Timeline for Example of function with a certain behavior.
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 15, 2011 at 14:00 | comment | added | de Araujo | @Emil: Your example is correct with regard to the condition $f(x)x\leq c(1 + x^2) $, that's right for the problem which i intend to apply the example you gave. What matters is growth on top. If you intend to use your example I referred to you as the author of it, if you do not mind. | |
| Aug 15, 2011 at 10:06 | comment | added | Emil Jeřábek | @Anderson: Yes, that’s what the example does. However, it exploits the fact that (4) does not require anything about the behaviour of $f(x)$ for negative $x$ as long as $f$ is nonnegative, which made me wonder whether that’s the intended meaning of the condition: the context suggests that you are considering properties which somehow restrict the growth rate of the function; the current (4) does not do that, one would expect the condition to read $|f(x)x|\le c(1+x^2)$ (and then the example no longer works). | |
| Aug 14, 2011 at 14:49 | comment | added | de Araujo | Aubrey, again Emil is correct. | |
| Aug 14, 2011 at 14:45 | comment | added | de Araujo | Dear Emil, If I understand correctly, you gave an example that: $(1) + (3) + (4)$ does not imply $(2)$ right? | |
| Aug 12, 2011 at 18:49 | comment | added | Emil Jeřábek | @Aubrey: (2) + (3) implies (4) by taking $y=0$ in (2). | |
| Aug 12, 2011 at 18:43 | comment | added | Aubrey da Cunha | I don't find it easy to see that $(1)+(2)+(3)$ implies $(4)$ considering $f(x)=-x^3$ appears to be a counterexample. I think Emil is right and you want absolute values in $(2)$ and $(4)$. | |
| Aug 12, 2011 at 17:11 | comment | added | Emil Jeřábek |
As given, $f(x)=\begin{cases}x^{42}(1+\sin x),&x<0,\\0,&x\ge0\end{cases}$ is a counterexample. Are you really sure you don’t want absolute values on the left-hand side of (4)?
|
|
| Aug 12, 2011 at 16:56 | comment | added | de Araujo | Thanks, i had forgotten, in (1) is |x-y| and not (x-y). | |
| Aug 12, 2011 at 16:52 | history | edited | de Araujo | CC BY-SA 3.0 |
edited body
|
| Aug 12, 2011 at 16:43 | comment | added | Emil Jeřábek | (1) can only hold for $a=0$ and $f$ constant. Did you mean to have $|x-y|$ on the right-hand side? Are there any other absolute values missing in (2) or (4)? And what does “$f(0)$ is limited” mean? | |
| Aug 12, 2011 at 16:40 | comment | added | Kenneth Hung | For $(1)$, When $x < y$, the right hand side is certainly negative, but the left hand side is non negative. Is there a typo? | |
| Aug 12, 2011 at 16:28 | history | asked | de Araujo | CC BY-SA 3.0 |