Find all $f:\mathbb{R}^+\to \mathbb{R}^+$ satisfies that:

Find all $f:\mathbb{R}^+\to \mathbb{R}^+$ satisfies that: $$f(x)\cdot f(yf(x))=f(y+f(x))$$ $\forall x,y \in \mathbb{R}^+$

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$f(x)=1$ for all $x\in\mathbb{R}^+$. –  Paul Dec 23 '12 at 10:31
I 've guess this but I need the exactly answer, can you show me –  Haruboy15 Dec 23 '12 at 10:32
What do you mean by "exactly answer"? Do you want to find all $f$ satisfies the functional equation? If yes, it's better you edit your question, because it seems to me that you just want to an $f$ satisfying that. –  Paul Dec 23 '12 at 10:34
thanks you so, sorry about my bad English –  Haruboy15 Dec 23 '12 at 10:43
Are there any other constraints? (e.g. $f$ continuous) –  Hurkyl Dec 23 '12 at 10:47

The following is a complete proof that $f\equiv 1$ is the only solution.

Suppose $u,v\in f(\mathbb R^+)$. Then $u = f(x)$ for some $x\in\mathbb R^+$. Therefore the functional equation tells you that $f(x) f(vf(x)) = f(v+f(x))$ or in other words $$u f(uv)=f(u+v).$$ (Note that we have used the fact that $f(x)\in\mathbb R^+$ for $x\in\mathbb R^+$, so that the functional equation holds for $u$ and $v$.)

But as we only assumed that $u,v\in f(\mathbb R^+)$, the same also holds if we interchange the roles of $u$ and $v$, therefore $$v f(uv)=f(u+v)$$ must also hold. Combining the two equations, we have $u f(uv)=v f(uv)$, but because $f(x)$ is positive for all $x\in\mathbb R^+$, we can divide by $f(uv)$. This gives us $u=v$.

What does this mean? We assumed $u,v\in f(\mathbb R^+)$ and proved that $u=v$. This means that $f(\mathbb R^+)$ only has a single element. In other words, $f$ is a constant function, i.e. $f(x)=c$ for all $x$, where $c$ is some positive constant. But from the functional equation we now have that $c$ must satisfy $c^2 = c$, and since $c$ is positive, this is only possible if $c=1$. This completes the proof.

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I can't see, can you show me more –  Haruboy15 Dec 23 '12 at 14:50
@Haruboy15: I have edited my hint into a complete answer. I hope that's ok. –  Dejan Govc Dec 23 '12 at 15:50
@Dejan Govc: good job! (+1) –  Chris's sis Dec 24 '12 at 15:48

Assume there exists an $x_0$ such that $f(x_0)>1$. Then with $y:=\frac{f(x_0)}{f(x_0)-1}$ (i.e. the solution of $yf(x_0)=y+f(x_0)$) we obtain the contradiction $f(x_0)=\frac{f(y+f(x_0))}{f(yf(x_0))}=1$. Therefore, $f(x)\le 1$ for all $x$.

Assume $q:=f(x_1)<1$. Then define recursively $x_{n+1}=\frac{x_n}q+q$ to find $$f(x_{n+1})=f\left(\frac{x_n}{q}+f(x_1)\right)=f(x_1)f\left(\frac{x_n}qf(x_1)\right)=qf(x_n),$$ hence by induction $f(x_n)=q^n$. Select $n\in\mathbb N$ with $q^n<f(1)$. Then with $y:=1-q^n$ we find $$f(1)=f(y+f(x_n))=f(x_n)f(yf(x_n))\le f(x_n)=q^n,$$ contradiction. Therefore the assumption $f(x_1)<1$ cannot hold. We conclude $f(x)=1$ for all $x$.

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von Etizen how can you think that $x_{n+1}=\frac{x_n}{q}+q$ help us to do this homework –  Haruboy15 Dec 23 '12 at 15:19