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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>.  
 
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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Let <math>x=y=1</math> and we have <math>f(f(1))=f(1)</math>.  Now, let <math>y=f(1)</math> and we have <math>f(xf(f(1)))=f(1)f(x)</math> which simplifies to <math>f(x)=f(1)f(x)</math> and since <math>f(x)\ne0</math> we have <math>f(x)=1</math>.   
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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>.   
  
 
Plug in <math>y=x</math> and we have <math>f(xf(x))=xf(x)</math>.  If <math>a=1</math> is the only solution to <math>f(a)=a</math> then we have <math>xf(x)=1\Rightarrow f(x)=\frac{1}{x}</math>.  We prove that this is the only function by showing that there does not exist any other <math>a</math>:
 
Plug in <math>y=x</math> and we have <math>f(xf(x))=xf(x)</math>.  If <math>a=1</math> is the only solution to <math>f(a)=a</math> then we have <math>xf(x)=1\Rightarrow f(x)=\frac{1}{x}</math>.  We prove that this is the only function by showing that there does not exist any other <math>a</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>.
 
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>.

Revision as of 09:56, 20 August 2008

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 $x\to \infty$.

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 $x\to \infty$.

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