Timeline for For a continuous function $f:\mathbb{R}^{+}\to\mathbb{R}^{+}$ does $(f(x)-f(y)) (f(\frac{x+y}{2}) - f(\sqrt{xy}))=0$ imply that $f$ is constant?
Current License: CC BY-SA 3.0
21 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 26, 2017 at 9:42 | comment | added | Fan Zheng | Just for the record, this is Problem 4 in Year 2015 of the notoriously difficult Hungarian mathematics competition for undergraduates (artofproblemsolving.com/community/c6h1224690p6149915). | |
| Aug 24, 2017 at 22:14 | comment | added | Ron P | You're right, I was not precise enough. I'm looking for a generalization of the theorem. Let me try to state one. Let "~" be an equivalence relation on the (positive) reals. Suppose 1) "~" has closed classes; 2) for all x < y, either x~y or there exists x <z < (x+y)/2 such that z~(x+y)/2. Then, "~" is trivial (namely, any two reals are equivalent). | |
| Aug 24, 2017 at 18:51 | comment | added | Aditya Guha Roy | Every constant function is continuous. So how can you ensure $f$ is constant claiming that $f$ is not continuous ? So, I don't see any point in claiming existence of discontinuous functions that obey this (means obeys the truth of the given and also they are constant functions). However it may be sensible to see if a proof needs something weaker than continuity, but remember that continuity will eventually be implied. Also initially the main interest was to investigate this property of "continuous" functions. | |
| Aug 24, 2017 at 13:47 | comment | added | Ron P | Nice example! Note that for the theorem to hold f need not be continuous. For example, you can replace it by $g\circ f$, where $g $ is any permutation of the reals. In fact, my proof relies only on the fact that the inverse image of any number is a closed set. My question is: Is any function with that property a composition of a continuous function and a permutation? | |
| Aug 24, 2017 at 8:42 | comment | added | Ian Morris | 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!) | |
| Aug 24, 2017 at 8:20 | comment | added | Ron P | I added a solution based on the "injective over some subinterval" argument. | |
| Aug 24, 2017 at 7:57 | answer | added | Ron P | timeline score: 0 | |
| Aug 24, 2017 at 1:01 | answer | added | Ron P | timeline score: 0 | |
| May 6, 2017 at 6:22 | vote | accept | Aditya Guha Roy | ||
| May 5, 2017 at 17:46 | answer | added | Terry Tao | timeline score: 43 | |
| S May 5, 2017 at 16:26 | history | suggested | the_fox | CC BY-SA 3.0 |
improved formatting
|
| May 5, 2017 at 16:20 | review | Suggested edits | |||
| S May 5, 2017 at 16:26 | |||||
| May 5, 2017 at 16:07 | comment | added | Aditya Guha Roy | Also an example which shows why the "injective over some subinterval" argument won't work is the Weierstrass function . | |
| May 5, 2017 at 16:05 | comment | added | Aditya Guha Roy | Yes you may think of the function $\lfloor x \rfloor$ then see that the condition is not valid "for all" positive reals x,y. Hence a contradiction !! | |
| May 5, 2017 at 16:02 | comment | added | Igor Rivin | Do you have a counterexample for a discontinuous function? | |
| May 5, 2017 at 15:37 | comment | added | Aditya Guha Roy | 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. | |
| May 5, 2017 at 15:21 | comment | added | Gro-Tsen | 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. | |
| May 5, 2017 at 15:06 | comment | added | Aditya Guha Roy | Actually there are some narrow gaps which if avoided makes the problem very simple. | |
| May 5, 2017 at 15:04 | answer | added | Igor Rivin | timeline score: -1 | |
| May 5, 2017 at 14:24 | review | First posts | |||
| May 5, 2017 at 14:26 | |||||
| May 5, 2017 at 14:20 | history | asked | Aditya Guha Roy | CC BY-SA 3.0 |