Solving the Functional Equation $f big( x + y f ( x ) big) = f ( x ) f ( y )$

I want to find all functions $f : \mathbb R \to \mathbb R$ satisfying the functional equation $f \big( x + y f ( x ) \big) = f ( x ) f ( y )$, for all real numbers $x$ and $y$.

An interesting fact about the functional equation is a symmetry that is not at first sight visible. If one substitutes $x + z f ( x )$ for $x$ in the functional equation, the equation $f \big( x + z f ( x ) + y f ( x ) f ( z ) \big) = f ( x ) f ( y ) f ( z )$ follows, which is also the result of substituting $z + y f ( z )$ for $y$ in the original equation.

This is my attempt:

Clearly, for every constant real number $a$, $f ( x ) = a x + 1$ is a solution. The constant zero function is another solution. I conjecture that these are the only solutions.

It's easy to see that $f \big( x + y f ( x ) \big) = f \big( y + x f ( y ) \big)$ and thus if $f$ is injective, one must have $x + y f ( x ) = y + x f ( y )$, which letting $y = 1$ yields $f ( x ) = \big( f ( 1 ) - 1 \big) x + 1$.

Letting $x = y = 0$ in the functional equation, one gets $f ( 0 ) = 0$ or $f ( 0 ) = 1$. If $f ( 0 ) = 0$ then letting $y = 0$ in the functional equation one can find out that $f$ is the constant zero function. So from now on it's assumed that $f ( 0 ) = 1$.

If $f$ is differentiable, differentiating the functional equation with respect to $y$, one gets $f ( x ) f ' \big( x + y f ( x ) \big) = f ( x ) f ' ( y )$. Letting $y = 0$ in the last equation, one gets $f ( x ) \big( f ' ( x ) - f ' ( 0 ) \big) = 0$. Now since $f$ is differentiable at $0$, it is also continuous at $0$ and thus there is a positive real number $\delta$ such that if $- \delta < x < \delta$ then $f ( x ) > 0$. Hence if $- \delta < x < \delta$ then $f ' ( x ) = f ' ( 0 )$ which shows that on this interval $f ( x ) = x f ' ( 0 ) + 1$. For any $x$ such that $f ( x ) \ne 0$, one can substitute $\frac y { f ( x ) }$ for $y$ in the functional equation and get $f ( x + y ) = f ( x ) f \Big( \frac y { f ( x ) } \Big)$. Therefore if $- \delta f ( x ) < y < \delta f ( x )$ then $f ( x + y ) = f ( x ) + y f ' ( 0 )$.

I couldn't go further and I also couldn't avoid the differentiability condition.

Answer

Here is another solution that classifies all continuous solution. During the proof we exclude two trivial solutions $f \equiv 0$ and $f \equiv 1$ to avoid unnecessary case-division.

(Disclaimer: I heavily borrowed Tob Ernack's injectivity argument but tried a simpler proof. Although I have an independent proof, it is neither elegant nor shorter.)

Step 1. $Z: = f^{-1}(\{0\})$ is a non-empty closed interval.

$Z$ is obviously a closed set. Next, whenever $f(x) \neq 1$ we have

$$f\left(\tfrac{x}{1 - f(x)}\right) = f\left(x + \tfrac{x}{1 - f(x)}f(x)\right) = f(x)f\left(\tfrac{x}{1 - f(x)}\right).$$

and hence $\frac{x}{1 - f(x)} \in Z$. Then the assumption $f \not\equiv 1$ shows that $Z \neq \varnothing$. Finally, if $a, b \in Z$, then

$$f(x+af(x)) = f(x)f(a) = 0 \qquad \forall x \in \Bbb{R}$$

and the function $g(x) = x+af(x)$ takes values only in $Z$. Since $g(a) = a$ and $g(b) = b$, it follows from intermediate value theorem that $[a, b] \subseteq Z$.

Step 2. $f(x) = f(y) \neq 0$ implies $x = y$.

Let $p = (y-x)/f(x)$. From

$$f(x) = f(y) = f(x+pf(x)) = f(x)f(p),$$

we have $f(p) = 1$. Now assume that $p \neq 0$ Then

$$f(t+p) = f(p+tf(p)) = f(t)f(p) = f(t) \qquad \forall t \in \Bbb{R}$$

and hence $f$ is periodic. By Step 1, this implies that $Z$ is an interval which is unbounded in both direction, so $Z = \Bbb{R}$, contradicting the assumption $f \not\equiv 0$.

Step 3. There exists $a \in \Bbb{R}$ such that $f(x) = 1 + ax$ for $x \in \Bbb{R}\setminus Z$.

Use the assumption $f \not\equiv 0$ to choose $c \neq 0$ such that $f(c) \neq 0$ and let $a = \frac{f(c) - 1}{c}$. Then for any $x \notin Z$,

$$f(x + cf(x)) = f(x)f(c) = f(c+xf(c)) \neq 0$$

and so $x + cf(x) = c + xf(c)$. Solving this equation gives $f(x) = 1 + ax$.

Step 4. Except for the trivial solution $f \equiv 0$, only 2 types of solutions are possible:

• Case 1. $f(x) = 1+ax$ for some $a \in \Bbb{R}$


• Case 2. $f(x) = \max\{1+ax, 0\}$ for some $a \in \Bbb{R}$

Assume that $f \not\equiv 0$. Then it follows from $f(x) = f(x+0\cdot f(x)) = f(x)f(0)$ that $f(0) = 1$. This together with Step 3 shows that only Case 1 and 2 are possible solutions to the functional equation. We check that both cases are indeed solutions.

Since Case 1 is easy to verify, we focus on Case 2. Also, in Case 2, notice that $x \notin Z$ if and only if $1+ax > 0$. Then

• If $x \in Z$, then $f(x+yf(x)) = f(x) = 0 = f(x)f(y)$.




• If $x \notin Z$, then $x+yf(x) = x+y(1+ax) = x+y+axy$. So

  
  
  

  $$\begin{align*} x+yf(x) \notin Z &\quad \Leftrightarrow \quad 1+a(x+y+axy) = (1+ax)(1+ay) > 0 \\ &\quad \Leftrightarrow \quad 1+ay > 0. \end{align*}$$

  
  
  
  

  From this it is easy to check that $f(x+yf(x)) = f(x)f(y)$ holds for all $y$.

Therefore $f(x) = \max\{0,1+ax\}$ is also a solution of the functional equation.