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Let $f : [0, 1] → \mathbb{R}$ be a continuous function. Prove that $$\lim_{n\rightarrow \infty }\frac{1}{n}\sum_{k=1}^{n}(-1)^kf\left ( \frac{k}{n} \right )=0$$

First, I observed that any pair of consecutive terms' input values have the same distance $\frac{1}{n}$, and this converges to $0$ as $n$ goes to infinity.

Proof:

Consider $f(\frac{a}{n})$ and $f(\frac{a+1}{n})$ for any $a\in \{1,2,3,...,n\}$.

Note that for any $a$, $\lim_{n\rightarrow \infty }(\frac{a+1}{n}-\frac{a}{n})=\lim_{n\rightarrow \infty }(\frac{1}{n})=0$.

So, for some $\delta_{1}>0,\; \; \; \exists N $ s.t $$\left | \frac{1}{n} \right |< \delta _{1}\; \; \; \forall n>N$$

And since $f$ is continuous at $\frac{a}{n}$, $\forall \varepsilon > 0, \; \; \; \exists 0<\delta<\delta _{1} $ s.t $$\forall x\in [0,1] \; \; \; \text {that satisfies}\; \; \;\left | x-\frac{a}{n} \right |< \delta < \delta _{1}, \text {we have}\; \; \;\left | f(x)-f(\frac{a}{n}) \right |< \varepsilon $$

Thus $\forall n>N$, we get $$\left |\sum_{k=1}^{n}(-1)^kf(\frac{k}{n}) \right |=\left | -f(\frac{1}{n})+f(\frac{2}{n})-\cdot \cdot \cdot +(-1)^{n-1}f(\frac{n-1}{n})+(-1)^nf(\frac{n}{n}) \right |< \frac{n}{2}\varepsilon $$

This is because we have at least $ \frac{2}{n}$ many pairs of consecutive terms, and we can bound the series by Triangle Inequality.

Then this implies, $$\frac{1}{n}\left |\sum_{k=1}^{n}(-1)^kf(\frac{k}{n}) \right |= \left |\frac{1}{n} \right |\left |\sum_{k=1}^{n}(-1)^kf(\frac{k}{n}) \right |<\frac{\varepsilon }{2}<\varepsilon $$.

Hence, for $n>N$ and for any $\varepsilon$, $\left |\frac{1}{n} \sum_{k=1}^{n}(-1)^kf(\frac{k}{n}) \right |<\varepsilon$, so $\lim_{n\rightarrow \infty }\frac{1}{n}\sum_{k=1}^{n}(-1)^kf\left ( \frac{k}{n} \right )=0$. $\blacksquare $

Am I missing something in the proof?

Also, I doubt if we really can find some $0<\delta<\delta _{1}$. In other words, if we are looking for some $\delta$ for a fixed $\varepsilon$, is $\delta$ always bounded?

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    $\begingroup$ I think there is some mistake. Note that the $n$ is running, and you said that $f$ is continuous at $a/n$, for then the $\delta$ you choose afterward all depend on $n$, it seems that all are running, this is weird. $\endgroup$
    – user284331
    May 7, 2021 at 12:47
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    $\begingroup$ Express $\int _0^1 f(x) \, dx$ as limit of a Riemann sum using. Combining this Riemann sum with the sum given in question gives another sum which is also a Riemann sum for $f$ on $[0,1]$. Thus the desired limit is $0$. You can that we only need Riemann integrability of $f$ and not necessarily continuity. $\endgroup$
    – Paramanand Singh
    May 7, 2021 at 12:54
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    $\begingroup$ math.stackexchange.com/q/2886467/42969 $\endgroup$
    – Martin R
    May 7, 2021 at 13:11
  • $\begingroup$ @user284331 Then, can I just change the order of the proof? I mean, we can find some delta for any epsilon by using the continuity first, and then we can find N for such delta, then this mean, for n>N, any consecutive terms satisfies the $|x-a/n|$ so they are all bounded by epsilon. Do you think this works? $\endgroup$
    – john
    May 7, 2021 at 13:34
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    $\begingroup$ I think you really need to use uniform continuity to get rid of the particular choice of $a/n\in[0,1]$. $\endgroup$
    – user284331
    May 7, 2021 at 13:35

2 Answers 2

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$1/n \sum_{k=1}^n (-1)^k f(k/n) + 1/n \sum_{k=1}^n f(k/n) = 2/n \sum_{k=1}^{n/2} f(2k/n) = 2/n \sum_{k=1}^{n/2} f(2k/n)$ Now limit $n \rightarrow \infty$ => $\lim_{n \rightarrow \infty }1/n \sum_k (-1)^k f(k/n) = \int_{0}^1 f(x) dx - \int_{0}^1 f(x) dx = 0$ because $\lim_{n \rightarrow \infty} 2/n \sum_{k=1}^{n/2} f(2k/n) = \lim_{n \rightarrow \infty} 1/n \sum_{k=1}^{n} f(k/n)=\int_{0}^1 f(x) dx $

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  • $\begingroup$ Please fix your latex for the limit notation. $\endgroup$
    – Paramanand Singh
    May 7, 2021 at 13:00
  • $\begingroup$ Thanks...fixed it. $\endgroup$
    – Balaji sb
    May 7, 2021 at 13:02
  • $\begingroup$ +1 there. This is what I had suggested in comments and is the most straightforward approach to the problem. $\endgroup$
    – Paramanand Singh
    May 7, 2021 at 13:03
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Since $f$ is uniformly continuous on $[0,1]$, for $\epsilon>0$, there is some $\delta>0$ such that $|f(x)-f(y)|<\epsilon$ for $|x-y|<\delta$.

We first look at the subsequence of even terms, that is, \begin{align*} \dfrac{1}{2n}\sum_{k=1}^{2n}(-1)^{k}f\left(\dfrac{k}{2n}\right) \end{align*} Choose an $N$ such that $1/N<\delta$. For $n\geq N$, we have \begin{align*} \left|\dfrac{1}{2n}\sum_{k=1}^{2n}(-1)^{k}f\left(\dfrac{k}{2n}\right)\right|\leq\dfrac{1}{2n}\sum_{k=1}^{n}\left|f\left(\dfrac{k+1}{2n}\right)-f\left(\dfrac{k}{2n}\right)\right|\leq\dfrac{1}{2n}\sum_{k=1}^{n}\epsilon<\dfrac{\epsilon}{2}. \end{align*} For the subsequence of odd terms, one sees that

\begin{align*} \dfrac{1}{2n+1}\sum_{k=1}^{2n+1}(-1)^{k}f\left(\dfrac{k}{2n}\right)=-\dfrac{1}{2n+1}f\left(\dfrac{1}{2n+1}\right)+\dfrac{1}{2n+1}\sum_{k=2}^{2n+1}(-1)^{k}f\left(\dfrac{k}{2n+1}\right), \end{align*} and perform the similar trick for the second expression above since there are even terms for that.

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  • $\begingroup$ Can I get some hint for dealing with $-\dfrac{1}{2n+1}f\left(\dfrac{1}{2n+1}\right)$? I cannot get rid of it at the end.. $\endgroup$
    – john
    May 7, 2021 at 14:52
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    $\begingroup$ That is easy since $f$ is continuous at $x=0$, so $f(1/(2n+1))\rightarrow 0$. $\endgroup$
    – user284331
    May 7, 2021 at 14:54
  • $\begingroup$ So, here we are observing two cases when n is even and when n is odd because we need to show the limit of the "$\frac{1}{n}$series" has to be bounded by epsilon for any n>N? $\endgroup$
    – john
    May 7, 2021 at 14:57
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    $\begingroup$ Not like that. It is merely because when you do the grouping inside the sum, you need even terms. $\endgroup$
    – user284331
    May 7, 2021 at 14:59

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