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Timeline for Exponential Convexity

Current License: CC BY-SA 3.0

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Feb 5, 2014 at 19:24 vote accept Shinning Star
Feb 5, 2014 at 17:22 review Close votes
Feb 6, 2014 at 10:46
Feb 5, 2014 at 13:48 comment added Mark Meckes @RSG: Yes, you're missing the square in $\zeta^2 h(x)$.
Feb 5, 2014 at 13:05 answer added UwF timeline score: 2
S Feb 5, 2014 at 12:53 history suggested UwF CC BY-SA 3.0
the x_i are also chosen from R
Feb 5, 2014 at 12:50 review Suggested edits
S Feb 5, 2014 at 12:53
Feb 5, 2014 at 12:00 comment added RSG I am slightly confused. if $n=1$ (i.e. there is only $x$), we can always choose a $\zeta\in\mathbb{R}$ to make the product $\zeta h(x)$ not positive, whatever the function $h$ may be. Is it that, I am missing something? You have mentioned for all $n$. (Even for $n=2$, we can choose accordingly).
Feb 5, 2014 at 9:30 history edited Shinning Star
edited tags
Feb 5, 2014 at 9:14 comment added Shinning Star Yes, it was mistyping. Its fine now.
Feb 5, 2014 at 9:13 history edited Shinning Star CC BY-SA 3.0
deleted 8 characters in body
Feb 5, 2014 at 9:07 comment added Mark Meckes In any case, something has to be wrong or missing here. In (ii) of the Proposition, $(x_i + x_j)/2$ isn't necessarily in the domain of $h$.
Feb 5, 2014 at 9:05 comment added Mark Meckes Okay, your phrasing is confusing because you refer to "all choices of $\xi_i$… such that" something is true about the $x_i$.
Feb 5, 2014 at 9:02 comment added Shinning Star its mentioned in definition. there is not any relationship between these two. $x_i\in(a,b)\subset\mathbb{R}$ and $\xi_i\in\mathbb{R}$. for all $1\leq i\leq n$.
Feb 5, 2014 at 8:48 comment added Mark Meckes What is the relationship between $\xi_i$ and $x_i$?
Feb 5, 2014 at 8:37 history asked Shinning Star CC BY-SA 3.0