$\lim_{n\rightarrow \infty } \int_{a}^{b}g_{n}(x)\sin (2n\pi x)dx=0$ where $g_{n}$ is uniformly Lipschitz Let {$g_{n}$}be a bounded sequence of functions on $[0,1]$ which is uniformly Lipschitz. That is, there is a constant $M$ (independent of $n$) such that for all $n$, $|g_{n}(x)-g_n(y)|\leq M|x-y|$
for all $x,y\in [0,1]$ and $|g_{n}(x)|\leq M$ for all $x\in [0,1]$.
Then I have the following two questions:
(a) prove for all any $0\leq a\leq b\leq 1$,
$$\lim_{n\rightarrow \infty } \int_{a}^{b}g_{n}(x)\sin (2n\pi x)\,dx=0. $$
(b) prove that for any $f\in L^{1}[0,1]$,
$$\lim_{n\rightarrow \infty } \int_{0}^{1}f(x)g_{n}(x)\sin (2n\pi x)\,dx=0.$$
 A: There is a discrete analog of "integrate by parts to kill the periodic term". Namely, "translate by half-period and cancel". Like this: 
$$\int_a^b g_n(x)\sin (2n \pi x)\,dx = - \int_a^b g_n(x)\sin (2n \pi (x+1/(2n))\,dx \\ 
=- \int_{a+1/(2n)}^{b+1/(2n)} g_n(y-1/(2n))\sin (2n \pi y)\,dy $$
The right hand side is nearly the same as $-\int_{a}^{b} g_n(y)\sin (2n \pi y)\,dy$: the discrepancy of intervals of integration contributes $O(1/n)$, and the difference of integrands is also $O(1/n)$, due to the Lipschitz condition. Conclusion: $\int_a^b g_n(x)\sin (2n \pi x)\,dx = O(1/n)$.
I'll leave it for you to adapt this to (b). You'll need the usual "estimate the difference of products" trick, plus the fact that translation is continuous in $L^1$: $\|f(\cdot)-f(\cdot+1/n)\|_{L^1}\to 0$ as $n\to \infty$.
A: Here's an alternative proof of (a):
Since $g_n$ is Lipschitz, it is absolutely continuous, hence differentiable a.e. with $g_n' \in L^1[0,1]$, and $g_n(x) = \int_0^x g_n'(t) dt$. Let $\phi(x) = g_n(x) \frac{\cos(2 \pi nx)}{2 \pi n}$. We have $\phi'(x) = -g_n(x)\sin(2 \pi nx)+g_n'(x) \frac{\cos(2 \pi nx)}{2 \pi n}$. Then $\phi(b)-\phi(a) = \int_a^b \phi'(t) dt$, which gives 
$\frac{1}{n}(g_n(b) \frac{\cos(2 \pi n b)}{2 \pi}-g_n(a) \frac{\cos(2 \pi n a)}{2 \pi}) = - \int_a^b g_n(x)\sin(2 \pi nx) dx + \int_a^b g_n'(x) \frac{\cos(2 \pi nx)}{2 \pi n} dx$. Rearranging, and using the fact that $|g_n(x)| \leq M$, gives
$|\int_a^b g_n(x)\sin(2 \pi nx) dx| \leq \frac{1}{2n\pi}(2M+\|g_n'\|_1)$, from which the desired limit (a) follows.
