Difference between revisions of "1983 IMO Problems/Problem 1"

Revision as of 07:29, 9 July 2025

Problem

Find all functions $f$ defined on the set of positive reals which take positive real values and satisfy the conditions: (i) $f(xf(y))=yf(x)$ for all $x,y$ ; (ii) $f(x)\to0$ as $x\to \infty$ .

Solution 1

Let $x=y=1$ and we have $f(f(1))=f(1)$ . Now, let $x=1,y=f(1)$ and we have $f(f(f(1)))=f(1)f(1)\Rightarrow f(1)=[f(1)]^2$ since $f(1)>0$ we have $f(1)=1$ .

Plug in $y=x$ and we have $f(xf(x))=xf(x)$ . If $a=1$ is the only solution to $f(a)=a$ then we have $xf(x)=1\Rightarrow f(x)=\frac{1}{x}$ . We prove that this is the only function by showing that there does not exist any other $a$ :

Suppose there did exist such an $a\ne1$ . Then, letting $y=a$ in the functional equation yields $f(xa)=af(x)$ . Then, letting $x=\frac{1}{a}$ yields $f(\frac{1}{a})=\frac{1}{a}$ . Notice that since $a\ne1$ , one of $a,\frac{1}{a}$ is greater than $1$ . Let $b$ equal the one that is greater than $1$ . Then, we find similarly (since $f(b)=b$ ) that $f(xb)=bf(x)$ . Putting $x=b$ into the equation, yields $f(b^2)=b^2$ . Repeating this process we find that $f(b^{2^k})=b^{2^k}$ for all natural $k$ . But, since $b>1$ , as $k\to \infty$ , we have that $b^{2^k}\to\infty$ which contradicts the fact that $f(x)\to0$ as $x\to \infty$ .

Solution 2

Let $x=1$ so $f(f(y))=yf(1).$ If $f(a)=f(b),$ then $af(1)=f(f(a))=f(f(b))=bf(1)\implies a=b$ because $f(1)$ goes to the real positive integers, not $0.$ Hence, $f$ is injective. Let $x=y$ so $f(xf(x))=xf(x)$ so $xf(x)$ is a fixed point of $f.$ Then, let $y=1$ so $f(xf(1))=f(x)\implies f(1)=1$ as $x$ can't be $0$ so $1$ is a fixed point of $f.$ We claim $1$ is the only fixed point of $f.$ Suppose for the sake of contradiction that $a,b$ be fixed points of $f$ so $f(a)=a$ and $f(b)=b.$ Then, setting $x=a,y=b$ in (i) gives $f(ab)=f(af(b))=bf(a)=ab$ so $ab$ is also a fixed point of $f.$ Also, let $x=\frac{1}{a},y=a$ so $1=f(1)=f(\frac{1}{a}\cdot a)=f(\frac{1}{a}\cdot f(a))=af(\frac{1}{a})\implies f(\frac{1}{a})=\frac{1}{a}$ so $\frac{1}{a}$ is a fixed point of $f.$ If $f(a)=a$ with $a>1,$ then $f(a^n)$ is a fixed point of $f$ , contradicting (ii). If $f(a)=a$ with $0<a<1,$ then $f(\frac{1}{a^n})=\frac{1}{a^n}$ so $\frac{1}{a^n}$ is a fixed point, contradicting (ii). Hence, the only fixed point is $1$ so $xf(x)=1$ so $f(x)=\boxed{\frac{1}{x}}$ and we can easily check that this solution works.

Solution 3

Let $x=1$ . So $f(f(y)) = y f(1)$ . This clearly implies $f$ is injective. Let $y = 1$ . Then $f(f(y)) = f(1)$ . Applying $f^{-1}$ yields $f(1) = 1$ , so $f(f(y)) = y$ . So $f$ is surjective. Applying $f^{-1}$ yields $f(y) = f^{-1}(y)$ .

Let $u=x$ and $y = f^{-1}(v)$ . (i) becomes $f(uv) = f^{-1}(v)f(u) = f(u)f(v)$ since $f = f^{-1}$ . So $f$ is a group isomorphism on the group $(0,\infty)$ with binary operation multiplication. One gets $f(u)^n = f(u^n)$ . Letting $v = u^n$ implies $f(v^{1/n}) = f(v)^{1/n}$ . Combining these identities $f(u)^{\frac{m}{n}} = f(u^{\frac{m}{n}}.$ for all $m,n \in \mathbb{N}$ . Suppose $u>1$ . Taking $\frac{m}{n} \rightarrow \infty$ , the RHS goes to $0$ by (ii). Considering the LHS, necessarily $0 < f(u) < 1$ . So $f(u) < 1$ for $u>1$ . We claim $f$ is monotonically decreasing. Indeed, if $0<x<y$ then $f(y) = f\Big(x \frac{y}{x}\Big) = f(x)f\Big(\frac{y}{x}\Big) < f(x)$ since $y/x > 1$ . Now suppose $r \in (0,\infty)$ and $m_1/n_1 < r < m_2/n_2$ . So $f(x)^{\frac{m_1}{n_1}} = f(x^{\frac{m_1}{n_1}}) < f(x^r) < f(x^{\frac{m_2}{n_2}}) = f(x)^{\frac{m_2}{n_2}}$ Letting $m_1/n_1, m_2/n_2 \rightarrow r$ implies $f(x^r) = f(x)^r$ by the squeeze theorem. Suppose now $f(e) = c$ . Then $f(x) = f(e^{\ln x}) = f(e)^{\ln x} = c^{\ln x}.$ Solving $y = c^{\ln x}$ for $x$ yields $x = (\ln y)/(\ln c)$ . Since $f = f^{-1}$ it follows $c^{\ln x} = \frac{\ln x}{\ln c},$ which implies $(\ln c)^2 = 1$ when $x \neq 1$ . So either $c = e$ or $c = 1/e$ . Since $f$ is monotonically decreasing, $c = f(e) < 1$ , so $c = 1/e$ . In which case $f(x) = c^{\ln x} = 1/x$ .

~detriti

Video Solution

https://youtu.be/Fu0jBKKa4Lc [Video Solution by little fermat]

1983 IMO (Problems) • Resources
Preceded by First question	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 2
All IMO Problems and Solutions

@@ Line 14: / Line 14: @@
 Let <math>x=1</math> so <cmath>f(f(y))=yf(1).</cmath>  If <math>f(a)=f(b),</math>  then <math>af(1)=f(f(a))=f(f(b))=bf(1)\implies a=b</math> because <math>f(1)</math> goes to the real positive integers, not <math>0.</math>  Hence, <math>f</math> is injective.  Let <math>x=y</math> so
 <cmath>  f(xf(x))=xf(x)</cmath> so <math>xf(x)</math> is a fixed point of <math>f.</math> Then, let <math>y=1</math> so <math>f(xf(1))=f(x)\implies f(1)=1</math> as <math>x</math> can't be <math>0</math> so <math>1</math> is a fixed point of <math>f.</math>  We claim <math>1</math> is the only fixed point of <math>f.</math> Suppose for the sake of contradiction that <math>a,b</math> be fixed points of <math>f</math> so <math>f(a)=a</math> and <math>f(b)=b.</math>  Then, setting <math>x=a,y=b</math> in (i) gives <cmath>f(ab)=f(af(b))=bf(a)=ab</cmath> so <math>ab</math> is also a fixed point of <math>f.</math> Also, let <math>x=\frac{1}{a},y=a</math> so <cmath>1=f(1)=f(\frac{1}{a}\cdot a)=f(\frac{1}{a}\cdot f(a))=af(\frac{1}{a})\implies f(\frac{1}{a})=\frac{1}{a}</cmath> so <math>\frac{1}{a}</math> is a fixed point of <math>f.</math> If <math>f(a)=a</math> with <math>a>1,</math> then <math>f(a^n)</math> is a fixed point of <math>f</math>, contradicting (ii).  If <math>f(a)=a</math> with <math>0<a<1,</math> then <math>f(\frac{1}{a^n})=\frac{1}{a^n}</math> so <math>\frac{1}{a^n}</math> is a fixed point, contradicting (ii).  Hence, the only fixed point is <math>1</math> so <math>xf(x)=1</math> so <math>f(x)=\boxed{\frac{1}{x}}</math> and we can easily check that this solution works.
+==Solution 3==
+Let <math>x=1</math>. So <math>f(f(y)) = y f(1)</math>. This clearly implies <math>f</math> is injective. Let <math>y = 1</math>. Then <math>f(f(y)) = f(1)</math>. Applying <math>f^{-1}</math> yields <math>f(1) = 1</math>, so <math>f(f(y)) = y</math>. So <math>f</math> is surjective. Applying <math>f^{-1}</math> yields <math>f(y) = f^{-1}(y)</math>.
+Let <math>u=x</math> and <math>y = f^{-1}(v)</math>. (i) becomes
+<cmath>
+f(uv) = f^{-1}(v)f(u) = f(u)f(v)
+</cmath>
+since <math>f = f^{-1}</math>. So <math>f</math> is a group isomorphism on the group <math>(0,\infty)</math> with binary operation multiplication. One gets <math>f(u)^n = f(u^n)</math>. Letting <math>v = u^n</math> implies <math>f(v^{1/n}) = f(v)^{1/n}</math>. Combining these identities
+<cmath>
+f(u)^{\frac{m}{n}} = f(u^{\frac{m}{n}}.
+</cmath>
+for all <math>m,n \in \mathbb{N}</math>. Suppose <math>u>1</math>. Taking <math>\frac{m}{n} \rightarrow \infty</math>, the RHS goes to <math>0</math> by (ii). Considering the LHS, necessarily <math>0 < f(u) < 1</math>. So <math>f(u) < 1</math> for <math>u>1</math>. We claim <math>f</math> is monotonically decreasing. Indeed, if <math>0<x<y</math> then
+<cmath>
+f(y) = f\Big(x \frac{y}{x}\Big) = f(x)f\Big(\frac{y}{x}\Big) < f(x)
+</cmath>
+since <math>y/x > 1</math>. Now suppose <math>r \in (0,\infty)</math> and <math>m_1/n_1 < r < m_2/n_2</math>. So
+<cmath>
+f(x)^{\frac{m_1}{n_1}} = f(x^{\frac{m_1}{n_1}}) < f(x^r) < f(x^{\frac{m_2}{n_2}}) = f(x)^{\frac{m_2}{n_2}}
+</cmath>
+Letting <math>m_1/n_1, m_2/n_2 \rightarrow r</math> implies <math>f(x^r) = f(x)^r</math> by the squeeze theorem. Suppose now <math>f(e) = c</math>. Then
+<cmath>
+f(x) = f(e^{\ln x}) = f(e)^{\ln x} = c^{\ln x}.
+</cmath>
+Solving <math>y = c^{\ln x}</math> for <math>x</math> yields <math>x = (\ln y)/(\ln c)</math>. Since <math>f = f^{-1}</math> it follows
+<cmath>
+c^{\ln x} = \frac{\ln x}{\ln c},
+</cmath>
+which implies <math>(\ln c)^2 = 1</math> when <math>x \neq 1</math>. So either <math>c = e</math> or <math>c = 1/e</math>. Since <math>f</math> is monotonically decreasing, <math>c = f(e) < 1</math>, so <math>c = 1/e</math>. In which case <math>f(x) = c^{\ln x} = 1/x</math>.
+~detriti
 == Video Solution ==

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