# When $\lim_{n\rightarrow\infty}\int_0^1f(x)\sin(nx)=0$

What is a good sufficient condition for $f:[0,1]\rightarrow\mathbb{R}$ such that:

$$\lim_{n\rightarrow\infty}\int_0^1f(x)\sin(nx)dx=0$$

If $f$ is differentiable then by integral by part we can easily prove. But I think differentiablity is a too strong condition. Can we make it less strict?

• Yes, use $$\left\lvert \int_0^1 f(x)\sin (nx)\,dx - \int_0^1 g(x)\sin (nx)\,dx\right\rvert \leqslant \int_0^1 \lvert f(x) - g(x)\rvert\,dx.$$ – Daniel Fischer Oct 11 '14 at 13:08
• Lebesgue integrability is enough. See this. – David Mitra Oct 11 '14 at 13:08
• @DanielFischer How do you proceed further? – Leaning Oct 11 '14 at 13:26
• You approximate an integrable function by a continuously differentiable function in the integral norm to obtain the conclusion for all integrable functions. – Daniel Fischer Oct 11 '14 at 13:37
• Yeah I understand that argument! Is there any other nice way to show directly by some calculations? – Leaning Oct 11 '14 at 13:38

To be sure that everything makes sense, we assume that $f$ is Lebesgue integrable.

If we show the result for the characteristic function of a Borel subset, it OK. And by approximation, we can do it for finite disjoint unions of intervals, for which the result is true.

Hence $f$ Lebesgue integrable is a sufficient condition.

• I really don't understand anything but thanks, I can see the beauty :) – Leaning Oct 11 '14 at 13:32
• I guess you did not study yet measure theory. – Davide Giraudo Oct 11 '14 at 13:40
• Yes I haven't. I hope to obtain a beautiful proof in the case of continuity maybe and in the context of normal Riemann integral. Hope I will get to measure theory in the future... – Leaning Oct 11 '14 at 13:41

I'm inspired by another answer here

Computing $\lim_{n \rightarrow\infty} \int_{a}^{b}\left ( f(x)\left | \sin(nx) \right | \right )$ with $f$ continuous on $[a,b]$

So I will write down a proof for $f$ continous.

Take $I_n=\int_0^1f(x)\sin(nx)dx=\int_0^1g(x)dx$

We have

$$I_n=\int_0^{\pi/n}g(x)dx+\int_{\pi/n}^{2\pi/n}g(x)dx+\dots+\int_{(k-1)\pi/n}^{k\pi/n}g(x)dx+\int_{k\pi/n}^1g(x)dx$$

where $1-\frac{k\pi}{n}<\frac{\pi}{n}$ or $k=[\frac{n}{\pi}]$.

Using mean value theorem we have

$$\int_{(i-1)\pi/n}^{i\pi/n}f(x)\sin(nx)dx=f(u_i)\int_{(i-1)\pi/n}^{i\pi/n}\sin(nx)dx=\frac{f(u_i)}{n}\int_{(i-1)\pi}^{i\pi}\sin ydy=2\frac{f(u_i)}{n}(-1)^i$$

with $u_i\in [(i-1)\pi/n,i\pi/n]$.

Therefore we have:

$$I_n=\sum_{i=1}^k2\frac{f(u_i)}{n}(-1)^i+\int_{k\pi/n}^1g(x)dx$$ $$I_n=\frac{1}{\pi}\sum_{i=1}^{[k/2]}\frac{2\pi f(u_{2i})}{n}-\frac{1}{\pi}\sum_{i=1}^{[(k+1)/2]}\frac{2\pi f(u_{2i-1})}{n}+\int_{k\pi/n}^1g(x)dx$$

Then using the definition of Riemann Integral we have:

$$\lim_{n\rightarrow\infty} I_n=\frac{1}{\pi}\int_0^1f(x)dx-\frac{1}{\pi}\int_0^1f(x)dx+\lim_{n\rightarrow\infty}\int_{k\pi/n}^1g(x)dx$$

$$=\lim_{n\rightarrow\infty}\int_{k\pi/n}^1g(x)dx=\lim_{n\rightarrow\infty}f(v)\int_{k\pi/n}^1\cos(nx)dx=0$$

where $v\in[k\pi/n,1]$ (since $|\int_{k\pi/n}^1\cos(nx)dx|\le 1-k\pi/n<\pi/n$ and $f$ is continous so $|f(v)-f(1)|$ is small enough)

Is this proof true?