Counterexample to Riemann sum limit

Solution 1:

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 $m\to\infty$.