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 $n \in \{1, 2, 3, ...\}$.


*$\lim_{n\rightarrow\infty} E[F(Y_n)] = F(1)$.


*$\lim_{n\rightarrow\infty} E[(Y_n-1)^2]=1$.
(This example is taken from here.)
The question is whether by forcing $Y_n$ to be continuous, the answer changes.
 A: This is the kind of problem where continuity alone is unlikely to help. If you have no quantitative control on the continuity (a modulus of continuity, for instance), the functions $Y_n$ can get pretty ugly. In particular, they can go very quickly from one value to another, so that its image distribution can be quite close to whatever you want. I'll give a counter-example on $[0,1]$; Just take a product by $[0,1]$ to get a counter-example in $\mathbb{R}^2$.
Take:
$$F(x) = e^{-x},$$
$$Y_n (x) = \left\{ \begin{align} 1-\frac{1}{\sqrt{n}} & \text{ if } 0 \leq  x \leq 1-\frac{2}{n}, \\ 
1-\frac{1}{\sqrt{n}}+\frac{2}{3}n\sqrt{n} \left(x-1+\frac{2}{n}\right) & \text{ if } 1-\frac{2}{n} \leq  x \leq 1-\frac{1}{n}, \\
1-\frac{1}{\sqrt{n}}+\frac{2}{3}\sqrt{n} & \text{ if } 1-\frac{1}{n} \leq x. \\\end{align}\right.$$
If I am not mistaken, $\mathbb{E} (Y_n) = 1$ for all $n$. By the dominated convergence theorem, $\lim_{n \to + \infty} \mathbb{E} (F(Y_n)) = F(1) = e^{-1}$. Finally,
$$\mathbb{E} ((Y_n-1)^2) \geq \int_{1-\frac{1}{n}}^1 (Y_n(x)-1)^2 \ dx = \frac{1}{n} \times \left(\frac{2}{3} \sqrt{n}-\frac{1}{\sqrt{n}}\right)^2 \sim \frac{4}{9}.$$
One may notice that the distribution of $Y_n$ is quite close (up to a factor $2/3$, here for convenience) to the counter example you already have; more generally, you can approximate (in the weak sense) as well as you want any distribution on $\mathbb{R}$ by the distribution of a continuous function on $[0,1]$, by generalizing this trick.
