Find all $f:\mathbb{R}^+\to \mathbb{R}^+$ satisfies that: $f(x)\cdot f(yf(x))=f(y+f(x))$ $\forall x,y \in \mathbb{R}^+$
Find all $f:\mathbb{R}^+\to \mathbb{R}^+$ satisfies that:
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0no that 's all, i think it clearly – 2012-12-23
2 Answers
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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0@Dejan Govc: good job! (+1) – 2012-12-24
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
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0von Etizen how can you think that $x_{n+1}=\frac{x_n}{q}+q$ help us to do this homework – 2012-12-23