THEOREM

THEOREM: 5.1.2
Let f:E_1?E_2 be a mapping from a normed vector space E_1 into banach space E_2 such that f(tx) is a continuous in t?R and for each fixed x?E_1 assume that there exist a constant ?>0 and p?0,1)? with subject to the inequality
?f(x+y)-f(x)-f(y)???(?x?^p+?y?^p ) (5.4)
For all x,y?E_1, then there exists a unique linear mapping T:E_1?E_2 such that
?f(x)-T(x)??2?/(2-2^p ) ?x?^p (5.5)
For all x?E_1.
THEOREM: 5.1.3
Let f:E_1?E_2 be a mapping from a normed vector space E_1 into banach space E_2 subject to the inequality
?f(x+y)-f(x)-f(y)???(?x?^p ?y?^p ) (5.6)
For all x,y?E_1, where ? and p are constants with ?>0 and 0?p?1/2. Then there exists a limit
L(x)=lim?(n??)??(f(2^n x))/2^n ? (5 .7)
For all x?E_1 and L:E_1?E_2 is the unique additive mapping which satisfies
?f(x)-L(x)???/(2-2^2p ) ?x?^2p (5.8)
For all x?E_1. If p>1/2 the inequality (6) holds for x,y?E_1 and the limit
A(x)=lim?(n??)??2^n f(x/2^n ) ?(5.9)
Exists for all x?E_1 and A:E_1?E_2 is a unique additive mapping which satisfies
?f(x)-A(x)???/(2^2p-2) ?x?^2p (5.10)
For all x?E_1.
K.W. Jun and H.M. Kim introduced the following cubic functional equation
f(2x+y)+f(2x-y)=2f(x+y)+f(x-y)+12f(x) (5.11)
x,y?X , where f is a mapping from a real vector space X into a real vector space Y. They established the general solution and the generalized Hyres-Ulam-Rassias stability for the functional equation(5.11). The function f(x)=x^3 satisfies the functional equation (5.11) which is thus called a cubic functional equation. Every solution of a cubic functional equation is said to be a cubic function. Jun and Kim proved that mapping f from real vector space X and Y is solution of (5.11) if and only if there exists a unique function C:X×X×X?Y such that f(x)=C(x,x,x) for all x?X and C is symmetric for fixed one variable an additive for fixed two variables.
W.G. Park and J.H. Bae considered the following functional equation
f(x+2y)+f(x-2y)=4f(x+y)+f(x-y)+24f(y)-6f(x)(5.12)
They proved that the function f between real vector space X and Y is a solution of (6.12) if and only if there exists a unique symmetric multi-additive function B:X×X×X×X?Y such that f(x)=B(x,x,x,x) for all x?X. It is easy to show that the function f(x)=x^4 satisfies the functional equation(5.12), which is called a quartic functional equation and every solution of the quartic functional equation is said to be a quartic function.In this paper, i deal with the new generalized functional equation deriving from cubic and quartic functions
f(x+ny)+f(x-ny)=n^2 f(x+y)+f(x-y)-2(n^2-1)f(x)
+2/n(n+1)f(ny)-n^3 f(y) (5.13)
For any real number n with n?0 in banach spaces. This equation is called mixed type cubic and quartic functional equation because the function f(x)=ax^3+bx^4 where a and b are arbitrary constants, becomes the solution of the equation (5.13). The general solution of the functional equation(5.13) is discussed in the generalized Hyres-Ulam-Rassias stability of the functional equation (5.13) is investigated in section 3.
It may be observed that the above cubic and quartic functional equation (5.13) is not been dealt so far by any of the authors. So it is of great interest that the authors of this paper is much interested in finding the general solution and generalized Hyres-Ulam-Rassias stability of the generalized cubic and quartic functional equation(5.13) in Banach Spaces.
In this paper, i consider the most general case of the functional equation (5.13) for any real number n with n?0 and proceed to find its general solution.
5.2.THE GENERAL SOLUTION OF THE FUNCTIONAL EQUATION:
In this section, I establish the general solution of functional equation (5.13).
THEOREM: 5.2.1
Let X,Y be vector Spaces and let f:X?Y be a function which satisfies
f(x+ny)+f(x-ny)=n^2 f(x+y)+f(x-y)-2(n^2-1)f(x)+
2/n (n+ 1) f(ny)-n^3 f(y) for any real number n with n?0 in Banach spaces then the following assertions hold
If f is odd function then the function f is cubic functional equation
If f is even function then the function f is quartic functional equation
PROOF:
Put x=y=0 in(5.13), get f(0)=0. Setting x=0 in (5.13) and by oddness of f obtain
f(ny)=n^3 f(y) (5.14)
for all y?Y. Replacing n by 1/n in (5.14), get
f(1/n y)=1/n^3 f(y) (5.15)
for all y?Y. Substituting (5.14) in (5.13), get
f(x+ny)+f(x-ny)=n^2 f(x+y)+f(x-y)-2(n^2-1)f(x) (5.16)
for all x,y?X. Replacing x by nx in (5.16), get
f(nx+ny)+f(nx-ny)=n^2 f(nx+y)+(nx-y)-2(n^2-1)f(nx)
(5.17)
for all x,y?X. Again replacing y by x in (5.14), get
f(nx)=n^3 f(x)(5.18)
for all x?X. Applying equation in (5.18) in (5.17), get
f(nx+y)+f(nx-y)=nf(x+y)+f(x-y)+2n(n^2-1)f(x)(5.19)
for all x,y?X. Replacing x by (x+y) in (5.16), get
f(x+(n+1)y)+f(x-(n-1)y)=n^2 f(x+2y)+f(x)
+2(n^2-1)f(x+y) (5.20)
for all x,y?X. Replacing x by (x-y) in (5.16), get
f(x-(n+1)y)+f(x+(n-1)y)=n^2 f(x-2y)+f(x)
-2(n^2-1)f(x-y) (5.21)
for all x,y?X . Adding (5.20) and (5.21), get
?(f(x+(n+1)y)+f(x-(n-1)[email protected](x+(n-1)y+f(x-(n-1)y)= ?(n^2 f(x+2y)+f(x-2y)+2n^2 f(x)@-2(n^2-1)f(x+y)+f(x-y) ) (5.22)
for all x,y?X. Further replacing y by (x+y) in (5.16), obtain
f((n+1)x+ny)+f((1-n)x-ny)=n^2 f(2x+y)-f(y)
-2(n^2-1)f(x) (5.23)
for all x,y?X and replacing y by (-x+y) in (5.16), get
f((1-n)x+ny)+f((1+n)x-ny)=n^2 f(y)+f(2x-y)
-2(n^2-1)f(x) (5.24)
for all x,y?X. Adding (5.23) and (5.24), get
?(f((n+1)x+ny)+f((n+1)x-ny)[email protected]((1-n)x+ny)+f((1-n)x-ny) )=n^2 f(2x+y)+f(2x-y- ¬4(n^2-1)f(x) (5.25)
for all x,y?X. Interchanging x by y in (5.25)
?(f(nx+(n+1)y)+f(-nx+(n+1)y)[email protected](nx+(1-n)y)+f(-nx+(1-n)y) )
=n^2 f(x+2y)-f(x-2y-4(n^2-1)f(y) (5.26)
for all x,y?X. Simplifying (5.26) and using oddness, get
?(f(nx+(n+1)y)-f(x-(n+1)y)[email protected](nx-(n-1)y)-f(nx+(n-1)y) )=n^2 f(x+2y)-f(x-2y
-4(n^2-1)f(y) (5.27)
for all x,y?X. Subtracting (5.20) from (5.21)get
?(f(x+(n+1)y)-f(x-(n+1)y)[email protected](x-(n-1)y)-f(x+(n-1)y) )=n^2 f(x+2y)-f(x-2y
-2(n^2-1)f(x+y)-f(x-y) (5.28)
for all x,y?X. Replacing x by nx in (5.28), get
?(f(nx+(n+1)y)-f(nx-(n+1)y)[email protected](nx-(n-1)y)-f(nx+(n-1)y) )=?(n^2 f(nx+2y)-f([email protected](n^2-1)f(nx+y)-f(nx-y) )
(5.29)
for all x,y?X. By comparing (5.27) and (5.29), get
n^2 f(x+2y)-f(x-2y-4(n^2-1)f(y)=?(n^2 f(nx+2y)-f([email protected](n^2-1)f(nx+y)-f(nx-y) )
(5.30)
for all x,y?X. Now, by interchanging x and y in (5.30), get
n^2 f(x+2y)-f(x-2y-4(n^2-1)f(y)=?(n^2 f(2x+ny)+f([email protected](n^2-1)f(x+ny)-f(x-ny) ) (5.31)

for all x,y?X. Substituting equation (5.16) in (5.31),
f(2x+y)+f(2x-y)=
?(f(2x+ny)+f(2x-ny)-2(n^2-1)@f(x+y)+f(x-y)+4(n^2-1)f(x)) (5.32)
for all x,y?X. Simplifying (5.32), get
f(2x+ny)+f(2x-ny)=
?(f(2x+y)+f(2x-y)-2(n^2-1)@f(x+y)+f(x-y)-4(n^2-1)f(x)) (5.33)
for all x,y?X. Replacing x by 2x in (5.16) get
f(2x+ny)+f(2x-ny)=n^2 f(2x+y)+f(2x-y)16(n^2-1)f(x) (5.34)
for all x,y?X. Equating (6.33) and (6.34), get
n^2 f(2x+y)+f(2x-y)-2(n^2-1)f(x)= ?(f(2x+y)+f(2x-y)+2(n^2-1)@f(x+y)+f(x-y)-4(n^2-1)f(x) )(5.35)
for all x,y?X. Simplifying (5.35) arrive a cubic functional equation
f(2x+y)+f(2x-y)=2f(x+y)+f(x-y)+12f(x)
which proves our first part of theorem.
(b) putting x=y=0 in (5.13), get f(0)=0, setting in x=0 in (5.13) and using evenness of f, we get
f(ny)=n^4 f(y) (5.36)
for all y?X. Substitute (5.36) in (5.13), we get
f(x+ny)+f(2x-ny)=n^2 f(x+y)+f(x-y)-2(n^2-1)f(x)
+2n^2 (n^2-1)f(y) (5.37)