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Difference between revisions of "1983 IMO Problems/Problem 1"
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Find all functions <math>f</math> defined on the set of positive reals which take positive real values and satisfy:  <math>f(xf(y))=yf(x)</math> for all <math>x,y</math>; and <math>f(x)\to0</math> as <math>x\to \infty</math>.  
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==Problem==
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Find all functions <math>f</math> defined on the set of positive reals which take positive real values and satisfy the conditions:   
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(i)  <math>f(xf(y))=yf(x)</math> for all <math>x,y</math>;  
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(ii)  <math>f(x)\to0</math> as <math>x\to \infty</math>.
  
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==Solution 1==
 
Let <math>x=y=1</math> and we have <math>f(f(1))=f(1)</math>.  Now, let <math>x=1,y=f(1)</math> and we have <math>f(f(f(1)))=f(1)f(1)\Rightarrow f(1)=[f(1)]^2</math> since <math>f(1)>0</math> we have <math>f(1)=1</math>.   
 
Let <math>x=y=1</math> and we have <math>f(f(1))=f(1)</math>.  Now, let <math>x=1,y=f(1)</math> and we have <math>f(f(f(1)))=f(1)f(1)\Rightarrow f(1)=[f(1)]^2</math> since <math>f(1)>0</math> we have <math>f(1)=1</math>.   
  
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Suppose there did exist such an <math>a\ne1</math>.  Then, letting <math>y=a</math> in the functional equation yields <math>f(xa)=af(x)</math>.  Then, letting <math>x=\frac{1}{a}</math> yields <math>f(\frac{1}{a})=\frac{1}{a}</math>.  Notice that since <math>a\ne1</math>, one of <math>a,\frac{1}{a}</math> is greater than <math>1</math>.  Let <math>b</math> equal the one that is greater than <math>1</math>.  Then, we find similarly (since <math>f(b)=b</math>) that <math>f(xb)=bf(x)</math>.  Putting <math>x=b</math> into the equation, yields <math>f(b^2)=b^2</math>.  Repeating this process we find that <math>f(b^{2^k})=b^{2^k}</math> for all natural <math>k</math>.  But, since <math>b>1</math>, as <math>k\to \infty</math>, we have that <math>b^{2^k}\to\infty</math> which contradicts the fact that <math>f(x)\to0</math> as <math>x\to \infty</math>.
 
Suppose there did exist such an <math>a\ne1</math>.  Then, letting <math>y=a</math> in the functional equation yields <math>f(xa)=af(x)</math>.  Then, letting <math>x=\frac{1}{a}</math> yields <math>f(\frac{1}{a})=\frac{1}{a}</math>.  Notice that since <math>a\ne1</math>, one of <math>a,\frac{1}{a}</math> is greater than <math>1</math>.  Let <math>b</math> equal the one that is greater than <math>1</math>.  Then, we find similarly (since <math>f(b)=b</math>) that <math>f(xb)=bf(x)</math>.  Putting <math>x=b</math> into the equation, yields <math>f(b^2)=b^2</math>.  Repeating this process we find that <math>f(b^{2^k})=b^{2^k}</math> for all natural <math>k</math>.  But, since <math>b>1</math>, as <math>k\to \infty</math>, we have that <math>b^{2^k}\to\infty</math> which contradicts the fact that <math>f(x)\to0</math> as <math>x\to \infty</math>.
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==Solution 2==
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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
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<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.
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==Solution 3==
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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>.
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Let <math>u=x</math> and <math>y = f^{-1}(v)</math>. (i) becomes
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<cmath>
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f(uv) = f^{-1}(v)f(u) = f(u)f(v)
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</cmath>
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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,
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<cmath>
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f(u)^{\frac{m}{n}} = f(u^{\frac{m}{n}}).
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</cmath>
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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 strictly monotonically decreasing. Indeed, if <math>0<x<y</math> then
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<cmath>
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f(y) = f\Big(x \frac{y}{x}\Big) = f(x)f\Big(\frac{y}{x}\Big) < f(x)
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</cmath>
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since <math>y/x > 1</math> thus <math>f(y/x) < 1</math>. Now suppose <math>r > 0</math>. We claim <math>f(x^r) = f(x)^r</math>. Indeed let <math>m_1/n_1 < r < m_2/n_2</math>. If <math>x > 1</math> then
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<cmath>
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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}}
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</cmath>
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and letting <math>\frac{m_1}{n_1}, \frac{m_2}{n_2} \rightarrow r</math> implies <math>f(x^r) = f(x)^r</math> by the squeeze theorem. The case <math>x < 1</math> is similar. The case <math>x=1</math> is trivial since <math>f(1) = 1</math>. Suppose now <math>f(e) = c</math>. Then
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<cmath>
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f(x) = f(e^{\ln x}) = f(e)^{\ln x} = c^{\ln x}.
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</cmath>
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Solving <math>y = c^{\ln x}</math> for <math>x</math> yields <math>x = \exp\big((\ln y)/(\ln c)\big)</math>. Since <math>f = f^{-1}</math> it follows
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<cmath>
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c^{\ln x} = \exp\bigg(\frac{\ln x}{\ln c}\bigg),
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</cmath>
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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>x>1</math> implies <math>f(x) < 1</math> we have <math>c = f(e) < 1</math>, so <math>c = 1/e</math>. In which case <math>f(x) = c^{\ln x} = 1/x</math>.
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~not_detriti
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== Video Solution ==
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https://youtu.be/Fu0jBKKa4Lc [Video Solution by little fermat]
  
 
{{IMO box|year=1983|before=First question|num-a=2}}
 
{{IMO box|year=1983|before=First question|num-a=2}}
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[[Category:Olympiad Algebra Problems]]
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[[Category:Functional Equation Problems]]

Latest revision as of 14:48, 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, \[f(u)^{\frac{m}{n}} = f(u^{\frac{m}{n}}).\] 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 strictly 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 $y/x > 1$ thus $f(y/x) < 1$. Now suppose $r > 0$. We claim $f(x^r) = f(x)^r$. Indeed let $m_1/n_1 < r < m_2/n_2$. If $x > 1$ then \[f(x)^{\frac{m_1}{n_1}} = f(x^{\frac{m_1}{n_1}}) > f(x^r) > f(x^{\frac{m_2}{n_2}}) = f(x)^{\frac{m_2}{n_2}}\] and letting $\frac{m_1}{n_1}, \frac{m_2}{n_2} \rightarrow r$ implies $f(x^r) = f(x)^r$ by the squeeze theorem. The case $x < 1$ is similar. The case $x=1$ is trivial since $f(1) = 1$. 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 $x = \exp\big((\ln y)/(\ln c)\big)$. Since $f = f^{-1}$ it follows \[c^{\ln x} = \exp\bigg(\frac{\ln x}{\ln c}\bigg),\] which implies $(\ln c)^2 = 1$ when $x \neq 1$. So either $c = e$ or $c = 1/e$. Since $x>1$ implies $f(x) < 1$ we have $c = f(e) < 1$, so $c = 1/e$. In which case $f(x) = c^{\ln x} = 1/x$.

~not_detriti


Video Solution

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

1983 IMO (Problems) • Resources
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