# New Properties of Complex Functions with Mean Value Conditions

Authors: Yuzhen Bai; Lei Wu
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

We apply mollifiers to study the properties of real functions which satisfy mean value conditions and present new equivalent conditions for complex analytic functions. New properties of complex functions with mean value conditions are given.


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## Body

## 1. Introduction

There are many good properties of complex analytic function. In the references on complex function theory (see [1] and the references therein), we see that analytic function satisfies mean value theorem but the converse is wrong. Hence, mean value condition is weaker than analytic condition.The mean value problem has been a very active area in recent years. The mean value theorem for real-valued differentiable functions defined on an interval is one of the most fundamental results in analysis. However, the theorem is incorrect for complex-valued functions even if the function is differentiable throughout the complex plane. Qazi [2] illustrated that by examples and presented three results of a positive nature. A mean value theorem for continuous vector functions was introduced by mollified derivatives and smooth approximations in [3]. Crespi et al. [4] and La Torre [5] gave some characterizations of convex functions by means of second-order mollified derivatives. Second-order necessary optimality conditions for nonsmooth vector optimization problems were given by smooth approximations in [6]. Eberhard and Mordukhovich [7] mainly concerned deriving first-order and second-order necessary (and partly sufficient) optimality conditions for a general class of constrained optimization problems via convolution smoothing. Eberhard et al. [8] demonstrated that second-order subdifferentials were constructed via the accumulation of local Hessian information provided by an integral convolution approximation of the function. In [9], Aimar et al. showed the parabolic mean value formula.In this paper, we will apply mollifiers to study the properties of real functions which satisfy mean value conditions and present new equivalent conditions for complex analytic functions. New properties of complex functions with mean value conditions will be given.We introduce the notations:z=x+iy, z¯=x-iy, p0=(x0,y0), p=(x,y), Br(p0)={p∣dist⁡(p,p0)≤r},  ∂Br(p0)={p∣dist⁡(p,p0)=r}. Using the chain rule of derivation, we have (1.1)∂∂z=(∂∂x-i∂∂y),∂∂z¯=(∂∂x+i∂∂y),Δ=∂∂z∂z¯=∂∂z¯∂z=∂2∂x2+∂2∂y2.
The Cauchy-Riemann equation of analytic function f(z)=u(x,y)+iv(x,y) can be written as (1.2)∂f(z)∂z¯=0.We will use the following classical definitions and results of functional analysis.Definition 1.1 (see [10]).
The functions(1.3)φɛ(x)={cɛnexp⁡(ɛ2|x|2-ɛ2),|x|<ɛ,0,|x|≥ɛ,
with c∈R such that ∫Rnφɛ(x)dx=1, are called standard mollifiers.From the definition, we see the functionsφɛ are C∞.Definition 1.2 (see [3]).
Give a locally integrable functionf:Rn→Rm and a sequence of bounded mollifiers, and define the functions fɛ by the convolution
(1.4)fɛ(x)∶=∫Rnf(x-y)φɛ(y)dy=∫Rnf(y)φɛ(x-y)dy.
The sequence fɛ(x) is said to be a sequence of mollified functions.Proposition 1.3 (Properties of mollifiers, see [10]).
Suppose thatΩ⊂Rn is open, ɛ>0, write Ωɛ={x∈Ω∣dist⁡(x,∂Ω)>ɛ}. Then,(i)
fɛ∈C∞(Ωɛ),(ii)
fɛ→f a.e. as ɛ→0,(iii)
iff∈C(Ω), then fɛ→f uniformly on compact subsets of Ω,(iv)
If1≤p<∞ and f∈Lloc⁡p(Ω) then fɛ→f in Lloc⁡p(Ω).This paper is organized as follows. In Section2, we give the definitions of mean value conditions and their equivalent forms. Applying mollifiers, we show some properties of real functions with mean value conditions in Section 3. Section 4 contains our main results for complex functions satisfying mean value condition, that is, the new equivalent condition of complex analytic function and the new properties of complex functions. At last, we present two problems with their answers.


## 2. Mean Value Conditions

Definition 2.1 (Mean value condition).
LetΩ be a domain in complex number field (bounded or unbounded) and f(z)=u(x,y)+iv(x,y) a continuous complex function defined in Ω. For any z0∈Ω and {z∣|z-z0|≤r}⊂Ω, if
(2.1)f(z0)=12πr∫|z-z0|=rf(z)ds,
we say that f(z) satisfies the mean value condition in domain Ω.Remark 2.2.
Iff(z) is an analytic function in domain Ω, then f(z) satisfies the mean value condition in domain Ω (see [1]), the converse is wrong. For example, f(z)=1+iy satisfies the mean value condition in the complex number field, but it is not analytic. Hence, mean value condition is weaker than analytic condition.Definition 2.3 (Mean value condition).
Setw(p)∈C(Ω).(i)
For anyBr(p0)⊂Ω, if
(2.2)w(p0)=12πr∫∂Br(p0)w(p)ds,
we say that w(p) satisfies the first mean value condition.(ii)
For anyBr(p0)⊂Ω, if
(2.3)w(p0)=1πr2∫Br(p0)w(p)dp,
we say that w(p) satisfies the second mean value condition.Proposition 2.4.
(1) The first and the second mean value conditions ofw(p) are equivalent.
(2)(i)
The first mean value condition ofw(p) can be written as
(2.4)w(p0)=12π∫|ω|=1w(p0+rω)ds.(ii)
The second mean value condition ofw(p) can be written as
(2.5)w(p0)=1π∫|ω|≤1w(p0+rω)dω.
(3) The mean value condition of complex functionf(z) can be written as
(2.6)f(z0)=12π∫02πf(z0+reiθ)dθ.
(4) The complex functionf(z)=u(x,y)+iv(x,y) satisfies mean value condition if and only if real functionsu(x,y) andv(x,y) satisfy mean value conditions.Proof.
(1) Differentiating both sides of(2.7)w(p0)=1πr2∫Br(p0)w(p)dp=1πr2∫0rdρ∫∂Bρ(p0)w(p)ds,
with respect to r, we have
(2.8)0=-2πr3∫0rdρ∫∂Bρ(p0)w(p)ds+1πr2∫∂Br(p0)w(p)ds,
that is,
(2.9)12π∫∂Br(p0)w(p)ds=1πr2∫Br(p0)w(p)dp=w(p0).
We write the first mean value condition as
(2.10)w(p0)ρ=12π∫∂Bρ(p0)w(p)ds
and get the second mean value condition by integrating the both sides of (2.10) with respect to ρ on [0,r].
(2)(i)
Letp=p0+rω. Then, by integral transform formula, we get
(2.11)w(p0)=12πr∫∂Br(p0)w(p)ds=12π∫|ω|=1w(p0+rω)ds.(ii)
In the same way, we get(2.12)w(p0)=1πr2∫Br(p0)w(p)dp=1π∫|ω|≤1w(p0+rω)dω.
(3) Letz=z0+reiθ. Then, by integral transform formula, we get
(2.13)f(z0)=12πr∫|z-z0|=rf(z)ds=12π∫02πf(z0+reiθ)dθ.
(4) Letz=z0+r(ω1+iω2) and ω=(ω1,ω2). Then, by integral transform formula, we see
(2.14)u(x0,y0)+iv(x0,y0)=f(z0)=12πr∫|z-z0|=r[u(x,y)+iv(x,y)]ds=12π∫|ω|=1[u(p0+rω)+iv(p0+rω)]ds,
which implies
(2.15)u(x0,y0)=12π∫|ω|=1u(p0+rω)ds,v(x0,y0)=12π∫|ω|=1v(p0+rω)ds.


## 3. Preliminaries

In this section, we give the properties of real functions satisfying the mean value conditions. These properties will be used to prove our main results in Section4.Lemma 3.1.
IfΔw(p)=0,p=(x,y)∈Ω, then,  w(x,y) satisfies the mean value condition in domain  Ω.Proof.
For anyBr(p0)⊂Ω, using Green formula, we have
(3.1)∫Bρ(p0)Δw(p)dp=∫∂Bρ(p0)∂w∂νds=ρ∫|ω|=1∂w∂ρ(p0+ρω)ds=ρ∂∂ρ∫|ω|=1w(p0+ρω)ds.
Since w is harmonic in Ω, we obtain from (3.1) that
(3.2)∂∂ρ∫|ω|=1w(p0+ρω)ds=0.
Integrating both sides of (3.2) with respect to ρ on [0,r], we get
(3.3)∫|ω|=1w(p0+rω)ds=∫|ω|=1w(p0)ds=2πw(p0),
that is,
(3.4)w(p0)=12π∫|ω|=1w(p0+rω)ds=12πr∫∂Br(p0)w(p)ds.Lemma 3.2.
Assume  (i)ψ(ρ)∈C[0,r] and A(r):=∫Br(p0)ψ(|p-p0|)dp≠0;  (ii)  w(p) satisfies the mean value condition in Br(p0) and w(p)∈C(B¯r(p0)). Then,
(3.5)w(p0)=1A(r)∫Br(p0)w(p)ψ(|p-p0|)dp.Proof.
From (i) and (ii), we have(3.6)A(r)∶=∫Br(p0)ψ(|p-p0|)dp=∫0rdρ∫∂Bρ(p0)ψ(|p-p0|)ds=2π∫0rψ(ρ)ρdρ,(3.7)w(p0)=12πρ∫∂Bρ(p0)w(p)ds.
Multiplying the both sides of (3.7) by 2πρψ(ρ) and integrating the result with respect to ρ on [0,r], we have
(3.8)2πw(p0)∫0rψ(ρ)ρdρ=∫0r(∫∂Bρ(p0)w(p)ψ(ρ)ds)dρ=∫Br(p0)w(p)ψ(|p-p0|)dp.
Combining (3.6) and (3.8), we obtain the conclusion.Lemma 3.3.
Ifw(p)∈C(Ω) satisfies the mean value condition, then, (i) w(p)∈C∞(Ω); (ii) Δw(p)=0.Proof.
(i) Method 1: chooseφ(p)∈C0∞(B1(0)) with(3.9)∫B1(0)φ(p)dp=1,φ(p)=ψ(|p|).
Using integral transform formulas, we have
(3.10)2π∫01rψ(r)dr=1.
Defineφɛ(p)=(1/ɛ2)φ(p/ɛ), with ɛ<dist⁡(p,∂Ω),p∈Ω. Using integral transform formulas and (3.10), we get(3.11)∫Ωw(p)φɛ(p-p0)dp=1ɛ2∫∣p∣<ɛw(p0+p)φ(pɛ)dp=∫∣p∣<1w(p0+ɛp)φ(p)dp=∫01dr∫∂Br(p0)w(p0+ɛp)φ(p)ds=∫01rdr∫∂B1(p0)w(p0+ɛrω)φ(rω)ds=∫01ψ(r)rdr∫|ω|=1w(p0+ɛrω)ds=2πw(p0)∫01ψ(r)rdr=w(p0),
that is,
(3.12)w(p0)=(φɛ*w)(p0),∀p0=(x0,y0)∈Ωɛ={p0∣p0∈Ω,d(p0,∂Ω)>ɛ}.
Applying (3.12) and Proposition 1.3, noticing the arbitrariness of ɛ, we conclude that w(p)∈C∞(Ω).
Method 2: chooseφ(p) as above. Define φɛ(p)=(1/ɛ2)φ(p/ɛ), with ɛ<dist⁡(p,∂Ω),p∈Ω, then,(3.13)∫Br(p0)φɛ(p)dp=1.
Using Lemma 3.2, we obtain
(3.14)w(p0)=(φɛ*w)(p0),∀p0=(x0,y0)∈Ωɛ={p0∣p0∈Ω,d(p0,∂Ω)>ɛ}.
Applying Proposition1.3, noticing the arbitrariness of ɛ, we conclude that w(p)∈C∞(Ω).
(ii)
Using (3.1) and Proposition 2.4, we get(3.15)∫Br(p0)Δw(p)dp=r∂∂r∫|ω|=1w(p0+rω)ds=r∂∂r(2πw(p0))=0,∀Br(p0)⊂Ω,
which implies Δw(p)=0,p=(x,y)∈Ω.


## 4. Main Results

In this section, we give the main results for the complex functions which satisfy the mean value conditions.Proposition 4.1.
f(z) satisfies the mean value condition in  Ω if and only if  Δf(z)=0 in  Ω.Proof.
Proposition2.4, Lemmas 3.1, and 3.3 yield the assertion.Theorem 4.2.
f(z) satisfies the mean value condition in  Ω and  ∂f(z)/∂z¯=0 if and only if  f(z) is analytic in  Ω.Proof.
Denotef(z)=u(x,y)+iv(x,y).
Firstly, we prove the necessary condition. Employing the assumption and Lemma3.3, we can assert u(x,y),v(x,y)∈C∞(Ω); hence, the partial derivatives of u(x,y) and v(x,y) are continuous in Ω. Combining with Cauchy-Riemann equation ∂f(z)/∂z¯=0, we conclude that f(z) is analytic in Ω.
Secondly, we prove the sufficient condition. The assumption thatf(z) is analytic in Ω implies(i)
ux(x,y)=vy(x,y),uy(x,y)=-vx(x,y), that is, ∂f(z)/∂z¯=0,(ii)
Δf(z)=0. Combining with Proposition 4.1 implies f(z) satisfies the mean value condition in Ω.Theorem 4.3.
Suppose thatf(z) satisfies the mean value condition in  Ω, and  |f(z)| is bounded. Then  f(z) is a constant in  Ω.Proof.
Sincef(z)=u(x,y)+iv(x,y) satisfies the mean value condition in Ω, using Proposition 2.4 and Lemma 3.3, we get Δu=0,Δv=0,(x,y)∈Ω.
Since|f(z)| is bounded, we obtain u and v are bounded, respectively, Without loss of generality, we assume that u≥0. For all M0∈ℝ2, one can choose BR(O) with M0∈BR(O). Denote R0=d(M0,O). The Harnack inequality (see [11]) implies
(4.1)R-R0R+R0u(O)≤u(M0)≤R+R0R-R0u(O).
Letting R→+∞, we conclude u(M0)=u(O). In the similar way, we conclude v(M0)=v(O). Since M0 is arbitrary, we conclude f(z)=u(x,y)+iv(x,y) is a constant in Ω.Remark 4.4.
This theorem may be proved by the local estimates for harmonic functions too. On the local estimates for harmonic functions, one can see [10].Theorem 4.5.
Suppose thatf(z) satisfies the mean value condition in  Ω, and  |f(z)| is a constant. Then,  f(z) is a constant in  Ω.Proof.
Sincef(z)=u+iv satisfies the mean value condition in Ω, using Proposition 2.4 and Lemma 3.3, we obtain
(4.2)Δu(x,y)=Δv(x,y)=0.
Since|f(z)| is a constant, we get the following in Ω(4.3)u2(x,y)+v2(x,y)≡constant.
From (4.3), we get 2uux+2vvx=0 and
(4.4)ux2+vx2+uuxx+vvxx=0.
In the similar way, we have
(4.5)uy2+vy2+uuyy+vvyy=0.
Adding (4.4) to (4.5) and noting (4.2), we obtain
(4.6)ux2+uy2+vx2+vy2=0,
which implies u and v are constants, that is, f(z) is a constant in Ω.Theorem 4.6.
Suppose (1)f(z)=u(x,y)+iv(x,y) satisfies the mean value condition in Ω; (2)  f(z) is continuous on Ω¯; (3)  f(z) is not a constant. Then, max⁡Ω¯⁡|f(z)| can be obtained only on the boundary of Ω.Proof.
DenotingM=max⁡Ω¯⁡|f(z)|, then, we have 0<M<+∞. Suppose there is z0∈Ω such that |f(z0)|=M. For any Bρ(z0)⊂Ω, the mean value condition implies that, for all z∈∂Bρ,|f(z)|=M. Hence |f(z)| is a constant in the neighborhood of M0. Theorem 4.3 implies f(z) is a constant in this neighborhood of M0. Applying the circular chain method, we have f(z) is a constant in Ω, which is a contradiction.In the following, we present two problems.Problem 1.
Suppose (1)f(z)=u(x,y)+iv(x,y) satisfies the mean value condition in Ω; (2)  f(z) is continuous on Ω¯; (3)  f(z) is not a constant; (4) for all z∈Ω,f(z)≠0. Can one confirm that min⁡Ω¯⁡|f(z)| is obtained only on the boundary of Ω?Answer
One can't confirm. For example,f(z)=1+iy in Ω={z∣|z|<1}. This example shows that the minimal module principle doesn't hold for complex function satisfying mean value condition. But analytic complex function has minimal module principle.Problem 2.
Iff(z) satisfies the mean value condition in Ω, can one confirm that f(z) is infinitely differentiable in Ω?Answer
One can not confirm. For example,f(z)=1+iy in Ω={z∣|z|<1} does not satisfy the Cauchy-Riemann equation. This example shows that mean value condition can not imply the differential property of complex function. But analytic complex function is infinitely differentiable.



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