Hard functional equation: $f \big ( x y + f ( x ) \big) = f \big( f ( x ) f ( y ) \big) + x$

Let $$\mathbb R _ { > 0 }$$ be the set of positive real numbers. Find all functions $$f : \mathbb R _ { > 0 } \to \mathbb R _ { > 0 }$$ such that $$f \big ( x y + f ( x ) \big) = f \big( f ( x ) f ( y ) \big) + x$$ for all positive real numbers $$x$$ and $$y$$.

What I thought: We could change $$x$$ by $$y$$, and then subtract.

Source: Brazil National Olympiad 2019 #3

• Does $f$ have to be continuous? Nov 14, 2019 at 16:09
• @MaximilianJanisch Typically you may not assume that $f$ is continuous unless they said so, esp for contest-math. Nov 14, 2019 at 16:15
• Hint. Assuming $f(x)$ continuous, as $$f(x y+f(x))-f(x y+ f(y)) = x - y$$ we have $$\frac{f(x y+f(x))-f(x y+ f(y))}{f(x)-f(y)}\frac{f(x)-f(y)}{x-y} = 1$$ now is $f(x)$ is continuous... Nov 14, 2019 at 17:29
• @Cesareo Can we (assuming continuity) conclude that $f(x)=x$ from here? Nov 14, 2019 at 22:44
• @MeuluElisson There are two solutions on this forum thread. If no one else posts one of those as an answer here perhaps I might do so in a few days. Nov 17, 2019 at 10:23

Exchanging $$x$$ and $$y$$ and substracting, it follows $$f(xy+f(x))-f(xy+f(y))=x-y$$. In particular, if $$f(x)=f(y)$$ then $$x=y$$.

The equation also tells us that if $$r > f(x)$$, we can find a $$y> 0$$ such that $$r=f(x)+xy$$, so $$f(r)=f(xy+f(x))=f(f(x)f(y))+x > x$$, ie that if $$r > f(x)$$, $$f(r) > x$$.

In particular, if $$x > f(x)$$, $$f(x) > x$$, so we have, for all $$x$$, $$f(x) \geq x$$.

Now, let us fix some $$x > 0$$ such that $$f(x)>x$$.

Define, for any $$y > 0$$, $$g(y)=\frac{f(x)}{x}(f(y)-1)$$. If $$g(y)>0$$, then note that $$xg(y)+f(x)=f(x)f(y)$$, thus $$f(xy+f(x))=f(xg(y)+f(x))+x$$.

Therefore, if $$y >0$$ and $$g^n(y)>0$$ is defined, $$0. As a consequence, $$n < \frac{f(xy+f(x))}{x}+1$$ (the precise estimate is irrelevant, just remember tha the RHS is explicit in $$x$$ and $$y$$).

In particular, there exists some $$n \geq 0$$ (depending on $$x,y$$) such that $$g^n(y) > 0$$ is defined and $$g^{n+1}(y) \leq 0$$.

Now, take $$y > \alpha$$, where $$f(x)(\alpha-1)=x\alpha$$. Then $$g(y)=\frac{f(x)}{x}(f(y)-1) \geq \frac{f(x)}{x}(y-1) > f(x)(\alpha-1)/x=\alpha$$.

We find that $$g^n(y)$$ is defined and positive for all $$n$$, a contradiction.

$$f(xy + f(x)) = f(f(x)\cdot f(y)) + x.$$

Let us substitute $$y = 1$$:

$$f(x + f(x)) = f(f(x)\cdot f(1)) + x.$$

Now, let us substitute into the initial equation $$x = 1$$:

$$f(y + f(1)) = f(f(1)\cdot f(y)) + 1.$$

In the latter equation, let us replace $$y$$ by $$x$$:

$$f(x + f(1)) = f(f(x)\cdot f(1)) + 1.$$

Now, we have

$$f(x + f(x)) = f(f(x)\cdot f(1)) + x \\ f(x + f(1)) = f(f(x)\cdot f(1)) + 1$$

Let $$g(x) = f(f(x)\cdot f(1))$$. Then, we have

$$f(x + f(x)) = g(x) + x \\ f(x + f(1)) = g(x) + 1$$

We see that a linear shift in the argument of the function $$f(x)$$ results in a linear shift in the values of function g(x).

1. Shifting by $$f(x)$$, i.e. $$f(x + f(x))$$ implies shifting by $$x$$.
2. Shifting by $$f(1)$$, i.e. $$f(x + f(1))$$ implies shifting by $$1$$.

This is true if both $$f(x)$$ and $$g(x)$$ are linear functions, particularly if $$f(x) = x.$$

Let us check that $$f(x) = x$$ is a solution:

$$f(xy + f(x)) = f(f(x)\cdot f(y)) + x \Leftrightarrow f(xy + f(x)) = xy + x \text{ and }f(f(x)f(y)) + x = xy + x \text{ (TRUE). }$$

We want to find all functions, continuous or not, $$\,f:\mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}\,$$ such that for all $$\,x\,$$ and $$\,y\,$$ positive reals $$f(x y + f(x)) = f(f(x) f(y)) + x. \tag{1}$$ Now use equation $$(1)$$ with $$\,y,x\,$$ instead of $$\,x,y\,$$ which gives $$f(x y + f(y)) = f(f(x) f(y)) + y. \tag{2}$$ Solving for $$\,f(f(x)f(y))\,$$ in both equations gives $$f(x y + f(y)) - y = f(x y + f(x)) - x.\tag{3}$$

Now suppose $$\,f(x) = f(y).\,$$ Equation $$(3)$$ implies that $$\,x = y\,$$ which proves $$\,f\,$$ is one-to-one.

Given $$\,x>0,\,$$ suppose $$\,f(x) Then we solve for $$\,y>0\,$$ in $$xy+f(x)=x. \tag{4}$$ Apply $$\,f\,$$ to both sides to get $$f(x y + f(x)) = f(x). \tag{5}$$ Combine with equation $$(1)$$ to get $$f(x) = f(f(x) f(y)) + x. \tag{6}$$ This implies that $$\,f(x) > x\,$$ which contradicts our assumption $$\,f(x) < x.\,$$ Thus $$\,f(x)\ge x\,$$ for all $$\,x>0.\,$$

The obvious solution is $$\,f(x)=x\,$$ for all $$\,x>0\,$$ so now the question is how to prove $$\,f(x)>x\,$$ is impossible.