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Source: Baltic Way 1995

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#1 h Aug 24, 2016, 7:34 AM • 1 Y

Y by Adventure10

$f:\mathbb R\setminus \{0\} \to \mathbb R$
(i) $f(1)=1$ ,
(ii) $\forall x,y,x+y \neq 0:f(\frac{1}{x+y})=f(\frac{1}{x})+f(\frac{1}{y}) : P(x,y)$
(iii) $\forall x,y,x+y \neq 0:(x+y)f(x+y)=xyf(x)f(y) :Q(x,y)$
$f=?$

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pco

#2 h Aug 24, 2016, 12:32 PM • 2 Y

Y by nicky-glass, Adventure10

nicky-glass wrote:

$f:\mathbb R\setminus \{0\} \to \mathbb R$
(i) $f(1)=1$ ,
(ii) $\forall x,y,x+y \neq 0:f(\frac{1}{x+y})=f(\frac{1}{x})+f(\frac{1}{y}) : P(x,y)$
(iii) $\forall x,y,x+y \neq 0:(x+y)f(x+y)=xyf(x)f(y) :Q(x,y)$
$f=?$

Let $x>0$ : $Q(\frac x2,\frac x2)$ $\implies$ $f(x)=\frac x4f(\frac x2)^2$ $\ge 0$

So $f(\frac 1x)$ is additive non negative over $\mathbb R_{>0}$ and so is linear.

So $f(x)=\frac 1x$ $\forall x>0$ (since $f(1)=1$ )

Let $x<0$ : $Q(\frac x2,\frac x2)$ $\implies$ $f(x)=\frac x4f(\frac x2)^2$ $\le 0$

So $f(\frac 1x)$ is additive non positive over $\mathbb R_{<0}$ and so is linear.

So $f(x)=\frac ax$ $\forall x<0$ (for some $a\ge 0$ )

Plugging this back in original equations, we easily get $a=1$
And so $\boxed{f(x)=\frac 1x\text{ }\forall x\ne 0}$ which indeed is a solution

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#3 h Apr 16, 2025, 2:44 PM

Y by

pco wrote:

nicky-glass wrote:

$f:\mathbb R\setminus \{0\} \to \mathbb R$
(i) $f(1)=1$ ,
(ii) $\forall x,y,x+y \neq 0:f(\frac{1}{x+y})=f(\frac{1}{x})+f(\frac{1}{y}) : P(x,y)$
(iii) $\forall x,y,x+y \neq 0:(x+y)f(x+y)=xyf(x)f(y) :Q(x,y)$
$f=?$

Let $x>0$ : $Q(\frac x2,\frac x2)$ $\implies$ $f(x)=\frac x4f(\frac x2)^2$ $\ge 0$

So $f(\frac 1x)$ is additive non negative over $\mathbb R_{>0}$ and so is linear.

So $f(x)=\frac 1x$ $\forall x>0$ (since $f(1)=1$ )

Let $x<0$ : $Q(\frac x2,\frac x2)$ $\implies$ $f(x)=\frac x4f(\frac x2)^2$ $\le 0$

So $f(\frac 1x)$ is additive non positive over $\mathbb R_{<0}$ and so is linear.

So $f(x)=\frac ax$ $\forall x<0$ (for some $a\ge 0$ )

Plugging this back in original equations, we easily get $a=1$
And so $\boxed{f(x)=\frac 1x\text{ }\forall x\ne 0}$ which indeed is a solution

Can you explain what $f(\frac 1x)$ is additive mean and how did u get that. Also i dont know why it implies $f(x)=\frac 1x$ ?

This post has been edited 1 time. Last edited by HuongToiVMO, Apr 16, 2025, 2:45 PM
Reason: wrong typed

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#4 h Apr 16, 2025, 4:48 PM

Y by

HuongToiVMO wrote:

Can you explain what $f(\frac 1x)$ is additive mean and how did u get that. Also i dont know why it implies $f(x)=\frac 1x$ ?

A function $g(x)$ is additive if $g(x+y)=g(x)+g(y)$ all over its domain.
If an additive function $g(x)$ is either overbounded, either lowerbounded over a non-empty open interval, then it is linear and $g(x)=g(1)x$

In our case, let $g(x)=f(\frac 1x)$ :
(ii) is $g(x+y)=g(x)+g(y)$ and so $g(x)$ is additive.
Over $\mathbb R_{>0}$ we get $f(x)=\frac x4f(\frac x2)^2$ and so $f(x)>0$ $\forall x>0$ and so $g(x)=f(\frac 1x)>0$ $\forall x>0$
So $g(x)$ is additive and lowerbounded over $(0,+\infty)$ and so $g(x)=g(1)x=f(\frac 11)x=x$ which is $f(\frac 1x)=x$ and so $f(x)=\frac 1x$

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#5 h Apr 20, 2025, 5:14 AM

Y by

nicky-glass wrote:

$f:\mathbb R\setminus \{0\} \to \mathbb R$
(i) $f(1)=1$ ,
(ii) $\forall x,y,x+y \neq 0:f(\frac{1}{x+y})=f(\frac{1}{x})+f(\frac{1}{y}) : P(x,y)$
(iii) $\forall x,y,x+y \neq 0:(x+y)f(x+y)=xyf(x)f(y) :Q(x,y)$
$f=?$

$f(1)=1$ unnecessary.

You can also reduce this to Cauchy over $\mathbb R$ .

Let $g:\mathbb R\to\mathbb R$ be defined by $g(x)=\begin{cases}0&\text{if }x=0\\f\left(\frac1x\right)&\text{if }x\ne0\end{cases}$ , we can easily check that $g(x+y)=g(x)+g(y)$ for all $x,y\in\mathbb R$ with $x,y,x+y\ne0$ . Also, it's true if $x=0$ or $y=0$ .

Now, for any $x,y\in\mathbb R$ we can choose some $z$ with $x+y+z\ne0$ and $y+z\ne0$ (such as $z=1+(x+y)^2+y^2$ ), then:
$g(x+y)+g(z)=g(x+y+z)=g(x)+g(y+z)=g(x)+g(y)+g(z).$ So $g$ is additive.

Now from the second equation we get (setting $x=y$ ) $2xf(2x)=x^2f(x)^2$ , so $f(x)\ge0$ for $x>0$ , so $g(x)\ge0$ for $x>0$ . Combined with additivity, this gives $g(x)=cx$ for all $x\in\mathbb R$ for some constant $c\in\mathbb R$ , so $f(x)=\frac cx$ for all $x\ne0$ . Testing, we get $\boxed{f(x)=0}$ or $\boxed{f(x)=\frac1x}$ .

If we want to restrict solutions to $f(1)=1$ , it's obviously just $f(x)=\frac1x$ .

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