# Stability of a Logarithmic Functional Equation in Distributions on a Restricted Domain

Authors: Jaeyoung Chung; Prasanna K. Sahoo
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

Letℝ be the set of real numbers, ℝ+={x∈ℝ∣x>0}, ϵ∈ℝ+, and f,g,h:ℝ+→ℂ. As classical and L∞ versions of the Hyers-Ulam stability of the logarithmic type functional equation in a restricted domain, we consider the following inequalities: |f(x+y)-g(xy)-h((1/x)+(1/y))|≤ϵ, and f(x+y)-g(xy)-h(1/x)+(1/y)L∞(Γd)≤ϵ in the sectors Γd={(x,y):x>0,y>0,(y/x)>d}. As consequences of the results, we obtain asymptotic behaviors of the previous inequalities. We also consider its distributional version u∘S-v∘Π-w∘RΓd≤ϵ, where u,v,w∈𝒟'(ℝ+), S(x,y)=x+y, Π(x,y)=xy, R(x,y)=1/x+1/y, x,y∈ℝ+, and the inequality ·Γd≤ϵ means that |〈·,φ〉|≤ϵ∥φ∥L1 for all test functions φ∈Cc∞(Γd).


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## Body

## 1. Introduction

The Hyers-Ulam stability problem of functional equations was originated in 1940 when Ulam proposed a question concerning the approximate homomorphisms from a group to a metric group (see [1]). A partial answer was given by Hyers et al. [2, 3] under the assumption that the target space of the involved mappings is a Banach space. It is possible to prove stability results similar to Hyers for functions that do not have bounded Cauchy difference. In 1950, Aoki [4] first proved such a result for additive functions. Bourgin [5, 6] and Aoki [4] studied the Ulam problem from 1949 to 1951. The area rested there for a while until 1978 when Rassias [7] published a generalized version of Hyers’ result on linear mappings, where the Cauchy difference was allowed to be unbounded. Rassias’ work provided an impetus for the study on the stability of functional equations (see [2, 7–31]).Letℝ be the set of real numbers, ℝ+ the set of positive real numbers, and ℂ the set of complex numbers. The subset, for fixed real number d>0,
(1)Γd={(x,y):x>0,y>0,yx>d}
of the plane, ℝ2, will be referred to as a sector. A function f:ℝ+→ℂ is said to be logarithmic if and only if it satisfies the logarithmic functional equation:
(2)f(xy)-f(x)-f(y)=0,∀x,y∈ℝ+,
for all x,y∈ℝ+. There are several variants of logarithmic functional equations (see [14–16]). It was shown by Heuvers and Kannappan [16] that the logarithmic functional equation is equivalent to the following functional equation:
(3)f(x+y)-f(xy)-f(1x+1y)=0,∀x,y∈ℝ+.
They have also studied the following pexiderized version of (3):
(4)f(x+y)-g(xy)-h(1x+1y)=0,∀x,y∈ℝ+.
The general solution of the functional equation (4) has the form (see [16])
(5)f(x)=L(x)+c1+c2,g(x)=L(x)+c1,h(x)=L(x)+c2,
where L:ℝ+→ℂ is a logarithmic function and c1,c2 are arbitrary constants.In this paper, we study Hyers-Ulam stability of the functional equation (4). In Section 2, we treat the Hyers-Ulam stability of the functional equation (4) in the classical sense and present its asymptotic behavior. In Section 3, we consider the stability of (4) in L∞-sense and its asymptotic behavior. Finally, in Section 4 we present the stability of (4) in Schwartz distributions.


## 2. Stability of (4) in Classical Sense and Its Asymptotic Behavior

In this section, we consider the classical Hyers-Ulam stability of the functional equation (4) on the sector Γd and then study its asymptotic behavior.The following theorem is a direct consequence of the Hyers’ result [3] (see also result of Forti [32]).Theorem 1.
Letϵ be a nonnegative real number. Suppose that f:ℝ+→ℂ satisfies
(6)|f(xy)-f(x)-f(y)|≤ϵ
for all x,y∈ℝ+. Then there exists a unique logarithmic function L:ℝ+→ℂ such that
(7)|f(x)-L(x)|≤ϵ,∀x∈ℝ+.Next, we establish the Hyers-Ulam stability of the functional equation (4) on the restricted domain Γd.Theorem 2.
Suppose thatϵ≥0, d>0, and f, g, h satisfy the functional inequality
(8)|f(x+y)-g(xy)-h(1x+1y)|≤ϵ
for all (x,y)∈Γd. Then there exists a unique logarithmic function L:ℝ+→ℂ such that
(9)|f(x)-L(x)-f(1)|≤4ϵ,|g(x)-L(x)-g(1)|≤4ϵ,|h(x)-L(x)-h(1)|≤4ϵ
for all x∈ℝ+.Proof.
For givent,s>0, choose a real number u>0 such that
(10)u≥max{2ts2,2ts,2s,2s,2dts2,2dts,2ds,2ds2},
and let
(11)x1=tsu-t2s2u2-4t2,y1=tsu+t2s2u2-4t2,x2=tsu-t2s2u2-4ts2,y2=tsu+t2s2u2-4ts2,x3=su-s2u2-4s2,y3=su+s2u2-4s2,x4=su-s2u2-42,y4=su+s2u2-42.
Then it is easy to check that xj,yj>0, yj/xj>d for all j=1,2,3,4. Replacing x, y by xj, yj in (8), respectively, for j=1,2,3,4 we have
(12)|f(tsu)-g(t)-h(su)|≤ϵ,(13)|f(tsu)-g(ts)-h(u)|≤ϵ,(14)|f(su)-g(s)-h(u)|≤ϵ,(15)|f(su)-g(1)-h(su)|≤ϵ.
From (12)–(15), using the triangle inequality we have
(16)|g(ts)-g(t)-g(s)+g(1)|≤4ϵ
for all t,s>0. Similarly, for given t,s>0, choose u>0 such that
(17)u≥max{4t2s,4t2s2,4s2,4s,4dt2s,4dt2s2,4ds2,4ds}
and let
(18)x1=tsu-t2s2u2-4su2,y1=tsu+t2s2u2-4su2,x2=tsu-t2s2u2-4u2,y2=tsu+t2s2u2-4u2,x3=su-s2u2-4u2,y3=su+s2u2-4u2,x4=su-s2u2-4su2,y4=su+s2u2-4su2.
Then it is easy to check that xj,yj>0, yj/xj>d for all j=1,2,3,4. Next, replacing x, y by xj, yj in (8), respectively, for j=1,2,3,4, we have
(19)|f(tsu)-h(t)-g(su)|≤ϵ,|f(tsu)-h(ts)-g(u)|≤ϵ,|f(su)-h(s)-g(u)|≤ϵ,|f(su)-h(1)-g(su)|≤ϵ.
From (19), using the triangle inequality, we have
(20)|h(ts)-h(t)-h(s)+h(1)|≤4ϵ
for all t,s>0. Now we prove that
(21)|f(ts)-f(t)-f(s)+f(1)|≤4ϵ
for all t,s>0. For given t,s>0, choose u>0 such that
(22)u≤min{t2s24,t2s4,s24,s4,t2s24d,t2s4d,s24d,s4d}
and let
(23)x1=ts-t2s2-4u2,y1=ts+t2s2-4u2,x2=t-t2-4u/s2,y2=t+t2-4u/s2,x3=s-s2-4u2,y3=s+s2-4u2,x4=1-1-4u/s2,y4=1+1-4u/s2.
Then xj,yj>0, yj/xj>d for all j=1,2,3,4. Replacing x, y by xj, yj in (8), respectively, for j=1,2,3,4, we have
(24)|f(ts)-g(u)-h(tsu)|≤ϵ,|f(t)-g(us)-h(tsu)|≤ϵ,|f(s)-g(u)-h(su)|≤ϵ,|f(1)-g(us)-h(su)|≤ϵ.
From (24), using the triangle inequality we get (21).
Now by Theorem1, there exist Lj:ℝ+→ℂ for j=1,2,3 satisfying the logarithmic functional equation
(25)Lj(ts)=Lj(t)+Lj(s),j=1,2,3,
for which
(26)|f(t)-L1(t)-f(1)|≤4ϵ,(27)|g(t)-L2(t)-g(1)|≤4ϵ,(28)|h(t)-L3(t)-h(1)|≤4ϵ.
Now we show that L1=L2=L3. Putting s=u=1 and t=u=1 in (12) separately, we have
(29)|f(t)-g(t)-h(1)|≤ϵfort≥max{4,4d},(30)|f(s)-h(s)-g(1)|≤ϵfors≥max{2,2d}.
From (26), (27), and (29), using the triangle inequality we have
(31)|L1(t)-L2(t)|≤9ϵ+|f(1)-g(1)-h(1)|∶=Mfort≥max{4,4d}.
Let t>1. Then we can choose a positive integer n0 such that tn≥max{4,4d} for all integers n≥n0. In view of (25), and (31) we have
(32)|L1(t)-L2(t)|=1n|L1(tn)-L2(tn)|≤Mn
for all integer n≥n0. Letting n→∞ in (32), we have L1(t)=L2(t) for all t>1. For 0<t<1, we have L1(t)=-L1(1/t)=-L2(1/t)=L2(t). Thus, we have L1(t)=L2(t) for all t>0. Similarly, using (26), (28), and (30) we can show that L1=L3. The uniqueness of the logarithmic function L is obvious. This completes the proof of the theorem.Lettingg=h=f in Theorem 2 and using the inequalities (12)–(14) together with the triangle inequality, we obtain
(33)|f(ts)-f(t)-f(s)|≤3ϵ
for all t,s>0. Thus, by Theorem 1 we have the following theorem.Theorem 3.
Letd>0. Suppose that f:ℝ+→ℂ satisfies the functional inequality
(34)|f(x+y)-f(xy)-f(1x+1y)|≤ϵ
for all (x,y)∈Γd. Then there exists a unique logarithmic function L:ℝ+→ℂ such that
(35)|f(x)-L(x)|≤3ϵ,∀x∈ℝ+.Now we prove the following asymptotic result concerning (8).Theorem 4.
Suppose thatf,g,h:ℝ+→ℂ satisfy the asymptotic condition
(36)|f(x+y)-g(xy)-h(1x+1y)|⟶0
as  (y/x)→∞. Then there exists a logarithmic function L:ℝ+→ℂ and c1,c2∈ℂ such that
(37)f(x)=L(x)+c1+c2,g(x)=L(x)+c1,h(x)=L(x)+c2
for all x>0.Proof.
By the condition (36), for any positive integer n, there exists dn>0 such that
(38)|f(x+y)-g(xy)-h(1x+1y)|≤1n
for all x,y>0 with (y/x)>dn. By Theorem 1, there exists a logarithmic function Ln:ℝ+→ℂ such that
(39)|f(x)-Ln(x)-f(1)|≤4n,(40)|g(x)-Ln(x)-g(1)|≤4n,(41)|h(x)-Ln(x)-h(1)|≤4n
for all x>0. Replacing n by m in (39) and using the triangle inequality, we have
(42)|Ln(x)-Lm(x)|≤4n+4m≤8
for all x>0. Thus, we obtain
(43)|Ln(x)-Lm(x)|=1k|Ln(xk)-Lm(xk)|≤8k
for all x>0 and k∈ℕ. Letting k→∞ in (43), we have Ln(x)=Lm(x):=L(x) for all x>0. Finally, letting n→∞ in (39), (40), and (41), we have
(44)f(x)=L(x)+f(1),g(x)=L(x)+g(1),h(x)=L(x)+h(1)
for all x>0. Finally, substituting (44) in (36) we get f(1)=g(1)+h(1). Letting c1=g(1) and c2=h(1) we obtain the asserted result.


## 3. Stability of (4) in L∞-Sense and Its Asymptotic Behavior

In this section, we consider the Hyers-Ulam stability of the functional equation (4) in L∞-sense on the sector Γd and then examine its asymptotic behavior. Consider the functional inequality
(45)∥f(x+y)-g(xy)-h(1x+1y)∥L∞(Γd)≤ϵ,
where Γd={(x,y):x>0,y>0,(y/x)>d} and d>1 is fixed, where ∥·∥L∞(Γd) denotes the essential supremum norm of D(x,y)=f(x+y)-g(xy)-h((1/x)+(1/y)) on the set Γd. We employ the function δ on ℝ defined by
(46)δ(x)={qe-(1-x2)-1,if|x|<1,0,if|x|≥1,
where
(47)q=(∫-11‍e-(1-x2)-1dx)-1.
It is easy to see that δ(x) is an infinitely differentiable function with support {x:|x|≤1}. Let f be a locally integrable function and δt(x):=t-1δ(x/t),  t>0. Then for each t>0,
(48)f*δt(x)=∫-∞∞‍f(y)δt(x-y)dy
is a smooth function of x∈ℝ and f*δt(x)→f(x) for almost every x∈ℝ as t→0+.Now we prove the Hyers-Ulam stability of the functional equation (4) in L∞-sense on the sector Γd.Theorem 5.
Letf, g, h be locally integrable functions satisfying (45). Then there exist constants c1,c2,c3,a∈ℂ such that
(49)∥f(x)-c1-alnx∥L∞(R+)≤4ϵ,∥g(x)-c2-alnx∥L∞(R+)≤4ϵ,∥h(x)-c3-alnx∥L∞(R+)≤4ϵ.Proof.
We will use the diffeomorphism(50)J(x,y)=(lnxy,lnx+yxy).
Let u=lnxy, v=ln((x+y)/xy) and y/x=t>1. Then, we have
(51)u+2v=lnxy+2lnx+yxy=ln(2+xy+yx)=ln(2+t+1t).
Thus, we have J(Γd):=Ud={(u,v):u+2v>ln(2+d+1/d)}. Consequently, (45) is converted to
(52)∥f(eu+v)-g(eu)-h(ev)∥L∞(Ud)≤ϵ.
Now, let
(53)F(u)=f(eu),G(u)=g(eu),H(u)=h(eu).
Then, we have
(54)∥F(u+v)-G(u)-H(v)∥L∞(Ud)≤ϵ.
For eachx,y∈ℝ and t,s>0, we have
(55)∬-∞∞F(u+v)δt(x-u)δs(y-v)dudv=∫-∞∞‍F(u)(∫-∞∞‍δt(x-u+v)δs(y-v)dv)du=∫-∞∞‍F(u)(∫-∞∞‍δt(v)δs(x+y-u-v)dv)du=∫-∞∞‍F(u)(δt*δs)(x+y-u)du=F*δt*δs(x+y).
We also have
(56)∬-∞∞G(u)δt(x-u)δs(y-v)dudv=∫-∞∞‍G(u)δt(x-u)(∫-∞∞‍δs(y-v)dv)du=∫-∞∞‍G(u)δt(x-u)du=G*δt(x).
Similarly, we have
(57)∬-∞∞H(v)δt(x-u)δs(y-v)dudv=H*δs(y).
On the other hand, let x+2y>3+ln(2+d+1/d) and 0<t<1, 0<s<1. Then, we have
(58)supp(δt(x-u)δs(y-v))={(u,v):x-t≤u≤x+t,y-s≤v≤y+s}⊂Ud.
Let d′=ln(2+d+1/d). Then it follows from (54)~(58) that
(59)|F*δt*δs(x+y)-G*δt(x)-H*δs(y)|=|∬-∞∞(F(u+v)-G(u)-H(v))×δt(x-u)δs(y-v)dudv∬-∞∞(F(u+v)-G(u)-H(v))|=|∫-∞∞‍∫d′-2v∞‍(F(u+v)-G(u)-H(v))×δt(x-u)δs(y-v)dudv∫d′-2v∞‍|=∫-∞∞‍∫d′-2v∞‍|F(u+v)-G(u)-H(v)|×|δt(x-u)δs(y-v)|dudv≤ϵ∬-∞∞|δt(x-u)δs(y-v)|dudv=ϵ.
Thus, we have the functional inequality
(60)|F*δt*δs(x+y)-G*δt(x)-H*δs(y)|≤ϵ
for all x+2y>d1:=3+ln(2+d+1/d) and 0<t<1, 0<s<1. From now on, we assume that 0<t<1, 0<s<1. From (60), we have
(61)|F*δt*δs(x+y+z)-G*δt(x+y)-H*δs(z)|≤ϵ
for x+y+2z>d1,
(62)|F*δt*δs(x+y+z)-G*δt(x)-H*δs(y+z)|≤ϵ
for x+2y+2z>d1,
(63)|F*δt*δs(y+z)-G*δt(y)-H*δs(z)|≤ϵ
for y+2z>d1,
(64)|F*δt*δs(y+z)-G*δt(0)-H*δs(y+z)|≤ϵ
for 2y+2z>d1.
For givenx,y∈ℝ, choose z>(1/2)(d1+|x|+2|y|). Then, using the triangle inequality with (61)~(64), we have
(65)|G*δt(x+y)-G*δt(x)-G*δt(y)+G*δt(0)|≤4ϵ
for all x,y∈ℝ. Replacing (x,t) by (y,s), (y,s) by (x,t) in (60) and changing the roles of G and H, we have
(66)|H*δt(x+y)-H*δt(x)-H*δt(y)+H*δt(0)|≤4ϵ
for all x,y∈ℝ. Now we prove that
(67)|F*δt(x+y)-F*δt(x)-F*δt(y)+F*δt(0)|≤4ϵ
for all x,y∈ℝ. From (60), we have
(68)|F*δt*δs(x+y)-G*δt(z)-H*δs(x+y-z)|≤ϵ,|F*δt*δs(x)-G*δt(z-y)-H*δs(x+y-z)|≤ϵ,|F*δt*δs(y)-G*δt(z)-H*δs(y-z)|≤ϵ,|F*δt*δs(0)-G*δt(z-y)-H*δs(y-z)|≤ϵ,
for all x, y, z such that 2x+2y-z>d1, 2x+y-z>d1, 2y-z>d1, and y-z>d1. For given x,y∈ℝ, choose z≤-d1-2|x|-2|y|. Using the triangle inequality with (68), we have
(69)|F*δt*δs(x+y)-F*δt*δs(x)-F*δt*δs(y)+F*δt*δs(0)|≤4ϵ.
Letting s→0+ in (69), we get (67).
Applying Hyers’ stability theorem from [3] for (65), (66), and (67), we obtain that for each 0<t<1 there exist functions Aj(·,t),  j=1,2,3, satisfying
(70)Aj(x+y,t)=Aj(x,t)+Aj(y,t),x,y∈ℝ,
for which
(71)|F*δt(x)-A1(x,t)-F*δt(0)|≤4ϵ,(72)|G*δt(x)-A2(x,t)-G*δt(0)|≤4ϵ,(73)|H*δt(x)-A3(x,t)-H*δt(0)|≤4ϵ,
for all x∈ℝ.
Now we prove thatA1=A2=A3. From (60), using the triangle inequality we have
(74)|G*δt(x)|≤ϵ+|F*δt*δs(x+y)|+|H*δs(y)|
for all x+2y>d1. Since F*δt*δs(x)→F*δs(x) as t→0+, in view of (74) it is easy to see that
(75)G~(x)∶=limsupt→0+G*δt(x)
exists for all x∈ℝ. Similarly, we can show that
(76)H~(x)∶=limsups→0+H*δs(x)
exists for all x∈ℝ. Putting y=0 in (60) and letting s→0+ so that H*δs(0)→H~(0) we have
(77)|F*δt(x)-G*δt(x)-H~(0)|≤ϵ
for all x>d1. Similarly, we have
(78)|F*δt(x)-H*δt(x)-G~(0)|≤ϵ
for all x>(d1/2). Using (71), (72), (77), and the triangle inequality, we have
(79)|A1(x,t)-A2(x,t)|≤9ϵ+|F*δt(0)-G*δt(0)-H~(0)|∶=M(t)
for all x>d1. From (71) and (80), we have
(80)|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>d1. Letting k→∞ if x>0 and letting k→-∞ if x<0 in (80), 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 (71), (73), and (78) we obtain that A1=A3.
Finally, we prove thatA1 is independent of t. Fixing x∈ℝ and letting t→0+ so that G*δt(x)→G~(x) in (60), we have
(81)|F*δs(x+y)-G~(x)-H*δs(y)|≤ϵ
for all x+2y>d1. From (81), using the same substitutions as in (61)~(64) we have
(82)|G~(x+y)-G~(x)-G~(y)+G~(0)|≤4ϵ.
By Hyers’ stability theorem [3], there exists a unique function A satisfying the Cauchy functional equation
(83)A(x+y)-A(x)-A(y)=0
for which
(84)|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 (81), we have
(85)|F*δs(x)-G~(x)-H*δs(0)|≤ϵ
for all x>d1. From (71), (84), and (85), using the triangle inequality we have
(86)|A1(x,t)-A(x)|≤9ϵ+|F*δt(0)-H*δt(0)-G~(0)|
for all x>d1. From (86), using the method of proving A1=A2 we can show that A1(x,t)=A(x) for all x∈ℝ and 0<t<1. Thus, we have A1=A2=A3:=A.
Lettingt→0+ in (72) so that G*δt(0)→G~(0), we have
(87)∥G(x)-A(x)-G~(0)∥L∞≤4ϵ.
Similarly, letting t→0+ in (73) so that H*δt(0)→H~(0), we have
(88)∥H(x)-A(x)-H~(0)∥L∞≤4ϵ.
Now we prove the inequality
(89)∥F(x)-A(x)-F~(0)∥L∞≤4ϵ.
For given x∈ℝ, choosing z such that x+z>d1 replacing x by x-z and y by z in (81), and using the triangle inequality, we have
(90)|F*δs(x)|≤ϵ+|G~(x-z)+H*δs(z)|.
From (90), it is easy to see that
(91)F~(x)∶=limsups→0+F*δs(x)
exists for all x∈ℝ. Letting t→0+ in (71) so that F*δt(0)→F~(0), we get (89). Replacing x by lnx in (87), (88), and (89), we have
(92)∥f(x)-A(lnx)-F~(0)∥L∞(ℝ+)≤4ϵ,∥g(x)-A(lnx)-G~(0)∥L∞(ℝ+)≤4ϵ,∥h(x)-A(lnx)-H~(0)∥L∞(ℝ+)≤4ϵ.
Finally, we show that the solution A of the Cauchy equation (83) has the form A(x)=cx for some c∈ℂ. Since G~ is the supremum limit of a collection of continuous functions G*δt,  0<t<1,  G~ is a Lebesgue measurable function. Also, as we see in the proof of Hyers-Ulam stability theorem (see [3]), the function A is given by
(93)A(x)=limn→∞2-nG~(2nx).
Thus, A is a Lebesgue measurable function since it is the limit of a sequence of Lebesgue measurable functions. It is well known that every Lebesgue measurable solution A of the Cauchy functional equation (83) has the form A(x)=ax for some a∈ℂ. Letting c1=F~(0),  c2=G~(0),  c3=H~(0) we get the asserted result.Now we discuss an asymptotic behavior of the inequality (45).Theorem 6.
Letf,g,h:ℝ+→ℂ,  j=1,2,3, be locally integrable functions satisfying
(94)∥f(x+y)-g(xy)-h(1x+1y)∥L∞(Γd)⟶0
as d→∞. Then there exist constants a,c1,c2,c3∈ℂ such that
(95)∥f(x)-c1-alnx∥L∞(ℝ+)=0,∥g(x)-c2-alnx∥L∞(ℝ+)=0,∥h(x)-c3-alnx∥L∞(ℝ+)=0.Proof.
By the condition (94), for any positive integer n there exists dn>1 such that
(96)∥f(x+y)-g(xy)-h(1x+1y)∥L∞(Γd)≤1n
for all x,y>0 with (y/x)>dn. Now by Theorem 5, there exist constants a,c1,c2,c3∈ℂ (which are independent of n) such that
(97)∥f(x)-c1-alnx∥L∞(ℝ+)≤4n,∥g(x)-c2-alnx∥L∞(ℝ+)≤4n,∥h(x)-c3-alnx∥L∞(ℝ+)≤4n.
Letting n→∞ in (97), we obtain the asserted result.As a direct consequence of the previous result we have found the solution of functional equation (4) in the L∞-sense.Corollary 7.
Letf,g,h:ℝ+→ℂ be locally integrable functions satisfying
(98)∥f(x+y)-g(xy)-h(1x+1y)∥L∞(Γd)=0
for all x,y∈ℝ+. Then there exist a,c1,c2,c3∈ℂ such that
(99)∥f(x)-c1-alnx∥L∞(ℝ+)=0,∥g(x)-c2-alnx∥L∞(ℝ+)=0,∥h(x)-c3-alnx∥L∞(ℝ+)=0.Finally, we discuss the locally integrable solutionf,g,h:ℝ+→ℂ of the functional equation (c.f. [16])
(100)f(x+y)-g(xy)-h(1x+1y)=0
for all (x,y)∈Γd. The following result is a direct consequence of Theorem 2. However, we introduce an alternative proof using Corollary 7. The following method of proof will be useful when we know only regular solution in L∞-sense.Corollary 8.
Every locally integrable solutionf,g,h:ℝ+→ℂ of the functional equation (100) has the form
(101)f(x)=c1+c2+alnx,(102)g(x)=c1+alnx,(103)h(x)=c2+alnx
for some constants a,c1,c2∈ℂ.Proof.
It follows from Corollary7 that (101), (102), and (103) hold in almost everywhere sense; that is, there exists a subset Ω⊂ℝ+ with Lebesgue measure m(Ωc)=0 such that (101), (102), and (103) 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,  (1/y)+(1/z)∈Ω. Thus, we can write
(104)f(x)=g(yz)+h(1y+1z)=c1+aln(yz)+c2+aln(1y+1z)=c1+c2+aln(y+z)=c1+c2+alnx,
which gives (101). 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 (101) we can write
(105)g(x)=f(y+z)-h(1y+1z)=c1+c2+aln(y+z)-c2-aln(1y+1z)=c1+aln(yz)=c1+alnx,
which gives (102). Finally, (103) follows from (100), (101), and (102). This completes the proof of the corollary.


## 4. Stability of (4) in Schwartz Distributions

LetΩ be an open subset of ℝn. We briefly introduce the space 𝒟′(Ω) of distributions. We denote 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 9.
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
(106)|〈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 10.
Letuj∈𝒟′(Ωj) and let 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 it is usually denoted by u∘λ.Ifλ is a diffeomorphism (a bijection with λ,λ-1 smooth functions) the pullback u∘λ can be written as
(107)〈u∘λ,φ〉=〈u,(φ∘λ-1)(x)|∇λ-1(x)|〉.For more details of distributions we refer the reader to [29, 33].In this section, we consider the Hyers-Ulam stability of the functional equation of (4) in Schwartz distributions, that is, the functional inequality
(108)∥u∘S-v∘Π-w∘R∥Γd≤ϵ,
where u,v,w∈𝒟′(ℝ+),  Π:ℝ2→ℝ,  and R:ℝ2→ℝ are defined by
(109)S(x,y)=x+y,Π(x,y)=xy,R(x,y)=1x+1y
and the inequality ∥·∥Γd≤ϵ in (108) means that |〈·,φ〉|≤ϵ∥φ∥L1 for all test functions φ∈Cc∞(Γd). 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
(110)limt→0+∫-∞∞‍(u*δt)(x)φ(x)dx=〈u,φ〉
for all φ∈Cc∞(ℝn).Theorem 11.
Letu,v,w∈𝒟′(ℝ+) satisfy (108). Then there exist constants a,c1,c2,c3∈ℂ such that
(111)∥u-c1-alnx∥≤4ϵ,∥v-c2-alnx∥≤4ϵ,∥w-c3-alnx∥≤4ϵ.Proof.
The idea of the following proof is essentially the same as that of Theorem5, only with different terminologies. For the reader we give a sketch of proof. Let ud and J:Γd→Ud be the set and mapping in the proof of Theorem 5, respectively. Then, J-1:Ud→Γd is given by
(112)J-1(x,y)=(ex+y+e2x+2y-4ex2,ex+y-e2x+2y-4ex2).
Taking pullback by J-1 in (108), we have
(113)∥u∘E∘S-v∘E∘P1-w∘E∘P2∥Ud≤ϵ,
where E:ℝ→ℝ,  S,P1,P2:ℝ2→ℝ are given by
(114)E(x)=ex,S(x,y)=x+y,P1(x,y)=x,P2(x,y)=y.
Thus, instead of (54) we have the inequality
(115)∥u~∘S-v~∘P1-w~∘P2∥Ud≤ϵ,
where u~=u∘E,  v~=v∘E,  and w~=w∘E. Using the same approach as in the proof of Theorem 5, we have
(116)∥u~-cx-c1∥≤4ϵ,∥v~-cx-c2∥≤4ϵ,∥w~-cx-c3∥≤4ϵ
for some c∈ℂ. Taking pullback by E-1(x)=lnx in (116), we have
(117)∥u-alnx-c1∥≤4ϵ,∥v-alnx-c2∥≤4ϵ,∥w-alnx-c3∥≤4ϵ,
for some constants a,c1,c2,c3∈ℂ. This completes the proof of the theorem.



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*Source: 751680-2013-09-03.xml*