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Let $f$ be an integrable function over $[a,b]$. Prove that: $$\lim_{n \rightarrow \infty } \int _a ^b f(x)|\sin(nx)| dx= \frac {2}{\pi} \int _a^b f(x) dx.$$

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  • $\begingroup$ Tried to use lower integral $\endgroup$ – Ben Jun 11 '12 at 16:04
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    $\begingroup$ Welcome to math.stackexchange BenLi. It is your first time on this website, so please take some time to read over the FAQ. When asking questions, please tell us what you have tried, you need to put in some effort into a problem and get stuck in a specific area so we can help you there. This time, I'll direct you to this relevant question. $\endgroup$ – Ragib Zaman Jun 11 '12 at 16:06
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Hint: Prove the theorem for characteristic functions of intervals, then prove the theorem for step functions. Approximate integrable functions with step functions and apply a convergence theorem.

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Based on the hints given by Alex Becker, we can obtain a more general result: If $f$ is integrable on $[a,b]$,which is a bounded closed interval, and $g$ is integrable periodic function with period $\rm T$, then we have the following:

$$\lim\limits_{n\rightarrow \infty} \int_a^b f(x)g(nx)dx={\rm T}^{-1}\int_0^{\rm T} g(t)dt \int_a^bf(x)dx$$

(1): I show that $$\lim\limits_{n\rightarrow \infty} \int_a^b {\rm g}(nx)dx={\rm T}^{-1}(b-a)\int_0^{\rm T}g(t)dt $$

PROOF: For each $n\in \mathbb{N}$, define $b_n=a+\frac{{\rm T}}{n} \left[\frac{n(b-a)}{{\rm T}} \right]$. It is not difficult to see that $0\le b-b_n <\frac{{\rm T}}{n}$.

$$\int_a^{{b_n}} g (nx)dx = \sum\limits_{k = 1}^{\left[ {\frac{{n(b - a)}}{{\rm T}}} \right]} {\int_{a + (k - 1)\frac{{\rm T}}{n}}^{a + k\frac{{\rm T}}{n}} g } (nx)dx$$

$$ = \frac{1}{n}\sum\limits_{k = 1}^{\left[ {\frac{{n(b - a)}}{{\rm T}}} \right]} {\int_{na + (k - 1){\rm T}}^{na + k{\rm T}} g } (y)dy = \frac{1}{n}\sum\limits_{k = 1}^{\left[ {\frac{{n(b - a)}}{{\rm T}}} \right]} {\int_0^{\rm T} g } (y)dy$$

(since the integral of a periodic function over a length of its period is the same)

$$\int_a^{b_n} g(nx) dx=\frac{1}{n}\sum_{k=1}^{\left[\frac{n(b-a)}{{\rm T}}\right]}\int_{0}^{{\rm T}} g(y)dy =\frac{1}{n}\left[\frac{n(b-a)}{{\rm T}}\right]\int_0^{\rm T}g(t)dt$$

Since $$\left|\int_a^bg(nt)dt-\int_a^{b_n}g(nt)dt \right|\le (b-b_n)M \le \frac{{\rm T}}{n} M$$ where $$M=\sup|g(x)|$$

so $$\lim_{x\rightarrow \infty} \int_a^bg(nx)dx=\lim_{x\rightarrow \infty} \int_a^{b_n}g(nx)dx={\rm T}^{-1}(b-a)\int_0^{\rm T} g(t)dt $$

(2): I show $$\lim_{n\rightarrow \infty} \int_a^b f(x)g(nx)dx={\rm T}^{-1}\int_0^{\rm T} g(t)dt \int_a^bf(x)dx$$ for any step function $f$.

PROOF: This result follows from (1) easily.

(3): I prove $$\lim_{n\rightarrow \infty} \int_a^b f(x)g(nx)dx={\rm T}^{-1}\int_0^{\rm T} g(t)dt \int_a^bf(x)dx$$ for any integrable function.

PROOF: Given $\delta >0$, since $f$ is integrable, we can find a step function $h$ such that $\displaystyle \int_a^b|f-h|<\delta$.

$$\eqalign{ & \left| {\int_a^b f (x)g(nx)dx - {T^{ - 1}}\int_0^T g (t)dt\int_a^b f (x)dx} \right| \leqslant \cr & \left| {\int_a^b f (x)g(nx)dx - \int_a^b h (x)g(nx)dx} \right| + \left| {\int_a^b h (x)g(nx)dx - {T^{ - 1}}\int_0^T g (t)dt\int_a^b h (x)dx} \right| \cr & + \left| {{T^{ - 1}}\int_0^T g (t)dt\int_a^b f (x)dx - {T^{ - 1}}\int_0^T g (t)dt\int_a^b h (x)dx} \right| < \left( {M + 1 + {T^{ - 1}}\int_0^T g (t)dt} \right)\delta \cr} $$

for all $n\ge N$,(such $N$ exist by (2)),and since $\delta>0$ is arbitrary so

$$\lim_{n\rightarrow \infty} \int_a^b f(x)g(nx)dx=T^{-1}\int_0^Tg(t)dt \int_a^bf(x)dx$$

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  • $\begingroup$ I dont understand what the definition of $b_n$ in the first part of the proof is accomplishing. In particular, everything cancels and I'm left with $b_n = b$. $\endgroup$ – yoshi May 18 at 15:49
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    $\begingroup$ @yoshi The square bracket notation $[\cdot]$ in the definition of $b_n$ represents the integer part. So $b_n$ is close to, but not exactly $b$. $\endgroup$ – grand_chat May 18 at 21:17

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