EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
2025, Vol. 18, Issue 2, Article Number 5750
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Flux at Infinity of Subharmonic Functions on R2

Amulya Smyrna C.1,, N. Nathiya1∗

1 Department of Mathematics, School of Advanced Sciences, Vellore Institute of Technology
Chennai, Tamil Nadu, India

Abstract. For a C2-function f(x) on a bounded domain ω in R2 the flux is defined by means of
outer normal derivative of f . In this paper, we introduce the notion of flux(f) for any real-valued
function on R2. We define flux on bounded domain ω and take limits when ω grows into R2 and
the limit is defined as ”at infinity”, the flux(f) at infinity denoted as flux∞f . This limit flux∞f
may or may not be finite. The related development is carried out by employing the notion of
inversion on R2 and the fact that a harmonic function defined outside a compact set in R2 is the
difference of two subharmonic functions on R2 that are harmonic outside a compact set.

2020 Mathematics Subject Classifications: 31A05, 31A10

Key Words and Phrases: Subharmonic Function, Harmonic Function, Flux

1. Introduction

In the Euclidean plane R2, let ω be a bounded domain with smooth boundary and
f(x) be a C2−function defined on a neighbourhood of ω. Then the exit flux (the flux
acting outwards from a closed surface) of f from ω is defined as

∫
∂ω

∂f
∂n+ds where ∂

∂n+ is

the outward normal derivative at boundary points [1]. By using Green’s theorem, the
function f(x) is harmonic on ω if and only if the exit flux of f from ω is 0. There is also
a theorem in (Brelot [2]), by using a series representation for harmonic functions defined
outside a compact set in R2, it states that given a harmonic function h outside a compact
set in R2, there exists a harmonic function H on R2 such that |h−H| is bounded if and
only if the flux at infinity of h is 0. Now, what is the relation between these two notions
of flux in R2? In this article, we find an answer to this question by using more generally,
subharmonic function on R2 which is locally Lebesgue integrable functions with a certain
mean value property.

First we extend the definition of flux to a subharmonic function defined on a bounded
domain in R2 by using the distributions. Then we introduce the notion of flux at infinity
for a subharmonic function defined outside a compact set in R2. For this we require the

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v18i2.5750

Email addresses: amulyasmyrna.c@gmail.com (Amulya Smyrna C.), nadhiyan@gmail.com (N. Nathiya)

https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0)



Amulya Smyrna C., N. Nathiya / Eur. J. Pure Appl. Math, 18 (2) (2025), 5750 2 of 9

measure associated with a subharmonic function in a local Riesz representation as an in-
tegral, (Helms [3], Ransford [4]). We prove that for a subharmonic function s(x) defined
on R2, the flux at infinity of s is finite if and only if the total measure associated with s in
R2 is finite or equivalently if and only if s(x) has a harmonic majorant outside a compact
set.

As an aside, we have also that if s(x) is a subharmonic function outside a compact set
in R2, then s(x) = v(x)− αs log |x| where v(x) is subharmonic on R2, αs ≥ 0 a constant.

We define the flux at infinity of s(x) as [flux∞s(x)] = [total measure associated with
v(x) on R2] − αs; this [flux∞s(x)] is independent of the representation of s(x) = v(x) −
αs log |x|, which may or may not be finite. We deduce that if h(x) is a function harmonic
outside a compact set then h(x) = H(x) + αh log |x| + b(x), where H(x) is harmonic on
R2, and αh is uniquely determined constant and b(x) is bounded harmonic tending to 0
at infinity. We conclude with the result that if the subharmonic function s(x) outside a
compact set in R2 has a harmonic majorant, then the flux at infinity of s(x) is αh where
h(x) is the least harmonic majorant of s(x) outside a compact set.

In the Brelot’s axiomatic potential theory without positive potentials, a superharmonic
function s(x) on the harmonic space X is termed as admissible in (Anandam [5, 6]), if it
has a harmonic minorant outside a compact set. Therein is a definition of flux at infinity
of a harmonic function outside a compact set in X. But only with the three axioms of
(Brelot [7]), it is not possible to introduce a comprehensive comparative study of the total
measure of a superharmonic function and its flux at infinity on a harmonic space (that is
locally compact) without positive potentials. See also the paper (Bajunaid et. al. [8]).

If f(x) is a C2-function defined on a bounded domain ω̄ with smooth boundary, then
the flux(f) is defined by means of the outer normal derivatives of f . In this note, we
introduce the notion of the linear function flux(f) for any real-valued function f defined
on R2. Since any real-valued function f on R2 is the difference of two subharmonic func-
tions, it is enough to define flux(s) for any subharmonic functions s on R2 with associated
Radon measure µ in a local Riesz representation as a sum of an integral with respect to µ
and a harmonic function.

If u is a C2-function on D = {x : |x| ≤ r} in R2, then
∫

|x|<r

∆u(x)dx =
∫

|z|=r

∂u
∂n+ (z)dσ

which is the flux(u) on D. If s is any subharmonic function on R2, then there exists
an increasing sequence of C2-subharmonic functions tending to s. Thus, we can write∫
|x|<r

∆s(x)dx =
∫

|z|=r

∂s
∂n+ (z)dσ. Since ∆s ≥ 0 then ∆s defines a Radon measure µ on R2

and we define flux(s) = µ(R2) = lim
r→∞

∫
∂s
∂n+dσ at infinity. Thus the flux(s) at infinity

may or may not be finite.
By using inversion, we show that flux(s) is finite if and only if it has a harmonic

majorant h outside a compact set. Any such harmonic function outside a compact set has
a unique representation h(x) = H(x) + α log |x|+ b(x) where H(x) is harmonic on R2, α
is a constant and b(x) → 0 at infinity. Finally, we prove that if h(x) is the least harmonic
majorant of s(x) outside a compact set, then flux(s) = α.



Amulya Smyrna C., N. Nathiya / Eur. J. Pure Appl. Math, 18 (2) (2025), 5750 3 of 9

2. Preliminaries

An upper and semi-continuous function u, −∞ ≤ u(x) < ∞, u ̸≡ −∞ on R2 is said to
be subharmonic if for any x = reiθ, where R is the distance of ∂ω from the origin.

u(x) ≤ 1

2π

2π∫
0

R2 − r2

R2 − 2Rr cos θ − ϕ+ r2
u(Reiθ)dϕ

Suppose v(x) is a locally Lebesgue integrable function in R2 such that ∆v ≥ 0, in the
sense of distributions, then there exists a unique subharmonic u(x) such that v(x) = u(x)
almost everywhere. For any locally integrable function f(x), there exists a function ϕ(x)
such that ∆ϕ(x) = f(x), (Brelot [2]); write ∆s1(x) = g+(x) and ∆s2(x) = g−(x) where
g(x) = ∆f(x), then s1(x) and s2(x) are subharmonic on R2 and ∆[f − (s1 − s2)] = 0 so
that f − (s1−s2) is a harmonic function h almost everywhere. Thus any locally integrable
function is the difference of two subharmonic functions almost everywhere.

For a subharmonic function u(x), ∆u ≥ 0 in the sense of distributions, therefore∫
2π

dµ(x) = ∆u(x) is a Radon measure µ on R2; furthermore, for any bounded domain

ω, u(x) =
∫

y∈ω
log |x − y|dµ(y) + h(x) where h(x) is a harmonic function on ω. The total

measure µ(R2) associated with u may or may not be finite; µ(R2) = 0 if and only if u(x)
is a harmonic function.

If f(x) is a C2−function defined on a neighbourhood of ω where ω is a bounded domain
with smooth boundary ∂ω, then by Green’s theorem

∫
ω
∆f(x)dx =

∫
∂ω

∂f
∂n+ds, where

∂
∂n+

is the outward normal derivative. In particular, for a C2−subharmonic function u in R2,
the total measure of u is lim

ω−→R2

∫
∂ω

∂u
∂n+ds = lim

m−→∞

∫
∂ωm

∂u
∂n+ds.

Suppose now that s(x) is a subharmonic function on R2. Then (see Brelot [2]) there ex-
ists an increasing sequence {sn} of C2−functions tending to s on R2. Then ∆sn −→ ∆s so
that the total measure associated with s equals to lim

n−→∞
total measure of sn. Remark that

this limit does not depend on the particular sequence {sn} of C2−subharmonic functions
that increases to s.

3. Inversion and associated measure of a subharmonic function

Let s(x) be subharmonic on R2. Then ∆s(x) ≥ 0 in the sense of distributions [9,
10], and hence a Radon measure µ(x), dµ(x) = 1

2π∆s(x)dx such that locally (i.e. in a
neighbourhood N of a point x) s(x) =

∫
N

log |x− y|dµ(y)+ a harmonic function on N .

Definition 1. (Inversion) [11] A map x → x∗ = x
|x|2 if x ̸= 0, ∞; 0 → ∞ and ∞ → 0 is

known as an inversion (see for example: Axler et al. pp 59–61).
If u(x) is a subharmonic (respectively harmonic) function defined outside a compact set



Amulya Smyrna C., N. Nathiya / Eur. J. Pure Appl. Math, 18 (2) (2025), 5750 4 of 9

then u(x∗) = u
(

x
|x|2

)
is a subharmonic (respectively harmonic) function in a neighbourhood

of 0, excluding 0. For a set E in R2 write E∗ = {x∗ : x ∈ E}.

Theorem 1. Let u(x) be a subharmonic function defined outside a compact set K with
associated measure λ. Then u(x) has a harmonic majorant h(x) outside a compact set K
if and only if

∫
R2\K

dλ(x) < ∞.

Proof. Write z = x∗. Then v(z) = u(z) − h(z) is an upper bounded subharmonic
function on 0 < |z| < ϵ; hence extends as a subharmonic function v(z) on |z| < ϵ whose
associated measure is denoted by σ(x) so that v(z) =

∫
|y|<ϵ

log |z − y|dσ(y)+ a harmonic

function on |z| < ϵ. Note that
∫

|y|≤ ϵ
2

dσ(y) is finite and σ = λ on any compact set in

0 < |z| < ϵ
2 .

Hence λ(E) where E = (z : 0 < |z| < ϵ
2) is finite. Take an inversion to conclude that λ(E∗)

is finite leading to the conclusion λ(R2\K) is finite. Note that E∗ is a neighbourhood of
the point at infiniy.
Converse: If

∫
R2\K

dλ(x) < ∞, define a Radon measure ν on N = {z : |z| < ϵ} by taking

for any compact set A ∈ R2, ν(A) = λ[(A ∩ N)∗] where (A ∩ N)∗ is the inversion set
of (A ∩ N). Note ν is a Radon measure on R2 with compact harmonic support taking
ν({0}) = 0.

Write z = x∗, x in a neighbourhood of infinity. Then s(x) =
∫
R2

log |x − y|dν(y)

is subharmonic on R2. Hence ∃ a harmonic function h(x) ≥ s(x) on N . Note that
s(z) on N\{0} is the same as u(x∗) = u(z) up to an additive harmonic function v(z).
Also note that on N\{0}, the harmonic functions h(z) and v(z) can also be written
as h(x∗) and v(x∗) so that we have h(x∗) ≥ u(x∗) + v(x∗) on N\{0}. An inversion
shows that h(x) ≥ u(x) + v(x). Thus u(x) ≤ h(x) − v(x) near infinity. That is u(x) is
majorised by a harmonic function near infinity if the associated measure λ of u is such
that

∫
R2\K

dλ(x) < ∞.

Corollary 1. Let s(x) be a subharmonic function on R2 with associated measure µ(x).
Then s(x) has a harmonic majorant outside a compact set if and only if

∫
X

dµ(x) < ∞.

Proof. Follows from the theorem 1

4. Flux at infinity

Suppose u(x) is a subharmonic function defined outside a compact set. Then there
exists a subharmonic function s(x) that is harmonic on a neighbourhood of 0 and a constant
α ≥ 0 such that u(x) = s(x)− α log |x| outside a disc (Anandam [12]).



Amulya Smyrna C., N. Nathiya / Eur. J. Pure Appl. Math, 18 (2) (2025), 5750 5 of 9

Definition 2. Let u(x) be a subharmonic function defined outside of a compact set. Let
u(x) = s(x)−α log |x|. Then define the flux∞u = flux at infinity of u(x) = [Total measure
of s(x)]−α.

Remark 1. If u(x) = s1(x) − α1 log |x| is another such decomposition outside a com-
pact set, then s(x) − α log |x| = s1(x) − α1 log |x| outside a compact set. Since v(x) =
s(x)+α1 log |x| = s1(x)+α log |x| outside a compact set represents two forms of the same
subharmonic function v(x), [total measure of v(x)] can be calculated by using the normal
derivative form in the Green’s theorem.
Hence,

total measure of s1 + α2 = total measure of s2 + α1,

so that
total measure of s1 − α1 = total measure of s2 − α2

Remark 2. If s(x) is a subharmonic function on R2, then total measure of s(x) =
flux∞s(x)

Theorem 2. Let h(x) be a harmonic function defined outside a compact set in R2.
Then h(x) = H(x) + α log |x| + b(x) where H(x) is harmonic on R2, α - a constant,
and b(x) bounded harmonic tending to 0 at infinity. This representation is unique and
[flux∞h(x)] = α.

Proof. Write h(x) = s(x)−β log |x|, β ≥ 0 and s(x) subharmonic on R2. The function
s(x) should be harmonic outside a compact set K. Hence s(x) =

∫
K

log |x − y|dµ(y) +

a harmonic function H(x) on R2.
Since the integral equals µ(K) log |x|+ b(x) outside a compact set where b(x) −→ 0 at

infinity, we have
h(x) = H(x) + [µ(K)− β] log |x|+ b(x)

= H(x) + α log |x|+ b(x),writing α = µ(K)− β

Uniqueness of the representation: Suppose h(x) = H1(x)+α1 log |x|+ b1(x), then H(x)+
α log |x|+ b(x) = H1(x) + α1 log |x|+ b1(x) outside a compact set.

Now for large R, take the integral 1
2πR

2π∫
0

on |x| = R, to see

H(0)−H1(0) = (α1 − α) logR+ [a bounded function]

. This is possible for all large R if and only if α1 = α in which case H(x) − H1(x) is
bounded, hence a constant tending to 0. Therefore H1 = H, proving the uniqueness of
the representation.
flux at infinity of h: Since h(x) = s(x)−β log |x|, then flux∞h = [total measure of s]−β =
∥µ∥ − β = α

Corollary 2. Let h be harmonic outside a compact set. Then there exists a harmonic
function H on R2 such that |h−H| is bounded near infinity if and only if [flux∞h] = 0.



Amulya Smyrna C., N. Nathiya / Eur. J. Pure Appl. Math, 18 (2) (2025), 5750 6 of 9

Proof. From the above theorem h(x) is represented as

h(x) = H(x) + α log |x|+ b(x), (1)

where H(x) is harmonic on R2, b(x) is bounded harmonic tending to 0 at infinity and α
is a constant. Let flux∞h = 0. By (1) we get |h − H| = b(x) at infinity. This shows
that |h − H| is bounded at infinity. Conversely, if |h − H| is bounded at infinity then
flux∞h = 0.

Proposition 1. If b(x) is a bounded harmonic function outside a compact set then [flux∞b(x)] =
0.

Proof. Write b(x) = s(x) − α log |x|. Then s(x) is harmonic near infinity so that
s(x) =

∫
log |x− y|dµ|y|+H(x), where H(x) is harmonic on R2.

Now
∫
log |x− y|dµ(y)−∥µ∥ log |x| = b1(x) which is harmonic tending to 0 at infinity.

Then b(x) = (∥µ∥ − α) log |x|+H(x) + b1(x) near infinity.
Since (b − b1) is bounded, if ∥µ∥ − α = β ̸= 0 then H(x) is bounded on one side and

hence a constant so that β log |x| should be bounded near infinity, not possible. Hence
∥µ∥ − α = 0, so that flux∞b(x)] = ∥µ∥ − α = 0

Proposition 2. Let f(x) be a real-valued function is on R2, such that
∫
|f(x)|dx < ∞.

There exists two admissible subharmonic functions s1 and s2 on R2 such that, if s = s1−s2,
then ∆s(x) = f(x) in the sense of distributions.

Proof. Let ∆s1 = f+ and ∆s2 = f−. Then s1 and s2 are subharmonic on R2. Since∫
f+(x)dx < ∞ and

∫
f−(x)dx < ∞, then s1 and s2 are admissible. Hence the proposition

is proved.

Theorem 3. Suppose s(x) is a subharmonic function on R2, having a harmonic majorant
outside a compact set K. Then the flux of s at infinity=flux at infinity of the least harmonic
majorant of s outside K.

Proof. Suppose s(x) has a harmonic majorant on |x| > r. Take an increasing sequence
of numbers rn such that r < rn for all n and rn −→ ∞. Let sn be the function on |x| > r
which is the Dirichlet solution on r < |x| < rn with boundary values s(x) and extended
by s(x) on |x| > rn.
Then note:

(a) sn tends to the least harmonic majorant h of s on |x| > r;

(b) for every n, flux of sn at infinity = flux of s at infinity;

(c) flux of h(x) at infinity = lim
n−→∞

flux at infinity of sn(x).

Consequently, the flux of s(x) at infinity = flux of the least harmonic majorant h(x) of
s(x).



Amulya Smyrna C., N. Nathiya / Eur. J. Pure Appl. Math, 18 (2) (2025), 5750 7 of 9

5. Total measures of subharmonic functions and least harmonic
majorants

Proposition 3. If h1, h2 are two harmonic functions outside of a compact set in R2, such
that h1 ≥ h2. Let hi = Hi+αi log |x|+bi(x), for i = 1, 2, · · · , be the unique representations
as above. Then α1 ≥ α2.

Proof. For H1(x) + α1 log |x| + b1(x) ≥ H2(x) + α2 log |x| + b2(x) near infinity. Then
for large r,

∫
|x|=r

H1(x) + α1 log |x|+ b1(x)dx ≥
∫

|x|=r

H2(x) + α2 log |x|+ b2(x)dx.

Then rH1(0)+ 2πrα1 log r+
∫

|x|=r

b1(x)dx ≥ rH2(0)+ 2πrα2 log r+
∫

|x|=r

b2(x)dx. Now,

if |bi(x)| ≤ mi, then |
∫

|x|=r

bi(x)dx ≤ mir2π. Hence, from the above, after dividing by r,

we get the inequality

2πα1 ≥ 2πα2 log r + (a bounded function in r).

Allow r −→ ∞ to conclude that α1 ≥ α2.

If u is a C2-function and ω is a bounded domain in R2
∫
ω
∆u(x)d(x) =

∫
∂ω

∂u
∂n+ds, in

the classical sense. When v is a subharmonic function on R2, there exists an increasing
sequence of C2-subharmonic functions tending to v [2]. In classical sense, where limits
are not permissible, the continuity of subharmonic function (upper continuous function)
is not always true, therefore the distribution sense is adopted to show the continuity of
the function, as limits exists in distribution case [9].
Consequently in the sense of distributions∫

ω

∆v(x)d(x) =

∫
∂ω

∂v

∂n+
ds.

Theorem 4. Let v(x) be a subharmonic function on R2 having a harmonic majorant
near infinity. Let h(x) = α log |x|+ b(x)+ (a harmonic function R2) be the least harmonic
majorant of v(x) outside a compact set. Then the total measure associated with v(x) equals
α.

Proof. Suppose v has a harmonic majorant H outside a compact set. v(x) ≤ H(x) if
|x| ≥ m. Then for large r > m let hr be the Dirichlet solution in r < |x| < r + 1 with
boundary values v(x). Write

vr(x) =

{
hr(x), in r < |x| < r + 1

v(x), otherwise in R2

Then vr(x) is a subharmonic function on R2 and∫
|x|<r+2

∆vr(x)dx =

∫
|x|=r+2

∂vr(x)

∂n+
ds =

∫
|x|=r+2

∂v(x)

∂n+
ds =

∫
|x|<r+2

∆v(x)dx



Amulya Smyrna C., N. Nathiya / Eur. J. Pure Appl. Math, 18 (2) (2025), 5750 8 of 9

That is, the total measure associated with vr is the same as the total measure associated
with v. Note that vr(x) is a subharmonic function on R2 increasing with r and f(x) =
lim
r→∞

vr(x) is subharmonic on R2 and is a harmonic function h(x) when |x| > r; also h(x)

is the least harmonic majorant of v(x) on |x| > r. Thus

f(x) =

{
v(x), if |x| ≤ r is subharmonic on R2

h(x), if |x| > r

Since ∆f(x) = lim
r→∞

∆vr(x) in the sense of distributions, the total measure associated with

f(x) is the total measure associated with v(x).
Since near infinity f(x) = h(x) = α log |x| + b(x) + (a harmonic function on) R2 total
measure associated with f(x) is α. Consequently, total measure associated with v(x)
equals the constant α in the expression of h(x) which is the least harmonic majorant of
v(x) near infinity.

Corollary 3. Let s1 and s2 be two subharmonic functions on R2, s1 ≤ s2. Then the (total
measure associated with s1) ≤(total measure associated with s2).

Proof. If (total measure associated with s1) = ∞, then s1 does not have a harmonic
majorant near infinity consequently, s2 cannot have a harmonic majorant near infinity, so
that (total measure associated with s2).

Hence let us assume that the total measure of s1 is finite. In this case, if total measure
of s2 is infinite, nothing to prove. So we have to consider only the case when the total
measures of s1 and s2 are finite. Then outside a compact set if h1, h2 are the least harmonic
majorants of s1, s2, then h1 ≤ h2 so that if we write hi = Hi + αi log |x| + bi(x), then
α1 ≤ α2. Then by the above theorem (total measure associated with s1) ≤(total measure
associated with s2).

Conclusion

In complex analysis, the relation between the zeros of an analytic function f(z) and
the measure representing the subharmonic function log |f(z)| as an integral is intriguingly
fascinating. In this article, we have attempted to answer the question: if s(x) is a subhar-
monic function on the complex plane with the associated Radon measure µ representing
s(x) as an integral in a local representation, when will the total measure ∥µ∥ be finite?

By introducing the notion of flux at infinity of a subharmonic function defined outside
a compact set, we characterise the case where ∥µ∥ is finite. We have also mentioned some
related researchers on subharmonic functions on locally compact harmonic spaces in the
Brelot axiomatic potential without positive potentials.

Declarations

(i) Author Contribution Both the authors contributed equally.



Amulya Smyrna C., N. Nathiya / Eur. J. Pure Appl. Math, 18 (2) (2025), 5750 9 of 9

The authors (Amulya Smyrna C. and N. Nathiya) of this manuscript titled ”Flux at
infinity of subharmonic functions on R2” have no competing interests to declare that are
relevant to the content of this article.

References

[1] David H Armitage and Stephen J Gardiner. Classical potential theory. Springer
Science & Business Media, 2012.

[2] Marcel Brelot. Éléments de la théorie classique du potentiel. (No Title), 1959.
[3] Lester La Verne Helms et al. Potential theory. Springer, 2009.
[4] Thomas Ransford. Potential theory in the complex plane. Number 28. Cambridge

university press, 1995.
[5] Victor Anandam. Espaces harmoniques sans potentiel positif. In Annales de l’institut

Fourier, volume 22, pages 97–160, 1972.
[6] Victor Anandam. Admissible superharmonic functions and associated measures.

Journal of the London Mathematical Society, 2(1):65–78, 1979.
[7] Marcel Brelot. Axiomatique des fonctions harmoniques. (No Title), 1966.
[8] Ibtesam Bajunaid, Joel M Cohen, Flavia Colonna, and David Singman. A riesz

decomposition theorem on harmonic spaces without positive potentials. Hiroshima
mathematical journal, 38(1):37–50, 2008.

[9] Walter Rudin. The lemma of the logarithmic derivative for subharmonic functions. In
Mathematical Proceedings of the Cambridge Philosophical Society, volume 120, pages
347–354. Cambridge University Press, 1996.

[10] Walter Kurt Hayman and Patrick Brendan Kennedy. Subharmonic functions / W.
K. Hayman and P. B. Kennedy. L.M.S. monographs; 9, 20. Academic Press, London
;, 1976.

[11] Sheldon Axler, Paul Bourdon, and Ramey Wade. Harmonic function theory, volume
137. Springer Science & Business Media, 2013.

[12] Victor Anandam. Subharmonic functions outside a compact set in Rn. Proceedings
of the American Mathematical Society, 84(1):52–54, 1982.