Lemma 2.1.
Let
and
be linear spaces and let
be real numbers with
and
for some
Assume that a mapping
satisfies the functional equation (1.9) for all
Then the mapping
is Cauchy additive. Moreover,
for all
and all 
Proof.
Since
putting
in (1.9), we get
Without loss of generality, we may assume that
Letting
in (1.9), we get
for all
Letting
in (2.1), we get
for all
Similarly, by putting
in (2.1), we get
for all
It follows from (2.1), (2.2) and (2.3) that
for all
Replacing
and
by
and
in (2.4), we get
for all
Letting
in (2.5), we get that
for all
So the mapping
is odd. Therefore, it follows from (2.5) that the mapping
is additive. Moreover, let
and
Setting
and
for all
in (1.9) and using the oddness of
we get that 
Using the same method as in the proof of Lemma 2.1, we have an alternative result of Lemma 2.1 when 
Lemma 2.2.
Let
and
be linear spaces and let
be real numbers with
for some
Assume that a mapping
with
satisfies the functional equation (1.9) for all
Then the mapping
is Cauchy additive. Moreover,
for all
and all 
We investigate the generalized Hyers-Ulam stability of a generalized Euler-Lagrange type additive mapping in Banach spaces.
Throughout this paper,
will be real numbers such that
for fixed 
Theorem 2.3.
Let
be a mapping satisfying
for which there is a function
such that
for all
and
Then there exists a unique generalized Euler-Lagrange type additive mapping
such that
for all
Moreover,
for all
and all 
Proof.
For each
with
let
in (2.8), tthen we get the following inequality
for all
For convenience, set
for all
and all
Letting
in (2.10), we get
for all
Similarly, letting
in (2.10), we get
for all
It follows from (2.10), (2.12) and (2.13) that
for all
Replacing
and
by
and
in (2.14), we get that
for all
Putting
in (2.15), we get
for all
Replacing
and
by
and
in (2.15), respectively, we get
for all
It follows from (2.16) and (2.17) that
for all
where
It follows from (2.6) that
for all
Replacing
by
in (2.18) and dividing both sides of (2.18) by
we get
for all
and all
Therefore, we have
for all
and all integers
It follows from (2.20) and (2.22) that the sequence
is Cauchy in
for all
and thus converges by the completeness of
Thus we can define a mapping
by
for all
Letting
in (2.22) and taking the limit as
in (2.22), we obtain the desired inequality (2.9).
It follows from (2.7) and (2.8) that
for all
Therefore, the mapping
satisfies (1.9) and
Hence by Lemma 2.2,
is a generalized Euler-Lagrange type additive mapping and
for all
and all 
To prove the uniqueness, let
be another generalized Euler-Lagrange type additive mapping with
satisfying (2.9). By Lemma 2.2, the mapping
is additive. Therefore, it follows from (2.9) and (2.20) that
So
for all 
Theorem 2.4.
Let
be a mapping satisfying
for which there is a function
satisfying (2.6), (2.7) and
for all
and all
Then there exists a unique
-linear generalized Euler-Lagrange type additive mapping
satisfying (2.9) for all
Moreover,
for all
and all 
Proof.
By Theorem 2.3, there exists a unique generalized Euler-Lagrange type additive mapping
satisfying (2.9) and moreover
for all
and all 
By the assumption, for each
, we get
for all
. So
for all
and all
Since
for all
and 
for all
and all 
By the same reasoning as in the proofs of [41, 43],
for all
and all
Since
for all
the unique generalized Euler-Lagrange type additive mapping
is an
-linear mapping.
Corollary 2.5.
Let 
and
be real numbers such that
and
for all
where
Assume that a mapping
with
satisfies the inequality
for all
and all
. Then there exists a unique
-linear generalized Euler-Lagrange type additive mapping
such that
for all
where
Moreover,
for all
and all 
Proof.
Define
and apply Theorem 2.4.
Corollary 2.6.
Let
with
Assume that a mapping
with
satisfies the inequality
for all
and all
Then there exists a unique
-linear generalized Euler-Lagrange type additive mapping
such that
for all
Moreover,
for all
and all 
Proof.
Define
Applying Theorem 2.4, we obtain the desired result.
Theorem 2.7.
Let
be a mapping satisfying
for which there is a function
such that
for all
and
Then there exists a unique generalized Euler-Lagrange type additive mapping
such that
for all
Moreover,
for all
and all 
Proof.
By a similar method to the proof of Theorem 2.3, we have the following inequality
for all
where
It follows from (2.36) that
for all
Replacing
by
in (2.40) and multiplying both sides of (2.40) by
we get
for all
and all
Therefore, we have
for all
and all integers
It follows from (2.42) and (2.44) that the sequence
is Cauchy in
for all
and thus converges by the completeness of
Thus we can define a mapping
by
for all
Letting
in (2.44) and taking the limit as
in (2.44), we obtain the desired inequality (2.39).
The rest of the proof is similar to the proof of Theorem 2.3.
Theorem 2.8.
Let
be a mapping with
for which there is a function
satisfying (2.36), (2.37) and
for all
and all
Then there exists a unique
-linear generalized Euler-Lagrange type additive mapping
satisfying (2.39) for all
Moreover,
for all
and all 
Proof.
The proof is similar to the proof of Theorem 2.4.
Corollary 2.9.
Let
and
be real numbers such that
and
for all
where
Assume that a mapping
with
satisfies the inequality
for all
and all
Then there exists a unique
-linear generalized Euler-Lagrange type additive mapping
such that
for all
where
Moreover,
for all
and all 
Proof.
Define
Applying Theorem 2.8, we obtain the desired result.
Corollary 2.10.
Let
with
Assume that a mapping
with
satisfies the inequality
for all
and all
Then there exists a unique
-linear generalized Euler-Lagrange type additive mapping
such that
for all
Moreover,
for all
and all 
Proof.
Define
Applying Theorem 2.8, we obtain the desired result.
Remark 2.11.
In Theorems 2.7 and 2.8 and Corollaries 2.9 and 2.10 one can assume that
instead of 
For the case
in Corollaries 2.5 and 2.9, using an idea from the example of Gajda [56], we have the following counterexample.
Example 2.12.
Let
be defined by
Consider the function
by the formula
It is clear that
is continuous and bounded by 2 on
. We prove that
for all
and all
If
or
then
Now suppose that
Then there exists a nonnegative integer
such that
Therefore
Hence
for all
From the definition of
and (2.56), we have
Therefore
satisfies (2.54). Let
be an additive mapping such that
for all
Then there exists a constant
such that
for all rational numbers
So we have
for all rational numbers
Let
with
If
is a rational number in
, then
for all
So
which contradicts with (2.61).