Find all functions $f:\mathbb{R^+}\rightarrow \mathbb{R^+}$ such that: $$f(1+xf(y))=yf(x+y)$$ for all $x,y \in \mathbb{R^+}$


I can only prove $f$ is a surjective function. I tried to prove $f$ is an injective function, but I can't. Do you have any ideas?

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    $\begingroup$ To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people are much more willing to help you if you show that you've tried the problem yourself. Also, many would consider your post rude because it is a command ("Find..."), not a request for help, so please consider rewriting it. $\endgroup$ Jan 4, 2013 at 5:45
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    $\begingroup$ Also, does $\mathbb{R}^+$ denote the set of positive reals or the set of nonnegative reals? $\endgroup$ Jan 4, 2013 at 5:54
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    $\begingroup$ I simply cannot upvote this question because you have been here long enough to know that you should give context and show your work. $\endgroup$
    – davidlowryduda
    Jan 4, 2013 at 5:57
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    $\begingroup$ @Isomorphism I understand the downvotes, this is more like an assignment then asking for a helping hand. Also, as Zev said, no own thoughts are given. $\endgroup$ Jan 4, 2013 at 9:48
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    $\begingroup$ Can you put your proof of the surjectivity of $f(x)$ in the OP? It would help with the downvoting problem if you showed some work, especially if you have partially solved the problem. $\endgroup$ Jan 4, 2013 at 11:23

2 Answers 2


Set $x=0$. We then get that $$f(1) = y f(y) \implies f(y) = \dfrac{f(1)}y$$ Hence, we have $$\dfrac{f(1)}{1+xf(y)} = \dfrac{yf(1)}{x+y} \implies 1 + xf(y) = 1 + \dfrac{x}y \implies f(y) = \dfrac1{y} \implies f(1) = 1$$ Hence, we get that $$f(x) = \dfrac1x$$

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    $\begingroup$ I think you can't set $x=0$ here because it says that for all $ x;y \in \mathbb{R^+}$ $\endgroup$
    – Haruboy15
    Jan 4, 2013 at 6:23
  • $\begingroup$ @Haruboy15 That depends on your convention for the definition of $\mathbb{R}^+$ $\endgroup$
    – user7530
    Jan 4, 2013 at 7:10
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    $\begingroup$ If you allow $0$ for $x$, you must also allow it for $y$. But with $x=y=0$, you get $f(1) = 0$. $\endgroup$ Jan 4, 2013 at 9:10
  • $\begingroup$ @RobertIsrael Provided $f(0)$ is defined. $\endgroup$
    – user17762
    Jan 4, 2013 at 16:37
  • $\begingroup$ @Haruboy15 I only saw that the domain and range are in $\mathbb{R}^+$. If $x,y \neq 0$, then if you assume continuity of $f$ then again it gives you $1/x$ as a solution. $\endgroup$
    – user17762
    Jan 4, 2013 at 16:39
  • Let $y=1$. Then $f(1+xf(1))=f(1+x)$, so either $f(1)=1$, or $f(1+e^x)$ is periodic (this part is explained in more detail later). Let's assume the first to start. If $f$ is continuous on $(0,\infty)$ (this constraint is relaxed later), then we can let $x\to0$ so that we get $f(1)=yf(y)=1$ whence $f(x)=1/x$. Checking this solution, we get $$f(1+xf(y))=yf(y+x)\Rightarrow \frac1{1+x/y}=\frac y{y+x},$$ which is true, so $f(x)=1/x$ is a solution.

  • If we fix some $y\ne1$ such that $f(y)\ne1$ (which exists, because $f(x)=1$ is not a solution) and choose $x$ so that $1+xf(y)=y+x$, then $$x=\frac{y-1}{f(y)-1}\Rightarrow f\Big(\!\frac{yf(y)-1}{f(y)-1}\!\Big)=yf\Big(\!\frac{yf(y)-1}{f(y)-1}\!\Big)\Rightarrow f\Big(\!\frac{yf(y)-1}{f(y)-1}\!\Big)=0\notin\mathbb R^+,$$ which is a contradiction, so either $x=\frac{y-1}{f(y)-1}\le0$ or $y+x=\frac{yf(y)-1}{f(y)-1}\le0$. The second condition implies the first, which is equivalent to $$f(y)\ge1\mbox{ if }y\le1\mbox{ and }f(y)\le1\mbox{ if }y\ge1.$$

  • If we suppose $a:=\max(f(1),f(1)^{-1})\ne1$, then $f(1+ax)=f(1+x)$ (setting $y=1$). Moreover, this generalizes to $$yf(x+y)=f(1+xf(y))=f(1+axf(y))=yf(ax+y)\Rightarrow f(y+x)=f(y+ax),$$ so letting $y=1-x$, we get $a=f((a-1)x+1)$, so $f(x)=f(1)$ on $[1,a)$ (assuming $a>1$), and by $f(1+ax)=f(1+x)$, it is also constant on $[1,a^2)$, $[1,a^3)$, etc. so that (by induction) it is constant for all $x\ge1$. But then, choosing $y=2$, $f(1+xf(2))=2f(x+2)\Rightarrow$ $f(1)=2f(1)$, since both arguments are greater than $1$, and this is a contradiction. Thus, $f(1)=1$.

  • If we assume $f$ is continuous at $1$, then if we let $x\to0$, $$\lim_{x\to0}f(y+x)=\lim_{x\to0}\frac{f(1+xf(y))}y=\frac{f(1)}y=\frac1y$$ which implies that $f$ is continuous almost everywhere (discontinuous on a nowhere-dense set) and equals $1/x$ where it is continuous. For the remaining points, let $x$ be small enough that $f(1+xf(y))=\frac1{1+xf(y)}$ and $f(y+x)=\frac1{y+x}$. Then $\frac1{1+xf(y)}=\frac y{y+x}$ so that $f(y)=1/y$ on $(0,\infty)$.

In summary: $f(x)=1/x$ is a solution, and $\operatorname{sgn}(y-1)=\operatorname{sgn}(1-f(y))$. Other than that, it is known that any other solution will have to be discontinuous at $1$. More work to be done, I think...


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