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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{{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 13:04, 27 December 2018

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.

1983 IMO (Problems) • Resources
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