Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 6, Issue 2, Article 32, 2005
ON HYERS-ULAM STABILITY OF A SPECIAL CASE OF O’CONNOR’S AND
GAJDA’S FUNCTIONAL EQUATIONS
BELAID BOUIKHALENE, ELHOUCIEN ELQORACHI, AND AHMED REDOUANI
U NIVERSITY OF I BN T OFAIL
FACULTY OF S CIENCES
D EPARTMENT OF M ATHEMATICS
K ENITRA , M OROCCO
bbouikhalene@yahoo.fr
U NIVERSITY OF I BN Z OHR
FACULTY OF S CIENCES
D EPARTMENT OF M ATHEMATICS
AGADIR , M OROCCO
elqorachi@hotmail.com
U NIVERSITY OF I BN Z OHR
FACULTY OF S CIENCES
D EPARTMENT OF M ATHEMATICS
AGADIR , M OROCCO
redouani_ahmed@yahoo.fr
Received 08 November, 2004; accepted 01 March, 2005
Communicated by K. Nikodem
A BSTRACT. In this paper, we obtain the Hyers-Ulam stability for the following functional equation
XZ
f (xkϕ(y)k −1 )dωK (k) = |Φ|a(x)a(y), x, y ∈ G,
ϕ∈Φ
K
where G is a locally compact group, K is a compact subgroup of G, ωK is the normalized Haar
measure of K, Φ is a finite group of K-invariant morphisms of G and f, a : G −→ C are
continuous complex-valued functions such that f satisfies the Kannappan type condition
Z Z
Z Z
−1
−1
(*)
f (zkxk hyh )dωK (k)dωK (h) =
f (zkyk −1 hxh−1 )dωK (k)dωK (h),
K
K
K
K
for all x, y, z ∈ G.
Key words and phrases: Functional equations, Hyers-Ulam stability, Gelfand pairs.
2000 Mathematics Subject Classification. 39B72.
ISSN (electronic): 1443-5756
c 2005 Victoria University. All rights reserved.
213-04
2
B ELAID B OUIKHALENE , E LHOUCIEN E LQORACHI ,
AND
A HMED R EDOUANI
1. I NTRODUCTION
Let G be a locally compact group. Let K be a compact subgroup of G. Let ωK be the normalized Haar measure of K. A mapping ϕ : G → G is a morphism of G if ϕ is a homeomorphism
of G onto itself which is either a group-homomorphism, (i.e. ϕ(xy) = ϕ(x)ϕ(y), x, y ∈ G),
or a group-antihomomorphism, (i.e. ϕ(xy) = ϕ(y)ϕ(x), x, y ∈ G). We denote by M or(G)
the group of morphisms of G and Φ a finite subgroup of M or(G) which is K-invariant (i.e.
ϕ(K) ⊂ K, for all ϕ ∈ Φ). The number of elements of a finite group Φ will be designated by
|Φ|. The Banach algebra of the complex bounded measures on G is denoted by M (G), it is the
topological dual of C0 (G): Banach space of continuous functions vanishing at infinity. Finally
the Banach space of all complex measurable and essentially bounded functions on G is denoted
by L∞ (G) and C(G) designates the space of all continuous complex valued functions on G.
The stability problem for functional equations are strongly related to the question of S.M.
Ulam concerning the stability of group homomorphisms [26], [16]. During the last decades,
the stability problems of several functional equations have been extensively investigated by a
number of mathematicians ([16], [2], [3], [22], [23], [24], [25], [20], ...). The main purpose
of this paper is to generalize the Hyers-Ulam stability problem for the following functional
equation
XZ
(1.1)
f (xkϕ(y)k −1 )dωK (k) = |Φ|a(x)a(y), x, y ∈ G,
ϕ∈Φ
K
where G is a locally compact group, and f, a ∈ C(G) with the assumption that f satisfies the
Kannappan type condition: (*)
Z Z
Z Z
−1
−1
f (zkxk hyh )dωK (k)dωK (h) =
f (zkyk −1 hxh−1 )dωK (k)dωK (h),
K
K
K
K
for all x, y, z ∈ G.
In the case where G is a locally compact abelian group, O’Connor [19], Gajda [14] and
Stetkær [21] studied respectively the functional equation
(1.2)
f (x − y) =
n
X
ai (x)ai (y),
x, y ∈ G, n ∈ N,
i=1
(1.3)
f (x + y) + f (x − y) = 2
n
X
ai (x)ai (y),
x, y ∈ G, n ∈ N,
i=1
and
Z
f (xh · y)dh = a(x)a(y),
(1.4)
x, y ∈ G̃,
H
where G̃ is a locally compact group and H is a compact subgroup of Aut(G̃).
In the case n = 1 equations (1.2) and (1.3) are special cases of (1.1). Moreover, taking
G = G̃ ×s H the semi direct product of G̃ and H, and K = {e} × H, we observe that equation
(1.4) is also a special case of (1.1).
This equation may be considered as a common generalization of functional equations
(1.5)
f (xy −1 ) = a(x)a(y),
(1.6)
f (xy) + f (xy −1 ) = 2a(x)a(y),
J. Inequal. Pure and Appl. Math., 6(2) Art. 32, 2005
x, y ∈ G,
x, y ∈ G.
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H YERS -U LAM
3
STABILITY
It is also a generalization of the equations
Z
(1.7)
f (xky −1 k −1 )dωK (k) = a(x)a(y),
x, y ∈ G,
K
Z
Z
−1
f (xky −1 k −1 )dωK (k) = 2a(x)a(y),
f (xkyk )dωK (k) +
(1.8)
K
x, y ∈ G,
K
Z
f (xky −1 )χ(k)dωK (k) = a(x)a(y),
(1.9)
x, y ∈ G,
K
Z
Z
f (xky)χ(k)dωK (k) +
(1.10)
K
f (xky −1 )χ(k)dωK (k) = 2a(x)a(y),
x, y ∈ G,
K
Z
f (xky −1 )dωK (k) = a(x)a(y),
(1.11)
x, y ∈ G,
K
Z
Z
f (xky)dωK (k) +
(1.12)
K
f (xkϕ(y −1 ))dωK (k) = 2a(x)a(y),
x, y ∈ G, ([10], [11]).
K
If G is a compact group, equation (1.1) may be considered as a generalization of the equations
Z
(1.13)
f (xty −1 t−1 )dt = a(x)a(y),
x, y ∈ G,
G
Z
−1
Z
f (xtyt )dt +
(1.14)
G
f (xty −1 t−1 )dt = 2a(x)a(y),
x, y ∈ G,
G
XZ
(1.15)
f (xtϕ(y −1 )t−1 )dt = |Φ|a(x)a(y),
x, y ∈ G.
G
ϕ∈Φ
Furthermore the following equations are also a particular case of (1.1).
X
(1.16)
f (xϕ(y −1 )) = |Φ|a(x)a(y),
x, y ∈ G,
ϕ∈Φ
(1.17)
XZ
ϕ∈Φ
(1.18)
XZ
ϕ∈Φ
f (xkϕ(y −1 ))dωK (k) = |Φ|a(x)a(y),
x, y ∈ G,
K
f (xkϕ(y −1 ))χ(k)dωK (k) = |Φ|a(x)a(y),
x, y ∈ G,
K
where χ is a character of K. For more information about the equations (1.1) – (1.18) (see [1],
[4], [6], [7], [11], [12], [14], [15], [19], [21]).
In the next section, we note some results for later use.
J. Inequal. Pure and Appl. Math., 6(2) Art. 32, 2005
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B ELAID B OUIKHALENE , E LHOUCIEN E LQORACHI ,
2. G ENERALIZED
STABILITY RESULTS OF
AND
A HMED R EDOUANI
C AUCHY ’ S AND W ILSON ’ S EQUATIONS
Let G, K and Φ be given as above. One can prove (see [4]) the following two propositions.
Proposition 2.1. For an arbitrary fixed τ ∈ Φ, the mapping
Φ 3 ϕ 7→ ϕ ◦ τ ∈ Φ
is a bijection and for all x, y ∈ G, we have
XZ
XZ
−1
f (xkϕ(τ (y))k )dωK (k) =
f (xkψ(y)k −1 )dωK (k).
ϕ∈Φ
K
ψ∈Φ
K
Proposition 2.2. Let ϕ ∈ Φ and f ∈ C(G). Then
i)
Z
−1
Z
f (xkϕ(hy)k )dωK (k) =
K
f (xkϕ(yh)k −1 )dωK (k),
x, y ∈ G, h ∈ K.
K
ii) Moreover, if f satisfies the Kannappan type condition (∗), then we have
Z Z
Z Z
−1 −1
f (zhϕ(xkyk −1 )h−1 )dωK (h)dωK (k),
f (zhϕ(ykxk )h )dωK (h)dωK (k) =
K
K
K
K
for all z, y, x ∈ G.
The next results extend the ones obtained in [4], [8], [9], [10] and [13].
Theorem 2.3. Let ε : G −→ R+ be a continuous function. Let f, g : G −→ C be continuous
functions such that f satisfies the Kannappan type condition (∗) and
X Z
−1
(2.1)
f (xkϕ(y)k )dωK (k) − |Φ|f (x)g(y) ≤ ε(y),
x, y ∈ G.
K
ϕ∈Φ
If f is unbounded, then g satisfies the functional equation
XZ
(2.2)
g(xkϕ(y)k −1 )dωK (k) = |Φ|g(x)g(y),
ϕ∈Φ
x, y ∈ G.
K
Proof. Let ε : G −→ R+ be a continuous function, and let f, g ∈ C(G) satisfying inequality
(2.1). Let Φ = Φ+ ∪ Φ− , where Φ+ (resp. Φ− ) is a set of group-homomorphisms (resp. of
group-antihomomorphisms). By using Propositions 2.1, 2.2 and the fact that f satisfies the
condition (*), for all x, y, z ∈ G, we get
X Z
g(xkϕ(y)k −1 )dωK (k) − |Φ|g(x)g(y)
|Φ||f (z)| ϕ∈Φ K
X Z
−1
2
|Φ|f (z)g(xkϕ(y)k )dωK (k) − |Φ| f (z)g(x)g(y)
=
K
ϕ∈Φ
J. Inequal. Pure and Appl. Math., 6(2) Art. 32, 2005
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H YERS -U LAM
STABILITY
5
X Z X Z
f (zhτ (xkϕ(y)k −1 )h−1 )dωK (k)dωK (h)
≤
ϕ∈Φ K τ ∈Φ K
XZ
−
|Φ|f (z)g(xkϕ(y)k −1 )dωK (k)
ϕ∈Φ K
X Z X Z
+
f (zhτ (xkϕ(y)k −1 )h−1 )dωK (k)dωK (h)
ϕ∈Φ K τ ∈Φ K
XZ
−|Φ|g(y)
f (zhτ (x)h−1 )dωK (h)
τ ∈Φ K
X Z
−1
+ |Φ||g(y)| f (zkτ (x)k )dωK (k) − |Φ|f (z)g(x)
τ ∈Φ K
X Z X Z
=
f (zhτ (xkϕ(y)k −1 )h−1 )dωK (k)dωK (h)
ϕ∈Φ K τ ∈Φ K
XZ
−1
−
|Φ|f (z)g(xkϕ(y)k )dωK (k)
ϕ∈Φ K
X Z X Z
f (zhτ (x)kψ(y)k −1 )h−1 )dωK (k)dωK (h)
+
ψ∈Φ K τ ∈Φ+ K
XZ X Z
f (zhk −1 ψ(y)kτ (x)h−1 )dωK (k)dωK (h)
+
ψ∈Φ
K τ ∈Φ−
−|Φ|g(y)
K
f (zhτ (x)h−1 )dωK (h)
K
XZ
τ ∈Φ
X Z
+ |Φ||g(y)| f (zkτ (x)k −1 )dωK (k) − |Φ|f (z)g(x)
τ ∈Φ K
X Z X Z
=
f (zhτ (xkϕ(y)k −1 )h−1 )dωK (k)dωK (h)
ϕ∈Φ K τ ∈Φ K
XZ
|Φ|f (z)g(xkϕ(y)k −1 )dωK (k)
−
ϕ∈Φ K
X Z X Z
+
f (zhτ (x)h−1 kψ(y)k −1 )dωK (k)dωK (h)
ψ∈Φ K τ ∈Φ+ K
XZ X Z
+
f (zhτ (x)h−1 kψ(y)k −1 )dωK (k)dωK (h)
ψ∈Φ
K τ ∈Φ−
−|Φ|g(y)
K
−1
f (zhτ (x)h )dωK (h)
K
XZ
τ ∈Φ
X Z
+ |Φ||g(y)| f (zkτ (x)k −1 )dωK (k) − |Φ|f (z)g(x)
K
τ ∈Φ
J. Inequal. Pure and Appl. Math., 6(2) Art. 32, 2005
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B ELAID B OUIKHALENE , E LHOUCIEN E LQORACHI ,
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A HMED R EDOUANI
X Z X Z
f (zhτ (xkϕ(y)k −1 )h−1 )dωK (k)dωK (h)
=
ϕ∈Φ K τ ∈Φ K
XZ
−
|Φ|f (z)g(xkϕ(y)k −1 )dωK (k)
ϕ∈Φ K
X Z X Z
+
f (zhτ (x)h−1 kψ(x)k −1 )dωK (k)dωK (h)
ψ∈Φ K τ ∈Φ K
XZ
−1
−|Φ|g(y)
f (zhτ (x)h )dωK (h)
τ ∈Φ K
X Z
−1
+ |Φ||g(y)| f (zkτ (x)k )dωK (k) − |Φ|f (z)g(x)
τ ∈Φ K
X Z X Z
−1 −1
−1 ≤
f (zhτ (xkϕ(y)k )h )dωK (h) − |Φ|f (z)g(xkϕ(y)k )dωK (k)
ϕ∈Φ K τ ∈Φ K
X Z X Z
−1
−1
−1
+
f (zhτ (x)h kψ(y)k ))dωK (h) − |Φ|f (zhτ (x)h )g(y)dωK (h)
τ ∈Φ K ψ∈Φ K
X Z
+ |Φ||g(y)| f (zhτ (x)h−1 )dωK (h) − |Φ|f (z)g(x)
τ ∈Φ K
Z
X
≤
ε(xkϕ(y)k −1 )dωK (k) + |Φ|ε(y) + |Φ||g(y)|ε(x).
ϕ∈Φ
K
Since f is unbounded, then it follows that g is a solution of (2.1). This ends the proof of our
theorem.
The next results extend the ones obtained in [8] and [13].
Theorem 2.4. Let ε : G −→ R+ . Let f, g : G −→ C be continuous functions such that f
satisfies the condition (∗) and
X Z
(2.3)
f (xkϕ(y)k −1 )dωK (k) − |Φ|f (x)g(y) ≤ ε(y),
x, y ∈ G.
K
ϕ∈Φ
Suppose furthermore there exists x0 ∈ G such that |g(x0 )| > 1. Then there exists exactly one
solution F ∈ C(G) of the equation
XZ
(2.4)
F (xkϕ(y)k −1 )dωK (k) = |Φ|F (x)g(y),
x, y ∈ G,
ϕ∈Φ
K
such that F − f is bounded and one has
(2.5)
|F (x) − f (x)| ≤
J. Inequal. Pure and Appl. Math., 6(2) Art. 32, 2005
ε(x0 )
,
|Φ|(|g(x0 )| − 1)
x ∈ G.
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H YERS -U LAM
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Proof. In the proof, we use the ideas and methods that are analogous to the ones used in [8],
[13] and [20]. Let β = |Φ|g(x0 ), for all x ∈ G, one has
(2.6)
X Z
−1
f
(xkϕ(x
)k
)dω
(k)
−
βf
(x)
≤ ε(x0 ),
0
K
K
x, y ∈ G.
ϕ∈Φ
We define the following functions sequence
(2.7)
G1 (x) =
XZ
ϕ∈Φ
(2.8)
Gn+1 (x) =
XZ
ϕ∈Φ
f (xkϕ(x0 )k −1 )dωK (k),
x ∈ G,
K
Gn (xkϕ(x0 )k −1 )dωK (k),
x ∈ G and n ∈ N.
K
Next, we will prove the uniform convergence of the function sequence (β −n Gn )n≥1 , therefore
we need to show by induction the following inequalities
(2.9)
(2.10)
|Gn+1 (x) − βGn (x)| ≤ |Φ|n ε(x0 ),
x ∈ G, n ≥ 1,
|Gn (x) − β n f (x)| ≤ ε(x0 )(|Φ|n−1 + |Φ|n−2 |β| + · · · + |β|n−1 ),
and
(2.11)
|β −(n+1) Gn+1 (x) − β −n Gn (x)| ≤ |β|−(n+1) |Φ|n ε(x0 ).
In view of (2.5) one has for all x ∈ G
|G2 (x) − βG1 (x)|
Z
X Z
X
G1 (xkϕ(x0 )k −1 )dωK (k) − β
f (xkϕ(x0 )k −1 )dωK (k)
=
ϕ∈Φ K
ϕ∈Φ K
X Z X Z
f (xkϕ(x0 )k −1 hτ (x0 )h−1 )dωK (h)dωK (k)
=
ϕ∈Φ K τ ∈Φ K
XZ
−β
f (xkτ (x0 )k −1 )dωK (k)
τ ∈Φ K
X Z X Z
≤
f (xkτ (x0 )k −1 hϕ(x0 )h−1 )dωK (h)
τ ∈Φ K ϕ∈Φ K
−1 − βf (xkτ (x0 )k )dωK (k)
≤ |Φ|ε(x0 ).
J. Inequal. Pure and Appl. Math., 6(2) Art. 32, 2005
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B ELAID B OUIKHALENE , E LHOUCIEN E LQORACHI ,
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A HMED R EDOUANI
Assume (2.8) holds for n ≥ 1, then for n + 1, one has
X Z
Gn+1 (xkϕ(x0 )k −1 )dωK (k)
|Gn+2 (x) − βGn+1 (x)| = K
ϕ∈Φ
−β
ϕ∈Φ
XZ
≤
ϕ∈Φ
Gn (xkϕ(x0 )k −1 )dωK (k)
K
XZ
|Gn+1 (xkϕ(x0 )k −1 )dωK (k)
K
− βGn (xkϕ(x0 )k −1 )|dωK (k)
≤ |Φ|n+1 ε(x0 ).
In view of (2.5) we have for all x ∈ G
X Z
−1
|G1 (x) − βf (x)| = f (xkϕ(x0 )k )dωK (k) − βf (x)
K
ϕ∈Φ
≤ ε(x0 ).
Suppose (2.9) is true for n ≥ 1. For n + 1 one has
|Gn+1 (x) − β n+1 f (x)| ≤ |Gn+1 (x) − βGn (x)| + |β||Gn (x) − β n f (x)|
≤ |Φ|n ε(x0 ) + |β|ε(x0 )(|Φ|n−1 + |Φ|n−2 |β| + · · · + |β|n−1 )
= ε(x0 )(|Φ|n + |Φ|n−1 |β| + · · · + |β|n ).
For inequality (2.10), using (2.8), for all x ∈ G we get
|β −(n+1) Gn+1 (x) − β −n Gn (x)| = |β −(n+1) ||Gn+1 (x) − βGn (x)|
≤ |β|−(n+1) |Φ|n ε(x0 ).
So by using inequality (2.10) we deduce the uniform convergence of the sequence (β −n Gn )n≥1 .
Let F be a continuous function defined by
lim β −n Gn (x),
F (x) =
n−→+∞
Since
β
−(n+1)
Gn+1 (x) = β
−1
XZ
ϕ∈Φ
then one has
βF (x) =
XZ
ϕ∈Φ
x ∈ G.
β −n Gn (xkϕ(x0 )k −1 )dωK (k),
K
F (xkϕ(x0 )k −1 )dωK (k),
x ∈ G.
K
In view of (2.9), one has for all x ∈ G
|β −n Gn (x) − f (x)| ≤ |β|−n ε(x0 )(|Φ|n−1 + |Φ|n−2 |β| + · · · + |β|n−1 ),
which proves that
ε(x0 )
,
x ∈ G.
|Φ|(|g(x0 )| − 1)
Now we are going to show that F satisfies the equation
XZ
F (xkϕ(y)k −1 )dωK (k) = |Φ|F (x)g(y),
x, y ∈ G.
|F (x) − f (x)| <
ϕ∈Φ
K
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H YERS -U LAM
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STABILITY
Thus we need to show by induction the inequality
X Z
ε(y)
−n
−1
−n
β Gn (xkϕ(y)k )dωK (k) − |Φ|β Gn (x)g(y) ≤
(2.12)
.
|g(x0 )|n
ϕ∈Φ K
For n = 1, one has, by using the fact that f satisfies the condition (∗)
Z
1 X
G1 (xkϕ(y)k −1 )dωK (k) − |Φ|G1 (x)g(y)
β ϕ∈Φ K
Z XZ
1 X
= f (xkϕ(y)k −1 hτ (x0 )h−1 )dωK (k)dωK (h)
β ϕ∈Φ K τ ∈Φ K
XZ
−|Φ|g(y)
f (xkτ (x0 )k −1 )dωK (k)
τ ∈Φ K
Z XZ
1 X
f (xkτ (x0 )k −1 hϕ(y)h−1 )dωK (k)dωK (h)
= β τ ∈Φ K ϕ∈Φ K
XZ
−|Φ|g(y)
f (xkτ (x0 )k −1 )dωK (k)
τ ∈Φ K
Z X Z
1X
f (xkτ (x0 )k −1 hϕ(y)h−1 )dωK (h)
≤
β τ ∈Φ K ϕ∈Φ K
−1
− |Φ|g(y)f (xkτ (x0 )k )g(y)dωK (k)
≤
|Φ|ε(y)
ε(y)
=
.
|β|
|g(x0 )|
Assume (2.12) holds for some n ≥ 1. For n + 1, one has by using the fact that f satisfies the
condition (∗)
X Z
−(n+1)
−1
−(n+1)
β
Gn+1 (xkϕ(y)k )dωK (k) − |Φ|β
Gn+1 (x)g(y)
ϕ∈Φ K
Z XZ
1 X
= β −n Gn (xkϕ(y)k −1 hτ (x0 )h−1 )dωK (h)dωK (k)
β ϕ∈Φ K τ ∈Φ K
XZ
− |Φ|g(y)
β −n Gn (xkτ (x0 )k −1 )dωK (k)|
τ ∈Φ
K
Z X Z
1X
≤
β −n Gn (xkτ (x0 )k −1 hϕ(y)h−1 )dωK (h)
β τ ∈Φ K ϕ∈Φ K
− |Φ|g(y)β −n Gn (xkτ (x0 )k −1 )|dωK (k)|
≤
|Φ|
ε(y)
ε(y)
=
.
|Φ| | g(x0 ) || |g(x0 )|n
|g(x0 )|n+1
J. Inequal. Pure and Appl. Math., 6(2) Art. 32, 2005
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B ELAID B OUIKHALENE , E LHOUCIEN E LQORACHI ,
AND
A HMED R EDOUANI
3. T HE M AIN R ESULTS
In the next proposition, we investigate the stability of the functional equation (1.1).
Proposition 3.1. Let δ > 0. Let f, a ∈ C(G) such that
X Z
(3.1)
f (xkϕ(y −1 )k −1 )dωK (k) − |Φ|a(x)a(y) ≤ δ,
K
x, y ∈ G.
ϕ∈Φ
Then
i) If f is bounded then a is bounded and one has,
s
δ
|a(x)| ≤ sup |f | +
,
x ∈ G,
|Φ|
s
δ
δ
|f (x)| ≤ |a(e)| sup |f | +
+
,
x ∈ G.
|Φ| |Φ|
ii) If f is unbounded then a(e) 6= 0. Furthermore there exists x0 ∈ G such that |a(x0 )| >
|a(e)|.
Proof. i) Let f be a continuous bounded solution of (3.1), then by taking x = y in (3.1) we get
|Φ||a(x)|2 ≤ |Φ| sup |f | + δ,
i.e.
s
|a(x)| ≤
sup |f | +
δ
,
|Φ|
x ∈ G,
x ∈ G.
For y = e in (3.1) we get
|f (x)| ≤ |a(x)||a(e)| +
i.e.
s
|f (x)| ≤ |a(e)| sup |f | +
δ
,
|Φ|
δ
δ
+
,
|Φ| |Φ|
x ∈ G.
δ
We will prove (ii) by contradiction. If a(e) = 0 then |f (x)| < |Φ|
.
If |a(x)| ≤ |a(e)|, for all x ∈ G, then by taking y = e in (3.1) one has
δ
|f (x)| ≤ |a(e)|2 +
,
x ∈ G,
|Φ|
i.e. f is bounded, which is the desired contradiction.
The main results are the following theorems.
Theorem 3.2. Let δ > 0. Assume that f, a ∈ C(G) satisfy inequality (3.1) and f fulfills (∗).
Then
i) f, a are bounded
or
ii) f is unbounded and
XZ
(3.2)
a(e)
ǎ(xkϕ(y)k −1 )dωK (k) = |Φ|ǎ(x)ǎ(y),
x, y ∈ G,
ϕ∈Φ
K
where ǎ(x) = a(x−1 ), for x ∈ G.
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H YERS -U LAM STABILITY
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Proof. ii) Since f is unbounded then by Proposition 3.1 we have a(e) =
6 0. By using the fact
that f and a satisfy inequality (3.1) one has
XZ
f (xkϕ(y)k −1 )dωK (k) = |Φ|a(x)ǎ(y) + θ(x, y),
x, y ∈ G,
ϕ∈Φ
K
where |θ(x, y)| < δ. By taking y = e we get for all x ∈ G
|Φ|f (x) = |Φ|a(x)a(e) + θ(x, e),
so
(f (x) − a(x)a(e)) =
1
θ(x, e),
|Φ|
then we get
X Z
−1
f
(xkϕ(y)k
)dω
(k)
−
|Φ|f
(x)g(y)
< ε(y),
K
K
x, y ∈ G,
ϕ∈Φ
where
ǎ(y)
, and ε(y) = δ(1 + |g(y)|).
a(e)
In view of Theorem 2.3, we deduce that
XZ
a(e)
ǎ(xkϕ(y)k −1 )dωK (k) = |Φ|ǎ(x)ǎ(y),
x, y ∈ G.
g(y) =
ϕ∈Φ
K
The cases of f bounded follows from Proposition 3.1.
Theorem 3.3. Let δ > 0. Assume that f, a ∈ C(G) satisfy inequality (3.1) and f fulfills (∗).
Then either
s
δ
(3.3)
|a(x)| ≤ sup |f | +
,
x ∈ G,
|Φ|
s
|f (x)| ≤ |a(e)| sup |f | +
(3.4)
δ
δ
+
,
|Φ| |Φ|
x ∈ G,
or there exist x0 ∈ G such that |a(x0 )| > |a(e)| and a unique continuous function F : G −→ C
such that
a)
XZ
a(e)
F (xkϕ(y)k −1 )dωK (k) = |Φ|F (x)ǎ(y),
x, y ∈ G,
ϕ∈Φ
K
b) F − f is bounded and one has
|F (x) − f (x)| ≤
δ(|a(e)| + |a(x0 )|)
,
|Φ|(|a(x0 )| − |a(e)|)
x ∈ G.
Proof. If f is bounded, by using Theorem 3.2 and Proposition 3.1, we obtain the first case of
the theorem.
Now, let f be unbounded. Since a(e) 6= 0 it follows that
X Z
−1
f (xkϕ(y)k )dωK (k) − |Φ|f (x)g(y) < ε(y),
x, y ∈ G,
K
ϕ∈Φ
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B ELAID B OUIKHALENE , E LHOUCIEN E LQORACHI ,
AND
A HMED R EDOUANI
where
ǎ(y)
, and ε(y) = δ(1 + |g(y)|).
a(e)
Finally, by using Proposition 3.1 and Theorem 2.4 we get the rest of the proof.
g(y) =
4. A PPLICATIONS
The following theorems are a particular case of Theorem 3.3.
If K ⊂ Z(G), then we have
Theorem 4.1. Let δ > 0. Let f, a be a complex-valued functions on G such that f satisfies the
Kannappan condition (see [18])
x, y ∈ G
f (zxy) = f (zyx),
(4.1)
and the functional inequality
X
(4.2)
f (xϕ(y)) − |Φ|a(x)a(y) ≤ δ,
x, y ∈ G.
ϕ∈Φ
Then either
s
(4.3)
|a(x)| ≤
sup |f | +
s
(4.4)
|f (x)| ≤ |a(e)| sup |f | +
δ
,
|Φ|
x ∈ G,
δ
δ
+
,
|Φ| |Φ|
x ∈ G,
or there exist x0 ∈ G such that |a(x0 )| > |a(e)| and a unique function F : G −→ C such that
a)
X
a(e)
F (xϕ(y)) = |Φ|F (x)ǎ(y),
x, y ∈ G,
ϕ∈Φ
b) F − f is bounded and one has
|F (x) − f (x)| ≤
δ(|a(e)| + |a(x0 )|)
,
|Φ|(|a(x0 )| − |a(e)|)
x ∈ G.
If G is abelian then condition (4.1) holds. By taking Φ = {I} (resp. Φ = {I, −I}), we get
the following corollaries.
Corollary 4.2. Let δ > 0. Let f, a be complex-valued functions on G such that
(4.5)
|f (x − y) − a(x)a(y)| ≤ δ,
x, y ∈ G.
Then either
(4.6)
(4.7)
|a(x)| ≤
p
sup |f | + δ,
p
|f (x)| ≤ |a(e)| sup |f | + δ + δ,
x ∈ G,
x ∈ G,
or there exist x0 ∈ G such that |a(x0 )| > |a(e)| and a unique function F : G −→ C such that
a)
a(e)F (x + y) = F (x)ǎ(y),
x, y ∈ G,
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H YERS -U LAM STABILITY
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b) F − f is bounded and one has
δ(|a(e)| + |a(x0 )|)
,
(|a(x0 )| − |a(e)|)
|F (x) − f (x)| ≤
x ∈ G.
Corollary 4.3. Let δ > 0. Let f, a be a complex-valued functions on G such that
|f (x + y) + f (x − y) − 2a(x)a(y)| ≤ δ,
(4.8)
x, y ∈ G.
Then either
r
δ
sup |f | + ,
x ∈ G,
2
r
δ δ
|f (x)| ≤ |a(e)| sup |f | + + ,
x ∈ G,
2 2
or there exist x0 ∈ G such that |a(x0 )| > |a(e)| and a unique function F : G −→ C such that
a)
a(e)F (x + y) + a(e)F (x − y) = 2F (x)ǎ(y),
x, y ∈ G,
b) F − f is bounded and one has
δ(|a(e)| + |a(x0 )|)
|F (x) − f (x)| ≤
,
x ∈ G.
2(|a(x0 )| − |a(e)|)
|a(x)| ≤
(4.9)
If f (kxh) = χ(k)f (x)χ(h), k, h ∈ K and x ∈ G, where χ is a character of K, then we have
Theorem 4.4. Let δ > 0 and let χ be a character of K. Assume that (f, a) ∈ C(G) satisfy
f (kxh) = χ(k)f (x)χ(h), k, h ∈ K, x ∈ G,
Z Z
(4.10)
f (zkxhy)χ(k)χ(h)dωK (k)dωK (h)
K K
Z Z
=
f (zkyhx)χ(k)χ(h)dωK (k)dωK (h)
K
K
and
(4.11)
X Z
f (xkϕ(y −1 ))χ(k)dωK (k) − |Φ|a(x)a(y) ≤ δ,
K
x, y ∈ G.
ϕ∈Φ
Then either
s
|a(x)| ≤
(4.12)
sup |f | +
s
|f (x)| ≤ |a(e)| sup |f | +
(4.13)
δ
,
|Φ|
x ∈ G,
δ
δ
+
,
|Φ| |Φ|
x ∈ G,
or there exist x0 ∈ G such that |a(x0 )| > |a(e)| and a unique continuous function F : G −→ C
such that
a)
XZ
a(e)
F (xkϕ(y))χ(k)dωK (k) = |Φ|F (x)ǎ(y),
x, y ∈ G,
ϕ∈Φ
K
b) F − f is bounded and one has
|F (x) − f (x)| ≤
J. Inequal. Pure and Appl. Math., 6(2) Art. 32, 2005
δ(|a(e)| + |a(x0 )|)
,
|Φ|(|a(x0 )| − |a(e)|)
x ∈ G.
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B ELAID B OUIKHALENE , E LHOUCIEN E LQORACHI ,
AND
A HMED R EDOUANI
Corollary 4.5. Let δ > 0 and let χ be a character of K. Assume that (f, a) ∈ C(G) satisfy
f (kxh) = χ(k)f (x)χ(h), k, h ∈ K, x ∈ G,
Z Z
Z Z
f (zkxhy)χ(k)χ(h)dωK (k)dωK (h) =
f (zkyhx)χ(k)χ(h)dωK (k)dωK (h)
K
K
K
K
and
Z
Z
−1
(4.14) f (xky)χ(k)dωK (k) +
f (xky ))χ(k)dωK (k) − 2a(x)a(y) ≤ δ,
K
x, y ∈ G.
K
Then either
r
δ
sup |f | + ,
2
|a(x)| ≤
(4.15)
x ∈ G,
r
δ δ
+ ,
x ∈ G,
2 2
or there exist x0 ∈ G such that |a(x0 )| > |a(e)| and a unique continuous function F : G −→ C
such that
a)
Z
Z
a(e)
F (xky)χ(k)dωK (k) + a(e)
F (xky −1 ))χ(k)dωK (k) = 2F (x)ǎ(y),
x, y ∈ G,
|f (x)| ≤ |a(e)| sup |f | +
(4.16)
K
K
b) F − f is bounded and one has
|F (x) − f (x)| ≤
δ(|a(e)| + |a(x0 )|)
,
2(|a(x0 )| − |a(e)|)
x ∈ G.
Corollary 4.6. Let δ > 0 and let χ be a character of K. Assume that (f, a) ∈ C(G) satisfy
f (kxh) = χ(k)f (x)χ(h), k, h ∈ K, x ∈ G, and
Z
−1
(4.17)
x, y ∈ G.
f (xky ))χ(k)dωK (k) − a(x)a(y) ≤ δ,
K
Then either
|a(x)| ≤
p
sup |f | + δ,
x ∈ G,
p
|f (x)| ≤ |a(e)| sup |f | + δ + δ, x ∈ G,
or there exist x0 ∈ G such that |a(x0 )| > |a(e)| and a unique continuous function F : G −→ C
such that
a)
Z
(4.18)
a(e)
F (xky)χ(k)dωK (k) = F (x)ǎ(y),
x, y ∈ G,
K
b) F − f is bounded and one has
|F (x) − f (x)| ≤
δ(|a(e)|+ | a(x0 ) |)
,
(|a(x0 )| − |a(e)|)
x ∈ G.
Remark 4.7. If the algebra χωK ∗ M (G) ∗ χωK is commutative then the condition (∗) holds
[4]. Furthermore in the case where Φ = {I}, we do not need the condition (∗).
In the next theorem we assume that f is bi-K-invariant (i.e. f (hxk) = f (x), h, k ∈ K, x ∈
G), then we have
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Theorem 4.8. Let δ > 0 and assume that (f, a) ∈ C(G) satisfy f (kxh) = f (x), k, h ∈ K,
x ∈ G,
Z Z
Z Z
(4.19)
f (zkxhy)dωK (k)dωK (h) =
f (zkyhx)dωK (k)dωK (h)
K
K
K
K
and
X Z
−1
f (xkϕ(y ))dωK (k) − |Φ|a(x)a(y) ≤ δ,
K
(4.20)
x, y ∈ G.
ϕ∈Φ
Then either
s
|a(x)| ≤
(4.21)
sup |f | +
s
|f (x)| ≤ |a(e)| sup |f | +
(4.22)
δ
,
|Φ|
x ∈ G,
δ
δ
+
,
|Φ| |Φ|
x ∈ G,
or there exist x0 ∈ G such that |a(x0 )| > |a(e)| and a unique continuous function F : G −→ C
such that
a)
XZ
a(e)
F (xkϕ(y))dωK (k) = |Φ|F (x)ǎ(y),
x, y ∈ G,
ϕ∈Φ
K
b) F − f is bounded and one has
δ(|a(e)| + |a(x0 )|)
,
|Φ|(|a(x0 )| − |a(e)|)
|F (x) − f (x)| ≤
x ∈ G.
Corollary 4.9. Let δ > 0 and assume that (f, a) ∈ C(G) satisfy f (kxh) = f (x), k, h ∈ K,
x ∈ G,
Z Z
Z Z
f (zkxhy)dωK (k)dωK (h) =
f (zkyhx)dωK (k)dωK (h)
K
K
K
K
and
(4.23)
Z
Z
−1
f (xky)dωK (k) +
≤ δ,
f
(xky
))dω
(k)
−
2a(x)a(y)
K
K
x, y ∈ G.
K
Then either
r
δ
sup |f | + ,
x ∈ G,
2
r
δ δ
|f (x)| ≤ |a(e)| sup |f | + + ,
x ∈ G,
2 2
or there exist x0 ∈ G such that |a(x0 )| > |a(e)| and a unique continuous function F : G −→ C
such that
a)
Z
Z
a(e)
F (xky)dωK (k) + a(e)
F (xky −1 )dωK (k) = 2F (x)ǎ(y),
x, y ∈ G,
|a(x)| ≤
(4.24)
K
K
b) F − f is bounded and one has
|F (x) − f (x)| ≤
J. Inequal. Pure and Appl. Math., 6(2) Art. 32, 2005
δ(|a(e)| + |a(x0 )|)
,
2(|a(x0 )| − |a(e)|)
x ∈ G.
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B ELAID B OUIKHALENE , E LHOUCIEN E LQORACHI ,
AND
A HMED R EDOUANI
Corollary 4.10. Let δ > 0 and assume that (f, a) ∈ C(G) satisfy f (kxh) = f (x), k, h ∈ K,
x ∈ G, and
Z
−1
(4.25)
x, y ∈ G.
f (xky ))dωK (k) − a(x)a(y) ≤ δ,
K
Then either
|a(x)| ≤
p
sup |f | + δ,
x ∈ G,
p
|f (x)| ≤ |a(e)| sup |f | + δ + δ,
x ∈ G,
(4.26)
or there exist x0 ∈ G such that |a(x0 )| > |a(e)| and a unique continuous function F : G −→ C
such that
a)
Z
a(e)
F (xky)dωK (k) = F (x)ǎ(y),
x, y ∈ G,
K
b) F − f is bounded and one has
|F (x) − f (x)| ≤
δ(|a(e)| + |a(x0 )|)
,
(|a(x0 )| − |a(e)|)
x ∈ G.
Remark 4.11. If the algebra ωK ∗ M (G) ∗ ωK is commutative then the condition (∗) holds [4].
In the next corollary, we assume that G = K is a compact group.
Theorem 4.12. Let δ > 0 and let f, a be complex measurable and essentially bounded functions
on G such that f is a central function and (f, a) satisfy the inequality
X Z
−1
(4.27)
f (xtϕ(y)t )dt − |Φ|a(x)a(y) ≤ δ,
x, y ∈ G.
G
ϕ∈Φ
Then
s
(4.28)
|a(x)| ≤
sup |f | +
δ
,
|Φ|
and
s
|f (x)| ≤ |a(e)| sup |f | +
δ
δ
+
,
|Φ| |Φ|
for all x ∈ G.
Proof. Let f, a ∈ L∞ (G). Since f is central, then it satisfies the condition (∗) ([4], [6]). If f is
unbounded then a is a solution of the functional equation (3.2). In view of [15], we get the fact
that a is continuous. Since G is compact then a is bounded. Consequently f is bounded, which
is the desired property.
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