# Hyperstability of the Fréchet Equation and a Characterization of Inner Product Spaces

Authors: Anna Bahyrycz; Janusz Brzdęk; Magdalena Piszczek; Justyna Sikorska
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

We prove some stability and hyperstability results for the well-known Fréchet equation stemming from one of the characterizations of the inner product spaces. As the main tool, we use a fixed point theorem for the function spaces. We finish the paper with some new inequalities characterizing the inner product spaces.


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## Body

## 1. Introduction

In the literature there are many characterizations of inner product spaces. The first norm characterization of inner product space was given by Fréchet [1] in 1935. He proved that a normed space (X,∥·∥) is an inner product space if and only if, for all x,y,z∈X,
(1)∥x+y+z∥2+∥x∥2+∥y∥2+∥z∥2=∥x+y∥2+∥x+z∥2+∥y+z∥2.
In the same year Jordan and von Neumann [2] gave the celebrated parallelogram law characterization of an inner product space. Since then numerous further conditions, characterizing the inner product spaces among the normed spaces, have been shown. More than 300 such conditions have been collected in the book of Amir [3]. Many geometrical characterizations are presented in the book by Alsina et al. [4]; for some other see, for example, [5–8].The results that we obtain are motivated by the notion of hyperstability of functional equations (see, e.g., [9–14]), which has been introduced in connection with the issue of stability of functional equations (for more details see, e.g., [15, 16]).The main tool in the proof of the main theorem is a fixed point result for function spaces from [17] (for related outcomes see [18, 19]). Similar method of the proof has been already applied in [11, 20].To present the fixed point theorem we introduce the following necessary hypotheses (ℝ+ stands for the set of nonnegative reals and AB denotes the family of all functions mapping a set B≠∅ into a set A≠∅).(H1)
S is a nonempty set, E is a Banach space, and functions f1,…,fk:S→S and L1,…,Lk:S→ℝ+ are given.(H2)
𝒯:ES→ES is an operator satisfying the inequality
(2)∥𝒯ξ(x)-𝒯μ(x)∥≤∑i=1k‍Li(x)∥ξ(fi(x))-μ(fi(x))∥,ξ,μ∈EX,x∈S.(H3)
Λ:ℝ+S→ℝ+S is defined by
(3)Λδ(x):=∑i=1k‍Li(x)δ(fi(x)),δ∈ℝ+S,x∈S.Now we are in a position to present the above mentioned fixed point theorem for function spaces (see [17]).Theorem 1.
Let hypotheses (H1)–(H3) be valid and functionsε:S→ℝ+ and φ:S→E fulfil the following two conditions:
(4)∥𝒯φ(x)-φ(x)∥≤ε(x),x∈S,ε*(x):=∑n=0∞‍Λnε(x)<∞,x∈S.
Then there exists a unique fixed point ψ of 𝒯 with
(5)∥φ(x)-ψ(x)∥≤ε*(x),x∈S.
Moreover,
(6)ψ(x):=limn→∞𝒯nφ(x),x∈S.We start our considerations from the functional equation(7)f(x+y+z)+f(x)+f(y)+f(z)=f(x+y)+f(x+z)+f(y+z),
that is patterned on (1) and therefore quite often named after Fréchet (see, e.g., [21]).Note that (7) can be written in the form
(8)Δx,y,zf(0)=0,
where Δ denotes the Fréchet difference operator defined (for functions mapping a commutative semigroup (S,+) into a group) by
(9)Δyf(x)=Δy1f(x):=f(x+y)-f(x),x,y∈S,Δt,z:=Δt∘Δz,Δt2:=Δt,t,t,z∈S,Δt,u,z:=Δt∘Δu∘Δz,Δt3:=Δt,t,t,t,u,z∈S.
It is easy to check that
(10)Δt,zf(x)=f(x+t+z)-f(x+t)-f(x+z)+f(x),x,t,z∈S,Δt,z,uf(x)=f(x+t+z+u)-f(x+t+z)-f(x+t+u)-f(x+z+u)+f(x+t)+f(x+z)+f(x+u)-f(x),x,t,z,u∈S.Such operators were first considered by Fréchet in [22, 23] (we refer to [24] for more information and further references concerning this subject); so, it is still another motivation for (7) to be called the Fréchet equation.Let us yet observe (see [25]) that, alternatively, (7) can be written in the form
(11)C2f(x,y,z)=0,
where C2f(x,y,z)=Cf(x,y+z)-Cf(x,y)-Cf(x,z) and Cf(x,y)=f(x+y)-f(x)-f(y); that is, C2f is the Cauchy difference of f of the second order.We prove the subsequent theorem, which corresponds to [26, Theorem 3.1], where the equation
(12)Δz3f(x)=0
has been investigated (the author has named it the superstability result, which is not a precise description, because according to the terminology applied in [10–14] it should be rather called the hyperstability result). For some analogous investigations see [27–29]. Let us mention yet that stability of (7) has been already studied in [30–33] and our results complement the outcomes included there.It is easy to show that every solutionf of (7), mapping a commutative group (G,+) into a real linear space X, must be of the form f=a+q with some additive a:G→X and quadratic q:G→X (see, e.g., [21]). Namely, with x=y=z=0 from (7) we deduce that f(0)=0, and, next, taking z=-y in (7), we obtain that the even part of f is quadratic while the odd part is a solution of the Jensen equation, whence it is additive.


## 2. Main Results

The next theorem and corollary are the main results of the paper (ℕ and ℤ stand, as usual, for the sets of all positive integers and integers, respectively; moreover, ℤ0:=ℤ∖{0}).Theorem 2.
Let(X,+) be a commutative group, X0:=X∖{0}, Y a Banach space, and f:X→Y, c:ℤ0→[0,∞) and L:X03→[0,∞) satisfying the following three conditions:
(13)ℳ≔{m∈ℤ0:c(-2m)+2c(m+1)+2c(-m)fm∈ℤ0:c(-2m)+2c(m+1)+2c(-m)+c(2m+1)<1}≠∅,(14)L(kx,ky,kz)≤c(k)L(x,y,z),x,y,z∈X0,k∈{-2m,m+1,-m,2m+1},m∈ℳ,(15)∥f(x+y+z)+f(x)+f(y)+f(z)f-f(x+y)-f(x+z)-f(y+z)∥≤L(x,y,z),x,y,z∈X0.
Then there is a unique function F:X→Y satisfying (7) for all x,y,z∈X and such that
(16)∥f(x)-F(x)∥≤ρL(x),x∈X0,
where
(17)ρL(x)≔infm∈ℳL((2m+1)x,-mx,-mx)1-c(-2m)-2c(m+1)-2c(-m)-c(2m+1),x∈X0.Proof.
Replacingx by (2m+1)x and taking y=z=-mx in (15) we get
(18)∥f(x)+f((2m+1)x)+2f(-mx)f-2f((m+1)x)-f(-2mx)∥≤L((2m+1)x,-mx,-mx)=:εm(x),x∈X0,m∈ℤ0.
Further, put
(19)𝒯mξ(x):=ξ(-2mx)+2ξ((m+1)x)-2ξ(-mx)-ξ((2m+1)x),ξ∈YX,x∈X,m∈ℤ0.
Then,
(20)𝒯mnf(0)=0,n∈ℕ,m∈ℤ0,
and inequality (18) takes the form
(21)∥𝒯mf(x)-f(x)∥≤εm(x),x∈X0,m∈ℤ0.
Define an operator Λm:ℝ+X0→ℝ+X0 for m∈ℤ0 by
(22)Λmη(x):=η(-2mx)+2η((m+1)x)+2η(-mx)+η((2m+1)x)
for η∈ℝ+X0 and x∈X0. Then it is easily seen that, for each m∈ℤ0, the operator Λ:=Λm has the form described in (H3) with k=4, S=X0, E=Y, and
(23)f1(x)=-2mx,f2(x)=(m+1)x,f3(x)=-mx,f4(x)=(2m+1)x,L1(x)=L4(x)=1,L2(x)=L3(x)=2,x∈X0.
Moreover, for every ξ,μ∈YX0, x∈X0, m∈ℤ0,
(24)∥𝒯mξ(x)-𝒯mμ(x)∥=∥ξ(-2mx)+2ξ((m+1)x)-2ξ(-mx)ffffff-ξ((2m+1)x)-μ(-2mx)-2μ((m+1)x)fffff+2μ(-mx)+μ((2m+1)x)∥≤∥(ξ-μ)(-2mx)∥+2∥(ξ-μ)((m+1)x)∥+2∥(ξ-μ)(-mx)∥+∥(ξ-μ)((2m+1)x)∥=∑i=14‍Li(x)∥(ξ-μ)(fi(x))∥,
where (ξ-μ)(y)=ξ(y)-μ(y) for y∈X0. It is easy to check that, in view of (14),
(25)Λmεk(x)≤(c(-2m)+2c(m+1)+2c(-m)+c(2m+1))εk(x),k,m∈ℤ0,x∈X0.
Therefore, since the operator Λm is linear, we have
(26)εm*(x):=∑n=0∞‍Λmnεm(x)≤∑n=0∞‍(c(-2m)+2c(m+1)ffffffff+2c(-m)+c(2m+1))nεm(x)=εm(x)1-c(-2m)-2c(m+1)-2c(-m)-c(2m+1),m∈ℳ,x∈X0.
Thus, by Theorem 1 (with S=X0 and E=Y), for each m∈ℳ there exists a function Fm′:X0→Y with
(27)Fm′(x)=Fm′(-2mx)+2Fm′((m+1)x)-2Fm′(-mx)-Fm′((2m+1)x),x∈X0,∥f(x)-Fm′(x)∥≤εm(x)1-c(-2m)-2c(m+1)-2c(-m)-c(2m+1),x∈X0.
Moreover,
(28)Fm′(x)=limn→∞𝒯mnf(x),x∈X0,m∈ℳ.
Define Fm:X→Y by Fm(0)=0 and Fm(x):=Fm′(x) for x∈X0 and m∈ℳ. Then it is easily seen that, by (20),
(29)Fm(x)=limn→∞𝒯mnf(x),x∈X,m∈ℳ.
Next, we show that(30)∥𝒯mnf(x+y+z)+𝒯mnf(x)+𝒯mnf(y)+𝒯mnf(z)f-𝒯mnf(x+y)-𝒯mnf(x+z)-𝒯mnf(y+z)∥≤(c(-2m)+2c(m+1)+2c(-m)+c(2m+1))n×L(x,y,z)
for every x,y,z∈X0, n∈ℕ0:=ℕ∪{0}, m∈ℳ.
Fixm∈ℳ. For n=0, the condition (30) is simply (15). So, take r∈ℕ0 and suppose that (30) holds for n=r and x,y,z∈X0. Then,
(31)∥𝒯mr+1f(x+y+z)+𝒯mr+1f(x)+𝒯mr+1f(y)+𝒯mr+1f(z)f-𝒯mr+1f(x+y)-𝒯mr+1f(x+z)-𝒯mr+1f(y+z)∥=∥𝒯mrf(-2m(x+y+z))ggggg+2𝒯mrf((m+1)(x+y+z))ggggg-2𝒯mrf(-m(x+y+z))ggggg-𝒯mrf((2m+1)(x+y+z))ggggg+𝒯mrf(-2mx)+2𝒯mrf((m+1)x)ggggg-2𝒯mrf(-mx)-𝒯mrf((2m+1)x)ggggg+𝒯mrf(-2my)+2𝒯mrf((m+1)y)ggggg-2𝒯mrf(-my)-𝒯mrf((2m+1)y)ggggg+𝒯mrf(-2mz)+2𝒯mrf((m+1)z)ggggg-2𝒯mrf(-mz)-𝒯mrf((2m+1)z)ggggg-𝒯mrf(-2m(x+y))-2𝒯mrf((m+1)(x+y))ggggg+2𝒯mrf(-m(x+y))+𝒯mrf((2m+1)(x+y))ggggg-𝒯mrf(-2m(x+z))-2𝒯mrf((m+1)(x+z))ggggg+2𝒯mrf(-m(x+z))+𝒯mrf((2m+1)(x+z))ggggg-𝒯mrf(-2m(y+z))-2𝒯mrf((m+1)(y+z))ggggg+2𝒯mrf(-m(y+z))+𝒯mrf((2m+1)(y+z))∥≤(c(-2m)+2c(m+1)+2c(-m)+c(2m+1))rfffff×(((2m+1)x,(2m+1)y,(2m+1)z)L(-2mx,-2my,-2mz)fffff+2L((m+1)x,(m+1)y,(m+1)z)fffff+2L(-mx,-my,-mz)fffff+L((2m+1)x,(2m+1)y,(2m+1)z))≤((2m+1)c(-2m)+2c(m+1)+2c(-m)fffff+c(2m+1))r+1L(x,y,z)
for every x,y,z∈X0, which completes the proof of (30).
Lettingn→∞ in (30), we obtain that
(32)Fm(x+y+z)+Fm(x)+Fm(y)+Fm(z)=Fm(x+y)+Fm(x+z)+Fm(y+z),x,y,z∈X0.
So, we have proved that for each m∈ℳ there exists a function Fm:X→Y satisfying (7) for x,y,z∈X0 and such that
(33)∥f(x)-Fm(x)∥≤εm(x)1-c(-2m)-2c(m+1)-2c(-m)-c(2m+1),x∈X0.
Now, we show thatFm=Fk for all m,k∈ℳ. So, fix m,k∈ℳ. Note that Fk satisfies (32) with m replaced by k. Hence, replacing x by (2m+1)x and taking y=z=-mx in (32), we obtain that 𝒯mFj=Fj for j=m,k and
(34)∥Fm(x)-Fk(x)∥≤εm(x)1-c(-2m)-2c(m+1)-2c(-m)-c(2m+1)+εk(x)1-c(-2k)-2c(k+1)-2c(-k)-c(2k+1),x∈X0,
whence, by the linearity of Λ and (25),
(35)∥Fm(x)-Fk(x)∥=∥𝒯mnFm(x)-𝒯mnFk(x)∥≤Λmnεm(x)1-c(-2m)-2c(m+1)-2c(-m)-c(2m+1)+Λmnεk(x)1-c(-2k)-2c(k+1)-2c(-k)-c(2k+1)≤(c(-2m)+2c(m+1)+2c(-m)+c(2m+1))nεm(x)1-c(-2m)-2c(m+1)-2c(-m)-c(2m+1)+(c(-2m)+2c(m+1)+2c(-m)+c(2m+1))nεk(x)1-c(-2k)-2c(k+1)-2c(-k)-c(2k+1)
for every x∈X0 and n∈ℕ. Now, letting n→∞ we get Fm=Fk=:F.
Thus, in view of (33), we have proved that
(36)∥f(x)-F(x)∥≤εm(x)1-c(-2m)-2c(m+1)-2c(-m)-c(2m+1),x∈X0,m∈ℳ,
whence we derive (16).
Since (in view of (32)) it is easy to notice that F is a solution to (7) (i.e., (7) holds for all x,y,z∈X), it remains to prove the statement concerning the uniqueness of F. So, let G:X→Y be also a solution of (7) and ∥f(x)-G(x)∥≤ρL(x) for x∈X0. Then,
(37)∥G(x)-F(x)∥≤2ρL(x),x∈X0.
Further, 𝒯mG=G for each m∈ℤ0. Consequently, with a fixed m∈ℳ,
(38)∥G(x)-F(x)∥=∥𝒯mnG(x)-𝒯mnF(x)∥≤2ΛmnρL(x)≤2Λmnεm(x)1-c(-2m)-2c(m+1)-2c(-m)-c(2m+1)
for x∈X0 and n∈ℕ. Next, analogously as (25), by induction we get
(39)Λmnεm(x)≤(c(-2m)+2c(m+1)+2c(-m)+c(2m+1))nεm(x)
for x∈X0, n∈ℕ. This implies that G=F.Theorem2 yields at once the following hyperstability result.Corollary 3.
Let(X,+) be a commutative group, X0:=X∖{0}, let Y be a normed space, f:X→Y, c:ℤ0→[0,∞), L:X03→[0,∞), and let the conditions (13), (14), and (15) be valid. Assume that
(40)infm∈ℳL((2m+1)x,-mx,-mx)=0,x∈X0.
Then f satisfies (7) for all x,y,z∈X.Proof.
Note that without loss of generality we may assume thatY is complete, because otherwise we can replace it by its completion. Next, in view of (40), ρL(x)=0 for each x∈X0, where ρL is defined by (17). Hence, from Theorem 2, we easily derive that f is a solution to (7).


## 3. Final Remarks

Remark 4.
Note that if, in Theorem2,
(41)liminfm→∞c(-2m)+2c(m+1)+2c(-m)+c(2m+1)=0
(this is the case when, e.g., lim|m|→∞c(m)=0), then (13) holds and
(42)ρL(x):=infm∈ℳεm(x),x∈X0.
Further, letX be a normed space and
(43)L(x,y,z)=ε(∥x∥p+∥y∥p+∥z∥p),x,y,z∈X0
with some reals ε>0 and p<0. Then, the condition (14) is valid, for instance, with c(k)=|k|p for k∈ℤ0. Obviously, (40) holds, and there exists m0∈ℕsuch that
(44)|2m|p+2|m+1|p+2|m|p+|2m+1|p<1,m≥m0;
so we obtain (13), as well. Consequently, by Corollary 3, every function f:X→Y, fulfilling the inequality (15), satisfies (7) for all x,y,z∈X. In this way we have obtained a hyperstability result that corresponds to the recent hyperstability outcomes in [11, 20] and some classical stability results concerning the Cauchy equation (see, e.g., [9, page 3], [15, page 15, 16], and [16, page 2]).Below, we provide two further simple and natural examples of functionsL and c satisfying the conditions (13) and (14). The first one, clearly, includes the case just described. (a)
L(x,y,z)=(α1∥x∥s1+α2∥y∥s2+α3∥z∥s3)w for x,y,z∈X0 with some w,αi,si∈ℝ such that αi>0 and wsi<0 for i=1,2,3, and c(m)≡|m|-v|w|, where v:=min{|s1|,|s2|,|s3|}.(b)
L(x,y,z)=α∥x∥s∥y∥t∥z∥u for x,y,z∈X0 with some reals α>0 and s, t, u such that s+t+u<0.Clearly, if two functionsL satisfy the condition (14), then so do their sum and product, with suitable functions c. Therefore, we can easily produce numerous examples of such functions. Of course there are some other such examples that are a bit more artificial; for instance, L(x,y,z)=B(∥A(x)∥-n) for x∈X0, where n∈ℕ, A:X→X and B:ℝ→ℝ are functions with A(xm)=mA(x) and B(tm)=mB(t) for x∈X, t∈ℝ and m∈ℤ.We end the paper with a simple example of applications of our main result.Corollary 5.
LetX be a normed space and X0:=X∖{0}. Write
(45)D(x,y,z)∶=|∥x+y+z∥2+∥x∥2+∥y∥2+∥z∥2f-∥x+y∥2-∥x+z∥2-∥y+z∥2|
for x,y,z∈X. Assume that one of the following two hypotheses is valid. (i)
There existw0,αi,si∈ℝ such that αi,w0si>0 for i=1,2,3 and
(46)supx,y,z∈X0D(x,y,z)(α1∥x∥s1+α2∥y∥s2+α3∥z∥s3)w0<∞.(ii)
There exist realss,t,u such that s+t+u>0 and
(47)supx,y,z∈X0D(x,y,z)∥x∥s∥y∥t∥z∥u<∞. Then X is an inner product space.Proof.
Writef(x)=∥x∥2 for x∈X. Then, with L and c of the forms described in Remark 4 (with w=-w0), from Corollary 3 we easily derive that f is a solution to (7), which yields the statement.



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