# On the Superstability of Lobačevskiǐ’s Functional Equations with Involution

Authors: Jaeyoung Chung; Bogeun Lee; Misuk Ha
Journal: Journal of Function Spaces
 (2016)

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## Abstract

LetG be a uniquely 2-divisible commutative group and let f,g:G→C and σ:G→G be an involution. In this paper, generalizing the superstability of Lobačevskiǐ’s functional equation, we consider f(x+σy)/22-g(x)f(y)≤ψ(x) or ψ(y) for all x,y∈G, where ψ:G→R+. As a direct consequence, we find a weaker condition for the functions f satisfying the Lobačevskiǐ functional inequality to be unbounded, which refines the result of Găvrută and shows the behaviors of bounded functions satisfying the inequality. We also give various examples with explicit involutions on Euclidean space.


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## Body

## 1. Main Results

Throughout this paper, we denote byG a uniquely 2-divisible commutative group (i.e., for each x∈G, there exists a unique y∈G such that y+y=x) and by R, R+, C, and Rn the set of real numbers, nonnegative real numbers, complex numbers, and the n-dimensional Euclidean space, respectively. A function m:G→C is calledan exponential function provided that m(x+y)=m(x)m(y) for all x,y∈G and σ:G→G is calledan involution provided that σ(x+y)=σ(x)+σ(y) and σ(σ(x))=x for all x,y∈G. We denote σ(x) by σx. An exponential function m:G→C is called σ-exponential function if m satisfies m(σx)=m(x) for all x∈G and is denoted by mσ. In [1] (see also [2, Theorem 6.5]), Găvrută investigated the superstability of Lobačevskiǐ’s functional equation: (1)fx+y22-fxfy≤δfor all x,y∈G and for some δ>0 (see Albert and Baker [3] and Baker [4] for the superstability of the exponential functional equation). As a result he proved the following.Theorem 1.
Assume thatf:G→C satisfies (1). Then, every bounded function f satisfies(2)fx≤f0+f02+4δ2≔Kf,δfor all x∈G, and if there exists y0∈G such that |f(y0)|>Kf,δ, then f satisfies (3)fx+y22=fxfyfor all x,y∈G.A careful observation shows that the upper boundKf,δ in (2) is not optimal and may be replaced by a better one; namely, Kf,δ∗=(1/2)(mf+mf2+4δ) with mf=infx∈G⁡|f(x)|. For example, let f(x)=δcos⁡x. Then, Kf,δ=(1/2)(1+5)δ and Kf,δ∗=δ. Using the better upper bound, we investigate the behavior of bounded functions satisfying (1) (see Remark 3).As main results in this paper, we generalize Găvrută’s result and consider the inequalities:(4)fx+σy22-gxfy≤ψx,(5)fx+σy22-gxfy≤ψyfor all x,y∈G, where ψ:G→R+ is an unbounded function.Note that the casesf(x+σy)/22-f(x)g(y)≤ψ(y) and f(x+σy)/22-f(x)g(y)≤ψ(x) can be reduced to the special case of form (4) or (5) when σ=I, the identity involution on G (replacing g(y) by g(σy) and ψ(y) by ψ(σy)).As a direct consequence of our results, we obtain the following.Theorem 2.
Assume thatf:G→C satisfies(6)fx+σy22-fxfy≤δfor all x,y∈G. Then, every bounded function f satisfies(7)fx≤mf+mf2+4δ2≔Kf,δ∗for all x∈G, and if there exists y0∈G such that |f(y0)|>Kf,δ∗, then there exist σ-exponential function mσ and λ∈C such that (8)fx=λmσxfor all x∈G.Remark 3.
Letσ=I be the identity involution in Theorem 2; we obtain a refined version of Theorem 1. Now, from (7), we have (9)Mf-mf≤-mf+mf2+4δ2≤2δmf+mf2+4δ≤2δ4δ=δ.Thus, if {fn},n=1,2,3,…, is a sequence of functions satisfying (10)fnx+y22-fnxfny≤1nfor all x,y∈G, then there exists a sequence rn∈R such that |fn(x)|-rn converges uniformly to 0 as n→∞ (i.e., |fn| tends to straight lines).We refer the reader to [5–7] for related functional equations and their stabilities. We also refer the reader to [8, 9] for some important recent developments on the issues of stability and superstability for functional equations.


## 2. Stability of (4)

In this section, we prove the superstability of (4). For the proof, we need the following.Lemma 4.
Assume thatm1,m2:G→C are unbounded exponential functions satisfying (11)m1x-λm2x≤δfor all x∈G and for some λ∈C. Then, one has λ=1 and m1=m2.Proof.
Sincem2 is unbounded exponential function, it is easy to see that m2(x)≠0 for all x∈G. Replacing x by kx, k=1,2,3,…, in (11) and dividing the result by |m2(kx)| we have (12)m1xm2xk-λ≤δm2xk.If |m2(x)|>1, we have (13)limk→∞⁡m1xm2xk=λ.Since λ≠0, from (13), we have (14)m1x=m2xfor all x∈G such that |m2(x)|>1 and hence λ=1. Let |m2(x)|≤1. Choose y∈G such that |m2(y)|>|m2(x)|-1. Then, we have |m2(y)|>1 and |m2(x+y)|=|m2(x)m2(y)|>1. Thus, it follows from (14) that m1(y)=m2(y) and m1(x+y)=m2(x+y) and hence (15)m1x=m1x+ym1y=m2x+ym2y=m2xfor all x∈G such that |m2(x)|≤1. This completes the proof.From now on, we denoteMf=supx∈G⁡|f(x)| and mf=infx∈G⁡|f(x)|.Theorem 5.
Assume thatf,g:G→C satisfy (4). Then, every bounded function f satisfies(16)Mf≤infx∈G⁡gx+gx2+4ψx2.If f is unbounded, then there exists an exponential function m such that(17)gx=g0mx,fx≤f0g0m2x+ψ2xfor all x∈G, and, in particular, if(18)infx∈G⁡ψxgx+1gx=0,then there exist σ-exponential function mσ and λ∈C such that(19)fx=gx=λmσxfor all x∈G.Proof.
Assume thatf is bounded. Replacing y by -σx+2σy in (4) and using the triangle inequality with the result, we have(20)fy2≤gxf-σx+2σy+ψx≤gxMf+ψxfor all x,y∈G. Taking the supremum of the left hand side of (20), we have(21)Mf2≤gxMf+ψxfor all x∈G. Fixing x and solving the quadratic inequality (21), we obtain(22)Mf≤gx+gx2+4ψx2for all x∈G. Taking the infimum of the right hand side of (22), we get (16). Now, assume that f is unbounded. Putting x=0 in (4), we have(23)fσy22-g0fy≤ψ0for all y∈G. Choosing a sequence yn∈G, n=1,2,3,…, such that |f(yn)|→∞ as n→∞, putting y=yn, n=1,2,3,…, in (4), dividing the result by |f(yn)|, letting n→∞, and using (23), we have (24)gx=limn→∞⁡fx+σyn/22fyn=limn→∞⁡g0fσx+ynfynfor all x∈G. Multiplying both sides of (24) by g(y) and using (4) and (24), we have (25)gygx=limn→∞⁡g0gyfσx+ynfyn=limn→∞⁡g0fy+x+σyn/22fyn=g0gy+xfor all x,y∈G. Thus, g0(x)=g(x)/g(0) is an exponential function and let g0=m; then, we get(26)gx=g0mxfor all x∈G. Putting y=0 in (4), replacing x by 2x, and using (26) and the triangle inequality, we have |f(x)|≤|f(0)g(0)m(x)2|+ψ(2x) for all x∈G, which gives (17). Assume that (18) holds. Then, we can choose a sequence xn, n=1,2,3,…, such that(27)limn→∞⁡ψxngxn=limn→∞⁡1gxn=0.Putting x=xn, n=1,2,3,…, in (4), dividing the result by |g(xn)|, letting n→∞, and using (23), we have (28)fy=limn→∞⁡fxn+σy/22gxn=limn→∞⁡g0fσxn+ygxnfor all y∈G. Multiplying both sides of (28) by g(x) and using (4) and (28), we have (29)gxfy=limn→∞⁡g0gxfσxn+ygxn=limn→∞g0fx+σy+xn/22gxn=g0fσx+yfor all x,y∈G. Putting y=0 in (29), replacing x by σx, using (26), and dividing the result by g(0), we have(30)fx=f0mσxfor all x∈G. Putting (30) in (23), we have(31)f02my-g0f0-1mσy≤ψ0for all y∈G. Applying Lemma 4 in (31), we have m(y)=m(σy) for all y∈G and g(0)=f(0). Letting g(0)=f(0)≔λ, we get (19). This completes the proof.In particular, ifg=f in Theorem 5, we obtain the following.Corollary 6.
Assume thatf:G→C satisfies (32)fx+σy22-fxfy≤ψxfor all x,y∈G. Then, every bounded function f satisfies(33)Mf≤infx∈G⁡fx+fx2+4ψx2,and if f is unbounded, then there exists σ-exponential function mσ such that(34)fx=f0mσxfor all x∈G.Proof.
By Theorem5, we get (33) and every unbounded function f has the form f(x)=f(0)m(x) for some exponential function m. Putting x=0 and f(y)=f(0)m(y) in (32), we have (35)f02mσy-my≤ψ0for all y∈G. Using Lemma 4, we have m(y)=m(σy) for all y∈G. Thus, we get (34). This completes the proof.Remark 7.
In particular, ifψ(x)=δ in Corollary 6, we obtain Theorem 2.


## 3. Stability of (5)

In this section, we prove the stability of (5). We exclude the trivial case when f=0 or g=0.Theorem 8.
Assume thatf,g:G→C satisfy (5). Then, every bounded function f satisfies(36)supy∈G⁡Mgfy-ψy≤Mf2≤infy∈G⁡Mgfy+ψy.If f is unbounded, then there exists an exponential function m such that(37)fx=f0mxfor all x∈G. In particular, if(38)infx∈G⁡ψxfx+1fx=0,then there exist σ-exponential function mσ and λ∈C such that(39)fx=gx=λmσxfor all x∈G.Proof.
Choosingy0∈G such that f(y0)≠0 and putting y=y0 in (5), we get(40)fx+σy022-gxfy0≤ψy0for all x∈G, and it follows that f is bounded if and only if g is bounded. Assume that f and g are bounded. Replacing x by 2x-σy in (5) and using the triangle inequality with the result, we have(41)fx2≤g2x-σyfy+ψy≤Mgfy+ψyfor all x,y∈G. Taking the supremum of the left hand side of (41), we have(42)Mf2≤Mgfy+ψyfor all y∈G. Taking the infimum of the right hand side of (42), we have(43)Mf2≤infy∈G⁡Mgfy+ψy.Now, using the triangle inequality with (5) again, we have(44)gxfy≤fx+σy22+ψy≤Mf2+ψyfor all x,y∈G. Taking the supremum of g(x) in the left hand side of (44) and subtracting ψ(y) from both sides of the result, we have(45)Mgfy-ψy≤Mf2for all y∈G. Taking the supremum of the left hand side of (45), we have(46)supy∈G⁡Mgfy-ψy≤Mf2.From (43) and (46), we get (36). Now, we assume that f and g are unbounded. Putting y=0 in (5), we have(47)fx22-gxf0≤ψ0for all x∈G. Choosing a sequence xn∈G, n=1,2,3,…, such that |g(xn)|→∞ as n→∞, putting x=xn, n=1,2,3,…, in (5), dividing the result by |g(xn)|, letting n→∞, and using (47), we have (48)fy=limn→∞⁡fxn+σy/22gxn=limn→∞⁡f0gxn+σygxnfor all y∈G. Multiplying both sides of (48) by f(z) and using (5) and (48), we have (49)fyfz=limn→∞⁡f0gxn+σyfzgxn=limn→∞⁡f0fxn+σy+σz/22gxn=f0fy+zfor all y,z∈G. Thus, f0(x)=f(x)/f(0) is an exponential function, say f0=m; then, we get(50)fx=f0mxfor all x∈G. Assume that (38) holds. Then, we can choose a sequence yn, n=1,2,3,…, such that (51)limn→∞⁡ψynfyn=limn→∞⁡1fyn=0.Putting y=yn, n=1,2,3,…, in (5), dividing the result by |f(yn)|, letting n→∞, and using (47), we have (52)gx=limn→∞⁡fx+σyn/22fyn=limn→∞⁡f0gx+σynfynfor all y∈G. Multiplying both sides of (52) by f(y) and using (5) and (52), we have (53)gxfy=limn→∞⁡f0gx+σynfyfyn=limn→∞⁡f0fx+σyn+σy/22fyn=f0gx+σyfor all x,y∈G. Putting x=0 in (53), replacing y by σy, and using (50), we have(54)gx=g0mσxfor all x∈G. Putting (50) and (54) in (47), we have(55)f02mx-g0f0-1mσx≤ψ0for all x∈G. Applying Lemma 4, we have m(x)=m(σx) for all x∈G and g(0)=f(0). Letting g(0)=f(0)≔λ, we get (39). This completes the proof.In particular, ifg=f in Theorem 8, we obtain the following.Corollary 9.
Assume thatf:G→C satisfies(56)fx+σy22-fxfy≤ψyfor all x,y∈G. Then, every bounded function f satisfies(57)Mf≤infx∈G⁡fx+fx2+4ψx2,and if f is unbounded, then there exists σ-exponential function mσ such that(58)fx=f0mσxfor all x∈G.Proof.
By Theorem8, every bounded function f satisfies(59)Mf2≤Mffy+ψyfor all y∈G. Solving (59), we have(60)Mf≤fy+fy2+4ψy2for all y∈G, which gives (57). If f is unbounded, then by Theorem 8 we have(61)fx=f0mxfor all x∈G. Putting (61) in (56), we have (62)f02mxmσy-my≤ψyfor all x,y∈G. Since m is unbounded, we have m(y)=m(σy) for all y∈G, which gives (58). This completes the proof.In particular, ifψ(y)=δ for all y∈G in Theorems 5 and 8, we obtain the following.Corollary 10.
Assume thatf,g:G→C satisfy (63)fx+σy22-gxfy≤δfor all x,y∈G. If f is bounded, then f satisfies(64)MgMf-δ≤Mf2≤min⁡Mgmf,Mfmg+δ.If f is unbounded, then there exist σ-exponential function mσ and λ∈C such that(65)fx=gx=λmσxfor all x∈G.Proof.
By Theorem5, every bounded function f satisfies(66)Mf2≤Mfmg+δand, by Theorem 8, f satisfies(67)MgMf-δ≤Mf2≤Mgmf+δ.From (66) and (67), we get (64). If f is unbounded, then condition (38) holds. Thus, by Theorem 8, we get (65). This completes the proof.


## 4. Examples

In this section, employing some involutions on Euclidean space, we give various examples satisfying inequalities (4) and (5). We denote by c·x the inner product of c=(c1,c2,…,cn)∈Cn and x=(x1,x2,…,xn)∈Rn defined as c·x=∑j=1ncjxj, |x|=x12+⋯+xn2, and Rc=(Rc1,…,Rcn), where Rcj are the real parts of cj,j=1,2,…,n. It is easy to see that if G is uniquely 2-divisible, then mσ is σ-exponential if and only if(68)mσx=mx+σx2,∀x∈Gfor some exponential function m:G→C.Example 1.
LetG=〈R2,+〉 in Theorem 5 and let(69)σ=ab1-a2b-abe a 2×2 matrix, where a,b∈R with b≠0. Then, σ defines an involution on R2. Let f,g:R2→C satisfy(70)ft+au+bv2,s+b-11-a2u-av22-gt,sfu,v≤ϵtp+spfor all t,s,u,v∈R and for some ϵ>0, p>0. Then, by Theorem 5, every bounded function f satisfies(71)fu,v≤inft,s∈R2⁡gt,s+gt,s2+4ϵtp+sp2for all u,v∈R, and if f is unbounded, then (72)gt,s=g0,0mt,sfor all (t,s)∈R2, where m:R2→C is an exponential function. In particular, if g is unbounded, then we can choose (t0,s0)∈R2 such that |m(t0,s0)|>1, and hence (73)limn→∞⁡ψnt0,ns0gnt0,ns0≤limn→∞⁡ϵt0p+s0pg0,0limn→∞⁡npmt0,s0n=0,which implies that condition (18) holds. Thus, by Theorem 5, g,f have the form (74)gt,s=ft,s=λmt,sfor all (t,s)∈R2, where λ∈C and m:R2→C is an exponential function satisfying(75)mt,s=mat+bs,b-11-a2t-asfor all (t,s)∈R2. In particular, if g is Lebesgue measurable, m is Lebesgue measurable and has the form m(t,s)=ec1t+c2s for some c1,c2∈C. From condition (75), we have c1=(1/b)(1+a)c2. Thus, g,f has the form(76)gt,s=ft,s=λed1+at+bsfor all (t,s)∈R2 and for some d,λ∈C with Rd≠0.Example 2.
Letσ be an n×n matrix such that σ2=E, the unit matrix. Suppose that f,g:Rn→C satisfy (77)fx+σy22-gxfy≤exqfor all x,y∈Rn and for some 0≤q<1. Then, by Theorem 5, every bounded function f satisfies(78)fy≤infx∈Rn⁡gx+gx2+4exq2for all y∈Rn, and if f is unbounded, then(79)gx=g0mxfor all x∈Rn. If g is unbounded, it is easy to see that condition (18) holds. Thus, by Theorem 5, there exist σ-exponential function mσ:Rn→C and λ∈C such that(80)gx=fx=λmσxfor all x∈Rn. In particular, if g is continuous, then by (68) there exist c∈Cn, R(c+cσ)≠0 and λ∈C such that(81)gx=fx=λe1/2c+cσ·xfor all x∈Rn.Example 3.
Suppose thatf,g:C→C are unbounded function satisfying (82)fz+w¯22-gzfw≤ϵwpfor all z,w∈C and for some p>0. Then, by Theorem 8, there exists an exponential function m:C→C such that (83)fz=f0mzfor all z∈C. Choose z0∈C such that |m(z0)|>1. Then, we have (84)limn→∞⁡nz0pfnz0=z0pf0limn→∞⁡npmz0n=0,which implies that condition (38) holds. Thus, again, by Theorem 8, there exist an exponential function m satisfying m(z)=m(z¯) and λ∈C such that f(z)=g(z)=λm(z) for all z∈C. Now, from m(z)=m(z¯), m is given by m(z)=m(Rz). Thus, we have (85)fz=gz=λmRzfor all z∈C.Remark 11.
Leta,b∈Rn be two nonzero vectors that are not parallel; that is, b≠ra for all r∈R. Then, the hyperplane b·x=0 is not parallel to (b-a)·x=0 and hence there exists x0∈Rn such that b·x0>0 and (b-a)·x0>0. If b=ta for some t∈R, then there exists x0∈Rn such that b·x0>0 and (b-a)·x0>0 if and only if t>1. Thus, if b≠ta for all t≤1, then there exists x0∈Rn such that b·x0>0 and (b-a)·x0>0.Example 4.
Leta∈Rn be fixed. Suppose that f,g:Rn→C are unbounded continuous functions satisfying (86)fx+σy22-gxfy≤ea·yfor all x,y∈Rn. By Theorem 8, we have (87)fx=f0ec·xfor all x∈Rn and for some c∈Cn, λ∈C. If Rc≠ta for all t≤1, then, by Remark 11, there exists x0∈Rn such that(88)Rc·x0>0,Rc·x0>a·x0.From (87) and (88), we have (89)limk→∞⁡1+ea·kx0fkx0=1f0limk→∞⁡1+ea·kx0eRc·kx0=1f0limk→∞⁡1+eka·x0ekRc·x0=0.Thus, condition (38) holds. By Theorem 8, we get (90)gx=fx=λe1/2c+cσ·xfor all x∈Rn.



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