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Difference between revisions of "2016 APMO Problems/Problem 5"

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This gives us the power to compute <math>f(1)</math>. From <math>P(1,1,1)</math> we get <math>f(f(1)+1)=f(2)</math> and injectivity gives <math>f(1)=1</math>.
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This gives us the power to compute <math>f(1)</math>. From <math>P(1,1,1)</math> we get <math>f(f(1)+1)=f(2)</math> and injectivity gives <math>f(1)=1</math>. Showing that <math>f</math> is unbounded above is also easy as we can fix <math>(x,y)</math> and let <math>z</math> blow up to <math>\infty</math> in the original Functional equation..
  
  
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<cmath>(z+1)f(2x)=2f(x+xf(z)) \iff \frac{(z+1)f(2x)}{2}=f(x+xf(z))</cmath>
 
<cmath>(z+1)f(2x)=2f(x+xf(z)) \iff \frac{(z+1)f(2x)}{2}=f(x+xf(z))</cmath>
  
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This gives that <math>\text{Im}(f) \in \left(\frac{f(2x)}{2},\infty\right)</math>. Putting <math>x=\frac{1}{2}</math>, we get  <math>\text{Im}(f) \in \left(\frac{1}{2},\infty\right)</math>.
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This gives that <math>\text{Im}(f) \in \left(\frac{f(2x)}{2},\infty\right)</math>. Putting <math>x=\frac{1}{2}</math>, we get  <math>\text{Im}(f) \in \left(\frac{1}{2},\infty\right)</math>. By induction, surjectivity is proved as <math>\lim_{m \to \infty}\frac{1}{2^m}=0</math> and we are essentially done. <math>\square</math>
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Now, we have claim that if <math>a+b=c+d</math> for some <math>a,b,c,d \in \mathbb{R}^+</math>, then <math>f(a)+f(b)=f(c)+f(d)</math>. This can be achieved by putting <math>a=xf(z)+y</math>, <math>b=yf(z)+x</math>, <math>c=x'f(z)+y'</math> and <math>d=y'f(z)+x'</math>. Let us calculate <math>f(a)+f(b)</math>
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<cmath>f(a)+f(b)=(z+1)f(x+y)=(z+1)f\left(\frac{a+b}{f(z)+1}\right)=(z+1)f\left(\frac{c+d}{f(z)+1}\right)=(z+1)f(x'+y')=f(c)+f(d)</cmath>
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Everything is fine, but we would have show that such numbers <math>x,y,x',y' \in \mathbb{R}^+</math> do exist. We will show this in the next lemma.
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<b>Lemma 1:</b> There exits <math>x,y,x',y',a,b,c,d \in \mathbb{R}^+</math> such that
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<cmath>\begin{align*}
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a=xf(z)+y \\
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b=yf(z)+x \\
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c=x'f(z)+y' \\
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d=y'f(z)+x'
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\end{align*}</cmath>
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<i>Proof:</i> Notice that <math>f</math> is surjective. So, we will solve for <math>x,y,x',y'</math> in the system of linear equation. For convinience, let <math>f(z)=k</math> for <math>k \in \mathbb{R}^+</math> as <math>f</math> is surjective. Solving we get:
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<cmath>\begin{align*}
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x=\frac{ka-b}{k^2-1}\\
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y=\frac{kb-a}{k^2-1} \\
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x'=\frac{kc-d}{k^2-1} \\
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y'=\frac{kd-c}{k^2-1}
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\end{align*}</cmath>
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Now, you can chose <math>z</math> such that <math>f(z)\ne 1</math> and <math>x,y,x',y'</math> is positive. This finishes the proof to the lemma. <math>\square</math>
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<b>Claim 3:</b> <math>f</math> obeys Jensen's Functional Equation,i.e.,
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<cmath>\frac{f(x)+f(y)}{2}=f\left(\frac{x+y}{2}\right)</cmath>
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<i>Proof:</i>

Latest revision as of 08:20, 14 July 2021

Problem

Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that \[(z + 1)f(x + y) = f(xf(z) + y) + f(yf(z) + x),\]for all positive real numbers $x, y, z$.

Solution

We claim that $f(x)=x$ is the only solution. It is easy to check that it works. Now, we will break things down in several claims. Let $P(x.y,z)$ be the assertion to the Functional Equation.

Claim 1: $f$ is injective.

Proof: Assume $f(a)=f(b)$ for some $a,b \in \mathbb{R}^+$. Now, from $P(x,y,a)$ and $P(x,y,b)$ we have:


\[(a+1)f(x+y)=f(xf(a)+y)+f(yf(a)+x)\] \[(b+1)f(x+y)=f(xf(b)+y)+f(yf(b)+x)\]

Now comparing, we have $a=b$ as desired. $\square$


This gives us the power to compute $f(1)$. From $P(1,1,1)$ we get $f(f(1)+1)=f(2)$ and injectivity gives $f(1)=1$. Showing that $f$ is unbounded above is also easy as we can fix $(x,y)$ and let $z$ blow up to $\infty$ in the original Functional equation..


Claim 2: $f$ is surjective.

Proof: $P(x,x,z)$ gives \[(z+1)f(2x)=2f(x+xf(z)) \iff \frac{(z+1)f(2x)}{2}=f(x+xf(z))\]

This gives that $\text{Im}(f) \in \left(\frac{f(2x)}{2},\infty\right)$. Putting $x=\frac{1}{2}$, we get $\text{Im}(f) \in \left(\frac{1}{2},\infty\right)$. By induction, surjectivity is proved as $\lim_{m \to \infty}\frac{1}{2^m}=0$ and we are essentially done. $\square$


Now, we have claim that if $a+b=c+d$ for some $a,b,c,d \in \mathbb{R}^+$, then $f(a)+f(b)=f(c)+f(d)$. This can be achieved by putting $a=xf(z)+y$, $b=yf(z)+x$, $c=x'f(z)+y'$ and $d=y'f(z)+x'$. Let us calculate $f(a)+f(b)$

\[f(a)+f(b)=(z+1)f(x+y)=(z+1)f\left(\frac{a+b}{f(z)+1}\right)=(z+1)f\left(\frac{c+d}{f(z)+1}\right)=(z+1)f(x'+y')=f(c)+f(d)\]

Everything is fine, but we would have show that such numbers $x,y,x',y' \in \mathbb{R}^+$ do exist. We will show this in the next lemma.


Lemma 1: There exits $x,y,x',y',a,b,c,d \in \mathbb{R}^+$ such that \begin{align*} a=xf(z)+y \\ b=yf(z)+x \\ c=x'f(z)+y' \\ d=y'f(z)+x' \end{align*} Proof: Notice that $f$ is surjective. So, we will solve for $x,y,x',y'$ in the system of linear equation. For convinience, let $f(z)=k$ for $k \in \mathbb{R}^+$ as $f$ is surjective. Solving we get: \begin{align*} x=\frac{ka-b}{k^2-1}\\ y=\frac{kb-a}{k^2-1} \\ x'=\frac{kc-d}{k^2-1} \\ y'=\frac{kd-c}{k^2-1} \end{align*} Now, you can chose $z$ such that $f(z)\ne 1$ and $x,y,x',y'$ is positive. This finishes the proof to the lemma. $\square$

Claim 3: $f$ obeys Jensen's Functional Equation,i.e., \[\frac{f(x)+f(y)}{2}=f\left(\frac{x+y}{2}\right)\] Proof:

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