# How prove this integral limit $=f(\frac{1}{2})$

Let $f$ be a continuous function on the unit interval $[0,1]$. Show that
$$\lim_{n\to\infty}\int_{0}^{1}\cdots\int_0^1\int_{0}^{1}f\left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right)dx_{1}dx_{2}\cdots dx_{n}=f\left(\dfrac{1}{2}\right)$$

This problem is from Selected Problems in Real Analysis. But the author doesn’t include a solution. Maybe there is a method for this sort of problem?

Maybe we can use this:
$$\int_0^1\cdots\int_{0}^{1}\int_{0}^{1}\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}dx_{1}dx_{2}\cdots dx_{n}=\dfrac{1}{2}?$$

I also found a similar problem in this question.

#### Solutions Collecting From Web of "How prove this integral limit $=f(\frac{1}{2})$"

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$
$\ds{\lim_{n \to \infty}\int_{0}^{1}\cdots\int_{0}^{1} \fermi\pars{x_{1} + x_{2} + \cdots + x_{n} \over n} \,\dd x_{1}\,\dd x_{2}\ldots\dd x_{n} = \fermi\pars{\half}}$

\begin{align}
&\color{#c00000}{%
\lim_{n \to \infty}\int_{0}^{1}\cdots\int_{0}^{1}
\fermi\pars{x_{1} + x_{2} + \cdots + x_{n} \over n}
\,\dd x_{1}\,\dd x_{2}\ldots\dd x_{n}}
\\[3mm]&=\lim_{n \to \infty}\int_{0}^{1}\cdots\int_{0}^{1}\int_{-\infty}^{\infty}\tilde{\fermi}\pars{k}
\exp\pars{\ic k\,{x_{1} + x_{2} + \cdots + x_{n}\over n}}\,{\dd k \over 2\pi}
\,\dd x_{1}\,\dd x_{2}\ldots\dd x_{n}
\end{align}
where $\ds{% \tilde{\fermi}\pars{k} \equiv \int_{-\infty}^{\infty}\fermi\pars{x} \expo{-\ic k x}\,\dd x}$ is the $\ds{\fermi\pars{x}}$ Fourier Transform.

\begin{align}
&\color{#c00000}{%
\lim_{n \to \infty}\int_{0}^{1}\cdots\int_{0}^{1}
\fermi\pars{x_{1} + x_{2} + \cdots + x_{n} \over n}
\,\dd x_{1}\,\dd x_{2}\ldots\dd x_{n}}
\\[3mm]&=\lim_{n \to \infty}\int_{-\infty}^{\infty}\tilde{\fermi}\pars{k}
\pars{\int_{0}^{1}\expo{\ic kx/n}\,\dd x}^{n}\,{\dd k \over 2\pi}
=\lim_{n \to \infty}\int_{-\infty}^{\infty}\tilde{\fermi}\pars{k}
\pars{\expo{\ic k/n} – 1 \over \ic k/n}^{n}\,{\dd k \over 2\pi}
\\[3mm]&=\lim_{n \to \infty}\int_{-\infty}^{\infty}\tilde{\fermi}\pars{k}
\braces{\exp\pars{\ic k \over 2n}\,
\bracks{\exp\pars{\ic k \over 2n} – \exp\pars{-\,{\ic k \over 2n}}}\,
{1 \over \ic k/n}}^{n}\,{\dd k \over 2\pi}
\\[3mm]&=\lim_{n \to \infty}\int_{-\infty}^{\infty}\tilde{\fermi}\pars{k}
\exp\pars{\ic k \over 2}
\braces{\sin\pars{k/\bracks{2n}} \over k/\bracks{2n}}^{n}\,{\dd k \over 2\pi}
\end{align}

$$\color{#c00000}{% \lim_{n \to \infty}\int_{0}^{1}\cdots\int_{0}^{1} \fermi\pars{x_{1} + x_{2} + \cdots + x_{n} \over n} \,\dd x_{1}\,\dd x_{2}\ldots\dd x_{n}} =\int_{-\infty}^{\infty} \fermi\pars{x}\lim_{n \to \infty}{\rm K}_{n}\pars{x – \half}\,\dd x$$
where
$${\rm K}_{n}\pars{x} \equiv \int_{-\infty}^{\infty} \exp\pars{-\ic k x} \braces{\sin\pars{k/\bracks{2n}} \over k/\bracks{2n}}^{n}\,{\dd k \over 2\pi}$$

Since $\ds{\lim_{n \to \infty}{\rm K}_{n}\pars{x} = \delta\pars{x}}$
( $\ds{\delta\pars{x}}$ is the Dirac Delta Function ),
we’ll have:
$$\color{#c00000}{% \lim_{n \to \infty}\int_{0}^{1}\cdots\int_{0}^{1} \fermi\pars{x_{1} + x_{2} + \cdots + x_{n} \over n} \,\dd x_{1}\,\dd x_{2}\ldots\dd x_{n}} =\int_{-\infty}^{\infty} \fermi\pars{x}\delta\pars{x – \half}\,\dd x$$

$$\color{#00f}{\large% \lim_{n \to \infty}\int_{0}^{1}\cdots\int_{0}^{1} \fermi\pars{x_{1} + x_{2} + \cdots + x_{n} \over n} \,\dd x_{1}\,\dd x_{2}\ldots\dd x_{n} =\fermi\pars{\half}}$$

We can take advantage of a probabilistic interpretation. Let $X_1,\ldots,X_n$ be uniformly distributed random variables on $[0,1]$ and $\bar X=\frac{1}{n}(X_1+\cdots+X_n)$. Then
$$\lim\limits_{n\to \infty}\int_0^1\cdots\int_0^1f\left(\frac{x_1+\cdots+x_n}{n}\right)dx_1\cdots dx_n =\lim\limits_{n\to\infty}E[f(\bar X)]=E\left[f\left(\frac12\right)\right]=f\left(\frac12\right)$$
since $\bar X$ converges in distribution to $\frac12$ as $n\to \infty$.

Starting with OP’s hint,

$\displaystyle \lim\limits_{n\to\infty}\int_{0}^{1}\cdots\int_{0}^{1} \left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right)dx_{1}dx_{2}\cdots dx_{n}=\dfrac{1}{2}$,

Now, we try to show, $\displaystyle \lim\limits_{n\to\infty}\int_{0}^{1}\cdots\int_{0}^{1} \left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right)^k dx_{1}dx_{2}\cdots dx_{n}=\dfrac{1}{2^k}$

If we count the number of terms in the multinomial expansion of $\bigg(\sum\limits_{i=1}^n x_i\bigg)^k$, that contains all variables with power not exceeding $1$, is $n(n-1)(n-2)\cdots(n-k+1) = n^k + O(n^{k-1})$,

and the number of terms that has atleast one $x_i$ term with powers exceeding $1$ is not more than $n.n^{k-2} = n^{k-1}$.

So, combining all the terms we get,
$\displaystyle \int_{0}^{1}\cdots\int_{0}^{1} \left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right)^k dx_{1}dx_{2}\cdots dx_{n}=\dfrac{1}{2^k} + O\bigg(\frac{1}{n}\bigg)$

Thus the above limit is true for polynomials.

Using Weierstrass Approximation theorem, we can choose a polynomial $P$, such that $|f(x) – P(x)| < \epsilon/3$ for all $x \in [0,1]$.

Thus, we can find an $N \in \mathbb{N}$, such that $\forall n >N$,

$\displaystyle \left| \int_{0}^{1}\cdots\int_{0}^{1} P\left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right) dx_{1}dx_{2}\cdots dx_{n} – P\left(\frac12\right)\right| < \epsilon/3$

Therefore, for $n > N$,

$\displaystyle \left| \int_{0}^{1}\cdots\int_{0}^{1} f\left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right) dx_{1}dx_{2}\cdots dx_{n} – f\left(\frac12\right)\right| < \left| \int_{0}^{1}\cdots\int_{0}^{1} f\left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right) – P\left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right) dx_{1}dx_{2}\cdots dx_{n}\right| + \left| \int_{0}^{1}\cdots\int_{0}^{1} P\left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right) dx_{1}dx_{2}\cdots dx_{n} – P\left(\frac12\right)\right| + \left|f\left(\frac12\right) – P\left(\frac12\right)\right|$

$\le \displaystyle \int_{0}^{1}\cdots\int_{0}^{1} \left| f\left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right) – P\left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right)\right| dx_{1}dx_{2}\cdots dx_{n} + 2\epsilon/3 < \epsilon$

Thus, $\displaystyle \lim\limits_{n\to\infty} \int_{0}^{1}\cdots\cdots\int_{0}^{1}f\left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right)dx_{1}dx_{2}\cdots dx_{n}=f\left(\dfrac{1}{2}\right)$.

Define
$$\psi_n(x)=\overbrace{\left(n\chi_{[0,1/n]}\right)\ast\cdots\ast\left(n\chi_{[0,1/n]}\right)}^{\text{convolution of n copies}}$$
Then
\begin{align} &\int_0^1\cdots\int_0^1f\left(\frac1n\sum_{k=1}^nx_k\right)\,\mathrm{d}x_1\dots\,\mathrm{d}x_n\\ &=n^n\int_0^{1/n}\cdots\int_0^{1/n}f\left(\sum_{k=1}^nx_k\right)\,\mathrm{d}x_1\dots\,\mathrm{d}x_n\\ &=\int_\mathbb{R}f(x)\psi_n(x)\,\mathrm{d}x \end{align}
The Central Limit Theorem says that since $n\chi_{[0,1/n]}$ has mean $\frac1{2n}$ and variance $\frac1{12n^2}$, the convolution of $n$ copies tends to a normal distribution with mean $\frac12$ and variance $\frac1{12n}$
$$\psi_n\sim\sqrt{\frac{6n}\pi}\ e^{-6n(x-1/2)^2}$$
which is an approximation of the Dirac delta function at $x=1/2$. That is,
$$\lim_{n\to\infty}\int_\mathbb{R}f(x)\psi_n(x)\,\mathrm{d}x=f(1/2)$$