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+.

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 each x,y∈ℝ and t,s>0, we have
(55)∬-∞∞F(u+v)δt(x-u)δs(y-v)du dv =∫-∞∞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)du dv =∫-∞∞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)du dv=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)du dv∬-∞∞(F(u+v)-G(u)-H(v))| =|∫-∞∞∫d′-2v∞(F(u+v)-G(u)-H(v)) × δt(x-u)δs(y-v)du dv∫d′-2v∞| =∫-∞∞∫d′-2v∞|F(u+v)-G(u)-H(v)| ×|δt(x-u)δs(y-v)|du dv ≤ϵ∬-∞∞|δt(x-u)δs(y-v)|du dv=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 given x,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 that A1=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 that A1 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.

Letting t→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.

As a direct consequence of the previous result we have found the solution of functional equation (4) in the L∞-sense.