Solution of $f\bigl(x+f(y)\bigr)+f\bigl(y+f(x)\bigr)=2f\bigl(xf(y)\bigr)$

Question

What is the solution of following functional equation for $f:\mathbb R\to\mathbb R$? $$f\bigl(x+f(y)\bigr)+f\bigl(y+f(x)\bigr)=2f\bigl(xf(y)\bigr)$$

I tried something, but I am totally stuck. Following is my try.

1. If $f$ is a surjection (I cannot prove that), then there exists $\omega$ such that $f(\omega)=0$, then $2f(\omega)=f\bigl(\omega+f(\omega)\bigr)+f\bigl(\omega+f(\omega)\bigr)=2f\bigl(\omega f(\omega)\bigr)=2f(0)$, so $f(\omega)=f(0)$.

2. If $f$ is injective (but I cannot prove that either), then $\omega=0$. And further, for arbitrary $x,y$, $yf(x)=xf(y)$!

The question did not gave any properties. So, I don't know $f$ is injection, or surjection, or even continuous map!

Please help me. I am totally stuck. I estimated $f=0$ may be only answer, but it is just intuition.

Answer 1

It's straightforward to verify that any constant function $ f : \mathbb R \to \mathbb R $ satisfies $$ f \bigl ( x + f ( y ) \bigr ) + f \bigl ( y + f ( x ) \bigr ) = 2 f \bigl ( x f ( y ) \bigr ) \tag 0 \label 0 $$ for all $ x , y \in \mathbb R $. Indeed, they are the only solutions. To see that, let $ a = f ( 0 ) $ and $ b = f ( 1 ) $. Interchanging $ x $ and $ y $ in \eqref{0} and comparing to \eqref{0} itself, we have $$ f \bigl ( x f ( y ) \bigr ) = f \bigl ( y f ( x ) \bigr ) \tag 1 \label 1 $$ for all $ x , y \in \mathbb R $. In case $ a \ne 0 $, substituting $ \frac x a $ for $ x $ and $ 0 $ for $ y $ in \eqref{1}, we get $ f ( x ) = a $ for all $ x \in \mathbb R $, which means $ f $ must be constant in this case. From now on, consider the case $ a = 0 $. Putting $ y = 0 $ in \eqref{0}, we get $$ f \bigl ( f ( x ) \bigr ) = - f ( x ) \tag 2 \label 2 $$ for all $ x \in \mathbb R $. \eqref{2} implies $ f ( b ) = - b $, and then $ f ( - b ) = f \bigl ( f ( b ) \bigr ) = - f ( b ) = b $. Thus, setting $ x = - b $ and $ y = 1 $ in \eqref{0}, we have $ f ( 1 + b ) = 2 f \left ( - b ^ 2 \right ) $, while letting $ x = y = b $ in \eqref{0}, we also have $ f \left ( - b ^ 2 \right ) = 0 $. Therefore, $ f ( 1 + b ) = 0 $, which by putting $ x = y = 1 $ in \eqref{0} implies $ f ( b ) = 0 $, and thus $ b = 0 $. Finally, setting $ y = 1 $ in \eqref{1} and using \eqref{2}, we get $ f ( x ) = 0 $ for all $ x \in \mathbb R $, which means that $ f $ must be constant in this case, too.