another proof of Jensen’s inequality
First of all, it’s clear that defining
so it will we enough to prove only the simplified version.
Let’s proceed by induction.
1) ; we have to show that, for any and in ,
But, since must be equal to 1, we can put , so that the thesis becomes
2) Taking as true that , where , we have to prove that
First of all, let’s observe that
and that if all , belongs to as well. In fact, being non-negative,
and, summing over ,
We have, by definition of a convex function:
But, by inductive hypothesis, since , we have:
which is the thesis.
|Title||another proof of Jensen’s inequality|
|Date of creation||2013-03-22 15:52:53|
|Last modified on||2013-03-22 15:52:53|
|Owner||Andrea Ambrosio (7332)|
|Last modified by||Andrea Ambrosio (7332)|
|Author||Andrea Ambrosio (7332)|