28
$\begingroup$

Suppose that $f: \mathbb{R}^+ \to \mathbb{R}^+$ is a continuous function such that for all positive real numbers $x,y$ the following is true : $$(f(x)-f(y)) \left ( f \left ( \frac{x+y}{2} \right ) - f ( \sqrt{xy} ) \right )=0.$$ Is it true that the only solution to this is the constant function ?

$\endgroup$
12
  • 2
    $\begingroup$ By "makes the problem very simple", do you mean to say that you have a solution and are asking this as a riddle? Because if so, it should be made clear that MathOverflow is for asking questions which you do not know how to answer (at the time when you are asking). Maybe you should clarify how you came across this problem. $\endgroup$
    – Gro-Tsen
    May 5, 2017 at 15:21
  • 1
    $\begingroup$ No I actually couldn't edit that comment . Ok but you read "makes the problem simple" but did not see why I wrote "there are some narrow gaps" anyways I meant that if we had a better function f (instead of just being a continuous one) which would have shown the property that if f took two different values at some places (say f(a) $\ne$ f(b) and $a<b$ ) then f would behave injective over some subinterval of $(a,b)$ . And this is not a fake exercise. $\endgroup$ May 5, 2017 at 15:37
  • 2
    $\begingroup$ Do you have a counterexample for a discontinuous function? $\endgroup$
    – Igor Rivin
    May 5, 2017 at 16:02
  • 7
    $\begingroup$ For a discontinuous example take $f$ to be the characteristic function of the transcendental numbers. If $x$ and $y$ are both algebraic or both transcendental then the first factor, $f(x)-f(y)$, vanishes. If one is algebraic and the other transcendental then $(x+y)/2$ and $\sqrt{xy}$ are both transcendental and the second factor vanishes. (Here the fact that $x$ and $y$ are not allowed to be zero is important!) $\endgroup$
    – Ian Morris
    Aug 24, 2017 at 8:42
  • 2
    $\begingroup$ Just for the record, this is Problem 4 in Year 2015 of the notoriously difficult Hungarian mathematics competition for undergraduates (artofproblemsolving.com/community/c6h1224690p6149915). $\endgroup$
    – Fan Zheng
    Aug 26, 2017 at 9:42

4 Answers 4

43
$\begingroup$

Yes. If $f$ were not constant, then (since ${\bf R}^+$ is connected) it could not be locally constant, thus there exists $x_0 \in {\bf R}^+$ such that $f$ is not constant in any neighbourhood of $x_0$. By rescaling (replacing $f(x)$ with $f(x_0 x)$) we may assume without loss of generality that $x_0=1$.

For any $y \in {\bf R}^+$, there thus exists $x$ arbitrarily close to $1$ for which $f(x) \neq f(y)$, hence $f((x+y)/2) = f(\sqrt{xy})$. By continuity, this implies that $f((1+y)/2) = f(\sqrt{y})$ for all $y \in {\bf R}^+$. Making the substitution $z := (1+y)/2$, we conclude that $f(z) = f(g(z))$ for all $z > 1/2$, where $g(z) := \sqrt{2z-1}$. The function $g$ attracts $[1, \infty)$ to the fixed point $z=1$, so on iteration and by using the continuity of $f$ we conclude that $f(z)=f(1)$ for all $z >1$. Similarly, $h = g^{-1}$ defined by $h(z) = (z^2 + 1)/2$ attracts $(0, 1]$ to the fixed point $z = 1$, so by the same argument $f(z) = f(1)$ for $z < 1$, making $f$ constant on all of $\bf R^+$.

$\endgroup$
2
  • 1
    $\begingroup$ Slick, very slick. $\endgroup$
    – user78249
    May 5, 2017 at 19:03
  • $\begingroup$ This proof is strong. Also shows the pair $(AM,GM)$ can be replaced by some other terms (some relaxations maybe possible) and yet the result will be true. $\endgroup$ Aug 25, 2017 at 13:13
0
$\begingroup$

Trying to make sense of Igor's argument. Assume further that $f$ is defined on 0 and relaxt the continuity assumption by only assuming that the set $A=\{x>0:f (x)\neq f (0)\}$ is open.

It is shown that $A=\emptyset $ by showing that

  1. $A $ is dense in $A+\mathbb R^+$, and
  2. $A\cap 2A=\emptyset $.

Proof. (1) Let $(a,b) $ be a connected component of $A $. For every $0 <r <b-a $, $b+r\in A $ since the arithmetic average of $b\pm r $ is not in $A $ whereas the geometric is. It follows that the next connected component contains $(b,2b-a) $ and, by applying the same argument on the next component, the elements of $[a,\infty )\setminus A $ are at least $b-a $ apart from one another.

(2) Let $a\in A $. $a $ is the arithmetic average of 0 and $2a $ and 0 is their geometric.

$\endgroup$
0
$\begingroup$

Yet another solution (to the original question).

Define an equivalence relation on $\mathbb R^+$ by $x\sim y$ iff $f(x)=f(y)$. Since $f$ is continuous,

  1. every equivalence class is closed.

Let $0<x<y$. Since $y=\frac {x +(2y-x)}{2}$, either $x\sim 2y-x$ or there exists $z\in (x,y)$ such that $z\sim y$. Therefore,

  1. for every $x<y$, if [$\forall z\in(x,y)$ $z\not\sim y$] then $x\sim 2y-x$.

It suffices to prove the following proposition.

Proposition 1. Every equivalence relation on $\mathbb R^+$ that satisfies (1) and (2) is trivial (i.e., has one equivalence class).

Proof. Suppose there were a non-trivial equivalence relation "$\sim$" satisfying (1) and (2).

Step 1. Every equivalence class is unbounded and its interior is either unbounded or empty.

Suppose $\langle x\rangle$ were an equivalence class which is either bounded or has a non-empty bounded interior. There exits $\epsilon\geq 0$ such that $\langle x\rangle$ contains an interval of length $\epsilon$, and there exists $x'=\max\{x'': [x'',x''+\epsilon]\subset \langle x\rangle\}$. Take some $y \in (x',\infty)\setminus \langle x\rangle$, and let $y'=\min ( \langle y\rangle\cap[x',\infty))$. By (2), $[x',x'+\epsilon]'\sim 2y'-[x',x'+\epsilon]$ contradicting the maximality of $x'$.

Step 2. Every equivalence class is an arithmetic sequence.

Let $\langle x\rangle$ be an equivalence class. If $x_0=\inf \langle x \langle >0$, then by (2) the elements of $(x_0,2x_0)$ are not in $\langle x\rangle$. Since $\langle x\rangle$ is unbounded and closed, there is a minimal $r_x\geq x_0$ such that $x_0+r_x\in\langle x\rangle$. By (2), the elements of $(x_0,x_0+r_x)$ are not in $\langle x\rangle$ and $x_0+r_x$ is in $\langle x\rangle$. Applying the same argument inductively on all intervals $[x_0+nr_x,x_0+(n+1)r_x]$, we get $\langle x\rangle=\mathbb R^+\cap (x+r_x\mathbb Z)$.

Suppose $\inf \langle x \rangle = 0$ then, the compliment of $\langle x \rangle$ contains an interval $(a,b)$ such that $a,b\in \langle x \rangle$ and starting from that interval $\langle x \rangle$ is an arithmetic sequence with period $b-a$. Since this is the case for any such intervals, they must all have the same length; therefore $\inf \mathbb R^+\setminus\langle x\rangle >0$. Since $\langle x\rangle$ has a bounded interior it must have an empty interior (by Claim 1); therefore it does not contain a prefix of $\mathbb R^+$, so $\inf \langle x \rangle >0$.

Step 3. For every $x$ and $y$, $r_y \leq r_x$ with strict inequality for some $x,y$. Contradiction!

Let $x\not\sim y$. There exists $x'\sim x$ such that $\langle y\rangle\cap(x',x'+r_x)\neq\emptyset$. By (2), since $\forall z\in(x',x'+r_x)\ z\not\sim x'+r_x$, $\langle y\rangle\cap(x'+r_x,x'+2r_x)\neq\emptyset$. By induction, $\langle y\rangle\cap(x'+nr_x,x'+(n+1)r_x)\neq\emptyset$, for all $n\in\mathbb N$; therefore $r_y\leq r_x$. Take $x$ and $y=x+\frac 2 3 r_x$. By (2), $y\sim y'=x+\frac 4 3 r_x$. It follows that $r_y\leq |y'-y|<r_x$.

$\endgroup$
-1
$\begingroup$

if $\mathbb{R}^+$ is the set of non-negative real numbers, your condition implies that $$(f(x)-f(0)) (f(\frac{x}2)-f(0)) = 0$$ for all positive $x,$ which is pretty convincing (setting $g(x) = f(x) - f(0),$ the function $g$ will be identically $0.$).

Addendum Since the map $(x, y) -> (x+y)/2, \sqrt{x y})$ is surjective onto the set $x>y,$ if your function were real analytic and nonconstant, then the set where your condition held would be a subvariety of $R^+ \times R^+,$ so assuming more regularity makes the problem easy, as pointed out in comments.

$\endgroup$
10
  • 2
    $\begingroup$ No firstly $\mathbb{R}^+$ is the set of all positive reals (so $0$ is not included in the domain) . Also if $0$ was included in the domain still I think there remains a flaw in your argument which you can easily see because to infer your conclusion of $f(x) \cdot f(x/2)=0$ we would need $f(0)=0$ and then by our condition (of $f(x) \cdot f(x/2)=0$ ) for every $x$ and the continuity we get $f \equiv 0$ as the "only" solution which is not true since any constant function (obviously obeying the positivity of codomain ) fits in. So I ask you to check your solution to find the flaw. $\endgroup$ May 5, 2017 at 15:10
  • 1
    $\begingroup$ @adityaguharoy You are right that what I wrote is not quite correct, but almost. $\endgroup$
    – Igor Rivin
    May 5, 2017 at 15:55
  • 1
    $\begingroup$ @adityaguharoy see edit. $\endgroup$
    – Igor Rivin
    May 5, 2017 at 15:56
  • 2
    $\begingroup$ @Igor $f(0)$ is not defined! $\endgroup$ May 5, 2017 at 15:58
  • 4
    $\begingroup$ @IgorRivin I must be missing something. Let $g(x)$ be nonzero on the interval $(0.9,1.1)$ and zero everywhere else. Then $g(x) g(x/2)=0$ for all $x$. $\endgroup$ May 5, 2017 at 16:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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