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

$$f$$ 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.

• 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. – r9m May 18 '20 at 19:12
• omg, that's a super helpful hint. Thank you! – Est Mayhem May 18 '20 at 19:14
• Cool! Let me know when you get it. – r9m May 18 '20 at 20:44
• The title is not coherent. Please rewrite. – zhw. May 18 '20 at 23:15

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 $$E[f(e^{-\frac{S_n}{n}})] \xrightarrow{n} E[f(e^{-1})]=f(e^{-1}).$$
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}
• @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 $$\int_{0}^{1} x^{\alpha}\,\mathrm{d}x=\frac{1}{1+\alpha}, \qquad \alpha > -1$$ – Sangchul Lee May 19 '20 at 13:38