# 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.