# Ulam-Hyers Stability of Trigonometric Functional Equation with Involution

Authors: Jaeyoung Chung; Chang-Kwon Choi; Jongjin Kim
Journal: Journal of Function Spaces
 (2015)

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## Abstract

LetS and G be a commutative semigroup and a commutative group, respectively, C and R+ the sets of complex numbers and nonnegative real numbers, respectively, and σ:S→S or σ:G→G an involution. In this paper, we first investigate general solutions of the functional equation f(x+σy)=f(x)g(y)-g(x)f(y) for all x,y∈S, where f,g:S→C. We then prove the Hyers-Ulam stability of the functional equation; that is, we study the functional inequality |f(x+σy)-f(x)g(y)+g(x)f(y)|≤ψ(y) for all x,y∈G, where f,g:G→C and ψ:G→R+.


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## Body

## 1. Introduction

The following four functional equations(1)fx+y=fxgy+gxfy,(2)gx+y=gxgy-fxfy,(3)fx-y=fxgy-gxfy,(4)gx-y=gxgy+fxfyare calledtrigonometric functional equations. The above functional equations have been investigated together or individually. For early historical developments of the above functional equations we refer the reader to [1, p. 177] and [2, pp. 209–217]. In particular, general solutions of (3) are described in [2, pp. 216-217] and general solutions of (4) and regular solutions of (1) and (2) are described in [1, pp. 177–180]. Several authors [3–5] have studied some functional equations with involutions, particular cases σ(y)=-y of which are the usual d’Alembert’s functional equation [6], Wilson’s functional equation [7], or some related classical functional equations with usual operations. In this paper, generalizing the functional equation (3) we first consider the functional equation(5)fx+σy=fxgy-gxfyfor all x,y∈S, where f,g:S→C. As a result we first find all general solutions of functional equation (5). Secondly, we investigate the Hyers-Ulam stability of functional equation (5). The Hyers-Ulam stability problems of functional equations go back to 1940 when Ulam proposed a question concerning the approximate homomorphisms from a group to a metric group (see [8]). Székelyhidi [9] investigated the Hyers-Ulam stability of trigonometric functional equations (1) and (2). As a particular case of the result in [9] he obtained the stability of functional inequality(6)fx+y-fxgy-gxfy≤ψy,fx+y-fxfy+gxgy≤ψyfor all x,y∈G, where f,g:G→C and ψ:G→R+. On the other hand, we obtained a Székelyhidi type stability of functional equations (3) and (4) (see [10, 11] and also [12–14] for related results). In this paper we obtain a parallel Hyers-Ulam stability to that of [9] for functional equation (3); that is, we obtain the Hyers-Ulam stability of the functional inequality(7)fx+σy-fxgy+gxfy≤ψyfor all x,y∈G, where f,g:G→C and ψ:G→R+. Unfortunately, we have no idea yet how to obtain all general solutions of equation(8)gx+σy=gxgy+fxfyfor all x,y∈S, where f,g:S→C and its stability, which are the parallel generalizations of the functional equation and stability of (4).


## 2. General Solutions of (5)

Throughout this section we denote byS a 2-divisible commutative semigroup with an identity element. A function σ:S→S is said to bean involution if σ(x+y)=σ(x)+σ(y) for all x,y∈S and σ(σ(x))=x for all x∈S. For simplicity we write σx instead of σ(x). A function m:S→C is calledan exponential function provided that m(x+y)=m(x)m(y) for all x,y∈S and a:S→C is calledan additive function provided that a(x+y)=a(x)+a(y) for all x,y∈S. In this section, we present the general solutions (f,g) of functional equation (5).Lemma 1 (see [15, Theorem 2]).
Letf,g:S→C satisfy the functional equation(9)fx+y+fx+σy=2fxgyfor all x,y∈S. Then either (g,f) has the form(10)gx=mx+mσx2,fx=λ1mx+λ2mσxfor all x∈S, where m:S→C is an exponential function satisfying m≠m∘σ and λ1,λ2∈C, or(11)gx=mx,fx=mxβ+ax,where m:S→C is an exponential function satisfying m=m∘σ, and a:S→C is an additive function on S∗={x∈S:m(x)≠0} satisfying a(x)=-a(σx) for all x∈S∗ and taking arbitrary values on S∖S∗, and β∈C.In the following, we exclude the trivial cases whenf(x)=0 for all x∈S or g(x)=0 for all x∈S.Theorem 2.
Letf:S→C satisfy the functional equation(12)fx+σy=fxgy-gxfyfor all x,y∈S. Then either (f,g) has the form(13)fx=λmx-mσx,gx=γmx+1-γmσxfor all x∈S, where m:S→C is an exponential function satisfying m≠m∘σ and λ,γ∈C, or(14)fx=axmx,gx=1+βaxmxfor all x∈S, where m:S→C is an exponential function satisfying m=m∘σ, and a:S→C is an additive function on S∗={x∈S:m(x)≠0} satisfying a(x)=-a(σx) for all x∈S∗ and taking arbitrary values on S∖S∗, and β∈C.Proof.
Letge and go be the even part and odd part of g with respect to σ; that is,(15)gex=gx+gσx2,gox=gx-gσx2for all x∈S. Then we have(16)gx=gex+gox,geσx=gex,goσx=-goxfor all x∈S. Replacing (x,y) by (y,x) in (12), we have(17)fy+σx=fygx-gyfxfor all x,y∈S. From (12) and (17) we have(18)fx+σy=-fy+σxfor all x,y∈S. Putting y=0 in (18) we have(19)fσx=-fxfor all x∈S. Substituting (16) into (12) we have(20)fx+σy=fxgey+fxgoy-gexfy-goxfyfor all x,y∈S. Replacing (x,y) by (σx,σy) in (20) and using (18) and (19) we have(21)-fx+σy=-fxgey+fxgoy+gexfy-goxfyfor all x,y∈S. Adding (21) and (20) we have(22)fxgoy=goxfyfor all x,y∈S. Choosing z0 such that f(z0)≠0, putting y=z0 in (22), and dividing the result by f(z0) we have(23)gox=βfxfor all x∈S, where β=g(z0)/f(z0). Thus, we have(24)gx=gex+gox=gex+βfxfor all x,y∈S. Thus, it remains to find ge and f. Replacing y by σy in (20) and using (19) we have(25)fx+y=fxgey-fxgoy+gexfy-goxfyfor all x,y∈S. Adding (25) and (20) we have(26)fx+y+fx+σy=2fxgeyfor all x,y∈S. By Lemma 1, (ge,f) has the form (10) or (11). First, we consider the case when (ge,f) has the form (10); that is, ge(x)=(1/2)(m(x)+m(σx)), f(x)=λ1m(x)+λ2m(σx) for all x∈S and m≠m∘σ. From (19) we have λ1m(x)+λ2m(σx)=-λ1m(σx)-λ2m(x) for all x∈S, which implies (λ1+λ2)m(x)=-(λ1+λ2)m(σx) for all x∈S. If λ1+λ2≠0, we have m(x)=-m(σx) for all x∈S, and hence m(x)2=m(2x)=-m(2σx)=-m(σx)2=-m(x)2 for all x∈S, which leads to the contradiction m(x)=m(σx)=0 for all x∈S. Thus, we have λ2=-λ1≔λ and(27)fx=λmx-mσxfor all x∈S. From (24) and (27) we have(28)gx=mx+mσx2+βfx=mx+mσx2+βλmx-mσx=γmx+1-γmσxfor all x∈S, where γ=1/2+βλ. Thus, we get (13). Secondly, we consider the case when (ge,f) has the form (11); that is, ge(x)=m(x), f(x)=m(x)(β+a(x)), m(x)=m(σx), and a(x)=-a(σx) for all x∈S. From (19) we have m(x)(β+a(x))=-m(σx)(β+a(σx))=-m(x)(β-a(x)) for all x∈S, which implies β=0. Thus, we have f(x)=m(x)a(x) for all x∈S. From (23) we have g(x)=ge(x)+go(x)=m(x)+βm(x)a(x) for all x∈S. Thus, we get (14). This completes the proof.In particular ifS=G a commutative group, then since S∗=S, we have the following.Corollary 3.
Letf:G→C satisfy the functional equation(29)fx-y-fxgy+gxfy=0for all x,y∈G. Then either (f,g) has the form(30)fx=λmx-m-x,gx=γmx+1-γm-xfor all x∈G, where m:G→C is an exponential function satisfying m(x)≠m(-x) for some x∈G and λ,γ∈C, or else(31)fx=axmx,gx=1+βaxmxfor all x∈G, where m:G→C is an exponential function satisfying m(x)=m(-x) for all x∈G and a:G→C is an additive function, and β∈C.In particular, ifG is a 2-divisible commutative group, then m(x)=m(-x) for all x∈G if and only if m(x)=1 for all x∈G. Thus, we have the following.Corollary 4.
Letf:G→C satisfy the functional equation(32)fx-y-fxgy+gxfy=0for all x,y∈G. Then either (f,g) has the form(33)fx=λmx-m-x,gx=γmx+1-γm-xfor all x∈G, where m:G→C is an exponential function with m≠1, or else(34)fx=ax,gx=1+βaxfor all x∈G, where a:G→C is an additive function, and β∈C.LetH,+ be a commutative semigroup, and f,g:H×H→C. As an application of Theorem 2, we determine all general solutions of the functional equation(35)fx1+y2,x2+y1=fx1,x2gy1,y2-gx1,x2fy1,y2for all x1,x2,y1,y2∈H.In the following we exclude the trivial cases whenf=0 or g=0.Theorem 5.
Letf,g:H×H→C satisfy the functional equation (35). Then either (f,g) has the form(36)fx1,x2=λm1x1m2x2-m1x2m2x1,(37)gx1,x2=γm1x1m2x2+1-γm1x2m2x1for all x1,x2∈H, where m1,m2:H→C are exponential functions satisfying m1≠m2 and λ,γ∈C, or else(38)fx1,x2=ax1-ax2mx1+x2,(39)gx1,x2=1+βax1-ax2mx1+x2for all x1,x2∈H, where m:H→C is an exponential function and a:H→C is an additive function on H∗≔{x∈H:m(x)≠0} taking arbitrary values on H∖H∗, and β∈C.Proof.
Letσ(x1,x2)=(x2,x1) for all x1,x2∈H. Then by Theorem 2, all general solutions (f,g) of functional equation (35) are given by(40)fx1,x2=λmx1,x2-mx2,x1,(41)gx1,x2=γmx1,x2+1-γmx2,x1for all x1,x2∈H, where m:H×H→C is exponential functions satisfying m(x1,x2)≠m(x2,x1) for some (x1,x2)∈H×H and λ,γ∈C, or else(42)fx1,x2=ax1,x2mx1,x2,(43)gx1,x2=1+βax1,x2mx1,x2for all x1,x2∈H, where m:H×H→C is exponential functions satisfying m(x1,x2)=m(x2,x1) for all x1,x2∈H, a:H×H→C is an additive function on (H×H)∗≔{(x1,x2):m(x1,x2)≠0}, and β∈C. Since m can be written in the form m(x1,x2)=m1(x1)m2(x2) for some exponential functions m1,m2:H→C, from (40) and (41) we get (36) and (37), respectively. Now, we consider the case when m=m∘σ. Since m(x1,x2)=m(x2,x1) for all x1,x2∈H we have(44)mx1,x2=m1x1m2x2=m1x2m2x1=mx2,x1for all x1,x2∈H. It is easy to see that equality (44) implies m1=m2. Thus, (x1,x2)∈(H×H)∗ if and only if x1,x2∈H∗≔{x∈H:m1(x)≠0}. On the other hand, the additive function a can be written in the form a(x1,x2)=a1(x1)+a2(x2) for all x1,x2∈H∗, where a1,a2:H∗→C are additive functions. Now, the equality a(x1,x2)=-a(x2,x1) implies a2(x)=-a1(x) for all x∈H∗. Thus we have m(x1,x2)=m1(x1+x2) for all x1,x2∈H and a(x1,x2)=a1(x1)-a1(x2) for all x1,x2∈H∗. Thus, by letting m1=m, a1=a we get (40) and (39) from (42) and (43), respectively, since if x1∉H∗ or x2∉H∗, then m(x1,x2)=0. This completes the proof.


## 3. Stability of (5)

Throughout this section we denote byG a 2-divisible commutative group and f,g:G→C. In this section we consider the stability of functional equation (5); that is, we deal with the functional inequality(45)fx+σy-fxgy+gxfy≤ψyfor all x,y∈G. For the proof of the stability of (45) we need the following.Lemma 6.
Letf,g:G→C satisfy the functional inequality (45) for all x,y∈G. Then there exist μ1,μ2∈C (not both zero) and M>0 such that(46)μ1fx-μ2gx≤Mfor all x∈G, or else(47)fx+σy-fxgy+gxfy=0for all x,y∈G.Proof.
Suppose thatμ1f(x)-μ2g(x) is bounded only when μ1=μ2=0. Let(48)kx,y=fx+y-fxgσy+gxfσyfor all x,y∈G. Choose y1 satisfying f(σy1)≠0. Then we have(49)gx=ω1fx+ω2fx+y1-ω2kx,y1for all x∈G, where ω1=g(σy1)/f(σy1) and ω2=-1/f(σy1). From (48) and (49) we have(50)fx+y+z=fx+ygσz-gx+yfσz+kx+y,z=fxgσy-gxfσy+kx,ygσz-ω1fx+y+ω2fx+y+y1-ω2kx+y,y1fσz+kx+y,z=fxgσy-gxfσy+kx,ygσz-ω1fxgσy-gxfσy+kx,yfσz-ω2fxgσy+y1-gxfσy+y1+kx,y+y1fσz+ω2kx+y,y1fσz+kx+y,zfor all x,y,z∈G. Also, from (48) we have(51)fx+y+z=fxgσy+z-gxfσy+z+kx,y+zfor all x,y,z∈G. Equating (50) and (51) we have(52)gσygσz-ω1gσyfσz-ω2gσy+y1fσz-gσy+zfx+-fσygσz+ω1fσyfσz+ω2fσy+y1fσz+fσy+zgx=-kx,ygσz+ω1kx,yfσz+ω2kx,y+y1fσz-ω2kx+y,y1fσz-kx+y,z+kx,y+zfor all x,y,z∈G. Fixing y,z in (52) we have(53)-kx,ygσz+ω1kx,yfσz+ω2kx,y+y1fσz-ω2kx+y,y1fσz-kx+y,z+kx,y+z≤kx,ygσz+kx,yω1fσz+kx,y+y1ω2fσz+kx+y,y1ω2fσz+kx+y,z+kx,y+z≤ψσygσz+ψσyω1fσz+ψσy+y1ω2fσz+ψσy1ω2fσz+ψσz+ψσy+zfor all x,y,z∈G. So by our assumption, the left hand side of (52) vanishes, so does its right hand side; that is,(54)-kx,ygσz+ω1kx,y+ω2kx,y+y1-ω2kx+y,y1fσz=kx+y,z-kx,y+zfor all x,y,z∈G. From (48) we can write(55)kx+y,z-kx,y+z=fx+y+z-fx+ygσz+gx+yfσz-fx+y+z+fxgσy+z-gxfσy+z=-fσx+y+z+fσzgx+y-gσzfx+y+fσx+y+z-fσy+zgx+gσy+zfx=-fσz+σx+y+fσzgx+y-gσzfx+y+fσy+z+σx-fσy+zgx+gσy+zfx=-kσz,σx+y+kσy+z,σxfor all x,y,z∈G. Using (55) and the triangle inequality we have(56)kx+y,z-kx,y+z=kσz,σx+y-kσy+z,σx≤kσz,σx+y+kσy+z,σx≤ψx+y+ψxfor all x,y,z∈G. Thus, if we fix x,y in (54) the left hand side of (54) is a bounded function of z. Hence by our assumption, we have k(x,y)≡0. This completes the proof.Lemma 7 (see [16, p. 104]).
Assume thatf,g:G→C satisfy the functional inequality; for each y∈G, there exists a positive constant My such that(57)fx+y-fxgy≤Myfor all x,y∈G; then either f is a bounded function or g is an exponential function.In the following, we assume that(58)Φ1x≔∑k=0∞2-k-1ψ2kx<∞for all x∈G, or(59)Φ2x≔∑k=0∞2kψ2-k-1x<∞for all x∈G.Lemma 8 (see [17]).
Assume thatf:G→C satisfies the functional inequality(60)fx+y-fx-fy≤ψyfor all x,y∈G. Then there exists a unique additive function a1 given by(61)a1x=limn→∞⁡2-nf2nxsuch that(62)fx-a1x≤Φ1xfor all x∈G provided that (58) holds, and there exists a unique additive function a2 given by(63)a2x=limn→∞⁡2nf2-nxsuch that(64)fx-a2x≤Φ2xfor all x∈G provided that (59) holds.Theorem 9.
Letf,g:G→C satisfy the functional inequality(65)fx+σy-fxgy+gxfy≤ψyfor all x,y∈G; then (f,g) satisfies one of the following:(i)
f=0, g is arbitrary,(ii)
bothf and g are bounded functions,(iii)
f(x)=λ(m(x)-m(σx)), g(x)=γm(x)+1-γm(σx) for all x∈G, where m:G→C is an exponential function satisfying m≠m∘σ and λ,γ∈C,(iv)
f(x)=a(x)m(x), g(x)=(1+βa(x))m(x) for all x∈G, where m:G→C is an exponential function satisfying m=m∘σ, and a:G→C is an additive function satisfying a(x)=-a(σx) for all x∈G, and β∈C.(v)
there existβ∈C and a bounded exponential function m satisfying m≠m∘σ such that(66)gx=βfx+mxfor allx∈G, and f satisfies the condition that there exist C, d≥0 satisfying(67)fx≤Cψx+dfor allx∈G,(vi)
there existβ∈C and a bounded exponential function m satisfying m=m∘σ such that(68)gx=βfx+mxfor allx∈G, and f satisfies one of the following; there exists an additive function a1:G→C such that(69)fx-a1x+f0mx≤2Φ1xfor allx∈G, or there exists an additive function a2:G→C such that(70)fx-a2x+f0mx≤2Φ2xfor allx∈G, where Φ1 and Φ2 are the functions given in (58) and (59).Proof.
In view of Lemma6, if f,g satisfy (47), then by Theorem 2 we get (iii) or (iv). Now, we consider the case when f,g satisfy (46). If f=0, g is arbitrary which is the case (i). If f is a nontrivial bounded function, in view of (65), g is also bounded which gives the case (ii). If f is unbounded, it follows from (46) that μ2≠0 and g can be written as(71)gx=βfx+mxfor all x∈G, where β∈C and m is a bounded function. Putting (71) in (65) we have(72)fx+σy-fxmy+mxfy≤ψyfor all x,y∈G. Replacing y by σy in (72) we have(73)fx+y-fxmσy+mxfσy≤ψσyfor all x,y∈G. Using the triangle inequality with (73) we have(74)fx+y-fxmσy≤Lfσy+ψσyfor all x,y∈G, where L=supx∈G|m(x)|. By Lemma 7, m∘σ is an exponential function and so is m. If m=0, putting y=0 in (72) we have(75)fx≤ψ0for all x∈G. Thus we have m≠0, since f is unbounded. Since m is a nonzero bounded exponential function, it follows from the equalities(76)mx=mx-ymy,mkx=mxk,x,y∈G,k=0,±1,±2,…that m(x)≠0 for all x∈G and mx=1 for all x∈G. Putting x=0 in (72) and multiplying |m(x)| in the result we have(77)mxfσy+mxfy-f0mxmy≤ψyfor all y∈G. From (73) and (77) we have(78)fx+y-fxmσy-mxfy+f0mxmy≤2ψyfor all x,y∈G. First, we consider the case m(y0)≠m(σy0) for some y0∈G. Replacing x by y and y by x in (78) we have(79)fy+x-fymσx-myfx+f0mymx≤2ψxfor all x,y∈G. From (78) and (79), using the triangle inequality and putting y=y0 we have(80)fx≤Cψx+dfor all x∈G, where C=2/my0-mσy0 and d=ψ(y0)+fy0, which gives (v). Now, we consider the case when m(x)=m(σx) for all x∈G. Dividing both the sides (78) by mxmy we have(81)Fx+y-Fx-Fy≤2ψyfor all x,y∈G, where F(x)=f(x)/m(x)-f(0). Using Lemma 8 with (81) and multiplying mx in the result we get (69) and (70). This completes the proof.In particular, letG=X×X with X a normed space, ψ(x1,x2)=ϵ(x1p+x2p), p≥0, and σ(x1,x2)=(x2,x1) for all x1,x2∈X. Then by Theorems 5 and 9 we have the following.Corollary 10.
Letf,g:X×X→C satisfy the functional inequality(82)fx1+y2,x2+y1-fx1,x2gy1,y2+gx1,x2fy1,y2≤ϵy1p+y2pfor all x1,x2,y1,y2∈X. Then (f,g) satisfies one of the following:(i)
f=0, g is arbitrary,(ii)
bothf and g are bounded functions,(iii)
there exist exponential functionsm1,m2:X→C satisfying m1≠m2 and λ,γ∈C such that(83)fx1,x2=λm1x1m2x2-m1x2m2x1,gx1,x2=γm1x1m2x2+1-γm1x2m2x1for allx1,x2∈X,(iv)
there exist an exponential functionm:X→C, an additive function a:X→C, and β∈C such that(84)fx1,x2=ax1-x2mx1+x2,gx1,x2=1+βax1-x2mx1+x2for allx1,x2∈X,(v)
there existβ∈C and bounded exponential functions m1,m2:X→C satisfying m1≠m2 such that(85)gx1,x2=βfx1,x2+m1x1m2x2for allx1,x2∈X, and f satisfies the condition that there exist C,d≥0 satisfying(86)fx1,x2≤Cψx1,x2+dfor allx1,x2∈X,(vi)
there existβ∈C and a bounded exponential function m:X→C such that(87)gx1,x2=βfx1,x2+mx1+x2for allx1,x2∈X, and f satisfies one of the following; there exist additive functions a1,a2:X→C such that(88)fx1,x2-a1x1+a2x2+f0,0mx1+x2≤2ϵx1p+x2p2-2pfor allx1,x2∈X provided that 0≤p<1, or there exist additive functions b1,b2:X→C such that(89)fx1,x2-b1x1+b2x2+f0,0mx1+x2≤2ϵx1p+x2p2p-2for allx1,x2∈X provided that p>1.



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