Cfrgtkky

Let $X_nsim N(0,1/n)$. Is there a continuous function $f$ such that

1. $E[f^2(X_n)]<infty$


2. 
   $lim_{nrightarrow infty} E[f^2(X_n)] neq f^2(0)$?

Also, what would happen if I add the condition

3. 
   $E[|f(X)f(Y)|]<infty$ for all jointly normal $X,Y$ such that $EX=EY=0$
   

I know that there is no such $f$ if we additionally require $f$ to be bounded since $X_n stackrel{d}{rightarrow}delta_0$.

However, I am totally clueless when it comes to proving (or disproving) the existence of such $f$ if we drop out boundedness condition.

I appreciate every hint!

No, there doesn't exist such a function $f$.

Theorem: Let $X_n sim N(0,1/n)$ and let $g geq 0$ be a continuous function. If $mathbb{E}g(X_1)<infty$ then $mathbb{E}g(X_n) to g(0)$ as $n to infty$.

For the proof of the statement we use the following auxiliary result.

Lemma: Let $X_n sim N(0,1/n)$ and let $g geq 0$ be a continuous function. If $mathbb{E}g(X_1)<infty$ then $$lim_{R to infty} sup_{n in mathbb{N}} mathbb{E}(g(X_n) 1_{{|X_n| geq R}}) = 0.$$

Proof of the lemma: If we set $R_n := sqrt{log(n)/(n-1)}$ for $n geq 2$, then a straight-forward computation shows that $$exp left(- frac{y^2}{2} (n-1) right) leq frac{1}{sqrt{n}} quad text{for all $|y| geq R_n$}.$$ (Equality holds for $|y|=R_n$ and the monotonicity of the function of left-hand side then gives the desired inequality for $|y| geq R_n$.) Hence, $$exp left(-n frac{y^2}{2} right) leq frac{1}{sqrt{n}} exp left(-frac{y^2}{2} right) quad text{for all $|y| geq R_n$.}$$ Since $R_n to 0$ as $n to infty$, we have $R_0 := sup_{n geq 2} R_n$ and $$exp left(-n frac{y^2}{2} right) leq frac{1}{sqrt{n}} exp left(-frac{y^2}{2} right) quad text{for all $|y| geq R_0$, $n in mathbb{N}$.} tag{1}$$ (Note that $(1)$ is trivially satisfied for $n=1$.) Hence, begin{align*} mathbb{E}(g(X_n) 1_{{|X_n| geq R}}) &= sqrt{frac{n}{2pi}} int_{|y| geq R} g(y) exp left(-n frac{y^2}{2} right) ,d y \ &leq frac{1}{sqrt{2pi}} int_{|y| geq R} g(y) exp left(- frac{y^2}{2} right) , dy \ &= mathbb{E}(g(X_1) 1_{{|X_1| geq R}}) tag{2} end{align*} for all $R geq R_0$. By assumption, $mathbb{E}g(X_1)<infty$, and therefore the monotone convergence theorem yields that $$sup_{n in mathbb{N}} mathbb{E}(g(X_n) 1_{{|X_n| geq R}}) stackrel{(2)}{leq} mathbb{E}(g(X_1) 1_{{|X_1| geq R}}) xrightarrow{R to infty} 0.$$

Proof of the theorem: Fix $epsilon>0$. According to the above lemma, we can choose $R>0$ such that

$$sup_{n in mathbb{N}} mathbb{E}(g(X_n) 1_{{|X_n| geq R}}) leq epsilon.$$

Without loss of generality, we may assume that $R$ is sufficiently large such that

$$sup_{n in mathbb{N}} mathbb{P}(|X_n| geq R) leq epsilon.$$
By the triangle inequality, this implies that

$$begin{align*} mathbb{E}(|g(X_n)-g(0)|) &leq mathbb{E}(|g(X_n)-g(0)| 1_{{|X_n| leq R}}) + mathbb{E}(|g(X_n)-g(0)| 1_{{|X_n| > R}}) \ &leq mathbb{E}(|g(X_n)-g(0)| 1_{{|X_n| leq R}}) + 2 epsilon. end{align*}$$

The random variable $X_n sim N(0,1/n)$ equals in distribution $U/sqrt{n}$ for any $U sim N(0,1)$. Hence,

$$mathbb{E}(|g(X_n)-g(0)|) leq mathbb{E} left[ left| g left( frac{U}{sqrt{n}} right) - g(0) right| 1_{|U|/sqrt{n} leq R } right]+ 2 epsilon tag{3}$$

Noting that

$$ left| g left( frac{U}{sqrt{n}} right) - g(0) right| 1_{|U|/sqrt{n} leq R } leq 2 sup_{|y| leq R} |g(y)| < infty$$

and

$$ left| g left( frac{U}{sqrt{n}} right) - g(0) right| 1_{|U|/sqrt{n} leq R } xrightarrow{n to infty} 0$$

by the continuity of $g$, we conclude from the dominated convergence theorem that

$$lim_{n to infty} mathbb{E} left[ left| g left( frac{U}{sqrt{n}} right) - g(0) right| 1_{|U|/sqrt{n} leq R } right] = 0,$$

and so $(3)$ gives

$$lim_{n to infty} mathbb{E}(|g(X_n)-g(0)|) =0.$$

This implies, in particular, that

$$|mathbb{E}g(X_n)-g(0)| leq mathbb{E}(|g(X_n)-g(0)|) to 0,$$

i.e. $mathbb{E}g(X_n) to g(0)$.