Finding all $f: \mathbb{R} \to \mathbb{R}$ such that $f\bigl(xf(y)+y\bigr)+f\bigl(-f(x)\bigr)=f\bigl(yf(x)-y\bigr)+y $

functional-equations

Find all $f: \mathbb{R} \to \mathbb{R}$ such that
$$f\bigl(xf(y)+y\bigr)+f\bigl(-f(x)\bigr)=f\bigl(yf(x)-y\bigr)+y $$
for all $x,y \in \mathbb{R}$.

Help me solving this. My expectation of the answer is $f(x) = x+1$.

My try:
$$
P(x, y): f\bigl(xf(y)+y\bigr)+f\bigl(-f(x)\bigr)=f\bigl(yf(x)-y\bigr)+y \text. \\
P(x, 0): f\bigl(xf(0)\bigr)+f\bigl(-f(x)\bigr) = f(0) \text. \\
P(0, y): f(y)+f\bigl(-f(0)\bigr) = f\Bigl(y\bigl(f(0)-1\bigr)\Bigr)+y \text. \\
P(0, 0): f(0)+f\bigl(-f(0)\bigr) = f(0) \implies f\bigl(-f(0)\bigr)=0 \text. \\
\text {Assume) } f(a)=1 \text. \\
P(x, a): f(x+a) + f\bigl(-f(x)\bigr)=f\bigl(af(x)-a\bigr)+a \text. \\
x = 0 \text; \ 1 = f\Bigl(a\bigl(f(0)-1\bigr)\Bigr)+a \text. \\
x = a \text; \ f(2a)+f(-1) = f(0)+a \text. \\
$$

Best Answer

$$f(xf(y)+y)+f(−f(x))=f(yf(x)−y)+y$$ Define $f(0)=c$ for some $c\in \mathbb{R}$.

  • $(x,y)\equiv (0,0)$ $$f(0)+f(-f(0))=f(0)\implies f(-c)=0$$
  • $(x,y)\equiv (-c,-c)$ $$f(-c)+c=f(c)-c\implies f(c)=2c$$
  • $(x,y)\equiv (0,c)$ $$f(c)+f(-c)=f(c^2-c)+c\implies c=f(c^2-c)$$
  • $(x,y)\equiv(0,-c)$ $$f(-c)+f(-c)=f(-c^2+c)-c\implies c=f(-(c^2-c))$$
  • $(x,y)\equiv(-c,c^2-c)$ $$f(-c)+c=f(-c^2+c)+c^2-c \implies c^2=c\implies c\in\{0,1\}$$

If $c=f(0)=0$,

  • $(x,y)\equiv(x,0)$ $$f(-f(x))=0\;, \;\;\; \forall x\in \mathbb{R}$$
  • $(x,y)\equiv(-x/f(x),x)$ $$-f(xf(-x/f(x))-x)=x \\ f(-f(xf(-x/f(x))-x))=f(x) \\ \;\;\;\therefore\; f(x)=0\;, \;\;\; \forall x\in \mathbb{R}\;\;$$ But $f(x)=0$ for real $x$ is not a solution of the original F.E.

If $c=f(0)=1$,

  • $(x,y)\equiv(0,x)$ $$f(x)=f(cx-x)+x\implies f(x)=x+1.$$ Plugging this back in the original F.E, we verify that $\boxed{f(x)=x+1}\; \forall x\in \mathbb{R}$ is the only solution.
Related Question