Rate of weak convergence of sin(nx) Since $\sin(n\cdot)$ converges weakly to zero, we know that
$$
\lim_{n\rightarrow\infty} \int_a^b g(x)\sin(nx)\mathrm{d}x = \int_a^b g(x)\cdot 0\,\mathrm{d}x = 0
$$
holds for all $g\in L^2([a,b])$.
Is there a way to find an explicit formula for the rate of convergence in the above equation, i.e. to determine a function $C$ depending on $n$ such that
$$
\left|\int_a^b g(x)\sin(nx)\mathrm{d}x\right| \le C(n), \qquad \lim_{n\rightarrow\infty}C(n) = 0
$$
holds for all $g\in A$, where $A$ is a certain subset of $L^2([a,b])$?
If, for example, $A$ is the set of constant functions with $||g||_{L^\infty} < M$ for all $g\in A$, then it is easy to show that $C(n) = \frac{2}{n}M$ is such an upper bound (by integrating $\sin(nx)$).
I am particularly interested in the case where $A$ is the set of continuously differentiable (or smooth) functions with $||g||_{L^\infty}<M_1$ and $||\frac{\mathrm{d}}{\mathrm{d}x}g||_{L^\infty}<M_2$ for $M_1,M_2>0$.

Consider the functions $f_k = sin(kx)$ for all integers $k$. 
for every $k$:
$||f_k||_{L^\infty } = 1$ 
$\int_0^{2\pi} f_ksin(kx) \,dx= \pi \le C(k)$
therefore, there is no function $C(n)$ that satisfies your conditions.
