Find all real-valued functions f(x)

The Problem:
	Find all real-valued functions f(x) such that f(x3+y3)=x2f(x)+yf(y2) for all real numbers x and y .

The solution:

It is easy to see that for any real number m , the function f(x)=mx satisfies our functional equation:
On the left we have

f(x3+y3)=m(x3+y3),

while on the right,

x2f(x)=x2⋅mx and yf(y2)=y⋅my2.

We must now show that no other function satisfies the given identity. To that end, suppose that f(x) is a function that satisfies

f(x3+y3)=x2f(x)+yf(y2).

(1)

Then f(x3)=f(x3+0)=(1)x2f(x) , and f(x3)=f(0+x3)=(1)xf(x2) . That is,

f(x3)=x2f(x)=xf(x2).

(2)

When x=0 equation (2) tells us that f(0)=0 ; when x≠0 we can divide through by x to get xf(x)=f(x2) . Thus, for all real x ,

f(x2)=xf(x).

(3)

Furthermore, equation (2), when put back into (1), tells us that

f(x3+y3)=x2f(x)+yf(y2)=(2)f(x3)+f(y3)

for all real numbers x and y . It follows that for any reals a and b we can set x=a1/3 and b=y1/3 , and deduce that

f(a+b)=f(a)+f(b).

(4)

We recognize (4) to be Cauchy's functional equation, which has infinitely many other solutions in addition to the linear solutions f(x)=mx . Several of our correspondents felt obliged to make further assumptions (such as continuity) to eliminate the unwanted nonlinear solutions. The majority of our solvers saw, however, that equation (3) already does the job without the help of further assumptions! Note on the one hand that for all x ,

f((x+1)2)=(3)(x+1)f(x+1)=(4)(x+1)(f(x)+f(1))=xf(x)+xf(1)+f(x)+f(1).

On the other hand,

f((x+1)2)=f((x2+x)+(x+1))=(4)=(4)=(3)f((x2+x))+f(x+1)f(x2)+f(x)+f(x)+f(1)xf(x)+f(x)+f(x)+f(1).

It follows that

xf(x)+xf(1)+f(x)+f(1)=xf(x)+f(x)+f(x)+f(1),

whence, f(x)=f(1)⋅x . Since f(1) is an arbitrary real number we have proved that if f(x) satisfies equation (1) for all real numbers x , then there exists a real number m for which f(x)=mx .

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