## Thursday, January 5, 2006

### Dysfunctional. Topic: Algebra. Level: Olympiad.

Problem: (1983 IMO - #1) Find all functions $f$ defined on the set of positive reals which take positive real values and satisfy:$f(xf(y))=yf(x)$ for all $x,y$; and $f(x)\to0$ as $x\to\infty$.

Solution: First, it would help a lot if we showed the function was one-to-one (that is, for each $y$ there is at most one $x$ such that $f(x) = y$).

So suppose $f(a) = f(b)$.

Then $f(af(b) = f(af(a)) = bf(a)$. But we also have $f(af(a)) = af(a)$ so $f(af(a)) = bf(a) = af(a)$. And since $f$ takes positive real values, we know can divide by $f(a)$ to get $a = b$. Hence $f$ is one-to-one and we take take its inverse.

Now let's experiment with some values.

Let $x = y = 1$. We have $f(f(1)) = f(1) \Rightarrow f(1) = 1$ (because we can take its inverse).

Now take $x = y$. We have $f(xf(x)) = xf(x)$. Suppose $k = xf(x)$.

Assume $k > 1$. We have $f(k) = k \Rightarrow f(kf(k)) = f(k^2) = k^2$, and, repeating this process, we have $f(k^{2n}) = k^{2n}$. Then if we take $n \to \infty$, we have $f(x) \to \infty$ as $x \to \infty$, which contradcits the given condition.

Now assume $k < 1$. We have $f(k) = k \Rightarrow f\left(\frac{1}{k}f(k)\right) = f(1) = 1 = kf\left(\frac{1}{k}\right) \Rightarrow f\left(\frac{1}{k}\right) = \frac{1}{k}$. But again $f\left(\frac{1}{k}\right) = \frac{1}{k} \Rightarrow f\left(\frac{1}{k}f\left(\frac{1}{k}\right)\right) = f\left(\frac{1}{k^2}\right) = \frac{1}{k^2}$ which, repeating the process, becomes $f\left(\frac{1}{k^{2n}\right) = \frac{1}{k^{2n}$. Then if we take $n \to \infty$, we have $f(x) \to \infty$ as $x \to \infty$, again giving a contradiction.

Therefore $k = 1 = xf(x) \Rightarrow f(x) = \frac{1}{x}$ is the only such function. QED.

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Practice Problem: (1986 IMO - #5) Find all functions $f$ defined on the non-negative reals and taking non-negative real values such that: $f(2)=0,f(x) \neq 0$ for $0\le x<2$, and$f(xf(y))f(y)=f(x+y)$ for all $x,y$.