3

I thought when let $f(x) $ be a continuous function on $[0, 1]$,

$$\lim_{n \to \infty} \int^1_0 f(x) |\sin(n \pi x)|\,dx = \frac{2}{\pi} \int^1_0 f(x)\,dx$$

With my intuition, the formula above is followed from $$ \int^1_0 \sin(\pi x)\,dx = \frac{2}{\pi}$$

I tried to give a proof of this with mean value theorems for definite integrals but failed.

Are there rigorous proofs that $\displaystyle \lim_{n \to \infty} \int^1_0 f(x) |\sin(n \pi x)|\,dx = \frac{2}{\pi} \int^1_0 f(x)\,dx$?

StubbornAtom
  • 17,052
Soling
  • 478
  • 1
    If $f$ is continuous on [0,1], a closed interval, then it is uniformly continuous. – Lapin Apr 12 '21 at 11:05
  • Use the fact that $|\sin (\pi x) |$ is periodic with period $1$ and further use the generalization at end of this answer : https://math.stackexchange.com/a/3527941/72031 You don't need continuity of $f$. Just any integrable $f$ works. – Paramanand Singh Apr 12 '21 at 11:12
  • This is a generalization of the Riemann Lebesgue lemma and is true for any integrable $f$ and bounded integrable $g$ on some finite interval (one can think of such as periodic functions on the line by the obvious extension); one proves the result first for $f$ a trigonometric polynomial ( this is pretty much the classical riemann lebesgue) and then one uses dominated convergence (here one needs boundness of $g$) to show that if $||f_m-f||_1 \to 0$ and the result holds for $f_m$, it holds for $f$ too – Conrad Apr 12 '21 at 12:13

3 Answers3

3

We may use the following result.

Lemma (Fejér): Suppose $g$ is a bounded measurable $T$-periodic function on $\mathbb{R}$ ($T>0$). For any integrable function $f$ (denoted by $f\in\mathcal{L}_1(\mathbb{R})$) and numeric sequence $a_n\in\mathbb{R}$,

$$ \lim_n\int_\mathbb{R} f(x)g(nx+a_n)\,dx=\Big(\frac{1}{T}\int^T_0 g\Big)\int_\mathbb{R} f \tag{1}\label{one} $$

The result is atributed to Fejér, I learned of it from a problem in a prelim when I was a student long time ago. A short proof of this result is here (based on basic facts of integration such as dominated convergence and density of step functions).

Edit: Similar problems to to OP have been posted in the past. Common to their solutions are variant of the result I presented above. One such variant (more restrictive, but simpler in the sense that only Riemann calculus is needed) is given here.

Now, for the problem in the OP, set $g(x)=\sin(\pi x)$, $a_n=0$. Function $g$ has period is $T=1$. By extending $f$ to all of $\mathbb{R}$ by setting $f(x)=0$ if $x\notin [0,1]$, we have all the ingredient of the Lemma which we can no ally directly: \begin{align} \lim_n\int^1_0f(x)|\sin(n\pi x)|\,dx&=\Big(\int^1_0\sin(\pi x)\,dx\Big)\int^1_0f\\ &=\Big(\frac{1}{\pi}\int^\pi_0\sin( x)\,dx\Big)\int^1_0f\ =\frac{2}{\pi}\int^1_0f \end{align}

Mittens
  • 39,145
3

Since $f$ is continuous on $[0,1]$, it is uniformly continuous. That is, for any $\epsilon\gt0$, there is a $\delta\gt0$ so that if $|x-y|\le\delta$, then $|f(x)-f(y)|\le\epsilon$.

Pick an $\epsilon\gt0$ and let $\delta$ satisfy the requirements above. Choose $n$ so that $\frac1n\le\delta$.

For $1\le k\le n$, define $I_k=\left[\frac{k-1}{n},\frac{k}{n}\right]$ and $$ f_k=n\int_{I_k}f(x)\,\mathrm{d}x\tag1 $$ Taking the mean over $I_k$ of $\inf\limits_{u\in I_k}f(u)-f_k\le f(x)-f_k\le\sup\limits_{u\in I_k}f(u)-f_k$, we get $$ \inf_{u\in I_k}f(u)-f_k\le0\le\sup_{u\in I_k}f(u)-f_k\tag2 $$ By the choice of $n$, the difference between the left and right side of $(2)$ is less than or equal to $\epsilon$. Thus, $$ \sup_{u\in I_k}|f(u)-f_k|\le\epsilon\tag3 $$ Therefore, $(3)$ and $\int_{I_k}|\sin(n\pi x)|\,\mathrm{d}x=\frac2{n\pi}$ ensure $$ \int_{I_k}|f(x)-f_k|\,|\sin(n\pi x)|\,\mathrm{d}x\le\frac2{n\pi}\epsilon\tag4 $$ Thus, we get $$ \begin{align} &\left|\,\int_0^1f(x)\,|\sin(n\pi x)|\,\mathrm{d}x-\frac2\pi\int_0^1f(x)\,\mathrm{d}x\,\right|\tag{5a}\\ &=\left|\,\sum_{k=1}^n\left(\int_{I_k}f(x)\,|\sin(n\pi x)|\,\mathrm{d}x-\frac2\pi\int_{I_k}f(x)\,\mathrm{d}x\right)\right|\tag{5b}\\ &=\left|\,\sum_{k=1}^n\left(\int_{I_k}f(x)\,|\sin(n\pi x)|\,\mathrm{d}x-\frac2\pi\int_{I_k}f_k\,\mathrm{d}x\right)\right|\tag{5c}\\ &=\left|\,\sum_{k=1}^n\left(\int_{I_k}f(x)\,|\sin(n\pi x)|\,\mathrm{d}x-\int_{I_k}f_k|\sin(n\pi x)|\,\mathrm{d}x\right)\right|\tag{5d}\\ &\le\sum_{k=1}^n\int_{I_k}|f(x)-f_k|\,|\sin(n\pi x)|\,\mathrm{d}x\tag{5e}\\[3pt] &\le\frac2\pi\epsilon\tag{5f} \end{align} $$ Explanation:
$\text{(5b)}$: $\bigcup\limits_{k=1}^nI_k=[0,1]$
$\text{(5c)}$: apply $(1)$
$\text{(5d)}$: $\int_{I_k}|\sin(n\pi x)|\,\mathrm{d}x=\frac2\pi\int_{I_k}\mathrm{d}x$
$\text{(5e)}$: triangle inequality
$\text{(5f)}$: apply $(4)$

Since $\epsilon\gt0$ was arbitrary, inequality $(5)$ says that $$ \lim_{n\to\infty}\int_0^1f(x)\,|\sin(n\pi x)|\,\mathrm{d}x=\frac2\pi\int_0^1f(x)\,\mathrm{d}x\tag6 $$

robjohn
  • 345,667
2

Some thoughts

By the identity $$\left|\sin x\right| = \frac{2}{\pi}-\sum_{k = 1}^\infty \frac{4}{\pi(4k^2-1)}\cos(2k x),$$ we have $$\int_0^1 f(x) |\sin (n\pi x)| \mathrm{d} x = \frac{2}{\pi}\int_0^1 f(x) \mathrm{d} x - \sum_{k=1}^\infty I(k, n)$$ where $$I(k, n) = \int_0^1 \frac{4 f(x)}{\pi(4k^2-1)}\cos(2k n\pi x) \mathrm{d} x.$$

By Riemann-Lebesgue Lemma, we have, for each fixed $k\ge 1$, $$\lim_{n\to \infty} I(k, n) = 0.$$

By Tannery's theorem https://en.wikipedia.org/wiki/Tannery%27s_theorem, we have $$\lim_{n\to \infty} \sum_{k=1}^\infty I(k, n) = 0.$$

Tannery's theorem: Let $S_n=\sum_{k=0}^\infty a_k(n)$ and $\lim_{n\to\infty}a_k(n)=b_k$. If $|a_k(n)|\le M_k$ and $\sum_{k=0}^\infty M_k < \infty$, then $\lim_{n\to\infty}S_n=\sum_{k=0}^\infty b_k$.

River Li
  • 37,323