6
$\begingroup$

Let $\theta\in[0,1]$ be a constant and $f\in C^1[0,1] $. Show that $$\sum_k^{n-1}f\left(\frac{k+\theta} n\right)-n\int_0^1 f(x)\,dx$$

converges to $(\theta-\frac{1}{2})(f(1)-f(0))$, as $n\to\infty.$

\begin{align} & S_n=\sum_k^{n-1}f\left(\frac{k+\theta} n\right)-n\int_0^1 f(x)\,dx \\[10pt] ={} & n\sum_k^{n-1} \int_{k/n}^{(k+1)/n} \left( f \left(\frac{k+\theta} n\right)-f(x)\right) \, dx \\[10pt] = {} & n\sum_k^{n-1} \int_{k/n}^{(k+1)/n} \int_x^{(k+\theta)/n} f'(s) \, ds \, dx \\[10pt] = {} & n\sum_k^{n-1} \left( \int_{k/n}^{(k+\theta)/n} \int_x^{(k+\theta)/n}-\int_{k+\theta/n}^{(k+1)/n} \int_{(k+\theta)/n}^x \right)f'(s)\,ds\,dx \end{align}

\begin{align} & n\sum_k^{n-1} \int_{k/n}^{(k+1)/n} \int_x^{(k+\theta)/n} f'(s) \, ds \, dx \\[10pt] = {} & n\sum_k^{n-1} \left( \int_{k/n}^{(k+\theta)/n}\int_x^{(k+\theta)/n}-\int_{k+\theta/n}^{(k+1)/n} \int_{(k+\theta)/n}^x \right) f'(s) \, ds \, dx \end{align}

\begin{align}\int_{k/n}^{(k+\theta)/n}\int_x^{(k+\theta)/n} f'(s) \, ds \, dx=\int_{k/n}^{(k+\theta)/n}\int_{(k)/n}^s f'(s) \, ds \, dx=\int_{k/n}^{(k+\theta)/n}f'(s)(s-\frac{k}{n}) \, ds \int_{k+\theta/n}^{(k+1)/n}\int_{(k+\theta)/n}^x f'(s) \, ds \, dx=\int_{k+\theta/n}^{(k+1)/n}\int_s^{(k+1)/n} f'(s) \, ds \, dx=\int_{k+\theta/n}^{(k+1)/n}f'(s)(\frac{k+1}{n}-s) \, ds\end{align}

Therefore \begin{align} \sum_k^{n-1} \int_{k/n}^{(k+1)/n} f'(s)\phi_k(s) \, ds \end{align}

where \begin{align} \phi_k(s)=ns-k-\begin{cases} 0, & \mbox{ if }(k\leqslant ns <k+\theta)\\1, & \mbox{ if }k+\theta\leqslant ns <k+1\end{cases}\end{align},

Denoting by $\{x\}$ the fractional part of $x$ and noticing that $k=[ns]$ for $k\leqslant ns\leqslant k+1$, we see that $\phi_k(s)=g(ns)$, where

\begin{align}g(x)=\{x\}-\begin{cases} 0, & \mbox{ if }(\{x\}<\theta)\\1, & \mbox{ if }(\{x\}\geqslant \theta)\end{cases}=\{x-\theta\}+\theta-1\end{align}

is a piecewise linear periodic function of period 1.Any piecewise continuous periodic function $g(x)$, we conclude that \begin{align}S_n=\int_0^1 f'(x)g(nx)dx\to \mathscr{M}_[0,1](g)\times \int_0^1 f'(x) dx=(\theta-\frac{1}{2})(f(1)-f(0))\end{align} as $n\to\infty$ because $f'\in\mathscr{C}_[0,1]$

Question:

I do not understand what the author does when it is defined $g(x)=\{x\}-\begin{cases} 0, & \mbox{ if }(\{x\}<\theta)\\1, & \mbox{ if }(\{x\}\geqslant \theta)\end{cases}=\{x-\theta\}+\theta-1$ Where does this come from? How did he derive it?

How does the author prove \begin{align}S_n=\int_0^1 f'(x)g(nx)dx\to \mathscr{M}_[0,1](g)\times \int_0^1 f'(x) dx=(\theta-\frac{1}{2})(f(1)-f(0))\end{align}

Where did $\sum_{k=0}^{n-1}$ go to?

Thanks in advance!

$\endgroup$
9
  • $\begingroup$ Do you see how the first part, $$g(x) = \lbrace x\rbrace - \begin{cases} 0 &\text{if } \lbrace x\rbrace < \theta \\ 1 &\text{if} \lbrace x\rbrace \geqslant \theta \end{cases}$$ (it must be $\lbrace x\rbrace$ instead of $\lvert x\rvert$ in the first branch, somebody at some point misread a brace for a vertical bar) comes from $\phi_k(s)$? The second equality just uses $$\lbrace x - y\rbrace = \begin{cases} \lbrace x\rbrace - y &\text{if } \lbrace x\rbrace \geqslant y \\ \lbrace x\rbrace + (1-y) &\text{if } \lbrace x\rbrace < y \end{cases}$$ for $0 \leqslant y < 1$. $\endgroup$ Aug 20, 2018 at 12:04
  • $\begingroup$ I do not understand how the first part comes from $\phi_k(s)$. I have no clue whatsoever of how the author proceed on that regard. Concerning $\{\}$ instead of $||$, you are right. I have just corrected it. That was a typo of mine. Thanks for your reply. I would be glad if you told me how the indicator function on this case is derived. $\endgroup$ Aug 20, 2018 at 12:17
  • $\begingroup$ For $\frac{k}{n} \leqslant s < \frac{k+1}{n}$ we have $\lbrace ns\rbrace = ns - k$, for $k = \lfloor ns\rfloor$ then. And $k \leqslant ns < k + \theta$ is equivalent to $0 \leqslant ns - k < \theta$, which is $\lbrace ns\rbrace < \theta$. But I'm not sure what exactly you don't understand, so I'm not sure whether what I've said before helps or misses the point. $\endgroup$ Aug 20, 2018 at 12:24
  • $\begingroup$ @DanielFischer I still do not understand where the $x$ comes from. $\endgroup$ Aug 20, 2018 at 17:59
  • $\begingroup$ @DanielFischer I do not know if I was clear. My doubt arises at the point the author begins to define the function $g$. $\endgroup$ Aug 20, 2018 at 18:38

1 Answer 1

3
+25
$\begingroup$

You can simplify things by noting the expression equals

$$\tag 1 \sum_{k=0}^{n-1}[f(k/n+\theta/n)- f(k/n)] + \sum_{k=0}^{n-1}f(k/n)- n\int_0^1 f.$$

By the MVT, the first sum in $(1)$ equals

$$\sum_{k=0}^{n-1}f'(c(k,n))\cdot (\theta/n) = \theta\sum_{k=0}^{n-1}f'(c(k,n))\cdot (1/n).$$

This converges to $\theta\int_0^1 f' =\theta(f(1)-f(0)).$

The rest of $(1)$ equals

$$\tag 2 n \left (\sum_{k=0}^{n-1}\int_{k/n}^{(k+1)/n}[f(k/n)-f(x)]\,dx\right ).$$

For $n$ large we expect $f(x) \approx f(k/n) +f'(k/n)(x-(k/n))$ for $x$ in the $k$th interval. Let's just brazenly insert that into $(2).$ We get

$$n \left (-\sum_{k=0}^{n-1}\int_{k/n}^{(k+1)/n}f'(k/n)(x-(k/n))\,dx \right )$$ $$ = n \left (-\sum_{k=0}^{n-1}f'(k/n)(1/(2n^2)) \right ) = -(1/2)\sum_{k=0}^{n-1}f'(k/n)(1/n). $$

The last expression $\to -(1/2)\int_0^1 f' = -(1/2)f(1)-f(0)).$ So this looks very good. All we need to verify is that the approximations $f(x) \approx f(k/n) +f'(k/n)(x-(k/n))$ are good enough. This follows from the uniform continuity of $f'.$ I'll leave this verification to the reader for now.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .