show that $\lim_{n\to+\infty}\frac{1}{n}\sum_{k=0}^{n}f(x_{k})$ is exists

Question

let $f:[0,1]\to R^{+}$ real-valued continuous functions,and such $$\int_{0}^{1}f(x)dx=2019,~~\int_{0}^{1}f^2(x)dx=20181027$$ (1):show that:there exists unique sequence $x_{0},x_{1},\cdots,x_{n}\in [0,1]$,such $x_{0}<x_{1}<\cdots<x_{n}$ and for any postive integer $k=1,2,\cdots,n$.such $$\int_{x_{k-1}}^{x_{k}}f(t)dt=\dfrac{1}{n}\int_{0}^{1}f(t)dt$$

(2):and show that $$\lim_{n\to+\infty}\dfrac{1}{n}\sum_{k=0}^{n}f(x_{k})$$ is exists.and find this value.

I can do it $(1)$ exists.I think it is First mean value theorem for integration。but How to this sequence is unique. How to solve this (2),if this limt has exists.it seem use Stolz-Cesaro's Lemma:$$\lim_{n\to+\infty}\dfrac{1}{n}\sum_{k=0}^{n}f(x_{k})=\lim_{n\to\infty}f(x_{n})$$

this problem is my teacher gave me the exercise, these two problems I can not solve all , so I want to ask the teacher here,Thanks

Answer 1

• First note that if the stated conditions hold, then \begin{align} \int_{x_0}^{x_k}f(x)dx&=\sum_{i=1}^k\int_{x_{i-1}}^{x_i}f(x)dx\\ &=\frac{k}{n}\int_0^1f(x)dx\ , \end{align} and, in particular, $$ \int_{x_0}^{x_n}f(x)dx=\int_0^1f(x)dx\ . $$
• Subtracting the left side of the final equation above from both sides gives \begin{align} 0&=\int_0^1f(x)dx-\int_{x_0}^{x_n}f(x)dx\\ &=\int_0^{x_0}f(x)dx+\int_{x_n}^1f(x)dx\ , \end{align} and since $\ 0\le x_0\le x_n\le1\ $, and $\ f\ $ is positive and continuous, it follows from this that $\ x_0=0\ $ and $\ x_n=1\ $ if the stated conditions hold
• Now, following the suggestion of Kelenner in the first of the comments, put $$ F(x)=\int_0^xf(t)dt\ . $$ Then $\ F\ $ is strictly increasing and continuous on $\ [0,1]\ $, and if the stated conditions hold, then the first equation above tells us that $\ x_k\ $ must satisfy the equation $$ \ F\big(x_k\big)=\frac{k}{n}\int_0^1f(x)dx\ . $$ But since $$ F(0)=0<\frac{k}{n}\int_0^1f(x)dx<\int_0^1f(x)dx=F(1)\ , $$ for $\ k=1,2,\dots,n-1\ $ , then by the intermediate value theorem, therefore, there must exist $\ x_{nk}\in(0,1)\ $ with $$ F\big(x_{nk}\big)=\frac{k}{n}\int_\limits{0}^1f(x)dx\ ,$$ and since $\ F\ $ is strictly increasing, $\ x_{nk}\ $ is uniquely determined. Moreover, \begin{align} F\big(x_{nk}\big)-F\big(x_{n\,(k-1)}\big)&=\int_{x_{n\,(k-1)}}^{x_{nk}}f(x)dx\\ &=\frac{1}{n}\int_0^1f(x)dx\\ &\ge m\big(x_{nk}-x_{n\,(k-1)}\big)\ , \end{align} where $\ m=\min_\limits{x\in[0,1]}f(x)>0\ $. Therefore, $\ 0<x_{nk}-x_{n\,(k-1)}\le$$\frac{1}{mn}\int_\limits{0}^1f(x)dx\ $, and so $\ x_{nk}-x_{n\,(k-1)}\rightarrow0\ $ as $\ n\rightarrow\infty\ $, as Paramanand Singh has observed in the comments. For notational convenience, put $\ x_{n0}=x_0=0\ $ and $\ x_{nn}=x_n=1\ $.
• We are now in a position to prove that $$ \lim_{n\rightarrow\infty}\sum_{k=0}^nf\big(x_{nk}\big)=\frac{\int_\limits{0}^1f(x)^2dx}{\int_\limits{0}^1f(x)dx}=\frac{20181027}{2019}\ . $$ Let $\ M=\max_{x\in[0,1]}f(x)\ $ and $\ \epsilon\ $ be any positive real number. Since $\ f\ $ must be uniformly continuous on $\ [0,1]\ $, there exists a positive real number $\ \delta\ $ such that $\ \left|f(x)-f(y)\right|\le\frac{\epsilon}{M}\ $ for all $\ x,y\in[0,1]\ $ with $\ |x-y|\le\delta\ $. Therefore, if $\ n>\frac{1}{m\delta}\int_\limits{x=0}^1f(x)dx\ $, then $\ \left|x-x_{nk}\right|\le\delta\ $, and hence $\ \left|f(x)-f\big(x_{nk}\big)\right|\le\frac{\epsilon}{M}\ $, for all $\ x\in\left[x_{nk}, x_{n\,(k+1)}\right]\ $. Hence, for all such $\ n\ $ we have \begin{align} \left|\,\int_\limits{0}^1f(x)^2dx-\right.&\left.\frac{1}{n}\int_\limits{0}^1f(x)dx \sum_{k=0}^{n-1}f\big(x_{nk}\big)\right|\\ &=\left|\,\int_\limits{0}^1f(x)^2dx- \sum_{k=0}^{n-1}\int_{x_{nk}}^{x_{n\,(k+1)}}f\big(x_{nk}\big)f(x)dx\right|\\ &=\left|\,\sum_{k=0}^{n-1}\int_{x_{nk}}^{x_{n\,(k+1)}}\left(f(x)-f\big(x_{nk}\big)\right)f(x)dx\right|\\ &\le\sum_{k=0}^{n-1}\epsilon\big(x_{n\,(k+1)}-x_{nk}\big)\\ &=\epsilon\ . \end{align} Thus, $$ \lim_{n\rightarrow\infty}\frac{1}{n}\int_\limits{0}^1f(x)dx \sum_{k=0}^{n-1}f\big(x_{nk}\big)=\int_\limits{0}^1f(x)^2dx\ , $$ from which, since $\ \int_\limits{0}^1f(x)dx\ne0\ $, and $\ \frac{f\left(x_{nn}\right)}{n}=\frac{f(1)}{n}\rightarrow0\ $ as $\ n\rightarrow\infty\ $, it follows that $$ \lim_{n\rightarrow\infty}\frac{1}{n} \sum_{k=0}^nf\big(x_{nk}\big)=\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=0}^{n-1}f\big(x_{nk}\big)=\frac{\int_\limits{0}^1f(x)^2dx}{\int_\limits{0}^1f(x)dx}\ . $$