Problem: Let $f:[0,1]\to\mathbb R$ be a bounded, three times continuously differentiable function. Evaluate the following limit
$$\lim_{n\to\infty}\int_0^1\int_0^1\cdots\int_0^1 n\left[f\left(\frac{x_1+\cdots+x_n}{n}\right)-f\left(\frac{1}{2}\right)\right]\,dx_1\,dx_2\cdots\,dx_n.$$
It has been suggested to use a probabilistic approach by writing the integral as an expectation and to also try to use Taylor's theorem.
In addition, there is the following rather similar question: How prove this integral limit $=f(\frac{1}{2})$, which only differs by the factor of $n$.
I do not see how to employ a probabilistic interpretation, so I will go with Taylor's theorem. First, write
$$\frac{x_1+\cdots+x_n}{n}=\frac{S_n}{n},$$
where $x_1,\dots,x_n$ are i.i.d. uniform random variables on $[0,1].$ Using this notation, we expand around $x=1/2$ to get
$$n\left[f\left(\frac{S_n}{n}\right)-f\left(\frac{1}{2}\right)\right]=n\left[f'\left(\frac{1}{2}\right)\left(\frac{S_n}{n}-\frac{1}{2}\right)+\frac{f''(1/2)}{2}\left(\frac{S_n}{n}-\frac{1}{2}\right)^2+\frac{f'''(c)}{6}\left(\frac{S_n}{n}-\frac{1}{2}\right)^3\right].$$
Now note that
$$\int_0^1\cdots\int_0^1 n\left[f'\left(\frac{1}{2}\right)\left(\frac{S_n}{n}-\frac{1}{2}\right)\right]\,dx_1\cdots\,dx_n=0.$$
Now since
$$\left\vert\left(\frac{S_n}{n}-\frac{1}{2}\right)\right\vert\leq1,$$
the other terms can be bounded and we can conclude that
$$\lim_{n\to\infty}\int_0^1\int_0^1\cdots\int_0^1 n\left[f\left(\frac{x_1+\cdots+x_n}{n}\right)-f\left(\frac{1}{2}\right)\right]\,dx_1\,dx_2\cdots\,dx_n=0.$$
Do you agree with my approach above?
Any feedback is most welcomed. Thank you very much for your time and help.
Best Answer
The probability approach makes things simpler. Let $X_1,X_2,\dots,X_n,\dots$ be independent random variables, which are uniform on $[0,1]$.
$X_1$ (or each other member in the family) has mean $\mu=\frac 12$ and variance $\sigma^2=\frac 1{12}$.
Let $Y_n$ be the mean $$Y_n=\frac 1n(X_1+X_2+\dots+X_n)\ .$$
Let $\mu$ be $1/2$ for short, and $$ f(s) = \frac 1{0!}f(\mu) + \frac 1{1!}f'(\mu)(s-\mu) + \frac 1{2!}f''(\mu)(s-\mu)^2 + g(s)\ , $$ where $g(s)$ has an estimation of the shape $|g(s)|\le M\;|s-\mu|^3$ for some multiplicative constant $M$. We denote by $\Bbb E$ the expectation on the probability space where all variables $(X_n)$ live in.
Then let us denote by $I_n$ the integral: $$ \begin{aligned} I_n&= \int_0^1\int_0^1\cdots\int_0^1 n \left[\ f\left(\frac{x_1+\cdots+x_n}{n}\right)-f\left(\frac 12\right) \ \right] \,dx_1\,dx_2\cdots\,dx_n \\ &=\Bbb E[\ n(f(Y_n)-f(\mu))\ ] \\ &=\Bbb E\left[\ n\left( \frac 1{1!}f'(\mu)(Y_n-\mu) + \frac 1{2!}f''(\mu)(Y_n-\mu)^2 + g(Y_n) \right)\ \right] \\ &= \frac 1{1!}f'(\mu)\cdot n\;\underbrace{\Bbb E[\ (Y_n-\mu)\ ]}_{=0} + \frac 1{2!}f''(\mu)\cdot n\;\underbrace{\Bbb E[\ (Y_n-\mu)^2\ ]}_{\operatorname{Var}[Y_n]=\sigma^2/n} + \underbrace{n\;\Bbb E[\ g(Y_n)\ ]}_{\to 0} \\ &\to\frac 1{2!}f''(\mu)\cdot\sigma^2=\frac 12f''(\mu)\cdot\frac 1{12}=\frac 1{24}f''(\mu)\ . \end{aligned} $$ Here, we have used the estimation for $g$, and the limit $$ n\Bbb E[\ |Y_n-\mu|^3\ ]= n\|\ (Y_n-\mu)^3\ \|_{L^1} \le n \cdot \|\ (Y_n-\mu)^1\ \|_{L^2} \cdot \|\ (Y_n-\mu)^2\ \|_{L^2} \to 0\ . $$ So the result is $$ \lim_{n\to\infty}I_n=\frac 1{12}f''\left(\frac 12\right)\ . $$