Timeline for A functional inequality
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 3, 2014 at 13:50 | comment | added | Ilya Bogdanov | Yes you can, but I do not think it was quite essential... | |
| Sep 3, 2014 at 1:16 | comment | added | Juanito | Your comment to reformulate the problem in additive format, was very helpful for my work. Could I acknowledge you in my work as Ilya Bogdanov? | |
| May 30, 2014 at 5:55 | comment | added | Juanito | Is there a way to ensure that the bump fn introduced does not violate $ f((t+1)x)<f(tx)+f(x)$? | |
| May 30, 2014 at 5:45 | comment | added | Juanito | If I use $f(x)=-\log g(e^{-x})$, I would get $g(x)=e^{-f(-\log x)}$. Given the boundary conditions $g(0)=0$ and $g(1)=1$, I would require, the candidate function $f$ to satisfy $f(0)=0$ and $f(\infty)=\infty $. I was thinking of following the steps you recommended by using $f(x)=x^{1/2}$. This satisfies the boundary conditions and $ f((t+1)x)<f(tx)+f(x) $ for all $x \in(0,+\infty)$ and positive integer $t$. I perturb the function at $5,5/2$ and in $(1,1.9)$ Someone mentioned using bump functions to smooth the slope discontinuities. I hope using a bump fn dont violate $ f((t+1)x)<f(tx)+f(x)$. | |
| May 29, 2014 at 15:29 | comment | added | Juanito | you are absolutely correct, I also require $g(r^2)>g(r)^2$, thus, the correct condition should be $t>0$. $g(r^2)>g(r)^2$ is also implied by $m=n=1$ on the right hand side. | |
| May 29, 2014 at 5:22 | comment | added | Ilya Bogdanov | @Juanito: Are you sure then that there are no other misses? I am particularly interested in $t>1$ (in your present formulation, you don't say that $g(r^2)>g(r)^2$ --- is it what you wanted?). | |
| May 28, 2014 at 21:39 | comment | added | Juanito | Hello @Ilya, thanks for your reply. I think this should work out. TO get back g(x) from a function $f(x)=-\log g(e^{-x})=x+\eps$ I could replace $x$ by $-logx$, and I end up getting $g(x)=x.e^{-\eps}$. Similarly, $f(x)=1.1x+\eps$ on $(0,\,1.7)$ would give $g(x)=x^{1.1}.e^{-\eps}$. This function would not be differentiable at the kink. I think I missed including the conditions $g(0)=0$ and $g(1)=1$ in my initial post, and I apologize for that. But, given this extra information, $g(1)=1.e^{-\eps} \neq 1.$ | |
| May 28, 2014 at 6:42 | history | answered | Ilya Bogdanov | CC BY-SA 3.0 |