Let function pair $f,g : \mathbb{R^+} \rightarrow \mathbb{R^+}$ satisfies \[ f(g(x)y + f(x)) = (y+2015)f(x) \]for every $x,y \in \mathbb{R^+} $ a. Prove that $f(x) = 2015g(x)$ for every $x \in \mathbb{R^+}$ b. Give an example of function pair $(f,g)$ that satisfies the statement above and $f(x), g(x) \geq 1$ for every $x \in \mathbb{R^+}$
Problem
Source: 2015 Indonesia Math Olympiad Day 1 Problem 4
Tags: function, algebra, functional equation
Vietnamisalwaysinmyheart
30.06.2017 16:58
Here is my solution
Part a Replace $y$ by $\frac{y}{g(x)}$, we have: $f(y+f(x))=\frac{yf(x)}{g(x)}+2015f(x)$ $(1)$, now, replace $y$ by $f(y)$ in last equation, we have: $f(f(x)+f(y))=\frac{f(x)f(y)}{g(x)}+2015f(x)$, now, divide bothside with $f(x)f(y)$, we have:
$\frac{f(f(x)+f(y))}{f(x)f(y)}=\frac{1}{g(x)}+\frac{2015}{f(y)}$, now, interchange role of $x,y$, we have:
$\frac{1}{g(x)}+\frac{2015}{f(y)}=\frac{1}{g(y)}+\frac{2015}{f(x)}$. Thus, $\frac{1}{g(x)}-\frac{2015}{f(x)}=c$, or: $\frac{1}{g(x)}=c+\frac{2015}{f(x)}$, note that, we must have: $2015+cf(x)>0$ forall $x\in\mathbb{R^+}$
Now, plugging back to $(1)$, we have: $f(y+f(x))=yf(x)(\frac{2015}{f(x)}+c)+2015f(x)=2015y+cyf(x)+2015f(x)$ $(2)$
Replace: $y$ by $\frac{y}{2015+cf(x)}$, we have:
$f(\frac{y}{2015+cf(x)}+f(x))=y+2015f(x)$. Now, fix a number: $x_0\in\mathbb{R^+}$, we have:
$f(\frac{y+2015f(x_0)+cf(x_0)^2}{2015+cf(x_0)})=y+2015f(x_0)+cf(x_0)^2-cf(x_0)^2$
And thus, this means that, for large number, $f$ is linear, or: $f(x)=ax+b$ for all $x>M$
By $(2)$, we consider large $x,y$, then we have:
$a(y+ax+b)+b=2015y+cy(ax+b)+2015(ax+b)$
Lastly, divide bothside by $xy$, and taking limit that: $x,y,\rightarrow +\infty$, we must have:
$ca=0$, since: $a\neq 0$, thus, $c=0$, and hence: $f(x)=2015g(x)$
Part b: Let $f(x)=2015$ forall $x\in(0;1)$, and: $f(x)=2015x$ forall $x\geq 1$, for $g$, we just need to divide $2015$
Of course that: $f,g\geq 1$, forall $x\in\mathbb{R^+}$
Since: $g(x)y+f(x)>f(x)\geq 2015>1$, hence:
$f(g(x)y+f(x))=2015g(x)y+2015f(x)=(y+2015)f(x)$, which is valid!
ThE-dArK-lOrD
30.06.2017 17:11
We have $f(g(x)y+f(x)+zg(x))-f(g(x)y+f(x))=zf(x)$ for all $x,y,z\in \mathbb{R}^+$. So $f(T+Z)-f(T)=\frac{Zf(x)}{g(x)}$ for all $x,T,Z\in \mathbb{R}^+$ that $T>f(x)$. For any $a,b\in \mathbb{R}^+$, there exist $K>f(a),f(b)$, so $f(K+Z)-f(K)=\frac{Zf(a)}{g(a)}=\frac{Zf(b)}{g(b)}$ for all $Z\in \mathbb{R}^+$. So $\frac{f(a)}{g(a)}=\frac{f(b)}{g(b)}$. In other words, $f(x)=cg(x)$ for all $x\in \mathbb{R}^+$ for some positive constant $c$. Plug in the condition give us $g((y+c)g(x))=(y+2015)g(x)$ for all $x,y\in \mathbb{R}^+$. We have $(y+2015)(z+2015)g(x)=(y+2015)g((z+c)g(x))=g((y+c)g((z+c)g(x)))=g((y+c)(z+2015)g(x))$ and $g((y+c)(z+2015)g(x)) =\Big( (y+c)(z+2015)-c+2015\Big) g(x)$ for all $x,y,z\in \mathbb{R}^+$. So $(1+2015)(1+2015)=(1+c)(1+2015)-c+2015\rightarrow c=2015$. And example of such function $g$ is $g(x)=\begin{cases} 1,& \text{for } 0<x<1 \\ x,& \text{for } 1\leq x \end{cases}$.