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Let $f :\mathbb{R}\to \mathbb{R}$ be a continuous function with period $1$ and $$\lim_{n\to\infty}\int_0^1\sin^2(\pi x)f(nx)dx= \frac{1}{k}\int_0^1f(x)dx.$$ Find $k$.

My approach till now:

Applying half angle formula $2\sin^2(x) = 1-\cos(2x)$ I got : $$\frac{1}{2}\int_0^1f(nx)dx- \frac{1}{2}\int_0^1\cos(2\pi x)f(nx)dx.$$

I can't think of a way forward from here without applying integration by parts but i don't know if its right to apply it as we don't know about the differentiability of the function $f$.

Please help me with this problem.

Robert Z
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OhMyGauss
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  • See https://math.stackexchange.com/a/3527941/72031 – Paramanand Singh Mar 30 '20 at 14:51
  • Thank you. I learned something really beautiful today from it. – OhMyGauss Mar 30 '20 at 15:55
  • Learning on MSE is more enjoyable than doing the same with books. I hope you have a great time here. – Paramanand Singh Mar 30 '20 at 16:08
  • @ParamanandSingh Exactly Sir . Sir when can we interchange summation and integral ? In what cases is it permissible ? All possible permutations like summation-integral interchange, limit-integral interchange, derivative-integral interchange. Can you please tell me about that ? I really want to know that. – OhMyGauss Mar 30 '20 at 16:13
  • Well interchange of two limit operations (remember both derivative and integral are deep down limits) is not allowed in general. Finding when this is allowed is one of the central themes in analysis. The definitive result regarding interchange of limits and integral was given by Lebesgue. Search Lebesgue DCT. – Paramanand Singh Mar 30 '20 at 16:20
  • @ParamanandSingh Can you please suggest me a book to start with analysis ? – OhMyGauss Mar 30 '20 at 16:22
  • You can first try Hardy's A Course of Pure Mathematics then follow it with Apostol's Mathematical Analysis. – Paramanand Singh Mar 31 '20 at 03:22

1 Answers1

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Let $g:[0,1]\to\mathbb{R}$ be a continuous function. Then \begin{aligned}\int_0^1 g(x)f(nx)dx&=\frac{1}{n}\int_0^n g\left(\frac{t}{n}\right)f(t)dt\\ &=\frac{1}{n}\sum_{k=0}^{n-1}\int_0^1 g\left(\frac{t+k}{n}\right)f(t)dt\\ &=\frac{1}{n}\sum_{k=0}^{n-1} g(t_k)\int_0^1 f(t)dt \end{aligned} where we used the $1$-periodicity of $f$ and the fact that by the Mean Value Theorem $\exists t_k\in [k,k+1]$ such that $$\int_0^1 g\left(\frac{t+k}{n}\right)f(t)dt=g(t_k)\int_0^1 f(t)dt.$$ Therefore $$\lim_{n\to\infty}\int_0^1 g(x)f(nx)dt=\int_0^1g(t)dt\cdot \int_0^1f(t)dt.$$ Hence, from your work, it follows that as $n\to\infty$, $$\int_0^1\sin^2(\pi x)f(nx)dx=\frac{1}{2}\int_0^1f(nx)dx- \frac{1}{2}\int_0^1\cos(2\pi x)f(nx)dx\to \frac{1}{2}\int_0^1f(x)dx$$ which means that $k=2$.

Robert Z
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