Timeline for Continuous functions with convex level sets
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 18, 2015 at 18:20 | comment | added | Włodzimierz Holsztyński | @JohannesHahn -- thank you, it's a relief for me. I could feel that my (:-) term was a bit off. | |
| Jan 18, 2015 at 17:39 | comment | added | Noah Schweber | @DenisSerre but you also have the assumption that $\phi$ is increasing - isn't it the case that any increasing continuous function is differentiable a.e.? | |
| Jan 18, 2015 at 14:30 | comment | added | Johannes Hahn | @WłodzimierzHolsztyński It is usually called "bounded variation" as far as I know. | |
| Jan 18, 2015 at 11:25 | comment | added | Włodzimierz Holsztyński | Also function of finite variation $\ f:(a;b)\rightarrow\mathbb R\ $ are differentiable a.e.--they are the differences of two non-decreasing functions (is the finite variation a correct term? :-) | |
| Jan 18, 2015 at 11:07 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
deleted 1 character in body
|
| Jan 18, 2015 at 10:56 | vote | accept | Ali Taghavi | ||
| Jan 18, 2015 at 7:48 | comment | added | Ali Taghavi | @DenisSerre I think this is a classic theorem that an increasing function is almost every where differentiable and we have $\int_{a}^{b}f'\leq f(b)-f(a)$. I learned it from Royden book | |
| Jan 18, 2015 at 4:53 | comment | added | Ali Taghavi | @AlexDegtyarev the usual convexity as in the plane geometry. | |
| Jan 18, 2015 at 4:34 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
added 250 characters in body
|
| Jan 17, 2015 at 21:19 | comment | added | Noah Schweber | @DenisSerre, only for continuous $\phi$; and such $\phi$ are differentiable a.e. Or am I missing something? | |
| Jan 17, 2015 at 21:06 | answer | added | Włodzimierz Holsztyński | timeline score: 10 | |
| Jan 17, 2015 at 20:35 | comment | added | Noah Schweber | I think a positive answer to (2) would follow from the statement: "If $A\subseteq\mathbb{R}^2$ is such that every set of the form $\{x: (x, y)\in A\}$ or $\{y: (x, y)\in A\}$ has measure 0, then $A$ has measure 0." However, I'm also fairly certain that statement is false. | |
| Jan 17, 2015 at 19:38 | answer | added | Noah Schweber | timeline score: 2 | |
| Jan 17, 2015 at 19:18 | comment | added | Alex Degtyarev | What do you mean by convex? | |
| Jan 17, 2015 at 19:08 | history | asked | Ali Taghavi | CC BY-SA 3.0 |