Derivative of unknown continuous function

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Source: UM Mathematical Olympiad 2024

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#1 h May 13, 2025, 1:05 AM

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Let $f: \mathbb{R} \to \mathbb{R}$ be a function whose derivative is continuous on $[0,1]$ . Show that
$\lim_{n \to \infty} \sum^n_{k = 1}\left[f\left(\frac{k}{n}\right) - f\left(\frac{2k - 1}{2n}\right)\right] = \frac{f(1) - f(0)}{2}.$

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#2 h May 13, 2025, 8:37 AM

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We have $f({k\over n})-f({{2k-1}\over {2n}})={{f'(t_{n,k})}\over {2n}}$ where ${{2k-1}\over {2n}}<t_{n,k}<{k\over n}$ .
Hence $\sum_{k=1}^n (f({k\over n})-f({{2k-1}\over {2n}}))={1\over {2n}}\sum_1^n f'(t_{n,k})$ .
This is not a regular Riemann sum since the intervals $[{1\over {2n}},{1\over n}],[{3\over {2n}}, {2\over n}],\cdots$ do not cover $[0,1]$ .
Fix $\epsilon>0$ . Since $f'$ is uc, we have $|f'(t_{n,k})-f'(k/n)|<\epsilon$ for $n$ large and $1\le k\le n$ .
It follows that we may also consider the limit ${1\over {2n}}\sum_1^n f'(k/n)$ .
But this, without the factor $1/2$ , is a Riemann sum hence for the limit we find ${1\over 2}\int_0^1 f'(x)dx={{f(1)-f(0)}\over 2}$ .

@solyaris: thank you, I didn't notice that.

This post has been edited 2 times. Last edited by alexheinis, May 13, 2025, 3:51 PM

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#3 h May 13, 2025, 12:43 PM

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@above: In fact $\frac 1 n \sum_{k=1}^n f'(t_{n,k})$ already is a Riemann sum. Note that $t_{n,k} \in [\frac{k-1} n , \frac k n]$ .
Thus by continuity of $f'$ the above tends to $\int_0^1 f'(x) dx = f(1) - f(0)$ .

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