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$f$ is continuous on $[0;1].$ Prove $$\lim_{n \to \infty} \underbrace{\int_0^1 \cdots \int_0^1}_{n} f(\sqrt[n]{x_1\cdots x_n})\mathrm \, dx_1\cdots \mathrm dx_n = f(\frac{1}{e}).$$

At first, I thought that we should get inside the function with the limit, but it's probably restricted due to $dx_1\dots dx_n$. I feel like I'm missing an important Theorem here. And yet it seems like the problem should be easy. Can somebody smart help me out here (at least with a hint) ?

Perhaps, it's an induction method problem.

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    $\begingroup$ You can proceed as here. If $f$ were a polynomial try to show that the limit is $f(1/e)$ and then pass to any continuous function $f$ using Weierstrass approximation. $\endgroup$ – r9m May 18 '20 at 19:12
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    $\begingroup$ omg, that's a super helpful hint. Thank you! $\endgroup$ – Est Mayhem May 18 '20 at 19:14
  • $\begingroup$ Cool! Let me know when you get it. $\endgroup$ – r9m May 18 '20 at 20:44
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    $\begingroup$ The title is not coherent. Please rewrite. $\endgroup$ – zhw. May 18 '20 at 23:15
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A short solution could be. Consider the random variables $u_1,...,u_n$ which they are i.i.d with density function $\rho(u)=e^{-u}$ (exponential distribution with $\lambda =1 $). So the expected value $E[u] = \int^{\infty}_{0}u\cdot e^{-u}du = 1$. By law of large numbers $\frac{S_n}{n}\to E[u]=1$. Where $S_n := u_1+..+u_n$. Observe that $$\int_{0}^1\dots\int_{0}^1f(\sqrt[n]{x_1...x_n})dx_1\dots dx_n = \int_{0}^{\infty}\dots\int_{0}^{\infty}f(e^{-\frac{S_n}{n}}) e^{-u_1}du_1...e^{-u_n}du_n =E[f(e^{-\frac{S_n}{n}})]$$ Finally by continuity and dominated convergence $$E[f(e^{-\frac{S_n}{n}})] \xrightarrow{n} E[f(e^{-1})]=f(e^{-1}).$$

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    $\begingroup$ very clever. Thank you! $\endgroup$ – Est Mayhem May 19 '20 at 13:03
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Here's an elementary proof. Suppose $f(x) = x^k$ then we see that \begin{align} \int^1_0\cdots \int^1_0 dx_1\cdots dx_n\ f(\sqrt[n]{x_1\cdots x_n})=&\ \int^1_0\cdots \int^1_0 dx_1\cdots dx_n\ \left(\sqrt[n]{x_1\cdots x_n} \right)^k\\ =&\ \left(\int^1_0 dx\ x^{\frac{k}{n}} \right)^n = \left(1+\frac{k}{n} \right)^{-n}. \end{align} In particular, it follows \begin{align} \lim_{n\rightarrow \infty}\int^1_0\cdots \int^1_0 dx_1\cdots dx_n\ f(\sqrt[n]{x_1\cdots x_n}) = \lim_{n\rightarrow \infty}\left(1+\frac{k}{n} \right)^{-n} = \left(\frac{1}{e}\right)^k. \end{align} If $f$ is a polynomial, then it follows \begin{align} \lim_{n\rightarrow \infty}\int^1_0\cdots \int^1_0 dx_1\cdots dx_n\ f(\sqrt[n]{x_1\cdots x_n}) = f\left( \frac{1}{e}\right). \end{align}

Next, if $f$ is continuous, then, by Wierestrass approximation theorem, there exists a sequence of polynomials $p_m$ such that $p_m \rightarrow f$ uniformly on $[0, 1]$. Finally, it follows \begin{align} \lim_{n\rightarrow \infty}\int^1_0\cdots \int^1_0 dx_1\cdots dx_n\ f(\sqrt[n]{x_1\cdots x_n}) =&\ \lim_{n\rightarrow \infty}\int^1_0\cdots \int^1_0 dx_1\cdots dx_n\ \lim_{m\rightarrow \infty}p_m(\sqrt[n]{x_1\cdots x_n})\\ =&\ \lim_{n\rightarrow \infty}\lim_{m\rightarrow \infty}\int^1_0\cdots \int^1_0 dx_1\cdots dx_n\ p_m(\sqrt[n]{x_1\cdots x_n})\\ =&\ \lim_{m\rightarrow \infty}\lim_{n\rightarrow \infty}\int^1_0\cdots \int^1_0 dx_1\cdots dx_n\ p_m(\sqrt[n]{x_1\cdots x_n})\\ =&\ \lim_{m\rightarrow \infty} p_m\left(\frac{1}{e} \right) = f\left(\frac{1}{e} \right). \end{align}

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  • $\begingroup$ I can't understand second and third = signs in your proof. You raise to power n as if f is out of integral and the next step you are taking an integral with f within the integral... $\endgroup$ – Est Mayhem May 19 '20 at 12:52
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    $\begingroup$ @EstMayhem, The second step is a special instance of the following observation: $$\int\limits_{[a_1,b_1]\times\dots\times[a_n,b_n]} g_1(x_1)\dots g_n(x_n) \, \mathrm{d}x_1\dots\mathrm{d}x_n=\prod_{k=1}^{n}\int_{a_k}^{b_k} g_k(x_k) \, \mathrm{d}x_k. $$ Of course, this is a simple application of the Fubini's Theorem. Then the third step follows from $$\int_{0}^{1} x^{\alpha}\,\mathrm{d}x=\frac{1}{1+\alpha}, \qquad \alpha > -1 $$ $\endgroup$ – Sangchul Lee May 19 '20 at 13:38

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