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Find all functions run over positive real numbers such that $f(1+xf(y))=yf(x+y)$ where $x,y\in R^+$

MY ANSWER: Putting $x=y=0$,we get, $f(1)=0$ Putting $x=0$ we get, $f(1)=yf(y)$

or,$yf(y)=0$

or,$f(y)=0$ (since $y\ne 0$., $y \in \mathbb R^+$)

Hence,$f(x)=0$ is the solution.

Is my answer and solution collect? If not then please tell me the proper answer and solution and where I have made the mistake!!

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  • $\begingroup$ The $x=y=0$ does not seem valid, since on right side you have $f(0)$ and $0 \not\in \mathbb{R}^+$. Or what is domain and codomain of the function? $\endgroup$ – Sil Jun 17 '18 at 14:13
  • $\begingroup$ Your solution does not seem to apply to $f(0)$. If I may, an alternative approach could be using $y=1$ which gives $f(1)=f(x+1)\ \forall x\in\mathbb{R^+}\Rightarrow f(x)=0\ \forall x\in\mathbb{R^+}$, using that $f(1)=0$ $\endgroup$ – AnotherJohnDoe Jun 17 '18 at 14:15
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    $\begingroup$ Btw here this is discussed on AoPS, there should be some solutions: artofproblemsolving.com/community/c6h323174 . $\endgroup$ – Sil Jun 17 '18 at 14:18
  • $\begingroup$ I am satisfied. Thank you! $\endgroup$ – Sufaid Saleel Jun 17 '18 at 14:27
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For reference, here is CanVQ's solution taken from here.

Replacing $x$ by $\frac{x}{f(y)}$ in $(1),$ we have $$f(1+x)=y\cdot f\left(\frac{x}{f(y)}+y\right),\quad \forall x ,\,y \in \mathbb R^+. \quad (2)$$ Next, we replace $y$ by $\frac{f(1+x)}{y}$ in $(2)$ to get $$y=f\left(\frac{x}{f\left(\frac{f(1+x)}{y}\right)}+\frac{f(1+x)}{y}\right),\quad \forall x,\,y \in \mathbb R^+. \quad (3)$$ From $(3),$ it follows that $f$ is surjective. Now, we will prove that $f$ is decreasing. Replacing $x$ by $x+z$ in $(1),$ we get $$f\big(1+(x+z)\cdot f(y)\big)=y\cdot f(x+y+z),\quad \forall x,\,y,\,z \in \mathbb R^+. \quad (4)$$ Replacing $y$ by $y+z$ in $(1),$ we also have $$f\big(1+x\cdot f(y+z)\big)=(y+z)\cdot f(x+y+z),\quad \forall x,\,y,\,z \in \mathbb R^+. \quad (5)$$ Dividing $(4)$ for $(5),$ side by side, we obtain $$\frac{f\big(1+(x+z)\cdot f(y)\big)}{f\big(1+x\cdot f(y+z)\big)}=\frac{y}{y+z},\quad \forall x,\,y,\,z \in \mathbb R^+. \quad (6)$$ Now, assume that there exists a pair $(y,\,z)$ such that $f(y+z)>f(y).$ In this case, by choosing $x=\frac{z\cdot f(y)}{f(y+z)-f(y)}$ vào $(6),$ we obtain $y=y+z,$ which is a contradiction. So we must have $$f(y+z) \le f(y),\quad \forall y,\,z \in \mathbb R^+. \quad (7)$$ Now, we will prove that $f$ is injective. Assume that there are two numbers $a,\,b$ such that $f(a)=f(b).$ Replacing $y=a$ and $y=b$ in $(1)$ respectively in $(1),$ we get $$a\cdot f(x+a)=b\cdot f(x+b),\quad \forall x\in \mathbb R^+. \quad (8)$$ From this, it follows that $$1+a(y-1)\cdot f(x+a)=1+b(y-1)\cdot f(x+b),\quad \forall x, \, y \in \mathbb R^+,\, y>1. \quad (9)$$ Plugging this into $f$ and using $(1),$ we get $$(x+a)\cdot f(x+ay)=(x+b)\cdot f(x+by),\quad \forall x,\,y \in \mathbb R^+ ,\, y>1. \quad (10)$$ From $(10),$ it follows that, for any $x,\,y,\,z \in \mathbb R^+,\, y>1,$ $$1+(xz+az)\cdot f(x+ay)=1+(xz+bz)\cdot f(x+by). \quad (11)$$ Again, we plug this into $f$ and using $(1).$ It follows that $$(x+ay)\cdot f(x+ay+az+xz)=(x+by)\cdot f(x+by+bz+xz)\quad (12)$$ for any $x,\,y,\,z \in \mathbb R^+$ and $ y>1.$ On the other hand, according to $(10),$ we also have $$\big[(x+xz)+a\big]\cdot f\big( (x+xz)+a(y+z)\big)=\big[(x+xz)+b\big] \cdot f\big((x+xz)+b(y+z)\big),$$ or $$(x+xz+a)\cdot f(x+ay+az+xz)=(x+xz+b)\cdot f(x+ay+az+xz). \quad (13)$$ Dividing $(12)$ for $(13),$ side by side, we obtain $$\frac{x+ay}{x+xz+a}=\frac{x+by}{x+xz+b},\quad \forall x,\,y,\,z \in \mathbb R^+,\, y>1. \quad (14)$$ It is easy to deduce that $a=b$ here, so $f $ is injective. Now, replacing $x=y=1$ in $(1)$ with notice that $f $ is injective, we have $f(1)=1.$ Since $f$ is strictly decreasing ($f$ is decreasing and injective), we have $$f(x)<1,\quad \forall x>1.\quad (15)$$ Now, let us consider the case $y>x.$ Replacing $y$ by $y-x$ in $(1)$ and using the above remark, we get $$f(y)=\frac{f\big(1+x\cdot f(y-x)\big)}{y-x}<\frac{1}{y-x},\quad \forall x,\,y \in \mathbb R^+,\, x<y. \quad (16)$$ In $(16),$ we let $x\to 0^+$ and obtain $$f(y) \le \frac{1}{y},\quad \forall y >0. \quad (17)$$ Next, replacing $x$ by $x-1$ and $y$ by $f(y)$ in $(3),$ we get $$y=\frac{x-1}{f\left(\frac{f(x)}{f(y)}\right)}+\frac{f(x)}{f(y)},\quad \forall x ,\, y \in \mathbb R^+,\, x >1. \quad (18)$$ Since $f\left(\frac{f(x)}{f(y)}\right) \le \frac{f(y)}{f(x)},$ from $(18),$ we deduce that $$y\ge \frac{(x-1)\cdot f(x)}{f(y)}+\frac{f(x)}{f(y)},$$ or $$y\cdot f(y) \ge x \cdot f(x),\quad \forall x,\,y \in \mathbb R^+,\, x>1. \quad (19)$$ Changin the position of $x$ and $y$ in $(19),$ we also have $$x\cdot f(x) \ge y\cdot f(y),\quad \forall x,\,y \in \mathbb R^+,\, y>1. \quad (20)$$ From the inequalities $(19)$ and $(20),$ we can easily deduce that $$f(x)=\frac{k}{x},\quad \forall x>1. \quad (21)$$ Now, taking $x=1$ in $(1)$ and using $(21),$ we have $$\frac{1}{1+f(y)}=\frac{y}{1+y},$$ or $$f(y)=\frac{1}{y},\quad \forall y\in \mathbb R^+. \quad (22)$$ Clearly, the function $f(x)=\frac{1}{x}$ satisfies our equation.

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