# 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.

• One comment: if $f$ is differentiable at $0$, then it is differentiable everywhere. May 28 '17 at 20:11
• @Batominovski I can't see why. May 28 '17 at 20:18
• Likewise for continuity May 28 '17 at 20:23
• Putting $y=0$ into the equation we infer that $f(x)=f(x)f(0)$, so either $f$ is constantly $0$, or $f(0)=1$. May 28 '17 at 20:30

Here's a different solution. $$f(x)=\begin{cases} 1\quad\text{if}\quad x\in\mathbb{Q}\\ 0\quad\text{otherwise} \end{cases}$$ I'm quite sure there is a number of variations of this theme.

My suspicion is that without requiring continuity or differentiability, the relation will not be enough to limit the available solutions for $f(x)$ for all $x\in\mathbb{R}$, and that there might thus be an uncountable number of solutions which will be almost nowhere continuous (possibly requiring the axiom of choice to construct).

Here are a few observation on the continuous/differential case. I haven't properly proof read them, so I may have missed something.

Let's first assume that $f(0)=1$, otherwise $f(x)=0$ for all $x$.

Let's take some $x$ with $f(x)\not=0$. Because of the relation $f(x+yf(x))=f(x)f(y)$, if $f(y)$ is continuous at $y=0$, it follows that $f$ is continuous at $f(x)$.

Similarly, if $f(y)$ is differentiable at $y=0$ with $f'(0)=a$, it follows that $f$ is differentiable at $x$. Differentiation of $f(x+yf(x))=f(x)f(y)$ with $y$ at $y=0$ yields $f'(x)=f'(0)=a$. Again, this requires that $f(x)\not=0$.

So, basically, if $f(0)=0$ and $f'(0)=a$, we know that $f'(x)=a$ whenever $f(x)\not=0$, from which follows that $f(x)=ax+1$ in the segment with $ax+1\ge0$.

If $a=0$, this makes $f(x)=1$ for all $x$. For any other $a$, $f(x)=ax+1$ is a possible solution.

A solution already mentioned by others is $f(x)=\max(ax+1,0)$.

A last option, assuming $a\not=0$, is that there is a $b>1$ so that $$f(x)=\begin{cases} ax+1\quad\text{if}\quad ax+1>0\\ ax+b\quad\text{if}\quad ax+b<0\\ 0\quad\text{otherwise} \end{cases}$$ since $f'(x)=a$ whenever $f(x)\not=0$. However, for any $y$ with $f(y)=0$, we have $f(x+yf(x))=0$ for all $x$. This means $x+yf(x)$ must be bound as $x\rightarrow\infty$ and $-\infty$, which can only happen if $y=-1/a$. So this alternative is ruled out.

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

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.

• Unfortunately... I reread my proof this morning and realized my proof of Lemma 1 was incorrect. I said that if $f(T) = 1$ then $1 = f(T) = f(T/2 + T/2) = f(T/2 + T/2f(T)) = f(T/2)f(T)$ but the functional equation only works if we had $f(T/2 + Tf(T/2))$ or $f(T + T/2f(T))$, none of which lead to the same result directly. I should probably delete my answer, at least until a correction is found. May 29 '17 at 13:30
• In fact, it seems like you proved things in the right order, first showing the zero set is a closed interval (my Lemma 2) and then using that to prove the injectivity (my Lemma 1). May 29 '17 at 13:41
• @TobErnack, That is indeed a subtle mistake. I hope you come up with a fix, as I really like your idea of using injectivity and want to give a genuine credit to your solution. May 29 '17 at 13:57
• By the way, I myself borrowed the injectivity idea, from the OP himself. May 29 '17 at 14:01
• Could we also have $\min\{0, 1 + ax\}$ as a solution? May 29 '17 at 16:43

This paper seems to show that the only continuous solutions are \begin{align} f(x) &= 0 \\ f(x) &= 1+ax \\ f(x) &= \begin{cases}1+ax & x>-\frac{1}{a} \\ 0&x\leq-\frac{1}{a} \end{cases} \\ f(x) &= \begin{cases}1+ax & x<\frac{1}{a} \\ 0&x\geq\frac{1}{a} \end{cases} \end{align} where $a$ is any real, and $a$ is positive for the final two cases. Unfortunately it's in French otherwise I would reproduce the argument here.

• It seems that the paper has made a typo since the last form of $f$ is not continuous. I think it should be $1 - a x$ for $x < \frac 1 a$. May 28 '17 at 22:19
• Yeah, it's certainly a typo. It may be $x < -\frac{1}{a}$ instead in the final case instead, though I haven't checked May 28 '17 at 22:22
• For $a > 0$, changing $1 + a x$ to $1 - a x$ gives a solution, indeed. May 28 '17 at 22:24