AAA Abstract and Applied Analysis 1687-0409 1085-3375 Hindawi Publishing Corporation 275915 10.1155/2013/275915 275915 Research Article On the Stability of Trigonometric Functional Equations in Distributions and Hyperfunctions Chung Jaeyoung 1 Chang Jeongwook 2 Kılıçman Adem 1 Department of Mathematics Kunsan National University Kunsan 573-701 Republic of Korea kunsan.ac.kr 2 Department of Mathematics Education Dankook University Yongin 448-701 Republic of Korea dankook.ac.kr 2013 8 5 2013 2013 06 02 2013 10 04 2013 2013 Copyright © 2013 Jaeyoung Chung and Jeongwook Chang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We consider the Hyers-Ulam stability for a class of trigonometric functional equations in the spaces of generalized functions such as Schwartz distributions and Gelfand hyperfunctions.

1. Introduction

Hyers-Ulam stability problems of functional equations go back to 1940 when Ulam proposed the following question .

Let f be a mapping from a group G1 to a metric group G2 with metric d(·,·) such that (1)d(f(xy),f(x)f(y))ϵ.

Then does there exist a group homomorphism h and δϵ>0 such that (2)d(f(x),h(x))δϵ for all xG1?

This problem was solved affirmatively by Hyers  under the assumption that G2 is a Banach space. After the result of Hyers, Aoki  and Bourgin [4, 5] treated with this problem; however, there were no other results on this problem until 1978 when Rassias  treated again with the inequality of Aoki . Generalizing Hyers' result, he proved that if a mapping f:XY between two Banach spaces satisfies (3)f(x+y)-f(x)-f(y)Φ(x,y),forx,yX with Φ(x,y)=ϵ(xp+yp)  (ϵ0,0p<1), then there exists a unique additive function A:XY such that f(x)-A(x)2ϵ|x|p/(2-2p) for all xX. In 1951 Bourgin [4, 5] stated that if Φ is symmetric in x and y with j=1Φ(2jx,2jx)/2j< for each xX, then there exists a unique additive function A:XY such that f(x)-A(x)j=1Φ(2jx,2jx)/2j for all xX. Unfortunately, there was no use of these results until 1978 when Rassias  treated with the inequality of Aoki . Following Rassias’ result, a great number of papers on the subject have been published concerning numerous functional equations in various directions [610, 1025]. In 1990 Székelyhidi  has developed his idea of using invariant subspaces of functions defined on a group or semigroup in connection with stability questions for the sine and cosine functional equations. We refer the reader to [9, 10, 18, 19, 25] for Hyers-Ulam stability of functional equations of trigonometric type. In this paper, following the method of Székelyhidi  we consider a distributional analogue of the Hyers-Ulam stability problem of the trigonometric functional inequalities (4)|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 f,g:n and ψ:n[0,) is a continuous function. As a distributional version of the inequalities (4), we consider the inequalities for the generalized functions u,v𝒢(n)  (resp.,𝒮(n)), (5)u(x-y)-uxvy+vxuyψ(y),v(x-y)-vxvy-uxuyψ(y), where and denote the pullback and the tensor product of generalized functions, respectively, and ψ:n[0,) denotes a continuous infraexponential function of order 2 (resp., a function of polynomial growth). For the proof we employ the tensor product Et(x)Es(y) of n-dimensional heat kernel (6)Et(x)=(4πt)-n/2exp(-|x|24t),xn,t>0.

For the first step, convolving Et(x)Es(y) in both sides of (5) we convert (5) to the Hyers-Ulam stability problems of trigonometric-hyperbolic type functional inequalities, respectively, (7)|U(x-y,t+s)-U(x,t)V(y,s)+V(x,t)U(y,s)|Ψ(y,s),|V(x-y,t+s)-V(x,t)V(y,s)-U(x,t)U(y,s)|Ψ(y,s), for all x,yn, t,s>0, where U,V are the Gauss transforms of u,v, respectively, given by (8)U(x,t)=u*Et(x)=uy,Et(x-y),(9)V(x,t)=v*Et(x), which are solutions of the heat equation, and (10)Ψ(y,s)=ψ(η)Es(η-y)dη=(ψ*Es)(y).

For the second step, using similar idea of Székelyhidi  we prove the Hyers-Ulam stabilities of inequalities (7). For the final step, taking initial values as t0+ for the results we arrive at our results.

2. Generalized Functions

We first introduce the spaces 𝒮 of Schwartz tempered distributions and 𝒢 of Gelfand hyperfunctions (see  for more details of these spaces). We use the notations: |α|=α1++αn, α!=α1!αn!, |x|=x12++xn2, xα=x1α1xnαn, and α=1α1nαn, for x=(x1,,xn)n, α=(α1,,αn)0n, where 0 is the set of nonnegative integers and j=/xj.

Definition 1 (see [<xref ref-type="bibr" rid="B29">29</xref>]).

One denotes by 𝒮 or 𝒮(n) the Schwartz space of all infinitely differentiable functions φ in n such that (11)φα,β=supx|xαβφ(x)|< for all α, β0n, equipped with the topology defined by the seminorms ·α,β. The elements of 𝒮 are called rapidly decreasing functions, and the elements of the dual space 𝒮 are called tempered distributions.

Definition 2 (see [<xref ref-type="bibr" rid="B13">26</xref>]).

One denotes by 𝒢 or 𝒢(n) the Gelfand space of all infinitely differentiable functions φ in n such that (12)φh,k=supxn,α,β0n|xαβφ(x)|h|α|k|β|α!1/2β!1/2< for some h,k>0. One says that φj0 as j if φjh,k0 as j for some h,k, and one denotes by 𝒢 the dual space of 𝒢 and calls its elements Gelfand hyperfunctions.

It is well known that the following topological inclusions hold: (13)𝒢𝒮,𝒮𝒢.

It is known that the space 𝒢(n) consists of all infinitely differentiable functions φ(x) on n which can be extended to an entire function on n satisfying (14)|φ(x+iy)|Cexp(-a|x|2+b|y|2),x,yn for some a, b, and C>0 (see ).

By virtue of Theorem 6.12 of [27, p. 134] we have the following.

Definition 3.

Let uj𝒢(nj) for j=1,2, with n1n2, and let λ:n1n2 be a smooth function such that, for each xn1, the Jacobian matrix λ(x) of λ at x has rank n2. Then there exists a unique continuous linear map λ*:𝒢(n2)𝒢(n1) such that λ*u=uλ when u is a continuous function. One calls λ*u the pullback of u by λ which is often denoted by uλ.

In particular, let λ:2nn be defined by λ(x,y)=x-y, x,yn. Then in view of the proof of Theorem  6.12 of [27, p. 134] we have (15)uλ,φ(x,y)=u,φ(x-y,y)dy.

Definition 4.

Let ux𝒢(n1), uy𝒢(n2). Then the tensor product uxuy of ux and uy, defined by (16)uxuy,φ(x,y)=ux,uy,φ(x,y) for φ(x,y)𝒢(n1×n2), belongs to 𝒢(n1×n2).

For more details of pullback and tensor product of distributions we refer the reader to Chapter V-VI of .

3. Main Theorems

Let f be a Lebesgue measurable function on n. Then f is said to be an infraexponential function of order 2 (resp., a function of polynomial growth) if for every ϵ>0 there exists Cϵ>0 (resp., there exist positive constants C,N, and d) such that (17)|f(x)|Cϵeϵ|x|2  [resp.  C|x|N+d] for all xn. It is easy to see that every infraexponential function f of order 2 (resp., every function of polynomial growth) defines an element of 𝒢(n)(resp.,𝒮(n)) via the correspondence (18)f,φ=f(x)φ(x)dx for φ𝒢(n)(resp.𝒮(n)).

Let u,v𝒢(n)  (resp.,𝒮(n)). We prove the stability of the following functional inequalities: (19)u(x-y)-uxvy+vxuyψ(y),(20)v(x-y)-vxvy-uxuyψ(y), where and denote the pullback and the tensor product of generalized functions, respectively, ψ:n[0,) denotes a continuous infraexponential functional of order 2 (resp. a continuous function of polynomial growth) with ψ(0)=0, and ·ψ means that |·,φ|ψφL1 for all φ𝒢(n)(resp.,𝒮(n)).

In view of (14) it is easy to see that the n-dimensional heat kernel (21)Et(x)=(4πt)-n/2exp(-|x|24t),t>0, belongs to the Gelfand space 𝒢(n) for each t>0. Thus the convolution (u*Et)(x):=uy,Et(x-y) is well defined for all u𝒢(n). It is well known that U(x,t)=(u*Et)(x) is a smooth solution of the heat equation (/t-Δ)U=0 in {(x,t):xn,t>0} and (u*Et)(x)u as t0+ in the sense of generalized functions that is, for every φ𝒢(n), (22)u,φ=limt0+(u*Et)(x)φ(x)dx.

We call (u*Et)(x) the Gauss transform of u.

A function A from a semigroup S,+ to the field of complex numbers is said to be an additive function provided that A(x+y)=A(x)+A(y), and m:S is said to be an exponential function provided that m(x+y)=m(x)m(y).

For the proof of stabilities of (19) and (20) we need the following.

Lemma 5 (see [<xref ref-type="bibr" rid="B16">15</xref>]).

Let S be a semigroup and the field of complex numbers. Assume that f,g:S satisfy the inequality; for each yS there exists a positive constant My such that (23)|f(x+y)-f(x)g(y)|My for all xS. Then either f is a bounded function or g is an exponential function.

Proof.

Suppose that g is not exponential. Then there are y,zS such that g(y+z)g(y)g(z). Now we have (24)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 (25)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 (23) the right hand side of (25) is bounded as a function of x. Consequently, f is bounded.

Lemma 6 (see [<xref ref-type="bibr" rid="B34">30</xref>, p. 122]).

Let f(x,t) be a solution of the heat equation. Then f(x,t) satisfies (26)|f(x,t)|M,xn,t(0,1) for some M>0, if and only if (27)f(x,t)=(f0*Et)(x)=f0(y)Et(x-y)dy for some bounded measurable function f0 defined in n. In particular, f(x,t)f0(x) in 𝒢(n) as t0+.

We discuss the solutions of the corresponding trigonometric functional equations (28)u(x-y)-uxvy+vxuy=0,(29)v(x-y)-vxvy-uxuy=0, in the space 𝒢 of Gelfand hyperfunctions. As a consequence of the results [8, 31, 32] we have the following.

Lemma 7.

The solutions u,v𝒢(n) of (28) and (29) are equal, respectively, to the continuous solutions f,g:n of corresponding classical functional equations (30)f(x-y)-f(x)g(y)+g(x)f(y)=0,(31)g(x-y)-g(x)g(y)-f(x)f(y)=0.

The continuous solutions (f,g) of the functional equation (30) are given by one of the following:

f=0 and g is arbitrary,

f(x)=c1·x, g(x)=1+c2·x for some c1, c2n,

f(x)=λ1sin(c·x) and g(x)=cos(c·x)+λ2sin(c·x) for some λ1, λ2, cn.

Also, the continuous solutions (f,g) of the functional equation (31) are given by one of the following:

g(x)=λ and f(x)=±λ-λ2 for some λ,

g(x)=cos(c·x) and f(x)=sin(c·x) for some cn.

For the proof of the stability of (19) we need the followings.

Lemma 8.

Let G be an Abelian group and let U,V:G×(0,) satisfy the inequality; there exists a nonnegative function Ψ:G×(0,) such that (32)|U(x-y,t+s)-U(x,t)V(y,s)+V(x,t)U(y,s)|Ψ(y,s) for all x, yG, t, s>0. Then either there exist λ1, λ2, not both are zero, and M>0 such that (33)|λ1U(x,t)-λ2V(x,t)|M, or else (34)U(x-y,t+s)-U(x,t)V(y,s)+V(x,t)U(y,s)=0 for all x, yG, t, s>0.

Proof.

Suppose that inequality (33) holds only when λ1=λ2=0. Let (35)K(x,y,t,s)=U(x+y,t+s)-U(x,t)V(-y,s)+V(x,t)U(-y,s), and choose y1 and s1 satisfying U(-y1,s1)0. Now it can be easily calculated that (36)V(x,t)=λ0U(x,t)+λ1U(x+y1,t+s1)-λ1K(x,y1,t,s1), where λ0=V(-y1,s1)/U(-y1,s1) and λ1=-1/U(-y1,s1). By (35) we have (37)U(x+(y+z),t+(s+r))=U(x,t)V(-y-z,s+r)-V(x,t)U(-y-z,s+r)+K(x,y+z,t,s+r).

Also by (35) and (36) we have (38)U((x+y)+z,(t+s)+r)=U(x+y,t+s)V(-z,r)-V(x+y,t+s)U(-z,r)+K(x+y,z,t+s,r)=(U(x,t)V(-y,s)-V(x,t)U(-y,s)+K(x,y,t,s))V(-z,r)-(λ0U(x+y,t+s)+λ1U(x+y+y1,t+s+s1)-λ1K(x+y,y1,t+s,s1))U(-z,r)+K(x+y,z,t+s,r)=(U(x,t)V(-y,s)-V(x,t)U(-y,s)+K(x,y,t,s))V(-z,r)-λ0(U(x,t)V(-y,s)-V(x,t)U(-y,s)+K(x,y,t,s))U(-z,r)-λ1(U(x,t)V(-y-y1,s+s1)-V(x,t)U(-y-y1,s+s1)+K(x,y+y1,t,s+s1))U(-z,r)+λ1K(x+y,y1,t+s,s1)U(-z,r)+K(x+y,z,t+s,r).

From (37) and (38) we have (39)=(V(-y,s)V(-z,r)-λ0V(-y,s)U(-z,r)=-λ1V(-y-y1,s+s1)U(-z,r)=-V(-y-z,s+r))U(x,t)=+(-U(-y,s)V(-z,r)+λ0U(-y,s)U(-z,r)=+λ1U(-y-y1,s+s1)U(-z,r)=+U(-y-z,s+r))V(x,t)=-K(x,y,t,s)V(-z,r)+λ0K(x,y,t,s)U(-z,r)+λ1K(x,y+y1,t,s+s1)U(-z,r)-λ1K(x+y,y1,t+s,s1)U(-z,r)-K(x+y,z,t+s,r)+K(x,y+z,t,s+r).

Since K(x,y,t,s) is bounded by Ψ(-y,s), if we fix y, z, r, and s, the right hand side of (39) is bounded by a constant M, where (40)M=Ψ(-y,s)|V(-z,r)|+Ψ(-y,s)|λ0U(-z,r)|+Ψ(-y-y1,s+s1)|λ1U(-z,r)|+Ψ(-y1,s1)|λ1U(-z,r)|+Ψ(-z,r)+Ψ(-y-z,r+s).

So by our assumption, the left hand side of (39) vanishes, so is the right hand side. Thus we have (41)(-λ0K(x,y,t,s)-λ1K(x,y+y1,t,s+s1)+λ1K(x+y,y1,t+s,s1))U(-z,r)+K(x,y,t,s)V(-z,r)=K(x,y+z,t,s+r)-K(x+y,z,t+s,r).

Now by the definition of K we have (42)K(x+y,z,t+s,r)-K(x,y+z,t,s+r)=U(x+y+z,t+s+r)-U(x+y,t+s)V(-z,r)+V(x+y,t+s)U(-z,r)-U(x+y+z,t+s+r)+U(x,t)V(-y-z,s+r)-V(x,t)U(-y-z,s+r)=U(-y-z-x,s+r+t)-U(-y-z,s+r)V(x,t)+V(-y-z,s+r)U(x,t)-U(-z-x-y,r+t+s)+U(-z,r)V(x+y,t+s)-V(-z,r)U(x+y,t+s)=K(-y-z,-x,s+r,t)-K(-z,-x-y,r,t+s). Hence the left hand side of (41) is bounded by Ψ(x,t)+Ψ(x+y,t+s). So if we fix x, y, t, and s in (41), the left hand side of (41) is a bounded function of z and r. Thus K(x,y,t,s)0 by our assumption. This completes the proof.

In the following lemma we assume that Ψ:n×(0,)[0,) is a continuous function such that (43)ψ(x):=limt0+Ψ(x,t) exists and satisfies the conditions ψ(0)=0 and (44)Φ1(x):=k=02-kψ(-2kx)< or (45)Φ2(x):=k=12kψ(-2-kx)<.

Lemma 9.

Let U,V:n×(0,) be continuous functions satisfying (46)|U(x-y,t+s)-U(x,t)V(y,s)+V(x,t)U(y,s)|Ψ(y,s) for all x,yn, t,s>0, and there exist λ1,λ2, not both are zero, and M>0 such that (47)|λ1U(x,t)-λ2V(x,t)|M.

Then (U,V) satisfies one of the followings:

U=0, V is arbitrary,

U and V are bounded functions,

V(x,t)=λU(x,t)+eic·x-bt for some λn,c(0)n, and b, and f(x):=limt0+U(x,t) is a continuous function; in particular, there exists δ:(0,)[0,) with δ(t)0 as t0+ such that (48)|U(x,t)-f(x)e-bt|δ(t)

for all xn, t>0, and satisfies the condition; there exists d0 satisfying (49)|f(x)|ψ(-x)+d

for all xn,

V(x,t)=λU(x,t)+e-bt for some λn, b, and f(x):=limt0+U(x,t) is a continuous function; in particular, there exists δ:(0,)[0,) with δ(t)0 as t0+ such that (50)|U(x,t)-f(x)e-bt|δ(t)

for all xn, t>0, and satisfies one of the following conditions; there exists a1n such that (51)|f(x)-a1·x|Φ1(x)

for all xn, or there exists a2n such that (52)|f(x)-a2·x|Φ2(x)

for all xn.

Proof.

If U=0, V is arbitrary which is case (i). If U is a nontrivial bounded function, in view of (46) V is also bounded which gives case (ii). If U is unbounded, it follows from (47) that λ20 and (53)V(x,t)=λU(x,t)+R(x,t) for some λ and a bounded function R. Putting (53) in (46) we have (54)|U(x-y,t+s)-U(x,t)R(y,s)+R(x,t)U(y,s)|Ψ(y,s) for all x,yn, t,s>0. Replacing y by -y and using the triangle inequality, we have, for some C>0, (55)|U(x+y,t+s)-U(x,t)R(-y,s)|C|U(-y,s)|+Ψ(-y,s) for all x,yn, t,s>0. By Lemma 5, R(-y,s) is an exponential function. If R=0, putting y=0,s0+ in (54) we have (56)|U(x,t)|ψ(0)=0. Thus we have R0 since U is unbounded. Given the continuity of U and V we have (57)R(x,t)=eic·x-bt for some cn, b with b0. Putting y=0 and s=1 in (54), dividing R(0,1), and using the triangle inequality we have (58)|U(x,t)||R(0,1)|-1(|U(x,t+1)|+C|U(0,1)|+Ψ(0,1)) for all xn, t>0.

From (58) and the continuity of U it is easy to see that (59)limsupt0+U(x,t):=f(x) exists. Putting x=y=0 and replacing s and t by t/2 in (54) we have (60)|U(0,t)|Ψ(0,t2) for all t>0.

Fixing x, putting y=0 letting t0+ so that U(x,t)f(x) in (54), and using the triangle inequality and (60) we have (61)|U(x,s)-f(x)e-bs|Ψ(0,s2)+Ψ(0,s):=δ(s) for all xn, s>0. Letting s0+ in (61) we have (62)lims0+U(x,s)=f(x) for all xn. From (61) the continuity of f can be checked by a usual calculus. Letting t0+ in (60) we see that f(0)=0. Letting t,s0+ in (54) we have (63)|f(x-y)-f(x)eic·y+eic·xf(y)|ψ(y) for all x,yn. Putting x=0 in (63) and replacing y by -y we have (64)|f(-y)+f(y)|ψ(-y) for all yn.

Replacing y by -y and using (64) and the triangle inequality we have (65)|f(x+y)-f(x)e-ic·y-eic·xf(y)|2ψ(-y) for all x,yn. Now we divide (65) into two cases: c=0 and c0. First we consider the case c0. Replacing x by y and y by x in (65) we have (66)|f(x+y)-f(y)e-ic·x-eic·yf(x)|2ψ(-x) for all x,yn. From (65) and (66), using the triangle inequality and dividing |eic·y-e-ic·y| we have (67)|f(x)|2(ψ(-x)+ψ(-y)+|f(y)|)|eic·y-e-ic·y| for all x,yn such that c·y0. Choosing y0n so that c·y0=π/2 and putting y=y0 in (67) we have (68)|f(x)|ψ(-x)+d, where d=ψ(π/2)+|f(π/2)|, which gives (iii). Now we consider the case c=0. It follows from (65) that (69)|f(x+y)-f(x)-f(y)|2ψ(-y) for all x,yn. By the well-known results in , there exists a unique additive function A1(x) given by (70)A1(x)=limn2-nf(2nx) such that (71)|f(x)-A1(x)|Φ1(x) if Φ1(x):=k=02-kψ(-2kx)<, and there exists a unique additive function A2(x) given by (72)A2(x)=limn2nf(2-nx) such that (73)|f(x)-A2(x)|Φ2(x) if Φ2(x):=k=02kψ(-2-kx)<. Now by the continuity of U and inequality (61), it is easy to see that f is continuous. In view of (70) and (72), Aj(x), j=1,2, are Lebesgue measurable functions. Thus there exist a1,a2n such that A1(x)=a1·x and A2(x)=a2·x for all xn, which gives (iv). This completes the proof.

In the following we assume that ψ satisfies (44) or (45).

Theorem 10.

Let u,v𝒢 satisfy (19). Then (u,v) satisfies one of the followings:

u=0, and v is arbitrary,

u and v are bounded measurable functions,

v(x)=λu(x)+eic·x for some λ, c(0)n, where u is a continuous function satisfying the condition; there exists d0(74)|u(x)|ψ(-x)+d

for all xn,

v(x)=λu(x)+1 for some λ, where u is a continuous function satisfying one of the following conditions; there exists a1n such that (75)|u(x)-a1·x|Φ1(x)

for all xn, or there exists a2n such that (76)|u(x)-a2·x|Φ2(x)

for all xn,

u=λsin(c·x), v=cos(c·x)+λsin(c·x), for some cn, λ.

Proof.

Convolving in (19) the tensor product Et(x)Es(y) of n-dimensional heat kernels in both sides of inequality (19) we have (77)[u(ξ-η)*(Et(ξ)Es(η))](x,y)=uξ,Et(x-ξ-η)Es(y-η)dη=uξ,(Et*Es)(x-y-ξ)=uξ,Et+s(x-y-ξ)=U(x-y,t+s).

Similarly we have (78)[(uv)*(Et(ξ)Es(η))](x,y)=U(x,t)V(y,s),[(vu)*(Et(ξ)Es(η))](x,y)=V(x,t)U(y,s), where U,V are the Gauss transforms of u,v, respectively. Thus we have the following inequality: (79)|U(x-y,t+s)-U(x,t)V(y,s)+V(x,t)U(y,s)|Ψ(y,s) for all x, yn, t,s>0, where (80)Ψ(y,s)=ψ(η)Et(x-ξ)Es(y-η)dξdη=ψ(η)Es(η-y)dη=(ψ*Es)(y).

By Lemma 8 there exist λ1,λ2, not both are zero, and M>0 such that (81)|λ1U(x,t)-λ2V(x,t)|M, or else U,V satisfy (82)U(x-y,t+s)-U(x,t)V(y,s)+V(x,t)U(y,s)=0 for all x,yn, t,s>0. Assume that (81) holds. Applying Lemma 9, case (i) follows from (i) of Lemma 9. Using (ii) of Lemma 9, it follows from Lemma 7 the initial values u,v of U(x,t),V(x,t) as t0+ are bounded measurable functions, respectively, which gives (ii). For case (iii), it follows from (50) that, for all φ𝒢(n), (83)|u,φ-f,φ|=|limt0+U(x,t)φ(x)dx-f(x)φ(x)dx|=|limt0+(U(x,t)-f(x)e-bt)φ(x)dx|limt0+|U(x,t)-f(x)e-bt||φ(x)|dxlimt0+δ(t)|φ(x)|dx=0. Thus we have u=f in 𝒢(n). Letting t0+ in (iii) of Lemma 9 we get case (iii). Finally we assume that (82) holds. Letting t,s0+ in (82) we have (84)u(x-y)-uxvy+vxuy=0. By Lemma 6 the solutions of (84) satisfy (i), (iv), or (v). This completes the proof.

Let ψ(x)=ϵ|x|p,p>0. Then ψ satisfies the conditions assumed in Theorem 10. In view of (44) and (45) we have (85)Φ1(x)=2ϵ|x|p2-2p if 0<p<1, and (86)Φ2(x)=2ϵ|x|p2p-2 if p>1. Thus as a direct consequence of Theorem 10 we have the following.

Corollary 11.

Let 0<p<1 or p>1. Suppose that u,v𝒢 satisfy (87)u(x-y)-uxvy+vxuyϵ|y|p. Then (u,v) satisfies one of the followings:

u=0, and v is arbitrary,

u and v are bounded measurable functions,

v(x)=λu(x)+eic·x for some λ, c(0)n, where u is a continuous function satisfying the condition; there exists d0(88)|u(x)|ϵ|x|p+d

for all xn,

v(x)=λu(x)+1 for some λ, where u is a continuous function satisfying the conditions; there exists an such that (89)|u(x)-a·x|2ϵ|x|p|2p-2|

for all xn,

u=λsin(c·x), v=cos(c·x)+λsin(c·x), for some cn, λ.

Now we prove the stability of (20). For the proof we need the following.

Lemma 12.

Let U,V:G×(0,) satisfy the inequality; there exists a Ψ:G×(0,)[0,) such that (90)|V(x-y,t+s)-V(x,t)V(y,s)-U(x,t)U(y,s)|Ψ(y,s) for all x,yn, t,s>0. Then either there exist λ1,λ2, not both are zero, and M>0 such that (91)|λ1U(x,t)-λ2V(x,t)|M, or else (92)V(x-y,t+s)-V(x,t)V(y,s)-U(x,t)U(y,s)=0 for all x,yG, t,s>0.

Proof.

As in Lemma 9, suppose that λ1U(x,t)-λ2V(x,t) is bounded only when λ1=λ2=0, and let (93)L(x,y,t,s)=V(x+y,t+s)-V(x,t)V(-y,s)-U(x,t)U(-y,s).

Since we may assume that U is nonconstant, we can choose y1 and s1 satisfying U(-y1,s1)0. Now it can be easily got that (94)U(x,t)=λ0V(x,t)+λ1V(x+y1,t+s1)-λ1L(x,y1,t,s1), where λ0=-V(-y1,s1)/U(-y1,s1) and λ1=1/U(-y1,s1). From the definition of L and the use of (94), we have the following two equations: (95)V((x+y)+z,(t+s)+r)=V(x+y,t+s)V(-z,r)+U(x+y,t+s)U(-z,r)+L(x+y,z,t+s,r)=(V(x,t)V(-y,s)+U(x,t)U(-y,s)+L(x,y,t,s))V(-z,r)+(λ0V(x+y,t+s)+λ1V(x+y+y1,t+s+s1)-λ1L(x+y,y1,t+s,s1))U(-z,r)+L(x+y,z,t+s,r)=(V(x,t)V(-y,s)+U(x,t)U(-y,s)+L(x,y,t,s))V(-z,r)+λ0(V(x,t)V(-y,s)+U(x,t)U(-y,s)+L(x,y,t,s))U(-z,r)+λ1(V(x,t)V(-y-y1,s+s1)+U(x,t)U(-y-y1,s+s1)+L(x,y+y1,t,s+s1))U(-z,r)-λ1L(x+y,y1,t+s,s1)U(-z,r)+L(x+y,z,t+s,r),(96)V(x+(y+z),t+(s+r))=V(x,t)V(-y-z,s+r)+U(x,t)U(-y-z,s+r)+L(x,y+z,t,s+r).

By equating (95) and (96), we have (97)V(x,t)(V(-y,s)V(-z,r)+λ0V(-y,s)U(-z,r)+λ1V(-y-y1,s+s1)U(-z,r)-V(-y-z,s+r))+U(x,t)(U(-y,s)V(-z,r)+λ0U(-y,s)U(-z,r)+λ1U(-y-y1,s+s1)U(-z,r)-U(-y-z,s+r))=-L(x,y,t,s)V(-z,r)-λ0L(x,y,t,s)U(-z,r)-λ1L(x,y+y1,t,s+s1)U(-z,r)+λ1L(x+y,y1,t+s,s1)U(-z,r)-L(x+y,z,t+s,r)+L(x,y+z,t,s+r).

In (97), when y, s, z, and r are fixed, the right hand side is bounded; so by our assumption we have (98)L(x,y,t,s)V(-z,r)+(λ0L(x,y,t,s)+λ1L(x,y+y1,t,s+s1)-λ1L(x+y,y1,t+s,s1))U(-z,r)=L(x,y+z,t,s+r)-L(x+y,z,t+s,r).

Here, we have (99)L(x,y+z,t,s+r)-L(x+y,z,t+s,r)=V(x+y+z,t+s+r)-V(x,t)V(-y-z,s+r)-U(x,t)U(-y-z,s+r)-V(x+y+z,t+s+r)+V(x+y,t+s)V(-z,r)+U(x+y,t+s)U(-z,r)=L(-y-z,-x,s+r,t)-L(-z,-x-y,r,t+s)Ψ(x,t)+Ψ(x+y,t+s).

Considering (98) as a function of z and r for all fixed x,y,t, and s again, we have L(x,y,t,s)0. This completes the proof.

In the following lemma we assume that Ψ:n×(0,)[0,) is a continuous function such that (100)ψ(x):=limt0+Ψ(x,t) exists and satisfies the condition ψ(0)=0.

Lemma 13.

Let U,V:n×(0,) be continuous functions satisfying (101)|V(x-y,t+s)-V(x,t)V(y,s)-U(x,t)U(y,s)|Ψ(y,s) for all x,yn,t,s>0, and there exist λ1,λ2, not both zero, and M>0 such that (102)|λ1U(x,t)-λ2V(x,t)|M.

Then (U,V) satisfies one of the followings:

U and V are bounded functions in n×(0,1),

±iU(x,t)=V(x,t)-eia·x-bt for some an, b, and g(x):=limt0+V(x,t) is a continuous function; in particular, there exists δ:(0,)[0,) with δ(t)0 as t0+ such that (103)|V(x,t)-g(x)e-bt|δ(t)

for all xn,t>0, and g satisfies (104)|g(x)-cos(a·x)|12ψ(x)

for all xn.

Proof.

If U is bounded, then in view of inequality (100), for each y,s, V(x+y,t+s)-V(x,t)V(-y,s) is also bounded. It follows from Lemma 5 that V is (101). If V is bounded, case (i) follows. If V is a nonzero exponential function, then by the continuity of V we have (105)V(x,t)=ec·x+bt for some cn, b. Putting (105) in (101) and using the triangle inequality we have for some d0(106)|ec·xeb(t+s)(e-c·y-ec·y)|Ψ(y,s)+d for all x,yn,t,s>0. In view of (106) it is easy to see that c=ia, an. Thus V(x,t) is bounded on n×(0,1). If U is unbounded; then in view of (101) V is also unbounded, hence λ1λ20 and (107)U(x,t)=λV(x,t)+R(x,t) for some λ0 and a bounded function R. Putting (107) in (101), replacing y by -y, and using the triangle inequality we have (108)|V(x+y,t+s)-V(x,t)((λ2+1)V(-y,s)+λR(-y,s))||(λV(-y,s)+R(-y,s))R(x,t)|+Ψ(-y,s).

From Lemma 5 we have (109)(λ2+1)V(y,s)+λR(y,s)=m(y,s) for some exponential function m. From (107) and (109), m is continuous, and we have (110)m(x,t)=ec·x+bt for some cn, b. If λ2-1, we have (111)U=λm+Rλ2+1,V=m-λRλ2+1.

Putting (111) in (101), multiplying |λ2+1| in the result, and using the triangle inequality we have, for some d0, (112)|m(x,t)(m(-y,s)-m(y,s))||λ2+1|Ψ(y,s)+d for all x,yn, t,s>0. Since m is unbounded, we have (113)m(y,s)=m(-y,s) for all y and s>0. Thus it follows that m(x,t)=ebt and that U,V are bounded in n×(0,1). If λ2=-1, we have (114)U=±i(V-m), where m is a bounded exponential function. Putting (114) in (101) we have (115)|V(x-y,t+s)-V(x,t)m(y,s)-V(y,s)m(x,t)+m(x,t)m(y,s)|Ψ(y,s) for all x,yn,t, s>0. Since m is a bounded continuous function, we have (116)m(x,t)=eia·x-bt for some an,b with b0.

Similarly as in the proof of Lemma 9, by (101) and the continuity of V, it is easy to see that (117)limsupt0+V(x,t):=g(x) exists. Putting x=y=0 in (115), multiplying |ebt| in both sides of the result, and using the triangle inequality we have (118)|V(0,s)-e-bs||ebt|(|V(0,t+s)-V(0,t)e-bs|+Ψ(0,s)) for all t,s>0. Letting s0+ in (118) we have (119)limt0+V(0,t)=1.

Putting y=0, fixing x, letting t0+ in (115) so that V(x,t)g(x), and using the triangle inequality we have (120)|V(x,s)-g(x)e-bs||V(0,s)-e-bs|+Ψ(0,s) for all xn,s>0. Letting s0+ in (120) we have (121)lims0+V(x,s)=g(x) for all xn. The continuity of g follows from (120). Letting t,s0+ in (115) we have (122)|g(x-y)-g(x)eia·y-g(y)eia·x+eia·(x+y)|ψ(y) for all x,yn. Replacing y by x in (122) and dividing the result by 2eia·x we have (123)|g(x)-cos(a·x)|12ψ(x).

From (114), (116), (120) and (123) we get (ii). This completes the proof.

Theorem 14.

Let u,v𝒢 satisfy (20). Then (u,v) satisfies one of the followings:

u and v are bounded measurable functions,

v(x)=cos(a·x)+r(x),±u(x)=sin(a·x)+ir(x) for some an, where r(x) is a continuous function satisfying (124)|r(x)|12ψ(x)

for all xn,

v(x)=cos(c·x) and u(x)=sin(c·x) for some cn.

Proof.

Similarly as in the proof of Theorem 10 convolving in (20) the tensor product Et(x)Es(y) we obtain the inequality (125)|V(x-y,t+s)-V(x,t)V(y,s)-U(x,t)U(y,s)|Ψ(y,s) for all x,yn,t,s>0, where U,V are the Gauss transforms of u,v, respectively, and (126)Ψ(y,s)=ψ(η)Et(x-ξ)Es(y-η)dξdη=ψ(η)Es(η-y)dη=(ψ*Es)(y). By Lemma 12 there exist λ1,λ2, not both zero, and M>0 such that (127)|λ1U(x,t)-λ2V(x,t)|M, or else U,V satisfy (128)V(x-y,t+s)-V(x,t)V(y,s)-U(x,t)U(y,s)=0 for all x,yn,t,s>0.

Firstly we assume that (127) holds. Letting t0+ in (i) of Lemma 13, by Lemma 6, the initial values u,v of U(x,t),V(x,t) as t0+ are bounded measurable functions, respectively, which gives case (i). Using the same approach of the proof of case (iii) of Theorem 10, we have v=g in 𝒢. It follows from (104) that (129)v(x)=cos(a·x)+r(x), where r(x) is a continuous function satisfying (130)|r(x)|12ψ(x) for all xn. Letting t0+ in (ii) of Lemma 13 we have (131)±iu(x)=v(x)-eia·x.

Putting (129) in (131) we have (132)±u(x)=sin(a·x)+ir(x).

Secondly we assume that (128) holds. Letting t,s0+ in (127) we have (133)v(x-y)-vxvy-uxuy=0.

By Lemma 7 the solution of (133) satisfies (i) or (iii). This completes the proof.

Every infraexponential function f of order 2 defines an element of 𝒢(n) via the correspondence (134)f,φ=f(x)φ(x)dx for φ𝒢. Thus as a direct consequence of Corollary 11 and Theorem 14 we have the followings.

Corollary 15.

Let 0<p<1 or p>1. Suppose that f,g are infraexponential functions of order 2 satisfying the inequality (135)|f(x-y)-f(x)g(y)+g(x)f(y)|ϵ|x|p for almost every (x,y)2n. Then (f,g) satisfies one of the following:

f(x)=0, almost everywhere xn, and g is arbitrary,

f and g are bounded in almost everywhere,

f(x)=f0(x), g(x)=λf0(x)+eic·x for almost everywhere xn, where λ, c(0)n, and f0 is a continuous function satisfying the condition; there exists d0(136)|f0(x)|ϵ|x|p+d

for all xn,

f(x)=f0(x), g(x)=λf0(x)+1 for a.e. xn, where λ and f0 is a continuous function satisfying the condition; there exists an such that (137)|f0(x)-a·x|2ϵ|x|p|2p-2|

for all xn,

f(x)=λsin(c·x), g(x)=cos(c·x)+λsin(c·x) for a.e. xn, where cn, λ.

Corollary 16.

Suppose that f, g are infraexponential functions of order 2 satisfying the inequality (138)|g(x-y)-g(x)g(y)-f(x)f(y)|ϵ|y|p for almost every (x,y)2n. Then (f,g) satisfies one of the followings:

f and g are bounded in almost everywhere,

there exists an such that (139)|g(x)-cos(a·x)|12ϵ|x|p,(140)|f(x)±sin(a·x)|12ϵ|x|p

for almost every xn,

g(x)=cos(c·x) and f(x)=sin(c·x) for a.e. xn, where cn.

Remark 17.

Taking the growth of u=ec·x as |x| into account, u𝒮(n) only when c=ia for some an. Thus Theorems 10 and 14 are reduced to the following:

Corollary 18.

Let u,v𝒮 satisfy (19). Then (u,v) satisfies one of the followings:

u=0, and v is arbitrary,

u and v are bounded measurable functions,

v(x)=λu(x)+eic·x for some λ, c(0)n, where u is a continuous function satisfying the condition; there exists d0(141)|u(x)|ψ(-x)+d

for all xn,

v(x)=λu(x)+1 for some λ, where u is a continuous function satisfying one of the following conditions; there exists a1n such that (142)|u(x)-a1·x|Φ1(x)

for all xn, or there exists a2n such that (143)|u(x)-a2·x|Φ2(x)

for all xn.

Corollary 19.

Let u,v𝒮 satisfy (20). Then (u,v) satisfies one of the followings:

u and v are bounded measurable functions,

v(x)=cos(a·x)+r(x), ±u(x)=sin(a·x)+ir(x) for some an, where r(x) is a continuous function satisfying (144)|r(x)|12ψ(x)

for all xn.

Acknowledgments

The first author was supported by Basic Science Research Program through the National Foundation of Korea (NRF) funded by the Ministry of Education Science and Technology (MEST) (no. 2012008507), and the second author was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science, and Technology (2011-0005235).

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