# On the Stability of Trigonometric Functional Equations in Distributions and Hyperfunctions

Authors: Jaeyoung Chung; Jeongwook Chang
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

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.


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## Body

## 1. Introduction

Hyers-Ulam stability problems of functional equations go back to 1940 when Ulam proposed the following question [1].Letf 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 homomorphismh and δϵ>0 such that
(2)d(f(x),h(x))≤δϵ
for all x∈G1?This problem was solved affirmatively by Hyers [2] under the assumption that G2 is a Banach space. After the result of Hyers, Aoki [3] and Bourgin [4, 5] treated with this problem; however, there were no other results on this problem until 1978 when Rassias [6] treated again with the inequality of Aoki [3]. Generalizing Hyers' result, he proved that if a mapping f:X→Y between two Banach spaces satisfies
(3)∥f(x+y)-f(x)-f(y)∥≤Φ(x,y),forx,y∈X
with Φ(x,y)=ϵ(∥x∥p+∥y∥p)(ϵ≥0,0≤p<1), then there exists a unique additive function A:X→Y such that ∥f(x)-A(x)∥≤2ϵ|x|p/(2-2p) for all x∈X. In 1951 Bourgin [4, 5] stated that if Φ is symmetric in ∥x∥ and ∥y∥ with ∑j=1∞‍Φ(2jx,2jx)/2j<∞ for each x∈X, then there exists a unique additive function A:X→Y such that ∥f(x)-A(x)∥≤∑j=1∞‍Φ(2jx,2jx)/2j for all x∈X. Unfortunately, there was no use of these results until 1978 when Rassias [7] treated with the inequality of Aoki [3]. Following Rassias’ result, a great number of papers on the subject have been published concerning numerous functional equations in various directions [6–10, 10–25]. In 1990 Székelyhidi [24] 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 [24] 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:ℝn→ℂ and ψ:ℝn→[0,∞) is a continuous function. As a distributional version of the inequalities (4), we consider the inequalities for the generalized functions u,v∈𝒢′(ℝn)(resp.,𝒮′(ℝn)),
(5)∥u∘(x-y)-ux⊗vy+vx⊗uy∥≤ψ(y),∥v∘(x-y)-vx⊗vy-ux⊗uy∥≤ψ(y),
where ∘ and ⊗ denote the pullback and the tensor product of generalized functions, respectively, and ψ:ℝn→[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),x∈ℝn,t>0.For the first step, convolvingEt(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,y∈ℝn, 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 [24] we prove the Hyers-Ulam stabilities of inequalities (7). For the final step, taking initial values as t→0+ for the results we arrive at our results.


## 2. Generalized Functions

We first introduce the spaces𝒮′ of Schwartz tempered distributions and 𝒢′ of Gelfand hyperfunctions (see [26–29] for more details of these spaces). We use the notations: |α|=α1+⋯+αn, α!=α1!⋯αn!, |x|=x12+⋯+xn2, xα=x1α1⋯xnαn, and ∂α=∂1α1⋯∂nαn, for x=(x1,…,xn)∈ℝn, α=(α1,…,αn)∈ℕ0n, where ℕ0 is the set of nonnegative integers and ∂j=∂/∂xj.Definition 1 (see [29]).
One denotes by𝒮 or 𝒮(ℝn) the Schwartz space of all infinitely differentiable functions φ in ℝn such that
(11)∥φ∥α,β=supx|xα∂βφ(x)|<∞
for all α, β∈ℕ0n, 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 [26]).
One denotes by𝒢 or 𝒢(ℝn) the Gelfand space of all infinitely differentiable functions φ in ℝn such that
(12)∥φ∥h,k=supx∈ℝn,α,β∈ℕ0n|xα∂βφ(x)|h|α|k|β|α!1/2β!1/2<∞
for some h,k>0. One says that φj→0 as j→∞ if ∥φj∥h,k→0 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)𝒢↪𝒮,𝒮′↪𝒢′.It is known that the space𝒢(ℝn) consists of all infinitely differentiable functions φ(x) on ℝn which can be extended to an entire function on ℂn satisfying
(14)|φ(x+iy)|≤Cexp(-a|x|2+b|y|2),x,y∈ℝn
for some a, b, and C>0 (see [26]).By virtue of Theorem 6.12 of [27, p. 134] we have the following.Definition 3.
Letuj∈𝒢′(ℝnj) for j=1,2, with n1≥n2, and let λ:ℝn1→ℝn2 be a smooth function such that, for each x∈ℝn1, the Jacobian matrix ∇λ(x) of λ at x has rank n2. Then there exists a unique continuous linear map λ*:𝒢′(ℝn2)→𝒢′(ℝn1) such that λ*u=u∘λ when u is a continuous function. One calls λ*u the pullback of u by λ which is often denoted by u∘λ.
In particular, letλ:ℝ2n→ℝn be defined by λ(x,y)=x-y, x,y∈ℝn. Then in view of the proof of Theorem  6.12 of [27, p. 134] we have
(15)〈u∘λ,φ(x,y)〉=〈u,∫‍φ(x-y,y)dy〉.Definition 4.
Letux∈𝒢′(ℝn1), uy∈𝒢′(ℝn2). Then the tensor product ux⊗uy of ux and uy, defined by
(16)〈ux⊗uy,φ(x,y)〉=〈ux,〈uy,φ(x,y)〉〉
for φ(x,y)∈𝒢(ℝn1×ℝn2), belongs to 𝒢′(ℝn1×ℝn2).For more details of pullback and tensor product of distributions we refer the reader to Chapter V-VI of [27].


## 3. Main Theorems

Letf be a Lebesgue measurable function on ℝn. 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 x∈ℝn. It is easy to see that every infraexponential function f of order 2 (resp., every function of polynomial growth) defines an element of 𝒢′(ℝn)(resp.,𝒮′(ℝn)) via the correspondence
(18)〈f,φ〉=∫‍f(x)φ(x)dx
for φ∈𝒢(ℝn)(resp.𝒮(ℝn)).Letu,v∈𝒢′(ℝn)(resp.,𝒮′(ℝn)). We prove the stability of the following functional inequalities:
(19)∥u∘(x-y)-ux⊗vy+vx⊗uy∥≤ψ(y),(20)∥v∘(x-y)-vx⊗vy-ux⊗uy∥≤ψ(y),
where ∘ and ⊗ denote the pullback and the tensor product of generalized functions, respectively, ψ:ℝn→[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〉|≤∥ψφ∥L1 for all φ∈𝒢(ℝn)(resp.,𝒮(ℝn)).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(ℝn) for each t>0. Thus the convolution (u*Et)(x):=〈uy,Et(x-y)〉 is well defined for all u∈𝒢′(ℝn). 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):x∈ℝn,t>0} and (u*Et)(x)→u as t→0+ in the sense of generalized functions that is, for every φ∈𝒢(ℝn),
(22)〈u,φ〉=limt→0+∫‍(u*Et)(x)φ(x)dx.We call(u*Et)(x) the Gauss transform of u.A functionA from a semigroup 〈S,+〉 to the field ℂ of complex numbers is said to be an additive function provided that A(x+y)=A(x)+A(y), and m:S→ℂ 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 [15]).
LetS be a semigroup and ℂ the field of complex numbers. Assume that f,g:S→ℂ satisfy the inequality; for each y∈S there exists a positive constant My such that
(23)|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
(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 [30, p. 122]).
Letf(x,t) be a solution of the heat equation. Then f(x,t) satisfies
(26)|f(x,t)|≤M,x∈ℝn,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 ℝn. In particular, f(x,t)→f0(x) in 𝒢′(ℝn) as t→0+.We discuss the solutions of the corresponding trigonometric functional equations(28)u∘(x-y)-ux⊗vy+vx⊗uy=0,(29)v∘(x-y)-vx⊗vy-ux⊗uy=0,
in the space 𝒢′ of Gelfand hyperfunctions. As a consequence of the results [8, 31, 32] we have the following.Lemma 7.
The solutionsu,v∈𝒢′(ℝn) of (28) and (29) are equal, respectively, to the continuous solutions f,g:ℝn→ℂ 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:(i)
f=0 and g is arbitrary,(ii)
f(x)=c1·x, g(x)=1+c2·x for some c1, c2∈ℂn,(iii)
f(x)=λ1sin(c·x) and g(x)=cos(c·x)+λ2sin(c·x) for some λ1, λ2∈ℂ, c∈ℂn.
Also, the continuous solutions(f,g) of the functional equation (31) are given by one of the following:(i)
g(x)=λ and f(x)=±λ-λ2 for some λ∈ℂ,(ii)
g(x)=cos(c·x) and f(x)=sin(c·x) for some c∈ℂn.For the proof of the stability of (19) we need the followings.Lemma 8.
LetG be an Abelian group and let U,V:G×(0,∞)→ℂ satisfy the inequality; there exists a nonnegative function Ψ:G×(0,∞)→ℝ 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, y∈G, t, s>0. Then either there exist λ1, λ2∈ℂ, 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, y∈G, 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).
SinceK(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 ofK 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 thatΨ:ℝn×(0,∞)→[0,∞) is a continuous function such that
(43)ψ(x):=limt→0+Ψ(x,t)
exists and satisfies the conditions ψ(0)=0 and
(44)Φ1(x):=∑k=0∞‍2-kψ(-2kx)<∞
or
(45)Φ2(x):=∑k=1∞‍2kψ(-2-kx)<∞.Lemma 9.
LetU,V:ℝn×(0,∞)→ℂ 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,y∈ℝn, t,s>0, and there exist λ1,λ2∈ℂ, 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:(i)
U=0, V is arbitrary,(ii)
U and V are bounded functions,(iii)
V(x,t)=λU(x,t)+eic·x-bt for some λ∈ℂn,c(≠0)∈ℝn, and b∈ℂ, and f(x):=limt→0+U(x,t) is a continuous function; in particular, there exists δ:(0,∞)→[0,∞) with δ(t)→0 as t→0+ such that
(48)|U(x,t)-f(x)e-bt|≤δ(t)for allx∈ℝn, t>0, and satisfies the condition; there exists d≥0 satisfying
(49)|f(x)|≤ψ(-x)+dfor allx∈ℝn,(iv)
V(x,t)=λU(x,t)+e-bt for some λ∈ℂn, b∈ℂ, and f(x):=limt→0+U(x,t) is a continuous function; in particular, there exists δ:(0,∞)→[0,∞) with δ(t)→0 as t→0+ such that
(50)|U(x,t)-f(x)e-bt|≤δ(t)for allx∈ℝn, t>0, and satisfies one of the following conditions; there exists a1∈ℂn such that
(51)|f(x)-a1·x|≤Φ1(x)for allx∈ℝn, or there exists a2∈ℂn such that
(52)|f(x)-a2·x|≤Φ2(x)for allx∈ℝn.Proof.
IfU=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 λ2≠0 and
(53)V(x,t)=λU(x,t)+R(x,t)
for some λ∈ℂ 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,y∈ℝn, 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,y∈ℝn, t,s>0. By Lemma 5, R(-y,s) is an exponential function. If R=0, putting y=0,s→0+ in (54) we have
(56)|U(x,t)|≤ψ(0)=0.
Thus we have R≠0 since U is unbounded. Given the continuity of U and V we have
(57)R(x,t)=eic·x-bt
for some c∈ℝn, b∈ℂ with ℜb≥0. 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 x∈ℝn, t>0.
From (58) and the continuity of U it is easy to see that
(59)limsupt→0+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.
Fixingx, putting y=0 letting t→0+ 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 x∈ℝn, s>0. Letting s→0+ in (61) we have
(62)lims→0+U(x,s)=f(x)
for all x∈ℝn. From (61) the continuity of f can be checked by a usual calculus. Letting t→0+ in (60) we see that f(0)=0. Letting t,s→0+ in (54) we have
(63)|f(x-y)-f(x)eic·y+eic·xf(y)|≤ψ(y)
for all x,y∈ℝn. Putting x=0 in (63) and replacing y by -y we have
(64)|f(-y)+f(y)|≤ψ(-y)
for all y∈ℝn.
Replacingy 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,y∈ℝn. Now we divide (65) into two cases: c=0 and c≠0. First we consider the case c≠0. 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,y∈ℝn. 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,y∈ℝn such that c·y≠0. Choosing y0∈ℝn 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,y∈ℝn. By the well-known results in [3], there exists a unique additive function A1(x) given by
(70)A1(x)=limn→∞2-nf(2nx)
such that
(71)|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
(72)A2(x)=limn→∞2nf(2-nx)
such that
(73)|f(x)-A2(x)|≤Φ2(x)
if Φ2(x):=∑k=0∞‍2kψ(-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,a2∈ℂn such that A1(x)=a1·x and A2(x)=a2·x for all x∈ℝn, which gives (iv). This completes the proof.In the following we assume thatψ satisfies (44) or (45).Theorem 10.
Letu,v∈𝒢′ satisfy (19). Then (u,v) satisfies one of the followings:(i)
u=0, and v is arbitrary,(ii)
u and v are bounded measurable functions,(iii)
v(x)=λu(x)+eic·x for some λ∈ℂ, c(≠0)∈ℝn, where u is a continuous function satisfying the condition; there exists d≥0(74)|u(x)|≤ψ(-x)+dfor allx∈ℝn,(iv)
v(x)=λu(x)+1 for some λ∈ℂ, where u is a continuous function satisfying one of the following conditions; there exists a1∈ℂn such that
(75)|u(x)-a1·x|≤Φ1(x)for allx∈ℝn, or there exists a2∈ℂn such that
(76)|u(x)-a2·x|≤Φ2(x)for allx∈ℝn,(v)
u=λsin(c·x), v=cos(c·x)+λsin(c·x), for some c∈ℂn, λ∈ℂ.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)[(u⊗v)*(Et(ξ)Es(η))](x,y)=U(x,t)V(y,s),[(v⊗u)*(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, y∈ℝn, 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 Lemma8 there exist λ1,λ2∈ℂ, 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,y∈ℝn, 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 t→0+ are bounded measurable functions, respectively, which gives (ii). For case (iii), it follows from (50) that, for all φ∈𝒢(ℝn),
(83)|〈u,φ〉-〈f,φ〉|=|limt→0+∫‍U(x,t)φ(x)dx-∫‍f(x)φ(x)dx|=|limt→0+∫‍(U(x,t)-f(x)e-bt)φ(x)dx|≤limt→0+∫‍|U(x,t)-f(x)e-bt||φ(x)|dx≤limt→0+δ(t)∫‍|φ(x)|dx=0.
Thus we have u=f in 𝒢′(ℝn). Letting t→0+ in (iii) of Lemma 9 we get case (iii). Finally we assume that (82) holds. Letting t,s→0+ in (82) we have
(84)u∘(x-y)-ux⊗vy+vx⊗uy=0.
By Lemma 6 the solutions of (84) satisfy (i), (iv), or (v). This completes the proof.Letψ(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.
Let0<p<1 or p>1. Suppose that u,v∈𝒢′ satisfy
(87)∥u∘(x-y)-ux⊗vy+vx⊗uy∥≤ϵ|y|p.
Then (u,v) satisfies one of the followings:(i)
u=0, and v is arbitrary,(ii)
u and v are bounded measurable functions,(iii)
v(x)=λu(x)+eic·x for some λ∈ℂ, c(≠0)∈ℝn, where u is a continuous function satisfying the condition; there exists d≥0(88)|u(x)|≤ϵ|x|p+dfor allx∈ℝn,(iv)
v(x)=λu(x)+1 for some λ∈ℂ, where u is a continuous function satisfying the conditions; there exists a∈ℂn such that
(89)|u(x)-a·x|≤2ϵ|x|p|2p-2|for allx∈ℝn,(v)
u=λsin(c·x), v=cos(c·x)+λsin(c·x), for some c∈ℂn, λ∈ℂ.Now we prove the stability of (20). For the proof we need the following.Lemma 12.
LetU,V:G×(0,∞)→ℂ 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,y∈ℝn, t,s>0. Then either there exist λ1,λ2∈ℂ, 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,y∈G, t,s>0.Proof.
As in Lemma9, 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 thatU 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 thatΨ:ℝn×(0,∞)→[0,∞) is a continuous function such that
(100)ψ(x):=limt→0+Ψ(x,t)
exists and satisfies the condition ψ(0)=0.Lemma 13.
LetU,V:ℝn×(0,∞)→ℂ 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,y∈ℝn,t,s>0, and there exist λ1,λ2∈ℂ, not both zero, and M>0 such that
(102)|λ1U(x,t)-λ2V(x,t)|≤M.
Then(U,V) satisfies one of the followings:(i)
U and V are bounded functions in ℝn×(0,1),(ii)
±iU(x,t)=V(x,t)-eia·x-bt for some a∈ℝn, b∈ℂ, and g(x):=limt→0+V(x,t) is a continuous function; in particular, there exists δ:(0,∞)→[0,∞) with δ(t)→0 as t→0+ such that
(103)|V(x,t)-g(x)e-bt|≤δ(t)for allx∈ℝn,t>0, and g satisfies
(104)|g(x)-cos(a·x)|≤12ψ(x)for allx∈ℝn.Proof.
IfU 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 c∈ℂn, b∈ℂ. Putting (105) in (101) and using the triangle inequality we have for some d≥0(106)|ec·xeb(t+s)(e-c·y-ec·y)|≤Ψ(y,s)+d
for all x,y∈ℝn,t,s>0. In view of (106) it is easy to see that c=ia, a∈ℝn. Thus V(x,t) is bounded on ℝn×(0,1). If U is unbounded; then in view of (101) V is also unbounded, hence λ1λ2≠0 and
(107)U(x,t)=λV(x,t)+R(x,t)
for some λ≠0 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 Lemma5 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 c∈ℂn, b∈ℂ. 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 d≥0,
(112)|m(x,t)(m(-y,s)-m(y,s))|≤|λ2+1|Ψ(y,s)+d
for all x,y∈ℝn, t,s>0. Since m is unbounded, we have
(113)m(y,s)=m(-y,s)
for all y∈ℝ and s>0. Thus it follows that m(x,t)=ebt and that U,V are bounded in ℝn×(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,y∈ℝn,t, s>0. Since m is a bounded continuous function, we have
(116)m(x,t)=eia·x-bt
for some a∈ℝn,b∈ℂ with ℜb≥0.
Similarly as in the proof of Lemma9, by (101) and the continuity of V, it is easy to see that
(117)limsupt→0+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 s→0+ in (118) we have
(119)limt→0+V(0,t)=1.
Puttingy=0, fixing x, letting t→0+ 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 x∈ℝn,s>0. Letting s→0+ in (120) we have
(121)lims→0+V(x,s)=g(x)
for all x∈ℝn. The continuity of g follows from (120). Letting t,s→0+ in (115) we have
(122)|g(x-y)-g(x)eia·y-g(y)eia·x+eia·(x+y)|≤ψ(y)
for all x,y∈ℝn. 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.
Letu,v∈𝒢′ satisfy (20). Then (u,v) satisfies one of the followings:(i)
u and v are bounded measurable functions,(ii)
v(x)=cos(a·x)+r(x),±u(x)=sin(a·x)+ir(x) for some a∈ℝn, where r(x) is a continuous function satisfying
(124)|r(x)|≤12ψ(x)for allx∈ℝn,(iii)
v(x)=cos(c·x) and u(x)=sin(c·x) for some c∈ℂn.Proof.
Similarly as in the proof of Theorem10 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,y∈ℝn,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∈ℂ, 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,y∈ℝn,t,s>0.
Firstly we assume that (127) holds. Letting t→0+ in (i) of Lemma 13, by Lemma 6, the initial values u,v of U(x,t),V(x,t) as t→0+ 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 x∈ℝn. Letting t→0+ 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,s→0+ in (127) we have
(133)v∘(x-y)-vx⊗vy-ux⊗uy=0.
By Lemma7 the solution of (133) satisfies (i) or (iii). This completes the proof.Every infraexponential functionf of order 2 defines an element of 𝒢′(ℝn) via the correspondence
(134)〈f,φ〉=∫‍f(x)φ(x)dx
for φ∈𝒢. Thus as a direct consequence of Corollary 11 and Theorem 14 we have the followings.Corollary 15.
Let0<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)∈ℝ2n. Then (f,g) satisfies one of the following:(i)
f(x)=0, almost everywhere x∈ℝn, and g is arbitrary,(ii)
f and g are bounded in almost everywhere,(iii)
f(x)=f0(x), g(x)=λf0(x)+eic·x for almost everywhere x∈ℝn, where λ∈ℂ, c(≠0)∈ℝn, and f0 is a continuous function satisfying the condition; there exists d≥0(136)|f0(x)|≤ϵ|x|p+dfor allx∈ℝn,(iv)
f(x)=f0(x), g(x)=λf0(x)+1 for a.e. x∈ℝn, where λ∈ℂ and f0 is a continuous function satisfying the condition; there exists a∈ℂn such that
(137)|f0(x)-a·x|≤2ϵ|x|p|2p-2|for allx∈ℝn,(v)
f(x)=λsin(c·x), g(x)=cos(c·x)+λsin(c·x) for a.e. x∈ℝn, where c∈ℂn, λ∈ℂ.Corollary 16.
Suppose thatf, 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)∈ℝ2n. Then (f,g) satisfies one of the followings:(i)
f and g are bounded in almost everywhere,(ii)
there existsa∈ℝn such that
(139)|g(x)-cos(a·x)|≤12ϵ|x|p,(140)|f(x)±sin(a·x)|≤12ϵ|x|pfor almost everyx∈ℝn,(iii)
g(x)=cos(c·x) and f(x)=sin(c·x) for a.e. x∈ℝn, where c∈ℂn.Remark 17.
Taking the growth ofu=ec·x as |x|→∞ into account, u∈𝒮′(ℝn) only when c=ia for some a∈ℝn. Thus Theorems 10 and 14 are reduced to the following:Corollary 18.
Letu,v∈𝒮′ satisfy (19). Then (u,v) satisfies one of the followings:(i)
u=0, and v is arbitrary,(ii)
u and v are bounded measurable functions,(iii)
v(x)=λu(x)+eic·x for some λ∈ℂ, c(≠0)∈ℝn, where u is a continuous function satisfying the condition; there exists d≥0(141)|u(x)|≤ψ(-x)+dfor allx∈ℝn,(iv)
v(x)=λu(x)+1 for some λ∈ℂ, where u is a continuous function satisfying one of the following conditions; there exists a1∈ℂn such that
(142)|u(x)-a1·x|≤Φ1(x)for allx∈ℝn, or there exists a2∈ℂn such that
(143)|u(x)-a2·x|≤Φ2(x)for allx∈ℝn.Corollary 19.
Letu,v∈𝒮′ satisfy (20). Then (u,v) satisfies one of the followings:(i)
u and v are bounded measurable functions,(ii)
v(x)=cos(a·x)+r(x), ±u(x)=sin(a·x)+ir(x) for some a∈ℝn, where r(x) is a continuous function satisfying
(144)|r(x)|≤12ψ(x)for allx∈ℝn.



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*Source: 275915-2013-05-08.xml*