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?
 A: 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. 
A: 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?
