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Difference between revisions of "2001 IMO Shortlist Problems/A4"
 
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Conversely, let <math>G</math> be a subgroup of the multiplicative group <math>\mathbb R^*</math>. Take <math>f(x) = \left\{\begin{array}{c}f(1)x,\ x\in G \\
 
Conversely, let <math>G</math> be a subgroup of the multiplicative group <math>\mathbb R^*</math>. Take <math>f(x) = \left\{\begin{array}{c}f(1)x,\ x\in G \\
−
0,\ x\not \in G\end{array}</math>. It's easy to check the condition <math>f(xy)[f(x) - f(y)] = (x - y)f(x)f(y)</math>.
+
0,\ x\not \in G\end{array}\right\}</math>. It's easy to check the condition <math>f(xy)[f(x) - f(y)] = (x - y)f(x)f(y)</math>.
  
 
== Resources ==
 
== Resources ==

Latest revision as of 11:39, 1 November 2020

Problem

Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$, satisfying

$f(xy)(f(x) - f(y)) = (x - y)f(x)f(y)$

for all $x,y$.

Solution

Assume $f(1) = 0$. Take $y = 1$. We get $f^2(x) = 0,\ \forall x$, so $f(x) = 0,\ \forall x$. This is a solution, so we can take it out of the way: assume $f(1)\ne 0$.

$y = 1\Rightarrow f(x)[f(x) - f(1)] = (x - 1)f(x)f(1)$. We either have $f(x) = 0$ or $f(x) = f(1)x$, so for every $x$, $f(x)\in\{0,f(1)x\}$. In particular, $f(0) = 0$.

Assume $f(y) = 0$. We get $f(x)f(xy) = 0,\ \forall x$. This means that $f(a),f(b)\ne 0\Rightarrow f\left(\frac ab\right)\ne 0\ (*)$ ($\frac ab$ is defined because $f(b)\ne 0\Rightarrow b\ne 0$). Assume now that $x\ne y$ and $f(x),f(y)\ne 0$. We get $f(x) = f(1)x,\ f(y) = f(1)y$, and after replacing everything we get $f(xy) = f(1)xy\ne 0$, so $x\ne y,\ f(x),f(y)\ne 0\Rightarrow f(xy)\ne 0\ (**)$. Assume now $f(x)\ne 0$. From $(*)$ we get $f\left(\frac 1x\right)\ne 0$, and after applying $(*)$ again to $a = x,b = \frac 1x$ we get $f(x^2)\ne 0\ (***)$. We can now see that $(**),(***)$ combine to $f(x),f(y)\ne 0\Rightarrow f(xy)\ne 0\ (\#)$.

Let $G = \{x\in\mathbb R|f(x)\ne 0\}$. $(*)$ and $(\#)$ simply say that $(G,\ \cdot)$ is a subgroup of $(\mathbb R^{*},\ \cdot)$.

Conversely, let $G$ be a subgroup of the multiplicative group $\mathbb R^*$. Take $f(x) = \left\{\begin{array}{c}f(1)x,\ x\in G \\ 0,\ x\not \in G\end{array}\right\}$. It's easy to check the condition $f(xy)[f(x) - f(y)] = (x - y)f(x)f(y)$.

Resources

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