Timeline for A new type of convexity related to the exponential function
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 2, 2024 at 21:43 | vote | accept | Hendrik Matamoros | ||
| Mar 2, 2024 at 19:51 | comment | added | Lorenzo Pompili | You could also aim at a direct proof in principle. But honestly the definition of exp type convexity sounds a bit artificial and cryptic, so if you want help on that you could maybe add some more information on where that definition comes from and some motivation on why it is useful | |
| Mar 2, 2024 at 19:47 | comment | added | Lorenzo Pompili | @HendrikMatamoros So maybe you should look for a proof of the usual convexity of $f(x)=|x|^p$. Once you find such a proof, you can simply add the extra step I wrote above and you get a proof for the exponential-type convexity. | |
| Mar 2, 2024 at 19:41 | comment | added | Hendrik Matamoros | @LorenzoPompili What you say is true, in fact it is a theorem that every non-negative and convex function $f$ is also a convex function of exponential type, my question is not exactly to prove this but rather to find a particular proof for the function $f (x)=x^p$ with $p>1$ but algebraically, that is, testing it for a fixed $t$ in $[0,1]$ and with $x,y$ arbitrary real numbers. If possible, of course. | |
| Mar 2, 2024 at 19:26 | history | edited | Lorenzo Pompili | CC BY-SA 4.0 | added 13 characters in body |
| Mar 2, 2024 at 19:25 | comment | added | Lorenzo Pompili | Right, thank you! | |
| Mar 2, 2024 at 19:24 | comment | added | Kurt G. | Nice. But I think this holds only for $f\ge 0\,.$ | |
| Mar 2, 2024 at 19:20 | history | answered | Lorenzo Pompili | CC BY-SA 4.0 |