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Let $(x_i)$ be a sequence of numbers in $(0,1)$ such that the $lim_{n→∞} (1/n \sum_{i=1}^{n} x_i^k)$ exists for every $k=0,1,2,....$ Show that $lim_{n→∞} (1/n \sum_{i=1}^{\infty} f(x_i))$ exists for every $f∈C[0,1]$

My attempt:

I'm trying to prove this using the Weierstrass Approximation Theorem, but I'm running into a few difficulties.

What I have so far: if $\{1/n \sum_{i=1}^{n} f(x_i)\}$ converges then it is Cauchy, so $∀\epsilon>0$, $∃N>1$ such that $|(1/n) \sum_{i=1}^{n} f(x_i) - (1/s) \sum_{i=1}^{s} f(x_i)| < \epsilon$ $∀s,n > N$.

Now using Weierstrass Approximation Theorem, since there exists a sequence of polynomials $(p_n)$ that converge to $f∈C[0,1]$ then $∀\epsilon>0$, $∃N>1$ such that $|f-p_n|<\epsilon$ $∀n > N$. Thus $|(1/n)\sum_{i=1}^{n} (f(x_i)-p_m(x_i))|<\epsilon/3$ $∀m > N$ and $∀x_i∈(0,1)$.

Now,

$|(1/n) \sum_{i=1}^{n} f(x_i) - (1/s) \sum_{i=1}^{s} f(x_i)|$

$ <|(1/n) \sum_{i=1}^{n} f(x_i) - (1/n) \sum_{i=1}^{n} p_m(x_i)+(1/n) \sum_{i=1}^{n} p_m(x_i)-(1/s) \sum_{i=1}^{s} f(x_i)+(1/s) \sum_{i=1}^{s} p_m(x_i)-(1/s) \sum_{i=1}^{s} p_m(x_i)|$

$ <|(1/n) \sum_{i=1}^{n} f(x_i) - (1/n) \sum_{i=1}^{n} p_m(x_i)|+|(1/s) \sum_{i=1}^{s} f(x_i)-(1/s) \sum_{i=1}^{s} p_m(x_i)|+|(1/n) \sum_{i=1}^{n} p_m(x_i)-(1/s) \sum_{i=1}^{s} p_m(x_i)|$

$<\epsilon/3 + \epsilon/3 + |(1/n) \sum_{i=1}^{n} p_m(x_i)-(1/s) \sum_{i=1}^{s} p_m(x_i)|$

This is where I get stuck. What would my next step be? Any help/suggestions greatly appreciated

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  • $\begingroup$ You were 99% of the way there. $\endgroup$ – RRL Mar 15 '16 at 19:35
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Hint:

For any polynomial of degree $m$

$$\lim_{n \to \infty} \frac{1}{n}\sum_{i=1}^np_m(x_i) = \lim_{n \to \infty} \frac{1}{n}\sum_{i=1}^n\sum_{k=0}^m c_kx_i^k = \sum_{k=0}^mc_k \left(\lim_{n \to \infty}\frac{1}{n}\sum_{i=1}^n x_i^k\right) = S$$

So the partial sum $\frac1{n} \sum p_m(x_i)$ is a Cauchy sequence.

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  • $\begingroup$ Thanks. Ya I don't know why that wasn't obvious to me in the first place. Stupid brain $\endgroup$ – user147686 Mar 15 '16 at 19:36

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