real analysis – Does an asymptotic vanishing of Jensen gap imply that the variance tend to zero?

Let $Omega subseteq mathbb{R}^2$ be a nice open bounded, connected domain, having Lebesgue measure $m(Omega)=1$.

Let $F:(0,infty) to (0,infty)$ be a $C^2$ strictly convex function. Suppose that $F”$ is an everywhere positive strictly decreasing function, and that $lim_{x to infty} F”(x)=0$.

Let $Y_n:Omega to mathbb (0,infty)$ be continuous, with constant expectations $int_{Omega} Y_n=c>0$, and suppose that

$$lim_{n to infty} int_{Omega} F(Y_n)-F(int_{Omega} Y_n)=0.$$

Is $lim_{n to infty} int_{Omega} (Y_n-c)^2=0$?


If we replace $Omega$ with an arbitrary probability space $X$, and only require $Y_n:X to (0,infty)$ to be measurable, then the answer can be negative, as the following example shows:

Set $F(x) = e^{-x}$. For $n in {1, 2, 3, …}$ define
$$ Y_n := left{begin{array}{ll}
1 – frac{1}{sqrt{n}} & mbox{ with prob $1-1/n$}\
1+ frac{n-1}{sqrt{n}} & mbox{ with prob $1/n$}
end{array}right.$$

Then

  • $E(Y_n)=1$ for all $n in {1, 2, 3, …}$.

  • $lim_{nrightarrowinfty} E(F(Y_n)) = F(1)$.

  • $lim_{nrightarrowinfty} E((Y_n-1)^2)=1$.

(This example is taken from here.)

The question is whether by forcing $Y_n$ to be continuous, the answer changes.