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Let $g:[0,1]\mapsto\mathbb{R}$ be a continuous function, and $\lim_{x\to0^+}g(x)/x$ exists and is finite. Prove that $\forall f:[0,1]\mapsto\mathbb{R}$,

$$\lim_{n\to\infty}n\int_0^1f(x)g(x^n)dx=f(1)\int_0^1\frac{g(x)}{x}dx$$

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Where exactly are you stuck? (Also, I think $f$ should be continuous as well). –  Jason DeVito Nov 16 '12 at 3:57
    
Or we can use the substitution $t=x^n$ to get a clearer view of what is going on as $n\to\infty$. –  Sangchul Lee Nov 16 '12 at 5:35

2 Answers 2

Let $f(x)$ be a function that possesses a power series expansion, such that $\forall f:[0,1]\mapsto\mathbb{R}$, then for $f(x) = \sum_{k \geq 0} a_{k} x^{k}$, \begin{align} L &= \lim_{n\to\infty}n\int_0^1f(x)g(x^n)dx = \sum_{k=0}^{\infty} a_{k} \, \lim_{n \to \infty} n \, \int_{0}^{1} x^{k} \, g(x^{n}) \, dx. \end{align} Make the substitution $t = x^{n}$ to obtain \begin{align} L &= \sum_{k=0}^{\infty} a_{k} \, \lim_{n \to \infty} \int_{0}^{1} t^{\frac{k+1}{n} - 1} \, g(t) \, dt \\ &= \sum_{k=0}^{\infty} a_{k} \, \int_{0}^{1} \frac{g(t) \, dt}{t} \\ &= f(1) \, \int_{0}^{1} \frac{g(t) \, dt}{t}. \end{align} This is the desired result.

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I'm a little late to the game and see that this suggestion has been made, but let's carry it out.

Let $y=x^n$, $x=y^{1/n}$, $dy = (1/n) y^{1/n} dy/y$. Then

$$n \int_0^1 dx \: f(x) g(x^n) = \int_0^1 \frac{dy}{y} y^{1/n} f(y^{1/n}) g(y)$$

As $n \rightarrow \infty$, $y^{1/n} \rightarrow 1 \forall y \in (0,1]$; that is, apart from $y=0$; however, since this is an isolated point (i.e., measure zero), then we have

$$\lim_{n \rightarrow \infty} n \int_0^1 dx \: f(x) g(x^n) = f(1) \int_0^1 dy \frac{g(y)}{y}$$

as was to be shown.

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