$\lim_{n\rightarrow \infty}\int_{(0,1)^n}f\left(\left(\frac1n\sum _{x=1}^n x_i^a\right)^\frac{1}{a}\right)dx_1\cdots dx_n$

Question

Given $x_1, \cdots, x_n>0$, $a\in\mathbb R$ and $f\in C[0,1]$, then can we compute $$\lim_{n\rightarrow \infty}\int_{(0,1)^n}f\left(\left(\frac1n \sum _{i=1}^n x_i^a\right)^\frac{1}{a}\right)dx_1\cdots dx_n$$ In terms of $f, a$? Any help will be appreciated.

Answer 1

Probability theory makes the answer very simple. The limit exists and equals $f((\frac 1 {a+1})^{1/a}) $ at least when $a >-1$.

Let $(U_i)$ be i.i.d random variables uniformly distributed on $(0,1)$. By SLLN $\frac 1n \sum\limits_{k=1}^{n} U_i^{a} \to \int_0^{1} x^{a}dx=\frac 1 {a+1}$ almost surely. Hence the given integral tends to $f((\frac 1 {a+1})^{1/a})$.