Prove that the limit $\lim_{n \to \infty} \frac{1}{n} \sum_{k = 1}^{n} f(x_k)$ exists given that $\lim_{n \to \infty} \frac{1}{n} \sum_{k = 1}^{n} x_k^i$ exists for every $i \in \{0,1,\dots \}$ and $x_k \in (0,1).$ and $f$ is a continuous function on $[0,1]$.

At first, the sums look like Riemann sums to he me, but I don't think it will help me prove the existence of the limit. I feel like this is a nested limit argument. The presence of the index $i$ also bothers me. Here are my thoughts,

\begin{align} \lim_{n \to \infty} \frac{1}{n} \sum_{k = 1}^{n} f(x_k) &= f(\lim_{n \to \infty} \frac{1}{n} \sum_{k = 1}^{n} x_k) \\ &=f(x) \end{align}

where $x = \lim_{n \to \infty} \frac{1}{n} \sum_{k = 1}^{n} x_k$ and the exchange of composition is justified by continuity. But this problem occurs in a chapter from the Weistrass Approximation thoerm, so I suspect something about him is needed? If I could get a hint it would be great.

Also it looks like $lim_{n \to \infty} \frac{1}{n} \sum_{k = 1}^{n} x_k^i$ is the polynomial I need.

  • $\begingroup$ Prove the result for polynomials, then use the Weierstrass approximation theorem to prove the general case. $\endgroup$ – anomaly Nov 17 '14 at 3:41
  • $\begingroup$ @anomaly, you mean prove $\lim \frac{1}{n} \sum p(x_k)$ exists? $p$ is a polynomial? $\endgroup$ – Hawk Nov 17 '14 at 3:41
  • $\begingroup$ Right, prove the result first for $f$ a polynomial. $\endgroup$ – anomaly Nov 17 '14 at 3:44
  • $\begingroup$ The limit doesn't commute with $f$ like that, because what that commutation says is that something kind of like $f(1)+f(2)+f(3)=f(1+2+3)$ holds. And yes, first for a polynomial. $\endgroup$ – Arthur Nov 17 '14 at 3:47

Your reasoning is wrong because it assumes that $f$ is linear (or almost): you assumed $$ f\left(\frac1n\sum_{k=1}^nx_k\right)=\frac1n\sum_{k=1}^nf(x_k); $$ This fails for most continuous functions and most choices of $x_1,\ldots,x_n$.

If $p$ is a polynomial, $p(t)=\sum_{j=0}^ma_jx^j$. Then, writing $y_j=\lim_{n\to\infty}\frac1n\sum_{k=1}^nx_k^j$, $$ \lim_{n\to\infty}\frac1n\sum_{k=1}^np(x_k)=\lim_{n\to\infty}\frac1n\sum_{k=1}^n\sum_{j=0}^ma_jx_k^j =\sum_{j=0}^ma_j\lim_{n\to\infty}\frac1n\sum_{k=1}^nx_k^j=\sum_{j=1}^na_jy_j $$ exists.

Now, by Weierstrass, for each $m$ there exists a polynomial $p_m$ with $|f(t)-p_m(t)|<1/m$ for all $t\in(0,1)$. Then $$ \left|\frac1n\sum_{k=1}^np_{m_1}(x_k)-\frac1n\sum_{k=1}^np_{m_2}(x_k)\right|\leq\frac1n\sum_{k=1}^n|p_{m_1}(x_k)-p_{m_2}(x_k)|\\ \leq\frac1n\sum_{k=1}^n|p_{m_1}(x_k)-f (x_k)|+|f (x_k)-p_{m_2}(x_k)|\leq\frac1n\,\sum_{k=1}^n\frac1 {m_1}+\frac1 {m_2}=\frac1 {m_1}+\frac1 {m_2}. $$ Thus the numbers $\frac1n\sum_{k=1}^np_{1/m}(x_k)$ converge, and an estimation similar to the above one shows that $\frac1n\sum_{k=1}^nf(x_k)$ converges to that number.

  • $\begingroup$ sorry but what are the polynomial $p_{\epsilon_1}$ and $p_{\epsilon}$ for? $\endgroup$ – Hawk Nov 17 '14 at 4:19
  • $\begingroup$ Please see if the new version is clearer. $\endgroup$ – Martin Argerami Nov 17 '14 at 4:59
  • $\begingroup$ So actually the $p_{\epsilon_i}$ is "cauchy" as well? $\endgroup$ – Hawk Nov 17 '14 at 15:19
  • $\begingroup$ Yes. The assumption is that the $p_m$ are polynomials that converge uniformly to $f$. Any convergent sequence is Cauchy. $\endgroup$ – Martin Argerami Nov 17 '14 at 16:43

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