Given non-zero reals $ a$, $ b$, find all functions $ f: \mathbb{R} \longmapsto \mathbb{R}$, such that for every $ x, y \in \mathbb{R}$, $ y \neq 0$, $ f(2x) = af(x) + bx$ and $ \displaystyle f(x)f(y) = f(xy) + f \left( \frac {x}{y} \right)$.
Problem
Source: China TST 2004 Quiz
Tags: function, quadratics, algebra unsolved, algebra
JoeBlow
01.02.2009 19:37
$ f(x)^2 = f(2x) + f\left(\frac {x}{2}\right) = af(x) + bx + \frac {f(x) - \frac {bx}{2}}{a}$, which is quadratic in $ f(x)$. One can check that the discriminant is not positive for all $ x$ except if $ a = \frac {1}{2}$, so there does not in general exist such a function well-defined on all of $ \mathbb{R}$.
shobber
01.02.2009 19:44
JoeBlow wrote:
$ f(x)^2 = f(2x) + f\left(\frac {x}{2}\right) = af(x) + bx + \frac {f(x) - \frac {bx}{2}}{a}$, which is quadratic in $ f(x)$. After solving it, one finds that the discriminant is not positive for all $ x$ for any nonzero $ a,b$, so there does not exist such a function well-defined on all of $ \mathbb{R}$.
Oh! There was a typo. It should be $ f(x)f(y)=f(xy)+f \left( \frac{x}{y} \right)$. Sorry for the trouble...
BG Yoda
10.02.2009 01:04
Trivial. 1. $ f(x)f(y)=f(xy)+f\left(\frac{x}{y}\right)=f(xy)+f\left(\frac{y}{x}\right)$ $ \Rightarrow f\left(\frac{x}{y}\right)=f\left(\frac{y}{x}\right)$. 2. $ x=y=1$ gives $ f(1)=2$ or $ f(1)=0$. But if $ f(1)=0$, we get $ f(x)=0$, which doesn't give us a solution. $ x=y$ gives $ f^2(x)=f(x^2)+2$. $ \Rightarrow |f(x)|>\sqrt{2}$, which give us $ |f(x)|>\sqrt{2+\sqrt{2+...+\sqrt{2}}}$. $ \Rightarrow |f(x)|\ge 2$. 3. $ f(x^2)+f(y^2)=f(xy)f\left(\frac{x}{y}\right)$, since $ x^2=xy\frac{x}{y}$ and $ y^2=\frac{xy}{\frac{x}{y}}$. $ f^2(x)+f^2(y)-4=f(xy)f\left(\frac{x}{y}\right)=f(xy)\left(f(x)f(y)-f(xy)\right)$. $ \Leftrightarrow f^2(x)+f^2(y)+f^2(xy)-f(x)f(y)f(xy)=4$. There is a lemma which tells us that if $ x^2+y^2+z^2-xyz=4$, then there exist $ a,b,c$ such that $ x=a+\frac{1}{a}$ and so on. Therefore + the fact $ |f(x)|\ge 2$ for every real $ x$, we get the following - $ f(x)=g(x)+\frac{1}{g(x)}$ for some function $ g$. 4. From $ f^2(x)=f(x^2)+2$ we get $ g(x^2)=g^2(x)$ or $ g(x^2)g^2(x)=1$ - which gives us nothing. $ \Rightarrow g^2(x)=g(x^2)$. bla-bla $ \Rightarrow g(x)=x^c$. and we just have to find $ c$ depending on $ a$ and $ b$.
bambaman
10.02.2009 02:35
BG Yoda wrote: $ x = y$ gives $ f^2(x) = f(x^2) + 2$. $ \Rightarrow |f(x)| > \sqrt {2}$ I might be missing something trivial, but how did you conclude this? We don't know that sign of $ f$.
BG Yoda
10.02.2009 16:03
Use both of the conditions. By the way, only from $ f(x) = f\left(\frac {1}{x}\right)$ and $ f(x) = af\left(\frac {x}{2}\right) + \frac {b}{2}x$, we can conclude that $ a = \frac {1}{2}$, which I didn't find yesterday so directly. $ f\left(\frac {2}{x}\right) = a.f\left(\frac {1}{x}\right) + \frac {b}{x} = a.f(x) + \frac {b}{x} = f\left(\frac {x}{2}\right) = \frac {f(x) - \frac {b}{2}x}{a}$ $ \Rightarrow f(x) = \frac { - b}{a - \frac {1}{a}}\left(\frac {x}{2a} + \frac {1}{x}\right)$ But since $ \frac { - b}{a - \frac {1}{a}}\left(\frac {x}{2a} + \frac {1}{x}\right) = \frac { - b}{a - \frac {1}{a}}\left(\frac {1}{2ax} + x\right)$ $ \Leftrightarrow f(x) = f\left(\frac {1}{x}\right)$, we find $ a = \frac {1}{2}$.