Functional equation $ f \big( f ( x ) - 2 y ) = 2 x - 3 y + f \big( f ( y ) - x \big) $ - not so trivial, or is it?

Question

Consider the following functional equation: $$ f \big( f ( x ) - 2 y ) = 2 x - 3 y + f \big( f ( y ) - x \big) $$ for all real $ x $ and $ y $. Find $ f $.

It is easy to observe that the only polynomial solution of the FE is $ f ( x ) = x $. However, I haven't been able to prove that $ f ( x ) = x $ is the only solution. How do I prove or disprove it? In fact, what's the best way to approach the above functional equation? A rather general solution and less guesswork would be appreciated.

Thanks a lot!

Answer 1

Here's something I figured out after posting the question; so decided to put it up as an answer.

Let $P(x,y)$ be the assertion $f(f(x)-2y)=2x-3y+f(f(y)-x)$

Let $a=f(0)$

(1) : $P(f(x),x)$ $\implies$ $f(f(f(x))-2x)=2f(x)-3x+a$

(2) : $P(x,0)$ $\implies$ $f(f(x))-2x=f(a-x)$

Using (2), equality (1) becomes (3) : $f(f(a-x))=2f(x)-3x+a$

$P(a-x,0)$ $\implies$ $f(f(a-x))=f(x)-2x+2a$ Plugging this in (3), we get $\boxed{f(x)=x+a\quad\forall x}$ Which indeed is a solution, whatever is $a\in\mathbb R$

P.S. I request fellow Math SE users to share their different methods for solving this problem.

Answer 2

An alternative approach:

Assume $f(x)=px+q$ for some constants $p$ and $q$. Then

$p(px+q-2y) + q = 2x - 3y + p(py+q-x) + q$

$\Rightarrow p^2x -2py +pq + q = (2-p)x + (p^2-3)y +pq + q$

$\Rightarrow (p^2+p-2)x -(p^2 + 2p-3)y =0$

Since this must be true for all $x$ and $y$, we have $p^2+p-2=0$ and $p^2+2p-3=0$. So $p=1$ and $q$ can take any value.

Answer 3

Let y=0, we have f(f(x)) =2x+f(f(0)-x), which is f(f(x))=2x+f(f(0)- x). Let x=0, we have f(f(0)-2y) =-3y+f(f(y)), which is f(f(0)-2y)=-3y+f(f(y)). Since both x and y are random, we change this y to x, which is f(f(0)-2x) +3x=f(f(x)). So, f(f(x)) = 2x+f(f(0)-x) = f(f(0)-2x)+3x, which is the same price as f(f(0)-x)= f(f(0)-2x)+x. Now, we let “f(0)-2x”equals to t, so f(t+x)=f(t)+x. we set t=o, we have f(x)=x+f(0). And once you put f(x)=x+f(0) to the original equation, it works.