Timeline for Local equality of functions implies global equality?
Current License: CC BY-SA 4.0
11 events
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Jun 26, 2023 at 23:46 | vote | accept | Amr | ||
Jun 26, 2023 at 18:03 | comment | added | Amr | @StevenGubkin the set of points where $f_2\circ f_1,f_4\circ f_3$ (i.e. $f_1^{-1}(dom(f_2))\cap f_3^{-1}(dom(f_4)) $) are both defined need not be connected to apply the identity theorem | |
Jun 26, 2023 at 13:39 | answer | added | Robert Israel | timeline score: 4 | |
Jun 26, 2023 at 11:30 | comment | added | Steven Gubkin | @Amr Your question appears to be answered in the affirmative by two theorems: that the composition of real-analytic functions is again real-analytic, and by the identity theorem for real-analytic functions. Please look these theorems up and see if they answer your question. If not, post an edit which makes your question more precise and addresses exactly why these two theorems do not solve your problem. | |
Jun 26, 2023 at 7:12 | comment | added | Amr | @AlexandreEremenko I edited my question to clarify that. Thank you for your time. | |
Jun 26, 2023 at 7:09 | history | edited | Amr | CC BY-SA 4.0 |
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Jun 26, 2023 at 5:30 | comment | added | Alexandre Eremenko | You should be more specific about the exact meaning of the condition that $f_1(f_2(x_0)))$ and $f_4(f_3(x_0)))$ exist. | |
Jun 26, 2023 at 2:06 | comment | added | Amr | @ThomasKojar I am not convinced, perhaps I am missing something too obvious. I don't think this will work since I require the function $f_k$ to be defined only on the the interval whose radius is the radius of convergence of the power series | |
Jun 26, 2023 at 1:41 | comment | added | Amr | @ThomasKojar But the bump function does not come from a power series or the composition of two power serieses , or does it ? | |
Jun 26, 2023 at 1:18 | comment | added | Amr | @ThomasKojar I can do that locally around 0. Do I understand your question correctly? | |
Jun 26, 2023 at 1:09 | history | asked | Amr | CC BY-SA 4.0 |