2
$\begingroup$

USAMO Question 2 (via artofproblemsolving.com):

Find all functions $f:(0,\infty)\to(0,\infty)$ such that $$f\left(x+\frac1y\right)+f\left(y+\frac1z\right)+f\left(z+\frac1x\right)=1$$ for all $x, y, z>0$ with $xyz=1$.

The link redirects to a forum on AoPS. Check USAMO 2's solution. The author claims that $g$ can (uniquely) be extended to an additive function $h$ on all of $\mathbb{R}$. I don't understand why this is true by the explanations below this statement. Please help me understand. Maybe define the function $h$ explicitely. I was thinking of something like $$h(x)=g\left(x-\left\lfloor x+\frac13 \right\rfloor\right)+3\left\lfloor x+\frac13 \right\rfloor g\left(\frac13 \right), \forall x\in \mathbb R$$ but I don't think this works.

If you can't explain the solution in the link above, but you have a solution to this problem which does not involve analysis (only algebra, and other than Evan Chen's solution / AoPS solutions which are very long and hard to find in a contest), please post it here. It will help. Thank you in advance!

$\endgroup$
1
+400
$\begingroup$

We have $g: (-\frac{1}{3},\frac{1}{3}) \to \mathbb{R}$ with $g(x+y) = g(x)+g(y)$. Define $G: \mathbb{R} \to \mathbb{R}$ by $G(x) = Ng(\frac{x}{N})$ where $N \in \mathbb{N}$ is large enough to ensure $|\frac{x}{N}| < \frac{1}{3}$. To see that the definition does not depend on $N$, i.e. to show $Ng(\frac{x}{N}) = Mg(\frac{x}{M})$ for any $M$ with $|\frac{x}{M}| < \frac{1}{3}$, it suffices to show both are equal to $NMg(\frac{x}{NM})$, which is clear from additivity. Let's show $G(x+y) = G(x)+G(y)$ for $x,y \in \mathbb{R}$. Fix $x,y \in \mathbb{R}$, and take $N$ large so that $|\frac{x}{N}|,|\frac{y}{N}|,|\frac{x+y}{N}| < \frac{1}{3}$; then $G(x+y) = Ng(\frac{x+y}{N})$ and $G(x)+G(y) = Ng(\frac{x}{N})+Ng(\frac{y}{N})$, so just use additivity of $g$. Finally, it is clear that $G$ extends $g$.

$\endgroup$
0
$\begingroup$

We can fix the extension part of the solution as follows. Put $U=\left( -\tfrac 13, \tfrac 23\right)$.

We claim that for each $x_1,\dots, x_k\in U$ with $x_1+\dots+x_k=0$ we have $g(x_1)+\dots+g(x_k)=0$. Let’s prove this claim by induction with respect to $k$. For $k\le 3$ the claim is given. Assume that the claim is proved for each $k\le n\ge 3$. Let $x_1,\dots, x_{n+1}\in U$ with $x_1+\dots+x_{n+1}=0$. Without loss of generality we can assume that $x_1\le 0\le x_2$, so $x_1+x_2\in [x_1, x_2]\subset U$. By the inductive hypothesis we have $$g(x_1+x_2)+g(x_3)+\dots+g(x_n)=0,$$ so it remains to prove that $g(x_1+x_2)=g(x_1)+g(x_2)$. It is easy to see that $-\tfrac{x_1+x_2}2\in U$, so $$g(x_1+x_2)+2g\left(-\tfrac{x_1+x_2}2\right)=0.$$ Similarly we have $$g(x_1)+2g\left(-\tfrac{x_1}2\right)=0\mbox{ and }g(x_2)+2g\left(-\tfrac{x_2}2\right)=0.$$ Moreover, $$g\left(-\tfrac{x_1+x_2}2\right)+ g\left(\tfrac{x_1}2\right)+ g\left(\tfrac{x_2}2\right)=0,$$
$$g\left(\tfrac{x_1}2\right)+ g\left(-\tfrac{x_1}2\right)=0,\mbox{ and } g\left(\tfrac{x_2}2\right)+ g\left(-\tfrac{x_1}2\right)=0.$$ It follows $$g(x_1+x_2)=$$ $$-2g\left(-\frac{x_1+x_2}2\right)=2 g\left(\frac{x_1}2\right)+2g\left(\frac{x_2}2\right)=-2 g\left(-\frac{x_1}2\right)-2g\left(-\frac{2}2\right)=$$ $$g(x_1)+g(x_2).$$

Let $x\in\Bbb R$ be any number, $x=x_1+\dots+x_n$ and $x=x’_1+\dots+x’_m$ be two representations of $x$ with $x_1,\dots, x_n, x’_1,\dots, x’_m\in U$. Then $\pm \tfrac {x_i}2$ and $\pm \tfrac {x’_j}2$ belong to $U$ for each $i$ and $j$. By the claim we have $$g(x_1)+\dots+g(x_n)=$$ $$-2\left(g\left(-\frac{x_1}2\right)+\dots+ g\left(-\frac{x_n}2\right) \right)=$$ $$2\left(g\left(\frac{x’_1}2\right)+\dots+ g\left(\frac{x’_m}2\right) \right)=$$ $$-2\left(g\left(-\frac{x’_1}2\right)+\dots+ g\left(-\frac{x’_m}2\right) \right)=$$ $$g(x’_1)+\dots+g(x’_n).$$

Put $h(x)=g(x_1)+\dots+ g(x_n)$. The definition of $h(x)$ implies that $h$ is additive and an extension of $g$. The uniqueness of such an $h$ follows from its additivity and the claim, but I guess it is not needed for the solution, since existence of any additive extension of $g$ on $\Bbb R$ implies $g(x)=kx$ for some $k\in\left[-\tfrac 12,1\right]$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.