Find all functions $f: \mathbb R \to \mathbb R$ satisfying $f(xf(y) + f(x+y)) = y(f(x)+1)+f(x)$

Question

Can you tell me if my solution is so far correct and help me complete the second case?

The statement of the problem :

Find all functions $f: \mathbb R \to \mathbb R$ with the following property: $$ f(xf(y) + f(x+y)) = y(f(x)+1) + f(x) , \forall x,y \in \mathbb R. $$

My approach :

For $x = y = 0$ we get $$ f(f(0)) = f(0). $$ For $x = 0$ we get

$$ f(f(y))=y(f(0)+1)+f(0) , \forall y \in \mathbb R. $$

Now the function $y \mapsto ay+c$ is bijective , where $a$ and $c$ real constants and $a \neq 0$ . So , for $f(0) \neq -1$ we get $$ \begin{align} &y \mapsto y(f(0)+1)+f(0) \text{ is bijective,}\\ \implies &y \mapsto f(f(y)) \text{ is bijective,}\\ \implies &f \text{ is injective and $f$ is surjective,}\\ \implies &f \text{ is bijective}. \end{align} $$ Making $y = 0$ in the main equation we get $$f(f(x))=f(x) ,\forall x \in \mathbb R. $$ And since $f$ is bijective we obtain that $$f(x)=x ,\forall x \in \mathbb R, $$ which satisfies the initial equation.

Now we have left the case where $y \mapsto y(f(0)+1)+f(0)$ is not bijective , more precisely when $$ \begin{align} f(0) &= -1 \\ \implies f(f(y)) &= f(0) = -1 ,\forall y \in \mathbb R. \end{align} $$ Making $y=0$ we get $$f(-x+f(x))=f(x),\forall x \in \mathbb R. $$ From here I didn't know what to do. I think that we either have to show that $f \equiv -1$ or reach a contradiction. I hope that at least the first part is correct.

I am open to all your suggestions and solutions. Thank you!

Answer 1

Note that $f(x) \equiv -1$ is also a solution of the functional equation

$$ f(xf(y) + f(x+y)) = y(f(x)+1) + f(x) , \qquad \forall x,y \in \mathbb{R} \tag{1}\label{fe}$$

To examine other possibilities, assume that $f$ is not identically $-1$.

1. Let $a \in \mathbb{R}$ be an arbitrary element satsifying $f(a) \neq -1$. Then plugging $x = a$ to $\eqref{fe}$ gives $$ f(af(y) + f(a+y)) = y(f(a) + 1) + f(a). $$ This shows that $f$ is surjective.

2. Now we plug $x = 0$ to $\eqref{fe}$. Then $$ f(f(y)) = y(f(0) + 1) + f(0). $$ Since $y \mapsto f(f(y))$ is surjective, the same is true for the right-hand side. In particular, $f(0) \neq -1$ and this in fact shows that $f$ is bijective.

3. Finally, we plug $y = 0$ to $\eqref{fe}$. Then $$ f(xf(0) + f(x)) = f(x), $$ hence by the injectivity of $f$ we get $$ x f(0) + f(x) = x, \qquad\text{i.e.,}\qquad f(x) = (1 - f(0)) x. $$ This then shows that $f(0) = 0$ and therefore $f(x) = x$.

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Conclusion. The functional equation $\eqref{fe}$ has two solutions: $f(x) \equiv -1$ or $f(x) = x$.