# On a Generalized Hyers-Ulam Stability of Trigonometric Functional Equations

Authors: Jaeyoung Chung; Jeongwook Chang
Journal: Journal of Applied Mathematics
 (2012)

---

## Abstract

LetG be an Abelian group, let ℂ be the field of complex numbers, and let f,g:G→ℂ. We consider the generalized Hyers-Ulam stability for a class of trigonometric functional inequalities, |f(x-y)-f(x)g(y)+g(x)f(y)|≤ψ(y),|g(x-y)-g(x)g(y)-f(x)f(y)|≤ψ(y), where ψ:G→ℝ is an arbitrary nonnegative function.


---

## Body

## 1. Introduction

The Hyers-Ulam stability problems of functional equations go back to 1940 when S. M. 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. After the result of Hyers, Aoki [4], and Bourgin [5, 6] dealt with this problem, however, there were no other results on this problem until 1978 when Rassias [7] dealt again with the inequality of Aoki [4]. Following the Rassias' result, a great number of papers on the subject have been published concerning numerous functional equations in various directions [2, 7–21]. The following four functional equations are calledtrigonometric functional equations.(1.1)f(x+y)-f(x)g(y)-g(x)f(y)=0(1.2)g(x+y)-g(x)g(y)+f(x)f(y)=0(1.3)f(x-y)-f(x)g(y)+g(x)f(y)=0(1.4)g(x-y)-g(x)g(y)-f(x)f(y)=0The four functional equations have been investigated separately. The general solutions and regular solutions of the above equations are introduced [22, 23]. In particular, the last equation (1.4) is most interesting in the sense that (1.4) alone characterizes the two trigonometric functions f(x)=cos(ax), g(x)=sin(ax) under some regularities of g, which none of the remaining equations are able to do.In [19], Székelyhidi developed his idea of using invariant subspaces of functions defined on a group or semigroup to obtain the Hyers-Ulam stability of the trigonometric functional equations (1.1) and (1.2). As results, he obtained the Hyers-Ulam stability when for each fixed y the difference(1.5)T1(x)∶=f(x+y)-f(x)g(y)-g(x)f(y)
is a bounded function of x and the Hyers-Ulam stability when for each fixed y the difference(1.6)T2(x)∶=g(x+y)-g(x)g(y)+f(x)f(y)
is a bounded function of x, where f,g are mappings from an Abelian (amenable) group G to the field ℂ of complex numbers.In this paper, we complete the parallel Hyers-Ulam stability to that of [19] for the functional equations (1.3) and (1.4). As results, we obtained the Hyers-Ulam stability when for each fixed y the difference(1.7)T3(x)∶=f(x-y)-f(x)g(y)+g(x)f(y)
is a bounded function of x and the Hyers-Ulam stability when for each fixed y the difference(1.8)T4(x)∶=g(x-y)-g(x)g(y)-f(x)f(y)
is a bounded function of x.In fact, the authors [10] obtained weaker versions of the Hyers-Ulam stability for the functional equations (1.3) and (1.4), that is, we proved the Hyers-Ulam stability of (1.3) when the difference(1.9)T3(x,y)∶=f(x-y)-f(x)g(y)+g(x)f(y)
is uniformly bounded for all x and y, and we proved the Hyers-Ulam stability of (1.4) when the difference(1.10)T4(x,y)∶=g(x-y)-g(x)g(y)-f(x)f(y)
is uniformly bounded for all x and y.So, the results in this paper would be generalizations of those in [10]. We refer the reader to [9, 15, 16, 20, 21] for some related Hyers-Ulam stability of functional equations of trigonometric type.


## 2. Main Theorems

A functiona from a semigroup 〈S,+〉 to the field ℂ of complex numbers is said to bean additive function provided that a(x+y)=a(x)+a(y) and m:S→ℂ is said to bean exponential function provided that m(x+y)=m(x)m(y). Throughout this paper, we denote by G an Abelian group, ℂ the set of complex numbers, and ψ:G→ℝ a fixed nonnegative function. For the proof of stabilities of (1.3) and (1.4), we need the following.Lemma 2.1 (see [2]).
LetS be a semigroup. Assume that f,g:S→ℂ satisfy the inequality; for each y∈S, there exists a positive constant My such that
(2.1)|f(x+y)-f(x)g(y)|≤My,
for all x∈S, then either f is a bounded function or g is an exponential function.Proof.
Suppose thatg is not exponential, then there are y,z∈S such that g(y+z)≠g(y)g(z). Now we have
(2.2)f(x+y+z)-f(x+y)g(z)=(f(x+y+z)-f(x)g(y+z))-g(z)(f(x+y)-f(x)g(y))+f(x)(g(y+z)-g(y)g(z)),
and hence,
(2.3)f(x)=(g(y+z)-g(y)g(z))-1×((f(x+y+z)-f(x+y)g(z))-(f(x+y+z)-f(x)g(y+z))+g(z)(f(x+y)-f(x)g(y))).
In view of (2.1), the right hand side of (2.3) is bounded as a function of x. Consequently, f is bounded.We discuss the general solutionsf,g:G→ℂ of the corresponding trigonometric functional equations
(2.4)f(x-y)-f(x)g(y)+g(x)f(y)=0,(2.5)g(x-y)-g(x)g(y)-f(x)f(y)=0.Lemma 2.2 (see [22, 23]).
The general solutions(f,g) of the functional equation (2.4) are given by one of the following:(i)
f=0 and g is arbitrary,(ii)
f(x)=λ1(m(x)-m(-x)) and g(x)=λ2f(x)+(1/2)(m(x)+m(-x)) for some λ1,λ2∈ℂ, where m is an exponential function,(iii)
f(x)=a(x)m(x),g(x)=(1+λa(x))m(x) for some λ∈ℂ, where a is an additive function and m is an exponential function satisfying m2≡1.
Also, the general solutions(g,f) of the functional equation (2.5) are given by one of the following:(i)
g(x)=λ and f(x)=±λ-λ2 for some λ∈ℂ,(ii)
g(x)=(1/2)(m(x)+m(-x)) and f(x)=(1/2i)(m(x)-m(-x)), where m is an exponential function.Proof.
The solutions of the functional equation (2.4) are given in [23, p. 217,  Theorem  11]. For the functional equation (2.5), combining the result of L. Vietoris [22, p. 177] and that of J. A. Baker [23, p. 220], we obtain that every nonconstant function g satisfying (2.5) has the form
(2.6)g(x)=12(m(x)+m(-x)),
for some exponential function m. Thus, using (2.5), we have
(2.7)f(x)=12i(m(x)-m(-x)).
This completes the proof.For the proof of the stability of (1.1), we need the following. Throughout this paper, we denote by ψ an arbitrary nonnegative function on G.Lemma 2.3.
Letf,g:G→ℂ satisfy the inequality
(2.8)|f(x-y)-f(x)g(y)+g(x)f(y)|≤ψ(y),
for all x,y∈G, then either there exist λ1,λ2∈ℂ, not both zero, and M>0 such that
(2.9)|λ1f(x)-λ2g(x)|≤M,
or else
(2.10)f(x-y)-f(x)g(y)+g(x)f(y)=0,
for all x,y∈G.Proof.
Suppose that the inequality (2.9) holds only when λ1=λ2=0. Let
(2.11)k(x,y)=f(x+y)-f(x)g(-y)+g(x)f(-y),
and choose y1 satisfying f(-y1)≠0. Now it can be easily calculated that
(2.12)g(x)=λ0f(x)+λ1f(x+y1)-λ1k(x,y1),
where λ0=g(-y1)/f(-y1) and λ1=-1/f(-y1). By (2.11), we have
(2.13)f(x+(y+z))=f(x)g(-y-z)-g(x)f(-y-z)+k(x,y+z).
Also by (2.11) and (2.12), we have
(2.14)f((x+y)+z)=f(x+y)g(-z)-g(x+y)f(-z)+k(x+y,z)=(f(x)g(-y)-g(x)f(-y)+k(x,y))g(-z)-(λ0f(x+y)+λ1f(x+y+y1)-λ1k(x+y,y1))f(-z)+k(x+y,z)=(f(x)g(-y)-g(x)f(-y)+k(x,y))g(-z)-λ0(f(x)g(-y)-g(x)f(-y)+k(x,y))f(-z)-λ1(f(x)g(-y-y1)-g(x)f(-y-y1)+k(x,y+y1))f(-z)+λ1k(x+y,y1)f(-z)+k(x+y,z).
From (2.13) and (2.14), we have
(2.15)(g(-y)g(-z)-λ0g(-y)f(-z)-λ1g(-y-y1)f(-z)-g(-y-z))f(x)+(-g(-y)g(-z)+λ0f(-y)g(-z)+λ1f(-y-y1)f(-z)+f(-y-z))g(x)=-k(x,y)g(-z)+λ0k(x,y)f(-z)+λ1k(x,y+y1)f(-z)-λ1k(x+y,y1)f(-z)-k(x+y,z)+k(x,y+z).
Sincek(x,y) is bounded by ψ(-y), if we fix y, z, the right hand side of (2.15) is bounded by a constant M, where
(2.16)M=ψ(-y)|g(-z)|+ψ(-y)|λ0f(-z)|+ψ(-y-y1)|λ1f(-z)|+ψ(-y1)|λ1f(-z)|+ψ(-z)+ψ(-y-z).
So by our assumption, the left hand side of (2.15) vanishes, so does the right hand side. Thus, we have
(2.17)(-λ0k(x,y)-λ1k(x,y+y1)+λ1k(x+y,y1))f(-z)+k(x,y)g(-z)=k(x,y+z)-k(x+y,z).
Now by the definition of k, we have
(2.18)k(x+y,z)-k(x,y+z)=f(x+y+z)-f(x+y)g(-z)+g(x+y)f(-z)-f(x+y+z)+f(x)g(-y-z)-g(x)f(-y-z)=f(-y-z-x)-f(-y-z)g(x)+g(-y-z)f(x)-f(-z-x-y)+f(-z)g(x+y)-g(-z)f(x+y)=k(-y-z,-x)-k(-z,-x-y).
Hence, the right hand side of (2.17) is bounded by ψ(x)+ψ(x+y). So if we fix x, y in (2.17), the left hand side of (2.17) is a bounded function of z. Thus, by our assumption, we conclude that k(x,y)≡0. This completes the proof.In the following theorem, we assume that(2.19)Φ1(x)∶=∑k=0∞2-kψ(-2kx)<∞,
or
(2.20)Φ2(x)∶=∑k=1∞2kψ(-2-kx)<∞.For the proof, we discuss the following property.Lemma 2.4.
Letm:G→ℂ be a bounded exponential function satisfying m(x)≠m(-x) for some x∈G, then there exists y∈G such that
(2.21)|m(y)-m(-y)|≥3.
Furthermore, the constant 3 is the best one.Proof.
Sincem is a bounded exponential, there exists C>0 such that |m(x)|k=|m(kx)|≤C for all k∈ℤ and x∈G, which implies |m(x)|=1 for all x∈G. Assume that m(x0)≠m(-x0), then we have m(x0)≠±1, and we may assume that m(x0)=eiθ,0<θ<π. If θ∈[π/3,2π/3], we have |m(x0)-m(-x0)|=|eiθ-e-iθ|≥3. If θ∈[0,π/3], there exists a positive integer k such that kθ∈[π/3,2π/3], and we have |m(kx0)-m(-kx0)|=|eik0θ-e-ik0θ|≥3. If θ∈[2π/3,5π/6], then 2θ∈[4π/3,5π/3], and we have |m(2x0)-m(-2x0)|=|ei2θ-e-i2θ|≥3. Finally, if θ∈[5π/6,π], there exists a positive integer k such that 2kθ∈[-π/3,-2π/3], and we have |m(2kx0)-m(-2kx0)|=|ei2kθ-e-i2kθ|≥3. Now define m:ℤ→ℂ by m(k)=eikπ/3. Then we have |m(3k+1)-m(-3k-1)|=3 for all k∈ℤ. Thus, 3 is the biggest one. This completes the proof.Theorem 2.5.
Letf,g:G→ℂ satisfy the inequality
(2.22)|f(x-y)-f(x)g(y)+g(x)f(y)|≤ψ(y),
for all x,y∈G, then (f,g) satisfies one of the following:(i)
f=0, g is arbitrary,(ii)
f and g are bounded functions,(iii)
f(x)=λ1(m(x)-m(-x)) and g(x)=λ2f(x)+(1/2)(m(x)+m(-x)) for some λ1,λ2∈ℂ, where m is an exponential function,(iv)
there existλ∈ℂ and a bounded exponential function m such that
(2.23)g(x)=λf(x)+m(x),
for all x∈G, and f satisfies the condition; there exists d≥0 satisfying
(2.24)|f(x)|≤23(ψ(-x)+d),
for all x∈G,(v)
there existλ∈ℂ and a bounded exponential function m satisfying m2≡1 such that
(2.25)g(x)=λf(x)+m(x),
for all x∈G, and f satisfies one of the following conditions; there exists an additive function a1:G→ℂ such that
(2.26)|f(x)-(a1(x)+f(0))m(x)|≤Φ1(x),
for all x∈G, or there exists an additive function a2:G→ℂ such that
(2.27)|f(x)-(a1(x)+f(0))m(x)|≤Φ2(x),
for all x∈G, where Φ1 and Φ2 are the functions given in (2.19) and (2.20).Proof.
In view of Lemma2.3, we first consider the case when f,g satisfy (2.9). If f=0, g is arbitrary which is the case (i). If f is a nontrivial bounded function, in view of (2.22), g is also bounded which gives the case (ii). If f is unbounded, it follows from (2.9) that λ2≠0 and
(2.28)g(x)=λf(x)+m(x),
for some λ∈ℂ and a bounded function m. Putting (2.28) in (2.22), we have
(2.29)|f(x-y)-f(x)m(y)+m(x)f(y)|≤ψ(y),
for all x,y∈G. Replacing y by -y and using the triangle inequality, we have, for some C>0,
(2.30)|f(x+y)-f(x)m(-y)|≤C|f(-y)|+ψ(-y),
for all x,y∈G. By Lemma 2.1, m is an exponential function. If m=0, putting y=0 in (2.29), we have
(2.31)|f(x)|≤ψ(0).
Thus, we have m≠0 since f is unbounded. Since m is a nonzero bounded exponential function, it follows from the equalities
(2.32)m(x)=m(x-y)m(y),x,y∈G
that m(0)=1 and m(x)≠0, for all x∈G. Putting x=0 in (2.29) and replacing y by -y multiplying |m(x)| in the result, we have
(2.33)|m(x)f(-y)+m(x)f(y)-f(0)m(x)m(-y)|≤ψ(-y),
for all y∈G. Replacing y by -y in (2.29) and using (2.33), we have
(2.34)|f(x+y)-f(x)m(-y)-m(x)f(y)+f(0)m(x)m(-y)|≤2ψ(-y).
First we consider the casem(x)≠m(-x) for some x∈G. Replacing x by y and y by x in (2.34), we have
(2.35)|f(y+x)-f(y)m(-x)-m(y)f(x)+f(0)m(y)m(-x)|≤2ψ(-x),
for all x,y∈G. From (2.34) and (2.35), using the triangle inequality, putting y=y0 such that |m(y0)-m(-y0)|≥3 and dividing |m(y0)-m(-y0)|, we have
(2.36)|f(x)|≤23(ψ(-x)+d),
for all x∈G, where d=ψ(-y0)+|f(y0)|+|f(0)|, which gives (iv). Now we consider the case m(x)=m(-x), for all x∈G. Dividing both the sides of (2.34) by m(x)m(y), we have
(2.37)|F(x+y)-F(x)-F(y)|≤2ψ(-y),
for all x,y∈G, where F(x)=f(x)/m(x)-f(0). By the well-known results in [4], there exists a unique additive function a1(x) given by
(2.38)a1(x)=lim⁡n→∞⁡2-nf(2nx)
such that
(2.39)|F(x)-a1(x)|≤Φ1(x)
if Φ1(x)∶=∑k=0∞2-kψ(-2kx)<∞, and there exists a unique additive function a2(x) given by
(2.40)a2(x)=lim⁡n→∞⁡2nf(2-nx)
such that
(2.41)|F(x)-a2(x)|≤Φ2(x)
if Φ2(x)∶=∑k=1∞2kψ(-2-kx)<∞. Multiplying |m(x)| in both sides of (2.39) and (2.41), we get (v). Now we consider the case when f,g satisfy (2.10). In view of Lemma 2.2, the solutions of (2.10) are given by (i), (iii), or contained in the case (v). This completes the proof.LetX be a real normed space, and let ψ:X→ℝ be given by ψ(x)=ϵ∥x∥p,p≥0,p≠1, then ψ satisfies the conditions assumed in Theorem 2.5. In view of (2.19) and (2.20), we have
(2.42)Φ1(x)=2ϵ‖x‖p2-2p
if 0<p<1,
(2.43)Φ2(x)=2ϵ‖x‖p2p-2
if p>1. Thus, as a direct consequence of Theorem 2.5, we have the following.Corollary 2.6.
Letf,g:X→ℂ satisfy the inequality
(2.44)|f(x-y)-f(x)g(y)+g(x)f(y)|≤ϵ‖y‖p,p≠1,p≥0,
for all x,y∈X, then (f,g) satisfies one of the following:(i)
f=0, g is arbitrary,(ii)
f and g are bounded functions,(iii)
f(x)=λ1(m(x)-m(-x)) and g(x)=λ2f(x)+(1/2)(m(x)+m(-x)) for some λ1,λ2∈ℂ, where m is an exponential function,(iv)
there existλ∈ℂ and a bounded exponential function m such that
(2.45)g(x)=λf(x)+m(x),
for all x∈X, and f satisfies the condition; there exists d≥0 satisfying
(2.46)|f(x)|≤23(ψ(-x)+d),
for all x∈X,(v)
there existλ∈ℂ and a bounded exponential function m satisfying m2≡1 such that
(2.47)g(x)=λf(x)+m(x),
for all x∈X, and f satisfies one of the following conditions; there exists an additive function a:X→ℂ such that
(2.48)|f(x)-(a(x)+f(0))m(x)|≤2ϵ‖x‖p|2-2p|,
for all x∈X.Now we prove the stability of (1.2). For the proof, we need the following.Lemma 2.7.
Letf,g:G→ℂ satisfy the inequality
(2.49)|g(x-y)-g(x)g(y)-f(x)f(y)|≤ψ(y),
for all x,y∈G, then either there exist λ1,λ2∈ℂ, not both zero, and M>0 such that
(2.50)|λ1f(x)-λ2g(x)|≤M,
or else
(2.51)g(x-y)-g(x)g(y)-f(x)f(y)=0,
for all x,y∈G.Proof.
Suppose thatλ1f(x)-λ2g(x) is bounded only when λ1=λ2=0, and let
(2.52)l(x,y)=g(x+y)-g(x)g(-y)-f(x)f(-y).
Since we may assume that f is nonconstant, we can choose y1 satisfying f(-y1)≠0. Now it can be easily get that
(2.53)f(x)=λ0g(x)+λ1g(x+y1)-λ1l(x,y1),
where λ0=-g(-y1)/f(-y1) and λ1=1/f(-y1). From the definition of l and the use of (2.53), we have the following two equations:
(2.54)g((x+y)+z)=g(x+y)g(-z)+f(x+y)f(-z)+l(x+y,z)=(g(x)g(-y)+f(x)f(-y)+l(x,y))g(-z)+(λ0g(x+y)+λ1g(x+y+y1)-λ1l(x+y,y1))f(-z)+l(x+y,z)=(g(x)g(-y)+f(x)f(-y)+l(x,y))g(-z)+λ0(g(x)g(-y)+f(x)f(-y)+l(x,y))f(-z)+λ1(g(x)g(-y-y1)+f(x)f(-y-y1)+l(x,y+y1))f(-z)-λ1l(x+y,y1)f(-z)+l(x+y,z),(2.55)g(x+(y+z))=g(x)g(-y-z)+f(x)f(-y-z)+l(x,y+z).
Equating (2.54) and (2.55), we have
(2.56)g(x)(g(-y)g(-z)+λ0g(-y)f(-z)+λ1g(-y-y1)f(-z)-g(-y-z))+f(x)(f(-y)g(-z)+λ0f(-y)f(-z)+λ1f(-y-y1)f(-z)-f(-y-z))=-l(x,y)g(-z)-λ0l(x,y)f(-z)-λ1l(x,y+y1)f(-z)+λ1l(x+y,y1)f(-z)-l(x+y,z)+l(x,y+z).
In (2.56), when y,z are fixed, the right hand side is bounded, so by our assumption, we have
(2.57)l(x,y)g(-z)+(λ0l(x,y)+λ1l(x,y+y1)-λ1l(x+y,y1))f(-z)=l(x,y+z)-l(x+y,z).
Also we can write
(2.58)l(x,y+z)-l(x+y,z)=g(x+y+z)-g(x)g(y+z)-f(x)f(y+z)-g(x+y+z)+g(x+y)g(z)+f(x+y)f(z)=l(y+z,x)-l(z,x+y)≤ψ(-x)+ψ(-x-y).
Thus, if we fix x,y in (2.57), the right hand side of (2.57) is bounded. By our assumption, we have l(x,y)≡0. This completes the proof.Theorem 2.8.
Letf,g:G→ℂ satisfy the inequality
(2.59)|g(x-y)-g(x)g(y)-f(x)f(y)|≤ψ(y),
for all x,y∈G, then (f,g) satisfies one of the following:(i)
f and g are bounded functions,(ii)
g(x)=(1/2)(m(x)+m(-x)) and f(x)=(1/2)(m(x)-m(-x)), where m is an exponential function,(iii)
f=±i(g-m) for a bounded exponential function m, and g satisfies
(2.60)|g(x)-12(g(0)m(-x)+m(x))|≤12ψ(x),
for all x∈G. In particular if ψ(0)=0, one has g(0)=1,f(0)=0.Proof.
In view of Lemma2.7, we first consider the case when f,g satisfy (2.51). If f is bounded, then in view of the inequality (2.59), for each y, g(x+y)-g(x)g(-y), is also bounded. It follows from Lemma 2.1 that g is bounded or a nonzero exponential function. If g is bounded, the case (i) follows. If g is a nonzero exponential function, from (2.59), using the triangle inequality, we have for some d≥0,
(2.61)|g(x)(g(-y)-g(y))|≤ψ(y)+d,
for all x,y∈G. Thus, it follows that
(2.62)g(y)=g(-y),
for all y∈G, or else g is bounded, and equality (2.62) implies g2≡1, which gives the case (i).
Iff is unbounded, then in view of (2.59), g is also unbounded, and hence, λ1λ2≠0 and
(2.63)f(x)=λg(x)+r(x),
for some λ≠0 and a bounded function r. Putting (2.63) in (2.59), replacing y by -y, and using the triangle inequality, we have
(2.64)|g(x+y)-g(x)((λ2+1)g(-y)+λr(-y))|≤|(λg(-y)+r(-y))r(x)|+ψ(-y).
From Lemma 2.1, we have
(2.65)(λ2+1)g(y)+λr(y)=m(y),
for some exponential function m. If λ2≠-1, we have
(2.66)f(x)=λm(x)+r(x)λ2+1,g(x)=m(x)-λr(x)λ2+1.
Putting (2.66) in (2.59), multiplying |λ2+1| in the result, and using the triangle inequality, we have for some d≥0,
(2.67)|m(x)(m(-y)-m(y))|≤|λ2+1|ψ(y)+d,
for all x,y∈G. Since m is unbounded, we have
(2.68)m(y)=m(-y),
for all y∈G, which implies m2≡1, contradicting to the fact that m is unbounded. Thus, it follows that λ2=-1, and we have
(2.69)f=±i(g-m),
where m is a bounded exponential function. Putting (2.69) in (2.59), we have
(2.70)|g(x-y)-g(x)m(y)-g(y)m(x)+m(x)m(y)|≤ψ(y),
for all x,y∈G. Replacing y by x in (2.70) and dividing the result by 2m(x), we have
(2.71)|g(x)-12(g(0)m(-x)+m(x))|≤12ψ(x),
for all x∈G. From (2.69), (2.71), we get (iii). Now we consider the case when f,g satisfy (2.51). In view of Lemma 2.2, the solutions of (2.51) are contained in (i) or given by (ii). Furthermore, if ψ(0)=0, then putting x=y=0 in (2.70), we have g(0)=1, and from (2.69), we also have f(0)=0. This completes the proof.In particular, iff,g:ℝn→ℂ is a continuous function and ψ(x)=ϵ|x|p,p>0,p≠1, then Theorem 2.8 is reduced as follows.Corollary 2.9.
Letf,g:ℝn→ℂ be a continuous function satisfying (2.59) for ψ(x)=ϵ|x|p, then (f,g) satisfies one of the following:(i)
f and g are bounded functions,(ii)
g(x)=cos(c·x) and f(x)=sin(c·x) for some c∈ℂn,(iii)
there existsa∈ℝn such that(2.72)|f(x)-sin⁡(a⋅x)|≤ϵ2|x|p,|g(x)-cos⁡(a⋅x)|≤ϵ2|x|p,
for all x∈ℝn.



---

*Source: 610714-2012-03-15.xml*