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For convergent Riemann sums of $f \in C^1([0,1])$ there is the property:

$$\tag{A}\lim_{n \to + \infty} \left[\, \sum_{k=0}^{n} f \left( \frac{k}{n+1} \right) - \sum_{k=0}^{n-1} f \left( \frac{k}{n} \right) \, \right] = \int_0^1 f (x) \ dx$$

Proof using the mean value theorem goes like

$$\, \sum_{k=0}^{n} f \left( \frac{k}{n+1} \right) - \sum_{k=0}^{n-1} f \left( \frac{k}{n} \right) = f \left( \frac{n}{n+1} \right) + \sum_{k=0}^{n-1}f'(\theta_k)\left(\frac{k}{n+1} - \frac{k}{n} \right)$$ and then shows the right side converges to $f(1) - \int_0^1 xf'(x) dx = \int_0^1f(x)dx.$ So it is very helpful that $f \in C^1$.

If we only have $f \in C[0,1]$, then equation (A) may not be true, and this was shown by Daniel Fischer using Banach-Steinhaus theorem in this answer. Here a related statement was considered:

$$\tag{B}\lim_{n\to\infty}\left[\, \sum_{k=0}^{n-1} f\left(\frac{k}{n}\right) - n\int_0^1 f(x)\,dx\, \right] = \frac{f(0) - f(1)}{2}$$

My question: Can someone please show me a concrete counterexample, i.e., a continuous but not continuously differentiable function where (A) or (B) is false?

YuiTo Cheng
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WoodWorker
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  • It is just an idea. Try to consider a real continuous function of unbounded variation on $[0,1]$, The right-hand side of (A) would be finite, however the left-hand side would be infinite. – Pavel R. Dec 11 '18 at 00:46
  • Banach Steinhaus (and the uniform boundedness principle, Baire category theorem, etc.) are constructive and can be unwound to provide counterexamples. Daniel Fischer's argument uses the fact that there are $f_n$ with $|T_n(f_n)|\to\infty.$ Banach-Steinhaus is a recipe that computes some coefficients $c_n$ such that the linear combination $f=\sum c_nf_n$ is uniformly convergent and satisfies $\sup_n|T_n(f)|=\infty.$ – Dap Jan 24 '19 at 15:20

1 Answers1

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Consider Weierstrass function $$ f(x)=\sum_{l=1}^\infty 2^{-l/2}\cos \pi 2^l x. $$ It is nowhere differentiable but is Holder continuous with exponent $1/2$. Also it is an example of a lacunary Fourier series.

Let's show that the expression under the limit sign in (B) is not bounded. The integral in the lhs is zero. For $n=2^m$ $$ \sum_{k=0}^{2^m-1}f\left(\frac{k}{2^m}\right)= \sum_{l=1}^\infty 2^{-l/2}\sum_{k=0}^{2^m-1}\cos \pi 2^{l-m} k. $$ It's straightforward to check, presenting cosine as a sum of imaginary exponents and evaluating the appearing geometric progression, that $$ \sum_{k=0}^{2^m-1}\cos \pi 2^{l-m} k= \left\{ \begin{array}{r,l} 2^m,&&m<l,\\ 0,&&m\ge l. \end{array} \right. $$ Therefore $$ \sum_{k=0}^{2^m-1}f\left(\frac{k}{2^m}\right)= \sum_{l=m+1}^\infty 2^{-l/2}2^m=\left(1+\sqrt{2}\right) 2^{m/2}\to \infty $$ as $m\to\infty$.

Andrew
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