Convergence of a sum of sines

Question

If $ s_N(x) := \sum_{n = 1}^N c_n \sin(n x) $ converges uniformly on $[0, \pi]$ as $N \to \infty$ then $c_n = o(n^{-1})$.

a) Is $c_n = o(n^{-1})$ sufficient for uniform convergence?

b) Is $\sum_n n c_n^2 < \infty$ sufficient?

Answer 1

This is really subtle. If we put $f_n(x)=c_n \sin(nx)$ and assume that $a)$ or $b)$ (that implies $c_n=o(n^{-1/2})$) holds, we have that $$\sum_{n\geq 1}\,f_n'(x)$$ converges pointwise (in $L^2$) to $g(x)$ for any $x\in(0,\pi)$ due to Dirichlet's test. For the same reason, $$\sum_{n\geq 1}\,f_n(x)$$ converges pointwise to $f(x)$ on $(0,\pi)$. The dominated convergence theorem hence gives: $$ f(x) = f(\pi/2)+\int_{\pi/2}^x g(t)\,dt$$ where $g$ is a continuous function or a square integrable function over $(0,\pi)$, depending on $a)/b)$. In both cases we have $$ \sum_{n= 1}^M f_n(x)\xrightarrow[]{L_2}f(x) $$ hence Egorov's theorem ensures almost uniform convergence. I think it is possible to "lift" this kind of convergence to uniform convergence by exploiting the regularity of $\sin(nx)$, summation by parts and the fact that for any $x\in(0,\pi)$ we have: $$\sum_{n=1}^{N}\sin(nx)=\frac{\sin\frac{Nx}{2}\sin\frac{(N+1)x}{2}}{\sin\frac{x}{2}}$$ and: $$\left|\frac{\sin(kx/2)}{\sin(x/2)}\right|\leq 1$$ for any $x\geq\frac{\pi}{k+1}$.