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How would you show that if $f : [0,1] \rightarrow \Bbb R$ is continuous, then

$$\lim_{n\rightarrow \infty}\int_0^1\int_0^1 \cdots \int_0^1 f\left( \frac{x_1+x_2+\cdots+x_n}{n} \right)~dx_1~dx_2\cdots dx_n = f\left( \frac{1}{2} \right) $$

and

$$\lim_{n\rightarrow \infty}\int_0^1\int_0^1 \cdots \int_0^1 f((x_1 x_2 \cdots x_n)^{\frac{1}{n}})~dx_1~dx_2\cdots dx_n = f\left(\frac{1}{e}\right) $$

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  • $\begingroup$ What's $x_k$?$\quad$ $\endgroup$ Apr 16, 2016 at 21:42
  • $\begingroup$ Is $f$ is continuous? $\endgroup$
    – zhw.
    Apr 16, 2016 at 23:05
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    $\begingroup$ @EricTowers: The integration limits directly say that $0\le x_k\le 1$. $\endgroup$ Apr 17, 2016 at 12:46

3 Answers 3

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Note that the LHS is in the form of expectation of some function.

Let $(X_n)_n$ be a sequence of i.i.d random variables, which follow uniform distribution on $[0,1]$.

For the first case: by strong law of large numbers, $\displaystyle S_n = \frac{1}{n}\sum_{i=1}^n X_i \overset{a.s.}{\longrightarrow} \mathbb E(X_1) = \frac{1}{2}$.

As $f$ is continuous, $f(S_n) \to f(\frac{1}{2})$. As $S_n \in [0,1]$ compact, $f(S_n)$ is bounded by a constant (thus integrable on $[0,1]$). So by dominated convergence theorem, $\mathbb E [f(S_n)] \overset{a.s.}{\longrightarrow} \mathbb E [f(\frac{1}{2})] = f(\frac{1}{2})$.

For the second case, $\displaystyle (x_1\cdots x_n)^{1/n} = \exp(-\frac{1}{n} \sum_{i=1}^n \ln\frac{1}{x_i})$. Let $Y_i = \ln\frac{1}{X_i}$, then by strong law of large numbers, $\displaystyle V_n = \frac{1}{n}\sum_{i=1}^n Y_i \overset{a.s.}{\longrightarrow} \mathbb E(Y_1) = 1$. Again, as $f(e^{-V_n})$ is continuous and bounded by a certain constant (thus integrable on $[0,1]$), $\mathbb E [f(e^{-V_n})] \overset{a.s.}{\longrightarrow} \mathbb E [f(e^{-1})] = f(\frac{1}{e})$.

BONUS: using the same approach, if moreover, we restrict $f: [0,1] \to [0,1]$, then $$\lim_{n \to \infty} \int_0^1 ... \int_0^1 \frac{f(x_1)+...f(x_n)}{n}dx_1...dx_n = \int_0^1 f(x) dx$$

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  • $\begingroup$ The 1st Q is equivalent to showing that for each $d>0$, the $n$-dimensional volume of $S_n(d)$ tends to $0$ as $n\to \infty,$ where $S_n(d)=$ $\{(x_1,...,x_n)\in [0,1]^n :$ $ |1/2-(1/n)\sum_{i=1}^nx_i|>d\}.$ Good answer. $\endgroup$ Apr 17, 2016 at 1:22
  • $\begingroup$ I do not understand your computation in the second case; are you sure you have the correct definition for $Y_i$? $\endgroup$
    – guest
    Apr 17, 2016 at 19:58
  • $\begingroup$ Ah yes, you are correct. It should be $\ln\frac{1}{X_i}$ instead. Corrected! $\endgroup$
    – SiXUlm
    Apr 17, 2016 at 20:08
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For the second problem, recall that Weierstrass tells us that polynomials are dense in $C[0,1].$ So it's enough to prove the result for polynomials, and for this it's enough to prove it for each $x^k.$ For $x^k$ the $n$-fold integral works out nicely to be $1/(1+k/n)^n.$ This has limit $1/e^k = (1/e)^k,$ which is the desired answer for this function.

Weiersrtass can also be used for the first problem as well, although it's not quite as simple.

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The first result can be shown introducing a Dirac delta (and its integral representation) into the game. We have

$$ \lim_{n\rightarrow \infty}\int_0^1\int_0^1 \cdots \int_0^1 f \left( \frac{x_1+x_2+\cdots +x_n}{n} \right)~dx_1~dx_2\cdots dx_n = $$ $$ \lim_{n\rightarrow \infty}\int_{\mathbb{R}} dt\ f(t)\int_0^1\int_0^1\cdots \int_0^1 \delta\left(t-\frac{1}{n}\sum_{k=1}^n x_k\right)~dx_1~dx_2\cdots dx_n= $$ $$ \lim_{n\rightarrow \infty}\int_{\mathbb{R}}dt\ f(t)\int_{\mathbb{R}}\frac{dz}{2\pi} e^{i tz} \int_0^1\int_0^1 \cdots \int_0^1\ e^{-iz\sum_{k=1}^n x_k/n} ~dx_1~dx_2\cdots dx_n= $$ $$ \lim_{n\rightarrow \infty}\int_{\mathbb{R}}dt\ f(t)\int_{\mathbb{R}}\frac{dz}{2\pi}e^{i tz}\left[\int_0^1 dx\ e^{-izx/n}\right]^n=\lim_{n\rightarrow \infty} \int_{\mathbb{R}}dt\ f(t)\int_{\mathbb{R}}\frac{dz}{2\pi}e^{i tz}\left[\frac{i n \left(-1+e^{-\frac{i z}{n}}\right)}{z}\right]^n $$ and using $$ \lim_{n\to\infty}\left[\frac{i n \left(-1+e^{-\frac{i z}{n}}\right)}{z}\right]^n=e^{-i z/2} $$ the result is $$ \int_{\mathbb{R}}dt f(t)\int_{\mathbb{R}}\frac{dz}{2\pi}e^{iz(t-1/2)} =\int_{\mathbb{R}}dt\ f(t)\delta(t-1/2)=f(1/2)\ . $$

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