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I came across this very nice Question so thought of sharing it!

Let $f(x)$ be a continuous function $f:R \to R$ with period $1$. Prove that $\displaystyle \lim_{n \to \infty} \int_{0}^{1} \sin^2(\pi x)f(nx)\,\mathrm{d}x = \frac{1}{2} \int_{0}^{1} f(x)\,\mathrm{d}x$.

Added the solution as an answer.

V.G
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user600016
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3 Answers3

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An alternative solution: Let $I = \int_0^1 f(t) \, dt$. Then $$ \int_0^1 \sin^2(\pi x)f(nx)\,dx = I \int_0^1 \sin^2(\pi x) \, dx + \int_0^1 \sin^2(\pi x)(f(nx)-I)\,dx \\ = \frac 12 I + \int_0^1 \sin^2(\pi x)(f(nx)-I)\,dx \, . $$ Define $G(x) = \int_0^x (f(t) - I) \, dt$. Then $G'(x) = f(x) - I$ and $G(0) = G(1) = 0$. In particular, $G$ is continuous and $1$-periodic and therefore bounded on $\Bbb R$. Now do integration by parts: $$ \int_0^1 \sin^2(\pi x)(f(nx)-I)\,dx = \int_0^1 \sin^2(\pi x)G'(nx) \, dx \\ = - \frac \pi n \int_0^1 \sin(2 \pi x) G(nx) \, dx $$ and that converges to zero for $n \to \infty$ because the integral is bounded independently of $n$.

Martin R
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Here is another solution based on this post and this solution which states that if $\phi\in \mathcal{L}_1(\mathbb{R})$, $f$ is a bounded $T$ periodic function and $a_n$ is any real numeric sequence, then

$$ \lim_n\int \phi(x)f(nx+a_n)\,dx=\Big(\frac{1}{T}\int^T_0f\Big)\int \phi \tag{1}\label{one} $$

In terms of the OP, $T=1$, $a_n=0$, and $\phi(x)=\mathbb{1}_{[0,1]}(x)\sin^2(\pi x)$. The rest follows by substituting $\phi$ on $\eqref{one}$.

Mittens
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$$\lim_{n \to \infty} \int_{0}^{1} \sin^2(πx)f(nx)dx$$

$$=\lim_{n \to \infty} \sum_{r=0}^{n-1} \int_{r/n}^{r/n+1/n} \sin^2(πx)f(nx)dx $$

$$=\lim_{n \to \infty} \sum_{r=0}^{n-1} \sin^2(πr/n) \int_{r/n}^{r/n+1/n} f(nx)dx$$ $nx=t$.

$$=\sum_{r=0}^{n-1}(\sin^2(πr/n))\frac{1}{n} \int_{r}^{r+1} f(t)dt$$

$$=\int_{0}^{1}f(t)dt \int_{0}^{1} \sin^2(πx)dx = \frac{1}{2} \int_{0}^{1} f(x)dx$$

user600016
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  • What is $x$ in the third and fourth ine? – Martin R Aug 30 '20 at 14:11
  • Thanks Corrected it – user600016 Aug 30 '20 at 14:15
  • Why are the second and third expression equal? – Martin R Aug 30 '20 at 14:17
  • As n is very large, the interval of the integration is very small, so we may approximate x to be a constant in this interval, and so we can treat sin²(πx) as a constant here, and it may be taken out of the integral – user600016 Aug 30 '20 at 14:27
  • Then the expressions are approximately equal, but not necessarily equal. – I am not saying that your end result is wrong, but you need to justify it better. – Martin R Aug 30 '20 at 14:32
  • But in an infidecimal interval wouldn't they be pretty much equal? I haven't learnt the epsilon delta method yet so not sure how to be more rigorous without it – user600016 Aug 30 '20 at 14:38
  • @zhw But this step concerns only with the size of any one interval and not the number of intervals isn't? – user600016 Aug 30 '20 at 15:05
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    Sorry, deleted my previous comment. You said the intervals are small, and I reminded you that the number of intervals is large. We have a sum of a large number of small numbers, so the the result is ...? It could be anything. For example $1/\sqrt n$ is small if $n$ is large, but a sum of $n$ of terms each equal to $1/\sqrt n$ is $n\cdot 1/\sqrt n = \sqrt n,$ which is large. What you need is an estimate on $|\sin(\pi x/n)-\sin(\pi k/n)|$ for $x\in [k-1,k].$ – zhw. Aug 30 '20 at 15:20
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    Your approach can be salvaged: Using the mean-value theorem for integrals you can show that $\int_{r/n}^{r/n+1/n} \sin^2(πx)f(nx)dx = \sin^2(π x_{n,j}) \int_{r/n}^{r/n+1/n} f(nx)dx$ for some $x_{n,j} $ in the interval $[r/j, (r+1)/j]$. That gives you Riemann sums converging to $\int_{0}^{1}\sin^2(πx)dx$. – Martin R Aug 30 '20 at 16:07