On separately subharmonic and harmonic functions

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On separately subharmonic andharmonic functionsJuhani Riihentaus a ba Department of Mathematical Sciences , University of Oulu , P.O.Box 3000, FI-90014 Oulun yliopisto , Finlandb Department of Physics and Mathematics , University of EasternFinland , P.O. Box 111, FI-80101 Joensuu , FinlandPublished online: 20 Sep 2013.

To cite this article: Juhani Riihentaus (2014) On separately subharmonic and harmonic functions,Complex Variables and Elliptic Equations: An International Journal, 59:2, 149-161, DOI:10.1080/17476933.2013.816845

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Complex Variables and Elliptic Equations, 2014Vol. 59, No. 2, 149–161, http://dx.doi.org/10.1080/17476933.2013.816845

On separately subharmonic and harmonic functions

Juhani Riihentaus∗

Department of Mathematical Sciences, University of Oulu, P.O. Box 3000, FI-90014 Oulunyliopisto, Finland; Department of Physics and Mathematics, University of Eastern Finland,

P.O. Box 111, FI-80101 Joensuu, Finland

Communicated by S. Plaksa

(Received 27 January 2013; accepted 7 June 2013)

We improve our previous generalizations to Arsove’s, and Kołodziej’s andThorbiörnson’s results concerning the subharmonicity of a function subharmonicwith respect to the first variable and harmonic with respect to the second.

Keywords: Separately subharmonic; harmonic; quasinearly subharmonic;generalized Laplacian

AMS Subject Classifications: 31C05; 31B25; 31B05

1. Introduction

Solving a long-standing problem, Wiegerinck [28, Theorem, p. 770], see also Wiegerinckand Zeinstra [29, Theorem 1, p. 246], showed that a separately subharmonic function neednot be subharmonic. On the other hand, it is an open problem, whether a function which issubharmonic in one variable and harmonic in the other is subharmonic. For older results onthis area, see e.g. Arsove [3, Theorem 2, p. 622], Imomkulov [8, Theorem, p. 9], Wiegerinckand Zeinstra [29, p. 248], Cegrell and Sadullaev [5, Theorem 3.1, p. 82] and Kołodziej andThorbiörnson [9, Theorem 1, p. 463]. The result of Kołodziej and Thorbiörnson (partially)includes the results of Arsove, of Cegrell and Sadullaev and of Imomkulov, and reads asfollows:

Theorem A Let � be a domain in Rm+n, m, n ≥ 2. Let u : � → R be such that

(a) for each y ∈ Rn the function

�(y) � x �→ u(x, y) ∈ R

is subharmonic and C2,(b) for each x ∈ R

m the function

�(x) � y �→ u(x, y) ∈ R

is harmonic.

Then u is subharmonic and continuous in �.

∗Emails: [email protected], [email protected] to the memory of Professor Promarz M. Tamrazov

© 2013 Taylor & Francis

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Above and below, we use the following notation: If � is a domain in Rm+n , m, n ≥ 2,

and if x ∈ Rm , y ∈ R

n , then write

�(y) := { x ′ ∈ Rm : (x ′, y) ∈ � }, �(x) := { y′ ∈ R

n : (x, y′) ∈ � }.We improved the result of Kołodziej and Thorbiörnson in a series of papers:

[15, Theorem 3, Theorem 4 and Corollary, p. 162–164], [16, Theorem 6, p. 234],[17, Theorem 1 and Corollary, p. 438, 444], [18, Theorem 5.1, Corollary 5.1 and Corol-lary 5.2, p. 67–68, 74], see also [20, Theorem 4.3.1, Corollary 4.3.3 and Corollary 4.3.4,p. e2625–e2626]. We will now return to the subject and improve our result still further, seeTheorem 2 below.

However, we begin with improving the above-cited results of Arsove, of Cegrell andSadullaev and our previous generalizations [18, Theorem 4.1, p. 64] and [20, Theorem 4.2.1,p. e2623]. Instead of subharmonic functions (resp. so called quasinearly subharmonicfunctions n.s.), we will now use quasinearly subharmonic functions. Observe that in certainsituations, such an approach is indeed useful. One example is the following. Armitageand Gardiner [1, Theorem 1, p. 256] gave a condition which ensures a separately sub-harmonic function to be subharmonic, and their condition was close to being sharp, see[1, p. 255–256]. With the aid of quasinearly subharmonic functions it was, nevertheless,possible to generalize and improve their result, see [19, Theorem 4.1 and Corollary 4.5,p. 8–9, 13] and [20, Theorem 3.3.1 and Corollary 3.3.3, p. e2621–e2622]. For our notationand definitions, see e.g. [7,18,20] and the references therein. For the convenience of thereader, we, however, recall the following. m N is the Lebesgue measure in the Euclideanspace R

N , N ≥ 2 and D is always a domain in RN .

We recall that an upper semicontinuous function u : D → [−∞,+∞) is subharmonicif for all closed balls B N (x, r) ⊂ D,

u(x) ≤ 1

νN r N

∫

B N (x,r)

u(y) dm N (y).

The function u ≡ −∞ is considered subharmonic.We say that a function u : D → [−∞,+∞) is nearly subharmonic, if u is Lebesgue

measurable, u+ ∈ L1loc(D), and for all B N (x, r) ⊂ D,

u(x) ≤ 1

νN r N

∫

B N (x,r)

u(y) dm N (y).

Observe that in the standard definition of nearly subharmonic functions, one uses the slightlystronger assumption that u ∈ L1

loc(D), see e.g. [7, p. 14]. However, our above, slightly moregeneral definition seems to be more practical, see e.g. [18, Proposition 2.1 (iii) and Propo-sition 2.2 (vi) and (vii), p. 54–55] and [20, Proposition 1.5.1 (iii) and Proposition 1.5.2 (vi)and (vii), p. e2615]. The following result emphasizes this fact still more.

Proposition ([18, Lemma, p. 52]) Let u : D → [−∞,+∞) be Lebesgue measurable.Then u is nearly subharmonic (in the sense defined above) if and only if there exists a functionu∗, subharmonic in D such that u∗ ≥ u and u∗ = u almost everywhere in D. Here u∗ isthe upper semicontinuous regularization of u:

u∗(x) = lim supx ′→x

u(x ′).

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Complex Variables and Elliptic Equations 151

The proof follows at once from [7, proof of Theorem 1, p. 14–15] (and referring also to[18, Proposition 2.1 (iii) and Proposition 2.2 (vii), p. 54–55]).

We say that a Lebesgue measurable function u : D → [−∞,+∞) is K -quasinearlysubharmonic, if u+ ∈ L1

loc(D) and if there is a constant K = K (N , u, D) ≥ 1 such thatfor all B N (x, r) ⊂ D,

uM (x) ≤ K

νN r N

∫

B N (x,r)

uM (y) dm N (y)

for all M ≥ 0, where uM := max{u,−M} + M . A function u : D → [−∞,+∞) isquasinearly subharmonic, if u is K -quasinearly subharmonic for some K ≥ 1.

2. Arsove’s result and its improvement

Arsove’s result is:

Theorem B ([3, Theorem 2, p. 622]) Let � be a domain in Rm+n, m, n ≥ 2. Let

u : � → R be such that


�(y) � x �→ u(x, y) ∈ R

is subharmonic,(b) for each x ∈ R

m the function

�(x) � y �→ u(x, y) ∈ R

is harmonic,(c) there is a nonnegative function ϕ ∈ L1

loc(�) such that −ϕ ≤ u.

Then u is subharmonic in �.

Arsove’s proof was based on mean value operators. Much later Cegrell and Sadullaev[5, Theorem 3.1, p. 82] gave a new proof using Poisson modification. Below in Theorem 1,we generalize the above result of Arsove, of Cegrell and Sadullaev and also our previousgeneralizations [18, Theorem 4.1 and Corollary 4.1, p. 64–65], see also [20, Theorem 4.2.1and Corollary 4.2.2, p. e2623]. Our proof is direct and short and it is different than theprevious proofs of Arsove and of Cegrell and Sadullaev. In the proof, we need the followingtwo lemmas.

Lemma 1 Let D be a domain in RN , N ≥ 2. Let E be a locally compact space, and let

μ be a positive measure on E. Let v : D × E → R be such that

(a) for each y ∈ E the function

D � x �→ v(x, y) ∈ R

is continuous,

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152 J. Riihentaus

(b) for each x ∈ D the function

E � y �→ v(x, y) ∈ R

is measurable.

Then v is measurable.

The idea of the proof is just to use dyadic cubes and approximate the function v(·, ·)with measurable functions of the type

D × E � (x, y) �→ v(a, y) ∈ R, a ∈ D.

We leave the details to the reader and just refer to a preliminary result [22, Exercise 8, p. 160]and to the special case results [3, Lemma 1, p. 624] and [13, Lemma 3.2, p. 103–104], say.

Lemma 2 (cf. [7, Theorem 2a), p. 15]) Let v be nearly subharmonic (now in the general-ized sense, defined above) in a domain D of R

N , N ≥ 2, ψ ∈ L∞(RN ), ψ ≥ 0, ψ(x) = 0when | x |≥ α and ψ(x) depends only on | x |. Then ψ � v ≥ v and ψ � v is subharmonicin Uα , provided

∫ψ(x)dm N (x) = 1, where Dα = {x ∈ D : B N (x, α) ⊂ D}.

Observe that the proof of the Lemma, see [7, proof of Theorem 2a), p. 15], works indeedalso in our slightly more general situation, where in the definition of nearly subharmonicfunctions, we use, instead of the standard condition v ∈ L1

loc(D), the slightly weakercondition u+ ∈ L1

loc(D).

Theorem 1 Let � be a domain in Rm+n, m, n ≥ 2, and K ≥ 1. Let u : � → R be such

that


�(y) � x �→ u(x, y) ∈ R

is K-quasinearly subharmonic,(b) for each x ∈ R

m the function

�(x) � y �→ u(x, y) ∈ R

is harmonic,(c) there is a nonnegative function ϕ ∈ L1

loc(�) such that −ϕ ≤ u.

Then u is K-quasinearly subharmonic in �.

Proof The Lebesgue measurability of u follows from Lemma 1. Thus, also the functionsuM := max{ u,−M } + M , M > 0, are Lebesgue measurable. We must show that u+ ∈L1

loc(�) and that each uM satisfies the generalized mean value inequality.

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To see that u+ ∈ L1loc(�), we proceed as follows. Observe first that 0 ≤ u+ ≤ uM ≤

vM := u + ϕ + M . To see that vM ∈ L1loc(�) requires only Fubini’s Theorem. As a matter

of fact, take Bm(a, R)× Bn(b, R) ⊂ � arbitrarily. Then,

0 ≤ K

mm+n(Bm(a, R)× Bn(b, R))

∫Bm(a,R)×Bn(b,R)

vM (x, y)dmm+n(x, y)

≤ K


∫Bm(a,R)×Bn(b,R)

[u(x, y)+ ϕ(x, y)+ M]dmm+n(x, y)

≤ K

νm Rm

∫Bm(a,R)

⎧⎪⎨⎪⎩

1

νn Rn

∫Bn(b,R)

[u(x, y)+ ϕ(x, y)+ M]dmn(y)

⎫⎪⎬⎪⎭ dmm(x)

≤ K

νm Rm

∫Bm(a,R)

⎡⎢⎣ 1

νn Rn

∫Bn(b,R)

u(x, y)dmn(y)

+ 1

νn Rn

∫Bn(b,R)

ϕ(x, y)dmn(y)+ M

⎤⎥⎦ dmm(x)

≤ K

νm Rm

∫Bm(a,R)

⎡⎢⎣u(x, b)+ 1

νn Rn

∫Bn(b,R)

ϕ(x, y)dmn(y)+ M

⎤⎥⎦ dmm(x)

≤ K

νm Rm

∫Bm(a,R)

u(x, b)dmm(x)

+ K

νm Rm

∫Bm(a,R)

⎡⎢⎣ 1

νn Rn

∫Bn(b,R)

ϕ(x, y)dmn(y)

⎤⎥⎦ dmm(x)+ K M

≤ K

νm Rm

∫Bm(a,R)

u(x, b)dmm(x)

+ K


∫Bm(a,R)×Bn(b,R)

ϕ(x, y)dmm+n(x, y)+ K M

<+ ∞.

It remains to show that for all (a, b) ∈ � and R > 0 such that Bm+n((a, b), R) ⊂ �,

uM (a, b) ≤ K

νm+n Rm+n

∫Bm+n((a,b),R)

uM (x, y)dmm+n(x, y).

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154 J. Riihentaus

To see this, we proceed in the following standard, direct and short way, see e.g.[7, Proposition 2(c) and proof of Theorem (a), p. 10–11, 32–33] and [18, p. 59]:

K

νm+n Rm+n

∫Bm+n((a,b),R)

uM (x, y)dmm+n(x, y)

= νm

νm+n Rm+n

∫Bn(b,R)

[(R2− | y − b |2)m

2K

νm(R2− | y − b |2)m2

∫

Bm(a,√

R2−|y−b|2)

uM (x, y)dmm(x)

]dmn(y)

≥ νm

νm+n Rm+n

∫Bn(b,R)

(R2− | y − b |2)m2 uM (a, y)dmn(y) ≥ uM (a, b).

Above we have used, in addition to the fact that, for every y ∈ Rm , the functions u(·, y) are

K -quasinearly, also the above Lemma 2. �

3. An improvement to the result of Kołodziej and Thorbiörnson

In our generalization to the cited result of Kołodziej and Thorbiornson, TheoremAabove, wewill use the generalized Laplacian defined with the aid of the Blaschke–Privalov operators,see e.g. [23, p. 451], [21, p. 278–279], [24, p. 498] and [26, p. 29]. Let D be a domain inR

N , N ≥ 2 and f : D → R, f ∈ L1loc(D). We write,

�∗ f (x) : = lim infr→0

2(N + 2)

r2·[

1

νN r N

∫

B N (x,r)

f (x ′)dm N (x′)− f (x)

],

�∗ f (x) : = lim supr→0

2(N + 2)

r2·[

1

νN r N

∫

B N (x,r)

f (x ′)dm N (x′)− f (x)

].

If �∗ f (x) = �∗ f (x), then write � f (x) := �∗ f (x) = �∗ f (x).If f ∈ C2(D), then

� f (x) =⎛⎝ N∑

j=1

∂2 f

∂x2j

⎞⎠ (x),

the standard Laplacian with respect to the variable x = (x1, x2, . . . , xN ). More generally,if x ∈ D and f ∈ t1

2 (x), i.e. f has an L1 total differential of order 2 at x , then� f (x) equalswith the pointwise Laplacian of f at x , i.e.

� f (x) =N∑

j=1

D j j f (x).

Here, D j j f represents a generalization to the usual ∂2 f∂x2

j, j = 1, 2, . . . , N . See e.g.

[25, p. 369] and [26, p. 29].

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Recall that there exist functions which are notC2, but for which the generalized Laplacianis, nevertheless, continuous, perhaps in the extended sense (in ([0,+∞], q), where q isthe spherical metric), see e.g. [26, p. 31], [18, Example 5 and Example 6, p. 66–67] and[20, Example 1 and Example 2, p. e2624–e2625].

If f is subharmonic on D, it follows from [23, p. 451] (see also [21, Lemma 2.2, p. 280]and [25, p. 376]) that � f (x) = �∗ f (x) = �∗ f (x) ∈ R for almost every x ∈ D.

Below, we use the following notation. Let � be a domain in Rm+n , m, n ≥ 2, and

u : � → R. If y ∈ Rn is such that the function

�(y) � x �→ f (x) := u(x, y) ∈ R

is in L1loc(�(y)), then we write �1∗u(x, y) := �∗ f (x), �∗

1u(x, y) := �∗ f (x), and�1u(x, y) := � f (x). Then we return to the generalization of our previous result [18,Theorem 5.1, p. 67–68] (or [20, Theorem 4.3.1, p. e2625] (where no proof is given!)) andthus, also the result of Kołodziej and Thorbiornson [9, Theorem 1, p. 463], Theorem Aabove. Though our proof will follow the main lines of [18, proof of Theorem 5.1, p. 67–72],it is different enough, nevertheless, to warrant that it be given in complete detail here: Now,our assumption (d) is essentially milder than our previous assumptions (d) and (e).

Theorem 2 Let � be a domain in Rm+n, m, n ≥ 2. Let u : � → R be such that

for each (x ′, y′) ∈ � there is (x0, y0) ∈ � and r1 > 0, r2 > 0 such that (x ′, y′) ∈Bm(x0, r1) × Bn(y0, r2) ⊂ Bm(x0, r1)× Bn(y0, r2) ⊂ � and such that the followingconditions are satisfied:

(a) For each y ∈ Bn(y0, r2) the function

Bm(x0, r1) � x �→ u(x, y) ∈ R

is continuous, and subharmonic in Bm(x0, r1).(b) For each x ∈ Bm(x0, r1) the function

Bn(y0, r2) � y �→ u(x, y) ∈ R

is continuous, and harmonic in Bn(y0, r2).(c) For each y ∈ Bn(y0, r2) one has �1∗u(x, y) < +∞ for each x ∈ Bm(x0, r1),

possibly with the exception of a polar set in Bm(x0, r1).(d) There are a set H1 ⊂ Bm(x0, r1), dense in Bm(x0, r1), and a set H2 ⊂ Bn(y0, r2),

dense in Bn(y0, r2), such that

(d1) for each y ∈ H2, for almost every x ∈ Bm(x0, r1) and for each x ∈ H1,

�1u(x ′, y) → �1u(x, y) ∈ R

as x ′ → x, x ′ ∈ H1, and(d2) for each y ∈ Bn(y0, r2) \ H2 and for almost every x ∈ Bm(x0, r1),

�1u(x, y′) → �1u(x, y) ∈ R

as y′ → y, y′ ∈ H2.


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Proof Choose r ′1 and r ′

2 such that 0 < r ′1 < r1 and 0 < r ′

2 < r2, and such that(x ′, y′) ∈ Bm(x0, r ′

1)× Bn(y0, r ′2). It is sufficient to show that u | Bm(x0, r ′

1)× Bn(y0, r ′2)

is subharmonic. For the sake of convenience of notation, we change the roles of r j and r ′j ,

j = 1, 2. We divide the proof into several steps.

Step 1 Construction of an auxiliar dense set G.For each k ∈ N, write

Ak := { x ∈ Bm(x0, r1) : −k ≤ u(x, y) ≤ k for each y ∈ Bn(y0, r2) }.Clearly Ak is closed and

Bm(x0, r1) =+∞⋃k=1

Ak .

Write

G :=+∞⋃k=1

int Ak .

It follows from Baire’s theorem that G is dense in Bm(x0, r1).

Step 2 The functions �1r u(x, ·) : Bn(y0, r2) → R (see the definition below), x ∈ G,0 < r < rx := dist(x, Bm(x0, r1) \ G), are nonnegative and harmonic.

For each (x, y) ∈ Bm(x0, r1) × Bn(y0, r2) and each r , 0 < r < dist(x, ∂Bm(x0, r ′1))

(observe that dist(x, ∂Bm(x0, r ′1)) > r ′

1 − r1 > 0), write

�1r u(x, y) := 2(m + 2)

r2·[

1

νm rm

∫Bm(x,r)

u(x ′, y) dmm(x′)− u(x, y)

]

= 2(m + 2)

r2· 1

νm rm

∫Bm(0,r)

[u(x + x ′, y)− u(x, y)

]dmm(x

′).

Since u(·, y) is subharmonic, �1r u(x, y) is defined and nonnegative. Suppose then thatx ∈ G and 0 < r < rx . Since Bm(x, r) ⊂ G and Ak ⊂ Ak+1 for all k = 1, 2, . . . ,Bm(x, r) ⊂ int AN for some N ∈ N. Therefore,

−N ≤ u(x ′, y) ≤ N for all x ′ ∈ Bm(x, r) and y ∈ Bn(y0, r2),

and hence,

−2N ≤ u(x + x ′, y)− u(x, y) ≤ 2N for all x ′ ∈ Bm(0, r) and y ∈ Bn(y0, r2). (1)

To show that �1r u(x, ·) is continuous, pick an arbitrary sequence y j → y0, y j , y0 ∈Bn(y0, r2), j = 1, 2, . . . . Using then (1), Lebesgue Dominated Convergence Theorem andthe continuity of u(x, ·), one sees easily that �1r u(x, ·) is continuous.

To show that �1r u(x, ·) satisfies the mean value equality, take y0 ∈ Bn(y0, r2) andρ > 0 arbitrarily such that Bn(y0, ρ) ⊂ Bn(y0, r2). Because of (1), we can use Fubini’s

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Theorem. Thus

1

νnρn

∫Bn(y0,ρ)

�1r u(x, y)dmn(y)

= 1

νnρn

∫Bn(y0,ρ)

⎧⎪⎨⎪⎩

2(m + 2)

r2· 1

νmrm

∫Bm(0,r)

[u(x + x ′, y)− u(x, y)

]dmm(x

′)

⎫⎪⎬⎪⎭ dmn(y)

= 2(m + 2)

r2· 1

νmrm

∫Bm(0,r)

⎧⎪⎨⎪⎩

1

νnρn

∫Bn(y0,ρ)

[u(x + x ′, y)− u(x, y)

]dmn(y)

⎫⎪⎬⎪⎭ dmm(x

′)

= 2(m + 2)

r2· 1

νm rm

∫Bm(0,r)

[u(x + x ′, y0)− u(x, y0)

]dmm(x

′)

= �1r u(x, y0).

Step 3 The functions �1u(x, ·) : Bn(y0, r2) → R, x ∈ G ∩ A, are defined, nonnegativeand harmonic. Here

A :=+∞⋂k=1

A(yk),

where H2 = { yk, k = 1, 2, . . . } (we may clearly suppose that H2 is countable), and, forarbitrary y ∈ Bn(y0, r2),

A(y) := { x ∈ Bm(x0, r1) : �1∗u(x, y) = �∗1u(x, y) = �1u(x, y) ∈ R }.

[21, Lemma 2.2, p. 280] (see also [23, p. 451] and [25, p. 376]) states that mm(Bm(x0, r1)\A(y)) = 0 for each y ∈ Bn(y0, r2).

Take x ∈ G ∩ A and a sequence r j → 0, 0 < r j < rx , j = 1, 2, . . . , arbitrarily.[7, Corollary 3 a), p. 6] (or [2, Lemma 1.5.6, p. 16]) gives that the family

�1r j u(x, ·) : Bn(y0, r2) → R, j = 1, 2, . . . ,

of nonnegative and harmonic functions is either uniformly equicontinuous and locallyuniformly bounded, or else

supj=1,2,...

�1r j u(x, ·) ≡ +∞.

On the other hand, since x ∈ G ∩ A, we know that for each yk ∈ H2, k = 1, 2, . . . ,

�1r j u(x, yk) → �1u(x, yk) ∈ R

as j → +∞. Therefore, by [27, Theorem 20.3, p. 68] and by [7, (c), p. 2](or [2, Theorem 1.5.8, p. 17]), the limit

�1u(x, ·) = limj→+∞�1r j u(x, ·)

exists and defines a harmonic function in Bn(y0, r2). Since the limit is clearly independentof the considered sequence r j , the claim follows.

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158 J. Riihentaus

Step 4 The function�1u(·, ·) | (G ∩ H1 ∩ A∩ B)× Bn(y0, r2) has a continuous extension�1u(·, ·) : (A ∩ B)× Bn(y0, r2) → R. Moreover, the functions �1u(x, ·) : Bn(y0, r2) →R, x ∈ A ∩ B, are nonnegative and harmonic. Here

B :=+∞⋂k=1

B(yk),

where, for arbitrary y ∈ Bn(y0, r2), we use the notation

B(y) := { x ∈ Bm(x0, r1) : �1u(x ′, y) → �1u(x, y) as x ′ → x, x ′ ∈ H1 }.Using the assumption (d1), one sees easily that G ∩ H1 ∩ A ∩ B is dense in A ∩ B.To show the existence of the desired continuous extension, it is clearly sufficient to show

that for each (x0, y0) ∈ (A ∩ B)× Bn(y0, r2), the limit

lim(x,y)→(x0,y0), (x,y)∈(G∩H1∩A∩B)×Bn(y0,r2)

�1u(x, y)

exists. (This is of course standard, see e.g. [6, (3.15.5), p. 54].) To see this, it is sufficient toshow that, for an arbitrary sequence (x j , y j ) → (x0, y0), (x j , y j ) ∈ (G ∩ H1 ∩ A ∩ B)×Bn(y0, r2), j = 1, 2, . . . , the limit

limj→+∞�1u(x j , y j )

exists.That this limit indeed exists, is seen as above, just using the facts:

− the functions�1u(x j , ·), j = 1, 2, . . . , are nonnegative and harmonic in Bn(y0, r2),by Step 3;

− for each yk ∈ H2, k = 1, 2, . . . , �1u(x j , yk) → �1u(x0, yk) ∈ R as j → +∞.

(See again [7, Corollary 3(a), p. 6] (or [2, Lemma 1.5.6, p. 16]) and [27, Theorem 20.3,p. 68]). That the functions �1u(x, ·) : Bn(y0, r2) → R, x ∈ A ∩ B, are harmonic, see[7, c), p. 2] (or [2, Theorem 1.5.8, p. 17]).

Step 5 For each x ∈ Bm(x0, r1) the functions

Bn(y0, r2) � y �→ v(x, y) :=∫

G Bm (x0,r1)(x, z)�1u(z, y)dmm(z) ∈ R

and

Bn(y0, r2) � y �→ h(x, y) := u(x, y)+ v(x, y) ∈ R

are harmonic. Above and below G Bm(x0,r1)(x, z) is the Green function of the ballBm(x0, r1), with x as a pole.

Using Fubini’s Theorem, one sees easily that for each x ∈ Bm(x0, r1), the functionv(x, ·) satisfies the mean value equality. To see that v(x, ·) is harmonic, it is sufficientto show that v(x, ·) ∈ L1

loc(Bn(y0, r2)). Using just Fatou’s Lemma, one sees that v(x, ·)

is lower semicontinuous, hence superharmonic. Therefore, either v(x, ·) ≡ +∞ or elsev(x, ·) ∈ L1

loc(Bn(y0, r2)). The following argument shows that the former alternative cannot

occur. Indeed, for each x ∈ A ∩ B and for each yk ∈ H2, k = 1, 2, . . . , we see, using thedefinition of the (continuous) function �1u(·, ·) and (d1), that

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�1u(x, yk) = limx ′→x, x ′∈G∩H1∩A∩B

�1u(x ′, yk) = limx ′→x, x ′∈G∩H1∩A∩B

�1u(x ′, yk)

= �1u(x, yk) ∈ R. (2)

Hence v(x, yk) = v(x, yk) ∈ R, for each x ∈ Bm(x0, r1) and yk ∈ H2, k = 1, 2, . . . .(See (3) in Step 6 below for the definition of v(·, ·) : Bm(x0, r1) × Bn(y0, r2) → R.)Therefore, for each x ∈ Bm(x0, r1), the function v(x, ·) and thus also the function h(x, ·) =u(x, ·)+ v(x, ·) are harmonic.

Step 6 For each y ∈ Bn(y0, r2) the function

Bm(x0, r1) � x �→ h(x, y) ∈ R

is harmonic.

With the aid of the version of Riesz’s Decomposition Theorem, given in [21, 1.3.Theorem II, p. 279, and p. 278, too] (see also [24, Theorem 1, p. 499]), for each y ∈Bn(y0, r2) one can write

u(x, y) = h(x, y)− v(x, y),

where

v(x, y) :=∫

G Bm (x0,r1)(x, z)�1u(z, y)dmm(z) (3)

and h(·, y) is the least harmonic majorant of u(·, y) | Bm(x0, r1). Here v(·, y) is continuousand superharmonic in Bm(x0, r1).

As shown above in (2), v(·, yk) = v(·, yk) for each yk ∈ H2, k = 1, 2, . . . . Therefore,h(·, yk) = h(·, yk) and thus h(·, yk) is harmonic for each yk ∈ H2, k = 1, 2, . . . .

To see that h(·, y) is harmonic also for y ∈ Bn(y0, r2) \ H2, take y0 ∈ Bn(y0, r2) \ H2arbitrarily and proceed in the following way. Take z ∈ A ∩ B ∩ A(y0) ∩ C(y0) arbitrarily,where, for arbitrary y ∈ Bn(y0, r2) \ H2,

C(y) := { z ∈ Bm(x0, r1) : �1∗u(z, y′) → �1∗u(z, y) as y′ → y, y′ ∈ H2 }.Since z ∈ A(y0), we have �1∗u(z, y0) = �1u(z, y0) ∈ R. Thus, we may also supposethat �1∗u(z, y′) = �1u(z, y′) ∈ R. Using then, our assumption (d2) and the continuity of�1u(·, ·), we see that

�1u(z, y0) = �1u(z, y0)

for every z ∈ A ∩ B ∩ A(y0) ∩ C(y0). Therefore, v(x, y0) = v(x, y0) and thus h(x, y0) =h(x, y0) for each x ∈ Bm(x0, r1).

Step 7 The use of the results of Lelong and of Avanissian.

By Steps 5 and 6, we know that h(·, ·) = h(·, ·) is separately harmonic in Bm(x0, r1)×Bn(y0, r2). Lelong’s result [11, p. 561] states that h(·, ·) is harmonic and thus locallybounded above in Bm(x0, r1)× Bn(y0, r2). Therefore, also u(·, ·) is locally bounded abovein Bm(x0, r1) × Bn(y0, r2). But then it follows from [4, Théoréme 9, p. 140] (see also[10, Théoréme 1 bis, p. 315], [3, Theorem 1, p. 622], [12, Proposition 3, p. 24],[14, Theorem 1, p. 69], [1, Theorem 1, p. 256] and [19, Corollary 4.5, p. 13]) that u(·, ·) issubharmonic on Bm(x0, r1)× Bn(y0, r2). �

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160 J. Riihentaus

Remark Observe that the assumption (d2) was needed only to see that

�1u(x, y) = �1u(x, y) for almost every x ∈ Bm(x0, r1) and for each y ∈ Bn(y0, r2)\H2.

(At this point, one might recall that the functions �1u(x, ·) : Bn(y0, r2) → R, x ∈Bm(x0, r1) are harmonic.)

From the above proof one sees easily that the assumption (d), that is (d1) and (d2), canbe replaced by:

(d∗) There is a set H∗1 ⊂ Bm(x0, r1), dense in Bm(x0, r1), such that for each y ∈

Bn(y0, r2), for almost every x ∈ Bm(x0, r1) and for each x ∈ H∗1 ,

�1u(x ′, y) → �1u(x, y) ∈ R

as x ′ → x, x ′ ∈ H∗1 .

Though our Theorem 2 might still be considered somewhat technical, it has, never-theless, the following concise corollary and already this corollary improves our previousresults [18, Corollary 5.1, p. 74], [20, Corollary 4.3.3, p. e2626], [17, Corollary, p. 444],[16, Theorem 6, p. 233] and thus also the result of Kołodziej and Thorbiornson, Theorem Aabove.

Corollary Let � be a domain in Rm+n, m, n ≥ 2. Let u : � → R be such that


�(y) � x �→ u(x, y) ∈ R

is continuous and subharmonic,(b) for each x ∈ R

m the function

�(x) � y �→ u(x, y) ∈ R

is harmonic,(c) for each y ∈ R

n one has �1∗u(x, y) < +∞ for each x ∈ �(y), possibly with theexception of a polar set in �(y),

(d) there is a set H3 ⊂ Rm dense in R

m such that for each y ∈ Rn and for almost every

x ∈ �(y) and for each x ∈ H3,

�1u(x ′, y) → �1u(x, y) ∈ R

as x ′ → x, x ′ ∈ H3.


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On separately subharmonic and harmonic functions