Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How to prove that if $f \in C[0,2 \pi]$, then $$ 2\pi \lim_{n \to \infty} \int_0^{2 \pi} |\sin(nx)-f(x)| \, dx = \int_0^{2 \pi} \int_0^{2 \pi} |\sin(y)-f(x)| \,dx \, dy\ ? $$ This is true for constant functions, by direct calculation.

share|cite|improve this question
What did you try? (Personally, I see no direct connectin between the two sides – but I would start by switching the order of the integrals on the right, then evaluating the inner integral explicitly. Perhaps as a warmup exercise, try to prove it when $f$ is a constant.) – Harald Hanche-Olsen Apr 24 '13 at 5:35
This statement is trivial for constant functions. – user64494 Apr 24 '13 at 5:46
Indeed. But what if $f$ is a constant times the characteristic function of an interval inside $[0,2\pi]$? (That's not continuous, but this should not matter.) Or a step function? We lack linearity here, so you have to be careful. But I still imagine you might get something out of these cases. – Harald Hanche-Olsen Apr 24 '13 at 10:05
up vote 3 down vote accepted

Start writing \begin{align} \int_0^{2\pi}|\sin(nx)-f(x)|dx&=\frac 1n\int_0^{2n\pi}|\sin(t)-f(tn^{-1})|dt\\ &=\frac 1n\sum_{k=0}^{n-1}\int_{2k\pi}^{2(k+1)\pi}|\sin t-f(tn^{-1})|dt\\ &=\frac 1n\sum_{k=0}^{n-1}\int_0^{2\pi}\left|\sin t-f\left(\frac tn+\frac{2\pi}nk\right)\right|dt. \end{align} It's a Riemann sum and the terms $tn^{-1}$ doesn't matter by uniform continuity.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.