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
15 events
| when toggle format | what | by | license | comment | |
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| Aug 24, 2017 at 1:02 | comment | added | Ron P | I tried to correct the proof. See answer below. | |
| May 5, 2017 at 16:39 | comment | added | Igor Rivin | @DavidSpeyer Guilty as charged. | |
| May 5, 2017 at 16:38 | comment | added | Aditya Guha Roy | @DavidSpeyer Yes you are correct. So we are not yet done. (In one of the earlier comments I wrote a flaw) . | |
| May 5, 2017 at 16:32 | comment | added | David E Speyer | @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 $x$. | |
| May 5, 2017 at 16:22 | history | edited | Igor Rivin | CC BY-SA 3.0 |
added a thought.
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| May 5, 2017 at 16:00 | comment | added | Igor Rivin | @DanPetersen If you look at the text it says that IF the domain includes zero THEN the argument works, so the fact that $f(0)$ is not defined (as clarified by the OP) means that the answer is to a slightly different question. | |
| May 5, 2017 at 15:58 | comment | added | Aditya Guha Roy | @Igor Rivin Yes now that is a nice work (really good), and we would have got a clean solution if the domain of f was non-negative reals but unfortunately 0 is not in the domain. | |
| May 5, 2017 at 15:58 | comment | added | Dan Petersen | @Igor $f(0)$ is not defined! | |
| May 5, 2017 at 15:56 | comment | added | Igor Rivin | @adityaguharoy see edit. | |
| May 5, 2017 at 15:56 | history | edited | Igor Rivin | CC BY-SA 3.0 |
corrected mistake.
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| May 5, 2017 at 15:55 | comment | added | Igor Rivin | @adityaguharoy You are right that what I wrote is not quite correct, but almost. | |
| May 5, 2017 at 15:54 | history | undeleted | Igor Rivin | ||
| May 5, 2017 at 15:54 | history | deleted | Igor Rivin | via Vote | |
| May 5, 2017 at 15:10 | comment | added | Aditya Guha Roy | 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. | |
| May 5, 2017 at 15:04 | history | answered | Igor Rivin | CC BY-SA 3.0 |