Timeline for On the average of continuous functions $f:\mathbb{R}^2\rightarrow[0,1]$
Current License: CC BY-SA 3.0
8 events
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| Jul 21, 2011 at 20:12 | history | edited | André Henriques | CC BY-SA 3.0 |
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| Jul 21, 2011 at 20:10 | comment | added | Marc Palm | Dear André Henriques, I misread that you had implied that the formula implies that $f$ is indeed harmonic, but you only stated that the formula holds for harmonic functions and answered the question. Sorry for this. But I guess now the question is more interesting, since there are no bounded nonconstant harmonic functions=) | |
| Jul 21, 2011 at 20:02 | history | edited | André Henriques | CC BY-SA 3.0 |
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| Jul 21, 2011 at 19:58 | comment | added | Nimr | Sorry, I mistyped my question. I meant f takes values in [0,1]! | |
| Jul 21, 2011 at 19:58 | comment | added | André Henriques | @Gerald: You're right I missed the simplest one. @pm: It is not true that $\int_0^1f(z+e^{2\pi\theta})d\theta=f(z)$ $\forall z$ implies $f$ is constant. I have provided a counterexample. | |
| Jul 21, 2011 at 19:49 | comment | added | Marc Palm | I think the question was, if $f: \mathbb{C} \rightarrow \mathbb{R}$ with $\int_0^{1} f( z + e^{2 \pi i \theta}) d \theta = f(z)$ implies that $f$ is constant. I am not seeing this being answered here. Can you please elaborate your answer? | |
| Jul 21, 2011 at 19:26 | comment | added | Gerald Edgar | He said unit circle. Harmonic functions have this property for all circles. Why do you say $xy$ is simplest? Shy not just $x$? | |
| Jul 21, 2011 at 19:20 | history | answered | André Henriques | CC BY-SA 3.0 |