Timeline for On the average of continuous functions $f:\mathbb{R}^2\rightarrow[0,1]$
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 3, 2011 at 15:08 | vote | accept | Nimr | ||
| Jul 22, 2011 at 23:38 | comment | added | Gerald Edgar | @George: although not solving the current problem, it is still quite interesting that $f(x,y)=e^{1.88044 x} \cos(6.947506 x)$ has the unit-circle averaging property, even though it is not harmonic! | |
| Jul 22, 2011 at 4:18 | comment | added | George Lowther | Sorry, it looks like the modified Bessel function $I_0(a)$ does hit 1, for $a\approx 1.88044+6.947506i$. Then, $f(x,y)=\Re[\exp(ax)]$ satisfies the required property (but is not bounded above or below). | |
| Jul 22, 2011 at 3:43 | comment | added | George Lowther | @Gerald: The Fourier transform gives the modified Bessel function of the first kind. I don't think it is equal to 1 except at the origin (?). If you take the delta function at the origin and transform back, then you just get a constant function. | |
| Jul 21, 2011 at 22:27 | comment | added | Gerald Edgar | @André: Your distribution is also called the (normalized) arc length measure on the unit circle. It is interesting that its Fourier transform can be written in terms of a Bessel function. After you get the set $A$ where that transform is 1, take something supported there, then inverse Fourier transform back. But can such a result be a bounded function? According to George's answer, no. But not requiring boundedness: we probably get a function with the unit circle averaging property that is not harmonic. | |
| Jul 21, 2011 at 21:55 | answer | added | George Lowther | timeline score: 16 | |
| Jul 21, 2011 at 21:20 | comment | added | Marc Palm | See jstor.org/pss/2034412 and en.wikipedia.org/wiki/Harmonic_function#Liouville.27s_theorem | |
| Jul 21, 2011 at 20:29 | comment | added | André Henriques | Say $f$ is a function that satisfies your condition. The function $f$ is an eigenfunction for the operator of convolution by the distribution $\delta_0(1-x^2-y^2)$ at the eigenvalue 1. After Fourier transforming, this operator becomes the operator of multiplication by $\widehat{\delta_0(1-x^2-y^2)}$. By looking at the solution of the equation $\widehat{\delta_0(1-x^2-y^2)}=1$ you'll find some restrictions on the possible support of the tempered distribution $\widehat f$... | |
| Jul 21, 2011 at 20:23 | comment | added | Nimr | I got it now @pm! However, I feel that claiming that f is harmonic because of the average property is somehow overstretched :-) | |
| Jul 21, 2011 at 20:13 | comment | added | Marc Palm | No, i just suggest. If the integral formula holds, then it is sufficient to show that this implies that $f$ is harmonic. Then you can argue that since $f$ is bounded and harmonic, it must be constant. | |
| Jul 21, 2011 at 20:09 | comment | added | Nimr | With the restriction to [0,1], I can't see the link with harmonic functions anymore. | |
| Jul 21, 2011 at 20:06 | comment | added | Marc Palm | Note there exists function with this poperty. Holomorphic function are suitable if you allow complex valued functions and real harmonic functions in the case before. There are no real valued holomorphic function except constant functions, there are no bounded harmonic functions, so you're finished if you can show that f is indeed harmonic. | |
| Jul 21, 2011 at 19:56 | comment | added | Nimr | Sorry, I mistyped my question. I meant f takes values in [0,1]! | |
| Jul 21, 2011 at 19:55 | history | edited | Nimr | CC BY-SA 3.0 |
deleted 5 characters in body; edited title
|
| Jul 21, 2011 at 19:20 | answer | added | André Henriques | timeline score: 4 | |
| Jul 21, 2011 at 18:53 | history | asked | Nimr | CC BY-SA 3.0 |