another proof of Jensen’s inequality
First of all, it’s clear that defining
we have
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
which is true by definition of a convex function.
2) Taking as true that , where , we have to prove that
where .
First of all, let’s observe that
and that if all , belongs to as well. In fact, being non-negative,
and, summing over ,
that is
We have, by definition of a convex function:
But, by inductive hypothesis, since , we have:
so that
which is the thesis.