# On a Stability of Logarithmic-Type Functional Equation in Schwartz Distributions

Authors: Jae-Young Chung
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We prove the Hyers-Ulam stability of the logarithmic functional equation of Heuvers and Kannappanf(x+y)-g(xy)-h(1/x+1/y)=0,x,y>0, in both classical and distributional senses. As a classical sense, the Hyers-Ulam stability of the inequality |f(x+y)-g(xy)-h(1/x+1/y)|≤ϵ,x,y>0 will be proved, where f,g,h:ℝ+→ℂ. As a distributional analogue of the above inequality, the stability of inequality ∥u∘(x+y)-v∘(xy)-w∘(1/x+1/y)∥≤ϵ will be proved, where u,v,w∈𝒟'(ℝ+) and ∘ denotes the pullback of distributions.


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## Body

## 1. Introduction

The stability problems of functional equations have originated with Ulam in 1940 (see [1]). One of the first assertions to be obtained is the following result, essentially due to Hyers [2], that gives an answer for the question of Ulam.Theorem 1.1.
Suppose thatG is an additive semigroup, B is a Banach space, f:G→B, ϵ≥0, and
(1.1)∥f(x+y)-f(x)-f(y)∥≤ϵ
for all x,y∈G. Then there exists a unique function A:G→B satisfying
(1.2)A(x+y)=A(x)+A(y)
for which
(1.3)∥f(x)-A(x)∥≤ϵ
for all x∈G.As a direct consequence of the result, we obtain the stability of the logarithmic functional equation (see also the result of Forti [3]) as follows:
(1.4)L(xy)-L(x)-L(y)=0,x,y>0.Theorem 1.2.
Letℝ+ be the set of positive real numbers. Suppose that f:ℝ+→C, ϵ≥0, and
(1.5)|f(xy)-f(x)-f(y)|≤ϵ
for all x,y>0. Then there exists a unique function L:ℝ+→ℂ satisfying (1.4) for which
(1.6)|f(x)-L(x)|≤ϵ
for all x>0.We call the functionL satisfying (1.4) logarithmic function. The logarithmic functional equation has been modified in various forms [4, 5] and Heuvers and Kannappan introduced the functional equation [6] as follows:
(1.7)f(x+y)-g(xy)-h(1x+1y)=0,x,y>0
for f,g,h:ℝ+→ℝ. In particular, it is shown that the general solution of (1.7) has the form
(1.8)f(x)=c1+c2+L(x),g(x)=c1+L(x),h(x)=c2+L(x),
where L:ℝ+→ℝ is a logarithmic function.In 1950, Schwartz introduced the theory of distributions in his monographThéorie des distributions [7]. In this book Schwartz systematizes the theory of generalized functions, basing it on the theory of linear topological spaces, relating all the earlier approaches, and obtaining many important results. After his elegant theory appeared, many important concepts and results on the classical spaces of functions have been generalized to the space of distributions.Making use of differentiation of distributions, several authors have dealt with functional equations in the spaces of Schwartz distributions, converting given functional equations to differential equations, and finding the solutions in the space of distributions (see [8–11]). However, when we try to consider the Hyers-Ulam stability problems of functional equations, the differentiation is not available for solving them in both the space of infinitely differentiable functions and the space of distributions. In the paper [12], using convolutional approach we initiated the following distributional version of the well-known Hyers-Ulam stability problem for the Cauchy functional equation:
(1.9)∥u∘A-u∘P1-u∘P2∥≤ϵ,
where ∘ is the pullback. See Section 3 for the pullback and see below the definition of the norm ∥·∥ in (1.9). Using the heat kernel
(1.10)Et(x)=(4πt)-n/2exp(-|x|24t),x∈ℝn,t>0,
we proved the stability problems (1.9) in the space of tempered distributions [7] by converting the inequality (1.9) to the classical stability problems
(1.11)|U(x+y,t+s)-U(x,t)-U(y,s)|≤ϵ
for all x,y∈ℝn,  t,s>0, where U is an infinitely differentiable function in ℝn×(0,∞) given by U(x,t)=u*Et(x). We also refer the reader to [13] for the stability of Pexider equations in the space of tempered distributions. In [14] we extend the stability problems in the space of tempered distributions to the space of distributions. Instead of the heat kernel, using the regularizing function δt(x):=t-nδ(x/t),  x∈ℝn,  t>0, where
(1.12)δ(x)={qe-(1-|x|2)-1,|x|<1,0,|x|≥1,(1.13)q=(∫|x|<1‍e-(1-|x|2)-1dx)-1,
we prove that the unknown distributions in the functional inequalities are tempered distributions and then use the same method as in [12, 13].In this paper, developing the previous method in [12–14], we consider a distributional version of the Hyers-Ulam stability of (1.7) in the space of distributions as
(1.14)∥u∘(x+y)-v∘(xy)-w∘(1x+1y)∥≤ϵ,
where u,v,w∈𝒟'(ℝ+), ∘ denotes the pullback of distributions and the inequality ∥·∥≤ϵ in (1.14) means |〈·,φ〉|≤ϵ∥φ∥L1 for all test functions φ(x,y) defined on ℝ+2. Since the tempered distributions are defined in whole real line or whole space ℝn, the methods used in [14] are not available for the inequality (1.14). For the proof of the above problem, we need some technical method than those employed in [12–14]. Indeed, we will show a method to control a functional inequality satisfied in a subset of ℝ2. As a direct consequence of the result, we obtain the Hyers-Ulam stability of (1.7) in L∞-sense, that is, the Hyers-Ulam stability of the inequality
(1.15)∥f(x+y)-g(xy)-h(1x+1y)∥L∞≤ϵ
will be obtained. Finally, we also find locally integrable solutions of (1.7) as a consequence of the stability of the inequality (1.15).


## 2. Stability in Classical Sense

In this section, we prove the Hyers-Ulam stability of the functional inequality(2.1)|f(x+y)-g(xy)-h(1x+1y)|≤ϵ,x,y>0,
where f,g,h:ℝ+→ℝ and ϵ≥0.Theorem 2.1.
Suppose thatf,g,h:ℝ+→ℂ satisfy (2.1). Then there exists a logarithmic function L:ℝ+→ℝ such that
(2.2)|f(x)-L(x)-f(1)|≤4ϵ,|g(x)-L(x)-g(1)|≤4ϵ,|h(x)-L(x)-h(1)|≤4ϵ
for all x>0.Proof.
Letxy=t,  1/x+1/y=s. Then we have
(2.3)|f(ts)-g(t)-h(s)|≤ϵ
for all t>0,  s>0 such that ts2≥4. For given t,s>0, choose a large u>0 so that ts2u2≥4,  tsu2≥4,  su2≥4,  s2u2≥2. Then in view of (2.3), we have
(2.4)|f(tsu)-g(t)-h(su)|≤ϵ,(2.5)|f(tsu)-g(ts)-h(u)|≤ϵ,(2.6)|f(su)-g(s)-h(u)|≤ϵ,(2.7)|f(su)-g(1)-h(su)|≤ϵ.
From (2.4)–(2.7), using the triangle inequality we have
(2.8)|g(ts)-g(t)-g(s)+g(1)|≤4ϵ
for all t,s>0. Changing the roles of g and h in (2.3), we can show that
(2.9)|h(ts)-h(t)-h(s)+h(1)|≤4ϵ
for all t,s>0. Now we prove that
(2.10)|f(ts)-f(t)-f(s)+f(1)|≤4ϵ
for all t,s>0. Replacing t by u and s by s/u in (2.3), we have
(2.11)|f(s)-g(u)-h(su)|≤ϵ,4u≤s2.
Similarly, we have
(2.12)|f(ts)-g(u)-h(tsu)|≤ϵ,4u≤t2s2,(2.13)|f(t)-g(us)-h(tsu)|≤ϵ,4u≤t2s,(2.14)|f(1)-g(us)-h(su)|≤ϵ,4u≤s.
For givent,s>0, let u=1/4min{s2,t2s2,t2s,s}. Then, from (2.11)–(2.14), using the triangle inequality we get the inequality (2.10).
Now by Theorem1.2, there exist functions Lj:ℝ+→ℝ,  j=1,2,3, satisfying the logarithmic functional equation
(2.15)Lj(ts)=Lj(t)+Lj(s),j=1,2,3,
for which
(2.16)|f(t)-L1(t)-f(1)|≤4ϵ,(2.17)|g(t)-L2(t)-g(1)|≤4ϵ,(2.18)|h(t)-L3(t)-h(1)|≤4ϵ.
Now we show thatL1=L2=L3. From (2.3), we have
(2.19)|f(t)-g(t)-h(1)|≤ϵ,t≥2,(2.20)|f(t)-h(t)-g(1)|≤ϵ,t≥4.
From (2.16), (2.17), and (2.19), using the triangle inequality we have
(2.21)|L1(t)-L2(t)|≤9ϵ+|f(1)-g(1)-h(1)|:=M,t≥2.
In view of (2.15) and (2.21), we have
(2.22)|L1(t)-L2(t)|=1|n||L1(tn)-L2(tn)|≤1|n|M
for all t>0,  t≠1 and tn≥2. Letting n→∞ for t>1 and letting n→-∞ for 0<t<1, we have L1(t)=L2(t) for t≠1. Since L1(1)=L2(1)=0, we have L1(t)=L2(t) for all t>0. Similarly we can show that L1=L3. This completes the proof.Lettingg=h=f in Theorem 2.1, in view of the inequalities (2.4), (2.5), and (2.6), using the triangle inequality we have
(2.23)|f(ts)-f(t)-f(s)|≤3ϵ
for all t,s>0. Thus by Theorem 1.2 we have the following.Theorem 2.2.
Letf:ℝ+→ℝ satisfy the inequality
(2.24)|f(x+y)-f(xy)-f(1x+1y)|≤ϵ
for all x,y>0. Then there exists a logarithmic function L:ℝ+→ℝ such that
(2.25)|f(x)-L(x)|≤3ϵ
for all x>0.


## 3. Schwartz Distributions

LetΩ be an open subset of ℝn. We briefly introduce the space 𝒟'(Ω) of distributions. We denote by α=(α1,…,αn)∈ℕ0n, where ℕ0 is the set of nonnegative integers, and |α|=α1+⋯+αn,  ∂α=∂1α1⋯∂nαn, ∂j=∂/∂xj,j=1,2,…,n.Definition 3.1.
LetCc∞(Ω) be the set of all infinitely differentiable functions on Ω with compact supports. A distribution u is a linear form on Cc∞(Ω) such that for every compact set K⊂Ω there exist constants C>0 and k∈ℕ0 for which
(3.1)|〈u,φ〉|≤C∑|α|≤k‍sup|∂αφ|
holds for all φ∈Cc∞(Ω) with supports contained in K. The set of all distributions is denoted by 𝒟'(Ω).LetΩj be open subsets of ℝnj for j=1,2, with n1≥n2.Definition 3.2.
Letuj∈𝒟'(Ωj) and λ:Ω1→Ω2 be a smooth function such that for each x∈Ω1 the derivative λ'(x) is surjective, that is, the Jacobian matrix ∇λ of λ has rank n2. Then there exists a unique continuous linear map λ*:𝒟'(Ω2)→𝒟'(Ω1) such that λ*u=u∘λ when u is a continuous function. We call λ*u the pullback of u by λ and is usually denoted by u∘λ.In particular, ifS(x,y)=x+y,P1(x,y)=x,P2(x,y)=y, the pullbacks u∘S,  u∘P1,  u∘P2 can be written as
(3.2)〈u∘S,φ(x,y)〉=〈u,∫‍φ(x-y,y)dy〉,(3.3)〈u∘P1,φ(x,y)〉=〈u,∫‍φ(x,y)dy〉,(3.4)〈u∘P2,φ(x,y)〉=〈u,∫‍φ(x,y)dx〉
for all test functions φ∈Cc∞(Ω).Also, ifλ is a diffeomorphism (a bijection with λ,  λ-1 smooth functions), the pullback u∘λ can be written as
(3.5)〈u∘λ,φ〉=〈u,(φ∘λ-1)(x)|∇λ-1(x)|〉.For more details of distributions we refer the reader to [7, 15].


## 4. Stability in Schwartz Distributions

We employ a functionδ on ℝn defined by
(4.1)δ(x)={qe-(1-|x|2)-1,|x|<1,0,|x|≥1,
where
(4.2)q=(∫|x|<1‍e-(1-|x|2)-1dx)-1.It is easy to see thatδ(x) is an infinitely differentiable function with support {x:|x|≤1}. Let δt(x):=t-nδ(x/t),t>0 and u∈𝒟'(ℝn). Then for each t>0,u*δt(x)=〈uy,δt(x-y)〉 is a smooth function of x∈ℝn and u*δt(x)→u as t→0+ in the sense that
(4.3)limt→0+∫‍(u*δt)(x)φ(x)dx=〈u,φ〉
for all φ∈Cc∞(ℝn). Here after we denote by S,P,P1,P2:ℝ2→ℝ,R:ℝ+2→ℝ,E:ℝ→ℝ+ by
(4.4)S(x,y)=x+y,P(x,y)=xy,P1(x,y)=x,P2(x,y)=y,R(x,y)=1x+1y,E(x)=2x.
Now we are in a position to prove the Hyers-Ulam stability of the inequality
(4.5)∥u∘S-v∘P-w∘R∥≤ϵ.
Recall that the inequality ∥·∥≤ϵ in (4.5) means that |〈·,φ〉|≤ϵ∥φ∥L1 for all test functions φ(x,y) defined on ℝ+2.Theorem 4.1.
Letu,v,w∈𝒟'(ℝ+) satisfy (4.5). Then there exist a,c1,c2,c3∈ℂ such that
(4.6)∥u-c1-alnx∥≤4ϵ,∥v-c2-alnx∥≤4ϵ,∥w-c3-alnx∥≤4ϵ.Proof.
LetU={(x,y):x+2y>2}, V={(x,y):x>y>0} and define J:U→V by
(4.7)J(x,y)=(2x+y+4x+y-2x+22,2x+y-4x+y-2x+22).
Then J is a diffeomorphism with J-1:V→U,  J-1(x,y)=(log2xy,log2(x+y)/xy). Taking pullback by J in (4.5) and using (3.5), we have
(4.8)∥u∘E∘S-v∘E∘P1-w∘E∘P2∥≤ϵ
in U. Thus it follows that
(4.9)∥u~∘S-v~∘P1-w~∘P2∥≤ϵ
in U, where u~=u∘E,v~=v∘E,w~=w∘E. Denoting by (δt⊗δs)(x,y)=δt(x)δs(y) and convolving δt⊗δs in the left hand side of (4.9) we have, in view of (3.2),
(4.10)[(u~∘S)*(δt⊗δs)](x,y)=〈u~∘(ξ+η),δt(x-ξ)δs(y-η)〉=〈u~ξ,∫‍δt(x-ξ+η)δs(y-η)dη〉=〈u~ξ,δt*δs(x+y-ξ)〉=u~*δt*δs(x+y).
Similarly we have, in view of (3.3) and (3.4),
(4.11)[(v~∘P1)*(δt⊗δs)](x,y)=v~*δt(x),[(w~∘P2)*(δt⊗δs)](x,y)=w~*δs(y).
Thus the inequality (4.9) is converted to the classical stability problem
(4.12)|u~*δt*δs(x+y)-v~*δt(x)-w~*δs(y)|≤ϵ∥δt⊗δs∥L1=ϵ
for all x+2y≥5 and 0<t<1,  0<s<1. From now on, we assume that 0<t<1,  0<s<1. From the inequality (4.12), we have
(4.13)|u~*δt*δs(x+y+z)-v~*δt(x+y)-w~*δs(z)|≤ϵ
for x+y+2z≥5,
(4.14)|u~*δt*δs(x+y+z)-v~*δt(x)-w~*δs(y+z)|≤ϵ
for x+2y+2z≥5,
(4.15)|u~*δt*δs(y+z)-v~*δt(y)-w~*δs(z)|≤ϵ
for y+2z≥5, and
(4.16)|u~*δt*δs(y+z)-v~*δt(0)-w~*δs(y+z)|≤ϵ
for  2y+2z≥5.
For givenx,y∈ℝ, choose z≥(1/2)(5+|x|+2|y|). Then in view of (4.13)–(4.16), using triangle inequality, we have
(4.17)|v~*δt(x+y)-v~*δt(x)-v~*δt(y)+v~*δt(0)|≤4ϵ
for all x,y∈ℝ. Replacing (x,t) by (y,s), (y,s) by (x,t) in (4.12) and changing the positions of v~ and w~, we have
(4.18)|w~*δt(x+y)-w~*δt(x)-w~*δt(y)+w~*δt(0)|≤4ϵ
for all x,y∈ℝ. Now we prove that
(4.19)|u~*δt(x+y)-u~*δt(x)-u~*δt(y)+u~*δt(0)|≤4ϵ
for all x,y∈ℝ. From the inequality (4.12), we have
(4.20)|u~*δt*δs(x+y)-v~*δt(z)-w~*δs(x+y-z)|≤ϵ,|u~*δt*δs(x)-v~*δt(z-y)-w~*δs(x+y-z)|≤ϵ,|u~*δt*δs(y)-v~*δt(z)-w~*δs(y-z)|≤ϵ,|u~*δt*δs(0)-v~*δt(z-y)-w~*δs(y-z)|≤ϵ
for all x,y,z such that 2x+2y-z≥5, 2x+y-z≥5, 2y-z≥5, and y-z≥5. For given x,y∈ℝ, choose z≤-5-2|x|-2|y|. Then in view of (4.20), using triangle inequality, we have
(4.21)|u~*δt*δs(x+y)-u~*δt*δs(x)-u~*δt*δs(y)+u~*δt*δs(0)|≤4ϵ.
Letting s→0+ in (4.21), we get the inequality (4.19).
Now in view of (4.17), (4.18), and (4.19), it follows from Theorem 1.1 that for each 0<t<1, there exist functions Aj(·,t),j=1,2,3, satisfying
(4.22)Aj(x+y,t)=Aj(x,t)+Aj(y,t),x,y∈ℝ,
for which
(4.23)|u~*δt(x)-A1(x,t)-u~*δt(0)|≤4ϵ,(4.24)|v~*δt(x)-A2(x,t)-v~*δt(0)|≤4ϵ,(4.25)|w~*δt(x)-A3(x,t)-w~*δt(0)|≤4ϵ
for all x∈ℝ.
Now we prove thatA1=A2=A3. From (4.12), using the triangle inequality, we have
(4.26)|v~*δt(x)|≤ϵ+|u~*δt*δs(x+y)|+|w~*δs(y)|
for all x+2y≥5. Since (u~*δt*δs)(x)→(u~*δs)(x) as t→0+, in view of (4.26) it is easy to see that
(4.27)g(x):=limsupt→0+v~*δt(x)
exists for all x∈ℝ. Similarly, we can show that
(4.28)h(x):=limsups→0+w~*δs(x)
exists for all x∈ℝ. Putting y=0 in (4.12) and letting s→0+ so that w~*δs(0)→h(0), we have
(4.29)|u~*δt(x)-v~*δt(x)-h(0)|≤ϵ
for all x≥5. Similarly, we have
(4.30)|u~*δt(x)-w~*δt(x)-g(0)|≤ϵ
for all x≥5/2. Using (4.23), (4.24), (4.29), and the triangle inequality, we have
(4.31)|A1(x,t)-A2(x,t)|≤9ϵ+|u~*δt(0)-v~*δt(0)-h(0)|:=M(t)
for all x≥5. From (4.22) and (4.31), we have
(4.32)|A1(x,t)-A2(x,t)|=1|k||A1(kx,t)-A2(kx,t)|≤1|k|M(t)
for all x∈ℝ,x≠0 and all integers k with kx≥5. Letting k→∞ if x>0 and letting k→-∞ if x<0 in (4.32) we have A1(x,t)=A2(x,t) for x≠0, which implies A1=A2 since A1(0,t)=A2(0,t)=0. Similarly, using (4.23), (4.25), and (4.30) we can show that A1=A3.
Finally we prove thatA1 is independent of t. Fixing x∈ℝ and letting t→0+ so that v~*δt(x)→g(x) in (4.12), we have
(4.33)|u~*δs(x+y)-g(x)-w~*δs(y)|≤ϵ
for all x+2y≥5. The same substitution as the inequalities (4.13)–(4.16) gives
(4.34)|g(x+y)-g(x)-g(y)+g(0)|≤4ϵ.
Using the stability Theorem [2], we obtain that there exists a unique function A satisfying the Cauchy functional equation
(4.35)A(x+y)-A(x)-A(y)=0
for which
(4.36)|g(x)-A(x)-g(0)|≤4ϵ.
Now we show that A1(x,t)=A(x) for all x∈ℝ and 0<t<1. Putting y=0 in (4.33), we have
(4.37)|u~*δs(x)-g(x)-w~*δs(0)|≤ϵ
for all x≥5. From (4.23), (4.36), and (4.37), using the triangle inequality, we have
(4.38)|A1(x,t)-A(x)|≤9ϵ+|u~*δt(0)-w~*δt(0)-g(0)|
for all x≥5. From (4.38), using the method of proving A1=A2, we can show that A1(x,t)=A(x) for all x∈ℝ and t. Thus we have A1=A2=A3=A.
Lettingt→0+ in (4.24) so that v~*δt(0)→g(0), we have
(4.39)∥v~-A(x)-c2∥≤4ϵ
for some c2∈ℂ. Similarly, letting t→0+ in (4.25) so that w~*δt(0)→h(0), we have
(4.40)∥w~-A(x)-c3∥≤4ϵ
for some c3∈ℂ. Now we prove the inequality
(4.41)∥u~-A(x)-c1∥≤4ϵ
for some c1∈ℂ. Putting x=-5,  y=5 in (4.33) and using the triangle inequality, we have
(4.42)|u~*δs(0)|≤ϵ+|g(-5)+w~*δs(5)|.
From (4.42), there exists a sequence sn→0+ such that u~*δsn(0) converges. Letting t=sn→0+ in (4.23), we get (4.41). Taking pullback by E-1(x)=log2x in (4.39), (4.40), and (4.41), we have
(4.43)∥u-A(log2x)-c1∥≤4ϵ,∥v-A(log2x)-c2∥≤4ϵ,∥w-A(log2x)-c3∥≤4ϵ.
Finally we show that the solution A of the Cauchy equation (4.35) has the form A(x)=ax for some a∈ℂ. Recall that g is the supremum limit of a collection of continuous functions v~*δt,0<t<1. Thus, if we let g=g1+ig2, then both g1 and g2 are Lebesgue measurable functions. Now, as we see in the proof of Hyers-Ulam stability Theorem [2], the function A is given by
(4.44)A(x)=limn→∞2-ng(2nx).
Thus, let A=A1+iA2. Then A1 and A2 are Lebesgue measurable functions as limits of sequences of Lebesgue measurable functions. It is well known that every Lebesgue measurable solution A of the Cauchy functional equation (4.35) has the form A(x)=cx for some c∈ℂ. This completes the proof.As a consequence of the above result we have the following.Corollary 4.2.
Letf,g,h:ℝ+→ℂ,j=1,2,3, be locally integrable functions satisfying
(4.45)∥f(x+y)-g(xy)-h(1x+1y)∥L∞≤ϵ.
Then there exist a,c1,c2,c3∈ℂ such that
(4.46)∥f(x)-c1-alnx∥L∞≤4ϵ,∥g(x)-c2-alnx∥L∞≤4ϵ,∥h(x)-c3-alnx∥L∞≤4ϵ.Proof.
Every locally integrable functionf defines a distribution via the equation
(4.47)〈f,φ〉=∫‍f(x)φ(x)dx,φ∈Cc∞(ℝ+).
Viewing f,g,h as distributions, the inequality (4.45) implies
(4.48)∥f∘S-g∘P-h∘R∥≤ϵ.
By Theorem 4.1, we have
(4.49)|∫‍(f(x)-c1-alnx)φ(x)dx|≤4ϵ∥φ∥L1,|∫‍(g(x)-c2-alnx)φ(x)dx|≤4ϵ∥φ∥L1,|∫‍(h(x)-c3-alnx)φ(x)dx|≤4ϵ∥φ∥L1
for all φ∈Cc∞(ℝ+). Viewing Cc∞ as a subspace of L1  (dense subspace) and using the Hahn-Banach theorem we obtain that the inequalities (4.49) hold for all φ∈L1. Now since L∞=(L1)′ we get the inequalities (4.46). This completes the proof.As a direct consequence of the above result we solve the functional equation(4.50)f(x+y)-g(xy)-h(1x+1y)=0
in L∞-sense, that is, we obtain the following.Corollary 4.3.
Letf,g,h:ℝ+→ℂ,j=1,2,3, be locally integrable functions satisfying
(4.51)∥f(x+y)-g(xy)-h(1x+1y)∥L∞=0.
Then there exist a,c1,c2,c3∈ℂ such that
(4.52)∥f(x)-c1-alnx∥L∞=0,∥g(x)-c2-alnx∥L∞=0,∥h(x)-c3-alnx∥L∞=0.Finally, we discuss the locally integrable solutionf,g,h:ℝ+→ℂ of (4.50) (c.f. [6]).Corollary 4.4.
Every locally integrable solutionf,g,h:ℝ+→ℂ of (4.50) has the form
(4.53)f(x)=c1+c2+alnx,(4.54)g(x)=c1+alnx,(4.55)h(x)=c2+alnx
for some a,c1,c2∈ℂ.Proof.
It follows from Corollary4.3 that the equalities (4.53), (4.54), and (4.55) hold in almost everywhere sense, that is, there exists a subset Ω⊂ℝ+ with Lebesgue measure m(Ωc)=0 such that the equalities (4.53), (4.54), and (4.55) hold for all x∈Ω. For given x>0, let p,q:(0,x)→ℝ by p(t)=1/t+(1/(x-t)),q(t)=t(x-t). Since m[(p-1(Ω)∩q-1(Ω))c]=m[p-1(Ωc)∪q-1(Ωc)]=0, we can choose y∈p-1(Ω)∩q-1(Ω). Let z=x-y. Then y+z=x and yz=y(x-y)=q(y)∈Ω, 1/y+1/z=p(y)∈Ω. Thus we can write
(4.56)f(x)=g(yz)+h(1y+1z)=c1+aln(yz)+c2+aln(1y+1z)=c1+c2+aln(y+z)=c1+c2+alnx,
which gives (4.53). For given x>0, let p:ℝ+→ℝ by p(t)=1/t+t/x. Then we have p-1(Ω)≠∅. Choose y∈p-1(Ω) and let z=x/y. Then yz=x,1/y+1/z∈Ω. Thus, using (4.53), we can write
(4.57)g(x)=f(y+z)-h(1y+1z)=c1+c2+aln(y+z)-c2-aln(1y+1z)=c1+aln(yz)=c1+alnx,
which gives (4.54). Finally, the equality (4.55) follows from (4.50), (4.53), and (4.54). This completes the proof.



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*Source: 435310-2012-11-21.xml*