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

Question

Find all continuous functions $ f : \mathbb R \to \mathbb R $ satisfying: $$ f ( x y ) + f ( x + y ) = f ( x y + x ) + f ( y ) $$ for all $ x , y \in \mathbb R $.

Let $ P ( x , y ) $ denote the functional equation. I have observed that:

$ P ( y , x ) - P ( x , y ) $: $ f ( x y + y ) + f ( x ) = f ( x y + x ) + f ( y ) $.

$ P ( x , 1 ) $: $ f ( x ) + f ( x + 1 ) = f ( 2 x ) + f ( 1 ) $.

$ P ( x + 1 , 1 ) $: $ f ( x + 1 ) + f ( x + 2 ) = f ( 2 x + 2 ) + f ( 1 ) $.
So $ f ( x + 2 ) - f ( x ) = f ( 2 x + 2 ) - f ( 2 x ) $.

Let $ g ( x ) = f ( x + 2 ) - f ( x ) $. We have $$ g ( 2 x ) = g ( x ) = g \left ( \frac x 2 \right ) = \dots = g ( 0 ) = c $$ because $ g $ is a continuous functions. After this, I have no idea towards solving the problem.

Answer 1

Let's denote the functional equation by $(*)$. Note that if $f(x)$ satisfies to $(*)$ then so do the functions of the form $af(x)+b$, for any real numbers $a$ and $b$. So we can assume that $f(0)=0$ by changing the function linearly, if needed. Now, as you got, we have $f(x+2)-f(x)=g(x)=g(0)=f(2)-f(0)=f(2)$.

On the other hand, if we take $y=2$ in $(*)$, we'll get $f(2x)+f(x+2)=f(3x)+f(2)$, and if we substitute the value of $f(2)$ from the above equation, we'll get the following

$f(2x)+f(x+2)=f(3x)+f(x+2)-f(x) \implies f(2x)+f(x)=f(3x)$. Now substitute $x+1$ for $x$: $f(3x+3)=f(3(x+1))=f(2(x+1))+f(x+1)=f(2x+2)+f(x+1)=f(2x)+f(2)+f(x+1)=f(2x)+f(x)-f(x)+f(2)+f(x+1)=f(3x)+f(x+1)-f(x)+f(2)$.

So, we'll have $f(x+1)-f(x)+f(2)=f(3x+3)-f(3x)=f(3x+1)+f(2)-f(3x) \implies f(x+1)-f(x)=f(3x+1)-f(3x)$. Therefore, in the same way as you did, we can get $f(x+1)-f(x)=f(1)-f(0)=f(1)$ using the continuity of the function $f$. Using simple induction, we'll get

$(1)\space f(x+n)-f(x)=nf(1)$ for any integer $n$.

In particular, we have $f(n)=nf(1)$ for all $n\in\mathbb Z$.

If we plug in $y=n$ in $(*)$ assuming $n$ is an integer, we get

$f(nx)+f(x+n)=f((n+1)x)+f(n)\implies f((n+1)x)=f(nx)+f(x)$. Therefore we can easily prove by induction that

$(2)\space f(nx)=nf(x)$ for all integer $n$ and real $x$.

Now, there are two cases:

$1. f(1)=0 \implies f(n)=0$ for all integer $n$ from $(1)$. And by $(2)$ we get $0=f(1)=f(m\frac{1}{m})=mf(\frac{1}{m})\implies f(\frac{1}{m})=0$ and therefore, again by $(2)\space f(\frac{n}{m})=0$. As the function is continuous and $0$ on rationals then it is $0$ everywhere.

$2. f(1)\neq0$. In this case we can change the function if needed to make $f(1)=1$ (by multiplying with a number). So, we'll have $f(n)=n$ for all integers, and in the same way as in the previous case, we show that $f(\frac{n}{m})=\frac{n}{m}$. Therefore,the function $f(x)-x$ is $0$ on rationals, so it is $0$ everywhere. That's why we get $f(x)=x$.

By taking the note at the beginning, we proved that all the solutions for this problem will be of the form $f(x)=ax+b$ for some real numbers $a$ and $b$.

Answer 2

You can prove that a function $ f : \mathbb R \to \mathbb R $ satisfies $$ f ( x y ) + f ( x + y ) = f ( x y + y ) + f ( x ) \tag 0 \label 0 $$ for all $ x , y \in \mathbb R $, iff there is an additive function $ A : \mathbb R \to \mathbb R $ and a constant $ b \in \mathbb R $ such that $ f ( x ) = A ( x ) + b $ for all $ x \in \mathbb R $. In case $ f $ is assumed to be continuous, $ A $ must be linear (see "Overview of basic facts about Cauchy functional equation"), and therefore there is also a constant $ a \in \mathbb R $ such that $ f ( x ) = a x + b $ for all $ x \in \mathbb R $. It's easy to check that functions of this form satisfy \eqref{0}. We try to prove the converse.

Assume $ f : \mathbb R \to \mathbb R $ satisfies \eqref{0} for all $ x , y \in \mathbb R $. Let $ b = f ( 0 ) $ and define $ A : \mathbb R \to \mathbb R $ with $ A ( x ) = f ( x ) - b $. Then by definition, we have $ A ( 0 ) = 0 $, and \eqref{0} yields $$ A ( x y ) + A ( x + y ) = A ( x y + y ) + A ( x ) $$ for all $ x , y \in \mathbb R $, and since the left-hand side is symmetric in $ x $ and $ y $, you'll get $$ A ( x y + x ) + A ( y ) = A ( x y + y ) + A ( x ) \tag 1 \label 1 $$ for all $ x , y \in \mathbb R $. Letting $ y = 1 $ in \eqref{1} we get $$ A ( 2 x ) + A ( 1 ) = A ( x + 1 ) + A ( x ) \tag 2 \label 2 $$ for all $ x \in \mathbb R $. Substituting $ y + 1 $ for $ y $ in \eqref{1} we have $$ A ( x y + x + y + 1 ) = A ( x y + 2 x ) + A ( y + 1 ) - A ( x ) \tag 3 \label 3 $$ for all $ x , y \in \mathbb R $. As the left-hand side of \eqref{3} is symmetric in $ x $ and $ y $, we conclude $$ A ( x y + 2 x ) + A ( y + 1 ) - A ( x ) = A ( x y + 2 y ) + A ( x + 1 ) - A ( y ) $$ for all $ x , y \in \mathbb R $, which using \eqref{2} simplifies to $$ A ( x y + 2 x ) + A ( 2 y ) = A ( x y + 2 y ) + A ( 2 x ) \tag 4 \label 4 $$ for all $ x , y \in \mathbb R $. Substituting $ 2 x $ for $ x $ and $ \frac y 2 $ for $ y $ in \eqref{1} we have $$ A ( x y + 2 x ) + A \left( \frac y 2 \right) = A \left( x y + \frac y 2 \right) + A ( 2 x ) $$ for all $ x , y \in \mathbb R $, which together with \eqref{4} show that $$ A ( 2 y ) - A \left( \frac y 2 \right) = A \left( x y + 2 y \right) - A \left( x y + \frac y 2 \right) \tag 5 \label 5 $$ for all $ x , y \in \mathbb R $. Putting $ x = - \frac 1 2 $ in \eqref{5} yields $$ A ( 2 y ) - A \left( \frac y 2 \right) = A \left( \frac 3 2 y \right) $$ for all $ y \in \mathbb R $, and hence using \eqref{5} we get $$ A \left( x y + 2 y \right) = A \left( x y + \frac y 2 \right) + A \left( \frac 3 2 y \right) \tag 6 \label 6 $$ for all $ x , y \in \mathbb R $. Substituting $ \frac 1 2 \left( \frac { 3 x } y - 1 \right) $ for $ x $ and $ \frac 2 3 y $ for $ y $ in \eqref{6} gives us $$ A ( x + y ) = A ( x ) + A ( y ) $$ forall $ x \in \mathbb R $ and all $ y \in \mathbb R \setminus \{ 0 \} $. As we also know that $ A ( x + 0 ) = A ( x ) + A ( 0 ) $ for all $ x \in \mathbb R $, $ A $ is additive, and therefore $ f $ is of the desired form.