Timeline for Are there functions satisfying the following integral condition?
Current License: CC BY-SA 2.5
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 11, 2011 at 20:22 | comment | added | Chulumba | @Todd, thanks. So,there are many types of counterexample functions. | |
| Feb 10, 2011 at 2:13 | vote | accept | Chulumba | ||
| Feb 9, 2011 at 17:00 | comment | added | Todd Trimble | Chulumba, it's not hard if you know how to construct bump functions. You can take $F(R) = G(R) = [0, 1]$, and say $supp(F) = [2, 3]$ and $supp(G) = [3, 4]$, as follows. Define f(x) = exp(-1/x)exp(1/(1-x)) for 1 > x > 0, then g(x) = f(x)/(1 + f(x)) for 1 > x > 0, then define a continuous function h(x) piecewise to be = 0 for 0 > x, = g(3x) for 1/3 > x > 0, = 1 for 2/3 > x > 1/3, g(3-3x) for 1 > x > 2/3, and 0 for x > 1. Finally, define F(x) = h(x-2), and G(x) = h(x-3). | |
| Feb 1, 2011 at 15:07 | comment | added | Chulumba | @Ady, thanks. I found it hard to find sample functions such that $supp\: F\cap G\left(\mathbb{R}\right)=supp\: G\cap F\left(\mathbb{R}\right)=\emptyset$. Could you come up with explicit two functions? | |
| Jan 30, 2011 at 11:20 | history | answered | Ady | CC BY-SA 2.5 |