# Solve the functional equation $f(1+xf(y))=yf(x+y)$

### Problem

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

### Progress

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?

• Also, does $\mathbb{R}^+$ denote the set of positive reals or the set of nonnegative reals? Jan 4, 2013 at 5:54
• I simply cannot upvote this question because you have been here long enough to know that you should give context and show your work. Jan 4, 2013 at 5:57
• @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. Jan 4, 2013 at 9:48
• 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. Jan 4, 2013 at 11:23

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 $$f(x) = \dfrac1x$$

• I think you can't set $x=0$ here because it says that for all $x;y \in \mathbb{R^+}$ Jan 4, 2013 at 6:23
• @Haruboy15 That depends on your convention for the definition of $\mathbb{R}^+$ Jan 4, 2013 at 7:10
• If you allow $0$ for $x$, you must also allow it for $y$. But with $x=y=0$, you get $f(1) = 0$. Jan 4, 2013 at 9:10
• @RobertIsrael Provided $f(0)$ is defined.
– user17762
Jan 4, 2013 at 16:37
• @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.
– 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...