arXiv:math/0106078v1 [math.AP] 11 Jun 2001 suenwthat now assume and (1) o al nastΩ epoeta,udrsial assumption suitable under that, prove We that Ω. implies set also a in balls for (3) o instance. for n uhroi ucin.W ei ycnieigteques the considering by inte when begin the said We t between the relations functions. that several subharmonic assumed consider and is we it paper when type this first in the of result [P converse i and a int [Ko] this all [Be], that particular, show for In to Ω PS]. possible Ku, set the is Ko, where a it ES, Be, Then, in [AG, satisfied ball Ω. are in defined equalities functions the that assumes that have we then Ω, oal nerbeo oanΩadstse 1 whenever (1) satisfies and Ω domain a on integrable locally nbt ae,teeeitcnes eut ttn hti t if that stating results converse exist there then cases, both In (2) radius HRCEIAINO AMNCADSUBHARMONIC AND HARMONIC OF CHARACTERIZATION oiae yti,adsnew ol o n n eeec in reference any find not could we since and this, by Motivated m these to properties converse of set known, lesser second, A h uhr eeprilyspotdb C,Prua,thro Portugal, FCT, by supported partially were authors The Date Let h r eray1 2008. 1, February : h utb amnc ntefis ae o ntne fw assume we if instance, for case, first the In harmonic. be must , Abstract. ntrso hi envle nblsado pee.Ti inc This spheres. func on subharmonic and these for between domains. balls type Beardon’s Harnack of in of inequalities integral values inequality obtain an mean of their converse of terms in eahroi ucini h lsdbl ete tapoint a at centred ball closed the in harmonic a be B surface h UCIN I ENVLEPROPERTIES MEAN–VALUE VIA FUNCTIONS ( ,r a, satisfies .Te ti elkonthat well–known is it Then ). h h ER RIA N JO AND FREITAS PEDRO and egv hrceiaino amncadsbamncfunc subharmonic and harmonic of characterization a give We shroi.Mr rcsl,w aetefollowing the have we precisely, More harmonic. is scniuu nΩ—frteecasclrsls e AR GT [ABR, see results, classical these for — Ω in continuous is h h ( h a ( volume a shroi nΩ iia euthlsfr()poie we provided (2) for holds result similar A Ω. in harmonic is = ) = ) 1. S nrdcinadresults and Introduction B a ( a vrgsaeequal. are averages r ( := ) r := ) B a | ( ∂B r | B = ) ( ( 1 1 ,r a, ,r a, 1 S OPLOOMATOS PALHOTO AO ˜ a ) ) ( | | h r Z Z ) aifistema–au properties mean–value the satisfies B ∂B ( a,r ( a,r ) ) h h g rga POCTI. program ugh ( x ( w en ngeneral in means two x ) ) dx rlmaso harmonic of means gral eeult ssatisfied, is equality he ds. B in.W also We tions. oaeae coincide, averages wo ( ino htcnbe can what of tion ,r a, ]cnie h case the consider S] ,ti atequality last this s, pista sa is Ω that mplies gal harmonic egrable a au results value ean scnandin contained is ) ue the ludes h ieaueto literature the tions a n with and that h is ], 2 PEDRO FREITAS AND JOAO˜ PALHOTO MATOS

Theorem 1.1. Let h be a in Ω. Then h is harmonic in Ω if and only if (3) is satisfied for all balls whose closure is contained in Ω.

Remark 1. Note that in the equality (3) it is not assumed a priori that the value of the averages remains fixed when a is a given point but the radius changes.

Remark 2. Continuity is essential, since otherwise any function coinciding with a on all but a set of finite points, for instance, would clearly satisfy (3) but not be harmonic. From the results in [Be], [Ko] and [PS] we have that the equality of the aver- ages (3) for all harmonic functions is possible only for balls. M. Rao has shown that in fact the averages of harmonic functions on general domains must satisfy a one– sided inequality of Harnack type [R]. Here we improve on this result establishing also a lower bound by using the methods of [Be] and [PS] to obtain the following Theorem 1.2. Let Ω be a domain satisfying the uniform exterior sphere condition. Then there exist constants 0

(4) Ba(r) ≤ Sa(r), so it makes sense to ask if this inequality characterizes subharmonic functions. In this respect we have the following Theorem 1.3. Let u be a continuous function in Ω. Then u is subharmonic in Ω if and only if (4) holds for all balls whose closure is contained in Ω. From the proof of Theorem 1.2 it follows that the second inequality in that result also holds for C2 subharmonic functions and this can easily be improved to C0 sub-harmonic functions. Also for particular domains or in dimension 1 one easily exhibits sequences of positive subharmonic functions for which the volume average is bounded and the surface average becomes unbounded. This can be improved for relatively general domains provided the geometry of the boundary allows a sufficiently rich supply of harmonic subfunctions, e.g., if the domain satisfies the uniform exterior sphere condition. CHARACTERIZATION OF HARMONIC AND SUBHARMONIC FUNCTIONS 3

Theorem 1.4. Let Ω be as above, and u be a non–negative subharmonic function in Ω and continuous in Ω. Then there exists a constant c2 as in Theorem 1.2 such that 1 c udx ≤ 2 u ds. |Ω| |∂Ω| ZΩ Z∂Ω On the other hand, there exists a sequence of positive subharmonic functions u for which the volume average in Ω remains bounded, while the surface average is unbounded. Finally, we consider a property of subharmonic functions proved by Beardon [B], which states that if u is subharmonic in a set Ω, then

(5) Sa(κr) ≤Ba(r) for all balls whose closure is contained in Ω and where 1 2, n =1  e−1/2, n =2 (6) κ =    2 1/(n−2) n , n ≥ 3.     We prove that this property again characterizes subharmonic functions. More pre- cisely, we have Theorem 1.5. Let u be a continuous function defined on a domain Ω and assume that condition (5) is satisfied for all balls whose closure is contained in Ω and for some κ satisfying

n 1 n2 0 <κ ≤ κ1 := − + +2n. 4 2s 4 Then u is subharmonic in Ω. Since the upper bound for κ in the theorem is larger than (or equal to, in the case where n is one) the values of κ given in (6), this yields a converse of Beardon’s property.

2. Harmonic functions 2.1. Proof of Theorem 1.1. Assume that (3) holds for a given fixed point a in Ω, and all r in (0, R), for some positive number R such that B(a, R) is contained n in Ω. Denoting by ωn the volume of the unit ball in R we have that 1 1 h(x)dx = h(x)ds ω rn nω rn−1 n ZB(a,r) n Z∂B(a,r) and thus r n h(x)ds dt = r h(x)ds. Z0 "Z∂B(a,t) # Z∂B(a,r) Letting

ϕ(r)= h(x)ds, Z∂B(a,r) 4 PEDRO FREITAS AND JOAO˜ PALHOTO MATOS we have that r n ϕ(t)dt = rϕ(r). Z0 Since h is continuous, we can differentiate this with respect to r which gives that ϕ satisfies the differential equation rϕ′(r)+(1 − n)ϕ(r)=0. Hence ϕ(r)= crn−1 for some constant c and c φ(r) = n−1 = Sa(r), nωn nωnr from which it follows that, for a and r as above, the average Ba(r) is independent of r. Since h is continuous, the value taken by this average must be attained by h for some point in the ball B(a, r). As this holds for arbitrarilly small r, we have that this value must coincide with h(a). Applying now the converse result to the mean–value equality (1) completes the proof of Theorem 1.1. 2.2. The integral Harnack inequality. Let v be the solution of the equation ∆v +1=0, x ∈ Ω (7) (v =0, x ∈ ∂Ω. We have that 1 1 1 ∂v h dx = − h∆v dx = h − ds. |Ω| |Ω| |Ω| ∂ν ZΩ ZΩ Z∂Ω   Since there exist constants c1 and c2 such that ∂v 0

3. Subharmonic functions 3.1. Proof of Theorems 1.3 and 1.4. We begin by proving the following

Lemma 3.1. If u is a continuous function in Ω such that Ba(r) ≤ Sa(r) holds for all balls whose closure is contained in Ω, then u satisfies the maximum principle in Ω. Proof. Let M be the maximum of u in Ω, and assume that this is attained at an point. Define A = {x ∈ Ω : u(x)= M} From the continuity of u we have that A is closed in Ω and that S is a continuous function of the radius. Let x0 be a point in A, and take r1 such that B(x0, r1) is in Ω. We have that either there exists r in (0, r1) such that Sx0 (r)

r∗ = inf {r ∈ (0, r) : Sx0 (r)= Sx0 (r)} .

Again by continuity, we have that r∗ is positive and also that Sx0 (r∗)= Sx0 (r)

n ωnr∗ Bx0 (r∗) = vdx B(x0,r∗) Z r∗ n−1 = nωn r Sx0 (r)dr 0 nZ > ωnr∗ Sx0 (r∗) and thus Bx0 (r∗) > Sx0 (r∗), contradicting the hypothesis. 6 PEDRO FREITAS AND JOAO˜ PALHOTO MATOS

Proof of Theorem 1.3: Assume that w satisfies the inequality Ba(r) ≤ Sa(r) but that it is not subharmonic. Then, there exists a ball B and a function h harmonic in B such that B is contained in Ω and h ≥ w on ∂B, but h(x0) < w(x0) for some x0 in B. Since the function w −h also satisfies the inequality but not the maximum principle, we have a contradiction and hence w must be subharmonic. Proof of Theorem 1.4: Let v be subharmonic. Let h be the harmonic solution of the with boundary values specified by v. Then v − h is subhar- monic and the maximum principle shows that v ≤ h. Hence Theorem 1.2 applied to h implies 1 1 c c v dx ≤ h dx ≤ 2 h ds = 2 v ds. |Ω| |Ω| |∂Ω| |∂Ω| ZΩ ZΩ Z∂Ω Z∂Ω For the second part of the statement consider first the trivial one dimensional case. Assume (without loss of generality) Ω = (0, 1) and let, for k a positive integer, 2 2 vk(x) = max{1, k−k x, k+k (x−1)}. Then the surface average is k and the volume average is bounded. To generalize the one dimensional case we will need for each element of the sequence a maximum of a finite number of harmonic functions which is large in a neighborhood of ∂Ω and small elsewhere. To deal with a fairly general geometry of ∂Ω requires using in the construction harmonic functions other than affine functions and will require a covering argument to control the oscillation of these on ∂Ω. For domains satisfying the uniform exterior sphere condition translating and rescaling fundamental solutions will be enough. Details follow below. k Let k > 2 be an integer and n > 1. For each y ∈ ∂Ω, let Bδk (y ) denote a k closed ball such that Bδk (y ) ∩ Ω = {y} and δk > 0. Here (δk) is a sequence of positive numbers smaller than the radius in the uniform exterior sphere condition and verifying δk → 0. Denote the fundamental solution of the by Ψ(|x|). Define, for each k> 2 and y ∈ ∂Ω

k 2 k 2 k vy (x)= −k Ψ(|x − y |/δk)+ k Ψ(|y − y |/δk)+ k.

k k Clearly vy (y)= k and vy < k on ∂Ω \{y}. By continuity and compactness there k is a finite number (dependent on k) of points yj ∈ ∂Ω and corresponding balls k k k Brjk (yj ) forming a covering of ∂Ω such that each vyj > k − 1 in Brjk (yj ). k k k k Let v = max{maxj {vyj }, 1}. Then v is subharmonic, k ≥ v ≥ 1 in Ω and vk > k − 1 on ∂Ω. k k Let Ωδ = {x ∈ Ω : dist(x, ∂Ω) ≥ δ}. If z ∈ Ωδk then |z − yj | ≥ 2δk for all yj k 2 2 k and vyj (z) ≤−k Ψ(2) + k Ψ(1) + k implying that, for sufficiently large k, v =1 in Ωδk . This allows, for sufficiently large k, the estimates

vk ds > |∂Ω|(k − 1), Z∂Ω k k v dx = 1+ v dx ≤ |Ω| + k|Ω \ Ωδk |. Ω Ω Ω\Ω Z Z δk Z δk

The result follows by choosing the sequence (δk)k>2 in such a way that it implies bundedness of the last expression. CHARACTERIZATION OF HARMONIC AND SUBHARMONIC FUNCTIONS 7

3.2. Beardon’s property. The converse of Beardon’s property follows from three facts: a standard mollification argument that shows the result holds if it holds for smooth functions, a computation for smooth functions that shows that Beardon’s inequalities holding for balls centered at a point x0 imply the Laplacian is non– negative at x0, provided κ<κ1, and a well known device to extend the latter result to κ ≤ κ1. We present these as three lemmas.

Lemma 3.2. Assume u is a continuous function for which (1.5) holds and is not subharmonic. Then a smooth subharmonic function exists for which the same prop- erties hold.

Proof. Let u be such a continuous function. Then, in some point x in the interior of some ball B ⋐ Ω, u is bigger then the solution h of Dirichlet’s problem for the the Laplace equation in B having u as boundary data. Denote γ ≡ u(x)−h(x) > 0. Let δ = dist(B, ∂Ω) and 0 <ǫ<δ/4 and consider the usual mollifiers ρǫ sup- ported in Bǫ(0), ϕ a continuous cut-off function which is 0 in Ω\Ωδ/4 and 1 in Ωδ/2 and extend u by 0 in the complement of Ω. Then Beardon’s inequality holds for ρǫ ∗ (uϕ) in Ω3δ/4. To check this statement let Br(a) ⋐ Ω3δ/4 and notice that 1 ρ ∗ (uϕ) dS rn−1κn−1ω ǫ n Z∂Bkr (a) 1 = ρ (y)u(x − y)ϕ(x − y) dy dS(x) rn−1κn−1ω ǫ n Z∂Bkr (a) ZBǫ(0) ! 1 = ρ (y) u(x − y) dS(x) dy rn−1κn−1ω ǫ n ZBǫ(0) Z∂Bkr (a) ! n ≤ ρ (y) u(x − y) dx dy rnω ǫ n ZBǫ(0) ZBr (a) ! n = ρ (y)u(x − y) dy dx rnω ǫ n ZBr (a) ZBǫ(0) ! n = ρ ∗ (uϕ) dx. rnω ǫ n ZBr (a)

As ρǫ ∗ (uϕ) → u uniformly on ∂B ∪{x} as ǫ → 0, we have, for sufficiently small ǫ> 0, ρǫ ∗ (uϕ) − γ/2 h(x). Hence ρǫ ∗ (uϕ) − γ/2 has all the desired properties in Ω3δ/4.

2 Lemma 3.3. Let u ∈ C (BR(a)) and assume (1.5) holds for u for some κ satisfy- ing 0 <κ<κ1 and all r satisfying 0

Proof. Assume a = 0. Beardon’s inequality can be rewritten as

n r 1 0 ≤ u(x)dS(x) dt − u(x) dS(x) rn rn−1κn−1 Z0 Z∂Bt(0) ! Z∂Bκr (0) n r = tn−1 u(tx)dS(x) dt − u(κrx) dS(x). rn Z0 Z∂B1(0) ! Z∂B1(0) 8 PEDRO FREITAS AND JOAO˜ PALHOTO MATOS

Let now r ψ(r)= n tn−1 u(tx)dS(x) dt − rn u(κrx) dS(x). Z0 Z∂B1(0) ! Z∂B1(0) As ψ ≥ 0 we have that ψ(r) lim r→0 ≥ 0, rn+2 which in turn equals ψ′(r) lim , r→0 (n + 2)rn+1 provided the latter limit exists. Now

ψ′(r)= nrn−1 u(tx) dS(x) − nrn−1 u(κrx) dS(x) Z∂B1(0) Z∂B1(0) − rn κ∇u(κrx) · x dS(x) Z∂B1(0) r d = nrn−1 (u(tx)) dt dS(x) − rn κ∇u(κrx) · x dS(x) dt Z∂B1(0) Zκr  Z∂B1(0) r = nrn−1 ∇u(tx) · xdt dS(x) − rn κ∇u(κrx) · x dS(x) Z∂B1(0) Zκr  Z∂B1(0) r = nrn−1 ∇u(tx) · x dS(x) dt − rn κ∇u(κrx) · x dS(x) Zκr "Z∂B1(0) # Z∂B1(0) r = nrn−1 t∆u(tx)dx dt − rn+1 κ2∆u(κrx)dx, Zκr "ZB1(0) # ZB1(0) where the last step follows by applying the divergence theorem. Thus, ψ′(r) n r lim = lim t∆u(tx)dx dt − κ2 ∆u(κrx)dx r→0 rn+1 r→0 r2 " Zκr "ZB1(0) # ZB1(0) # n r = lim t∆u(tx)dx dt − κ2 ∆u(0)dx. r→0 r2 " Zκr "ZB1(0) # # ZB1(0) Since n r n lim t∆u(tx)dx dt = (1 − κ) ∆u(0)dx, r→0 r2 2 " Zκr "ZB1(0) # # ZB1(0) we finally obtain that n (1 − κ) − κ2 ∆u(0) ≥ 0, 2 h i and thus if κ ∈ (0,κ1), it follows that ∆u(0) ≥ 0. The next lemma is only relevant in establishing the converse of Beardon’s prop- erty in the case n = 1.

Lemma 3.4. The previous lemma holds if κ = κ1. CHARACTERIZATION OF HARMONIC AND SUBHARMONIC FUNCTIONS 9

Proof. Again assume a = 0. Assume Beardon’s inequality holds for some smooth 2 function u with κ = κ1. Let v(x)= |x| . Then 1 v(x) dS = κ2r2 ω |κ|n−1rn−1 n Z∂Bκr and n r n v(x) dx = |ρ|2 dS dρ = r2. ω rn n +2 n ZBr Z0 Z∂Bρ ! Hence Beardon’s property holds for v with κ = κ0, hence it holds for u + ǫv for any ǫ> 0 and with κ = κ0 and consequently ∆u(0)+2nǫ = ∆(u + ǫv)(0) ≥ 0 for any ǫ> 0. But then ∆u(0) ≥ 0.

From the proof of Lemma 3.3 we have that if Ba(r) ≤ Sa(κr) for κ>κ1 then n (1 − κ) − κ2 ∆u(a) ≤ 0, 2 h i and we again conclude that ∆u(a) ≥ 0. This of course already followed from Theorem 1.3 (of which this argument gives a different proof). On the other hand, we have that for u(x)= |x|2p (p integer) nr2p B (r)= , while S (r)= κ2pr2p, 0 2p + n 0 giving that B0(r) ≤ S0(κr) provided n 1/(2p) κ ≥ , 2p + n   which converges to one as p → ∞. Thus we see that the best possible value of κ for this property of subharmonic functions is one.

References

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Departamento de Matematica,´ Instituto Superior Tecnico,´ Av.Rovisco Pais, 1049-001 Lisboa, Portugal E-mail address: [email protected], [email protected]