Timeline for Uniform approximation of increasing function in $C^{\infty}$
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 30, 2014 at 14:22 | vote | accept | Leffe | ||
| Oct 30, 2014 at 14:20 | answer | added | Willie Wong | timeline score: 1 | |
| Oct 30, 2014 at 14:17 | comment | added | Willie Wong | Let $f$ be your two-piece linear function. Let $\varphi\in C^\infty_0((-\epsilon,\epsilon))$ for some small $\epsilon$, such that $\varphi$ is even, with integral $\int \varphi = 1$, and $x\varphi' \leq 0$. Then you can check that the convolution $\varphi*f$ is increasing, smooth, and agrees with $f$ outside $(-2\epsilon,2\epsilon)$. | |
| Oct 30, 2014 at 14:14 | comment | added | Leffe | @Petya Thanks for your comment. For the linear function with two pieces, how can this smoothing be done? It appears that one could use the same linear functions. | |
| Oct 30, 2014 at 14:08 | comment | added | Leffe | @WillieWong Uniformly approximating the linear function with two pieces with a smooth function is enough for me. | |
| Oct 30, 2014 at 13:53 | comment | added | Willie Wong | Is your function actually piecewise linear with only two pieces? For that convolution against a compactly supported even mollifier will get you immediately your uniform approximation. So I assume there are some additional technicalities? | |
| Oct 30, 2014 at 13:46 | comment | added | Petya | One can approximate the given continuous function by an increasing piece-wise affine function and then smooth it. | |
| Oct 30, 2014 at 12:55 | history | edited | Leffe | CC BY-SA 3.0 |
added 81 characters in body
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| Oct 30, 2014 at 12:51 | review | First posts | |||
| Oct 30, 2014 at 12:59 | |||||
| Oct 30, 2014 at 12:50 | history | asked | Leffe | CC BY-SA 3.0 |