensional, and eigenfunctions for different eigenvalues are linearly independent. Proof. Let β Γ µ, ν () and let L be the positive operator (I + K µ + βν) 1 K ν on b (). It is compact, since K ν is compact, by Proposition 6.3. We note that I + K µ + αkν = (I + K µ + βkν )(I + (α β)l). So I + K µ + αkν is invertible (injective) if and only if I + (α β)l is invertible (injective) and, for every α 0, α is a ν-eigenvalue of µ if and only if 1/(α β) is an eigenvalue of L. So the result follows from the general theory of Fredholm operators. REMARK 7.6. In fact, it can be shown that the eigenspaces E α µ, ν(), α E µ, ν (), are pairwise orthogonal in L 2 (, ν) and that their (direct) sum is dense in L 2 (, ν). In particular, E µ, ν () is countable (see [?]). y our means, we are able to prove the following (where (1) and (4) could as well be proved using L 2 -theory). THEOREM 7.7. The following holds. 1. For the greatest ν-eigenvalue α 0 of µ, the space of all ν-eigenfunctions of µ is one-dimensional and consists of multiples of a strictly positive function h S (µ+α 0ν) + () = R + h If α < α 0, then S (µ+αν) + () = {0}. 4. For every α E α µ, ν() \ {α 0 }, there is no positive ν-eigenfunction of µ. 21

22 Proof. Let (γ n ) be a sequence in Γ µ, ν () which is decreasing to α 0. For every n N, let g n := H (µ+γnν) 1 and c n := g n. y Proposition 4.9, for every n N, 1 g n g n+1 and (7.2) g n + (K µ + γ nk ν )g n = 1. If sup c n <, then g := lim n g n is bounded and g + (K µ + α 0K ν )g = 1. So g H +µ+α0ν () and inf g() > 0, by Proposition Therefore α 0 Γ µ, ν (), by Theorem 6.4, a contradiction. Thus sup c n =. Since K µ, Kν are compact operators on ( b(), ) mapping b () into C 0 (), there is a subsequence (h n ) of (c 1 n g n ) such that (K µ h n + γ n K ν h n) converges uniformly to a function h 0 C 0 + (). y (7.2), the sequence (h n ) itself converges uniformly to h 0 and h 0 + K µ h 0 + α 0 K h 0 = 0, that is, h 0 E α 0 µ, ν(). Of course, h 0 = 1, since h n = 1 for every n N. Since h 0 0, we hence see, by Proposition 4.14, that h 0 > 0 on. Let µ 0 := µ+α 0 ν, α α 0, and s S µ αν ()\{0}, s 0. Then s is ( µ 0 )- superharmonic, by (4.6). sing Lemma 2.4, we may find a compact set A in such that W := \ A is regular and K µ 0 1 W < 1. Then K µ 0 W 1 < 1, and hence W is µ 0 -admissible. Let a := sup{α 0: αh 0 s on A} and t := s ah 0. Then t S µ 0 (), t 0 on A, and there exists a point x 0 A such that t(x 0 ) = 0. Clearly, lim inf y z t(y) 0 for every z W (recall that h 0 0 at ). Therefore, by Proposition 5.3, t 0 on W. So t 0 on. Since is connected and t(x 0 ) = 0, we conclude, by Proposition 4.14, that t = 0, that is, s = ah 0. Since h 0 S µ αν (), by (4.6), we obtain that s αν (), and hence s Eµ, α ν() E α 0 µ, ν(). Thus s = 0, if α < α 0. Finally, let h E α 0 µ, ν. There exists b > 0 such that g := bh 0 h 0 on A. Since h C 0 () and hence g C 0 (), we obtain, by Proposition 5.3, that g 0 on W as well. y the preceding considerations, there exists c > 0 such that bh 0 h = ch 0 and therefore h Rh 0. 8 Harnack s inequalities For the moment, let us suppose that d 3 and let x, y, z R d. Then, by the triangle inequality, x y x z + z y, hence 2 x z x y or 2 z y x y. So x z 2 d z y 2 d 2 d 2 x y 2 d and therefore G(x, z)g(z, y) 2 d 2 G(x, y) ( G(x, z) + G(z, y) ). For every regular set, we have a corresponding Green function G, defined by G (, y) := G(, y) H G(, y), y, and then, for all f b ( ) and x, K µ f(x) = G (x, y)f(y) dµ(y). 22

23 If is an open ball (x 0, r), then 1 { } 1 (1 + Φ(x, y)) 1 d/2 x y 2 d, if d 3, (d 2)ω d 1 G (x, y) = 4π ln{ 1 + Φ(x, y) }, if d = 2, 1{ r 2 (x x 0 )(y x 0 )} x y +, if d = 1, 2 r where Φ(x, y) := (r2 x x 0 2 )(r 2 y y 0 2 ) r 2 x y 2 (see [?]). Note that G is symmetric, that is, G (x, y) = G (y, x), for all x, y (this is true as well if is not a ball). Given d 1, there exists C > 1, such that, for all open balls in R d and x, y, z, the following similar inequality holds (8.1) G (x, z)g (z, y) C G (x, y)(g (x, z) + G (y, z)), where denotes the open ball which has the same center as, but double radius. (It is easily seen that cannot be replaced by, but it requires considerable efforts to obtain (8.1).) PROPOSITION 8.1. Let ν be a positive Radon measure on R d and let be an open ball in R d. Then, for every s S + ( ), K ν s 2C K ν 1 s. Proof. For all x, y, K ν G (, y)(x) = G (x, z)g (z, y) dν(z) C G (x, y) (G (x, z) + G (y, z)) dν(z) = C G (x, y)(k ν 1(x) + Kν 1(y)) 2C K ν 1 G (x, y). Integrating with respect to a Radon measure ρ on, we obtain that K ν (G ρ) 2C K ν 1 G ρ. The proof is finished using the fact that every s S + ( ) is the increasing limit of a sequence (G ρ n ). COROLLARY 8.2. Let µ be a signed Radon measure on R d and an open ball in R d such that K µ 1 1/(6C ). Then, for every s S + b ( ), (8.2) In particular, 1 2 s (I + Kµ ) 1 s 3 2 s. (8.3) 1 2 H 3 2 H. 23

24 Proof. y Proposition 8.1, K µ 1/3 and, for every s S+ b Therefore (I + K µ ) 1 s = s + ( K µ )n+1 s, where ( K µ )n+1 s n=0 n=1 (K µ n=0 )n s n=1 This proves (8.2), and (8.3) is an immediate consequence. (1) ns 1 = 3 2 s. ( ), K µ s 1 3 s. Looking at the Poisson kernel, it is easily verified that there exists c 0 > 0 such that, for every ball = (x 0, r) in R d and all x, y (x 0, r/2), (8.4) H (x, ) c 0 H (y, ). THEOREM 8.3. For every connected open set W in X and every compact set A in W, there exists c > 0 such that, for every positive h (W ), (8.5) sup h(a) c inf h(a). Proof. Let h (W ), h 0. Let x 0 W and r > 0 such that the closure of := (x 0, r) is contained in W and K µ 1 1/(6C ). y Corollary 8.2 and (8.5), for all x, y (x 0, r/2), h(x) = h(x) 3 2 H h(x) 3c 0 2 H h(y) 3c 0 h(y) = 3c 0 h(y). The proof is finished by a standard (and straightforward) covering argument. The following strong convergence property is now an immediate consequence of Theorem 8.3 and Corollary 2.9. COROLLARY 8.4. Let (h n ) be an increasing sequence of ( µ)-harmonic functions on a connected open set W in X such that h := sup h n is finite at some point. Then h is ( µ)-harmonic on W. 24