Convergence of $\sum\limits^\infty _{k=0} a_k \sin(kx)+b_k \cos(kx)$

Question

Ok, for the infinite series:

$$\sum^\infty _{k=0} a_k \sin(kx)+b_k \cos(kx)$$

How do I show that this converges on any finite interval if $\sum^\infty _{k=0} k(|a_k|+|b_k|)<\infty$?

Also, do the hack do I show that this function differentiable, and also finding its derivative?

Thanks!

Answer 1

Edit: as observed by @J.M. this is a Fourier series.

Actually, $\sum_{k\ge 0}|a_k|+|b_k|<\infty$ suffices for convergence, as it implies normal convergence on $\mathbb{R}$ (some people will rather invoke the Weierstrass M-test, which boils down to the same thing, but simply concludes that there is uniform convergence), i.e. $$ \sum_{k\geq 0}\sup_{x\in\mathbb{R}}|a_k\sin(kx)+b_k\cos(kx)|\leq \sum_{k\geq 0}|a_k|+|b_k|\leq \sum_{k\geq 0}k(|a_k|+|b_k|)<\infty. $$

Of course, normal convergence is a very strong form of convergence which implies in particular pointwise convergence. So your series converges on $\mathbb{R}$. A fortiori on any finite interval.

But your assumption is probably here to justify the normal convergence of the series obtained after term-by-term differentiation. Indeed $$ \sum_{k\geq 0}\sup_{x\in\mathbb{R}}|ka_k\cos(kx)-kb_k\sin(kx)|\leq \sum_{k\geq 0}k(|a_k|+|b_k|)<\infty. $$

This proves that your series is differentiable (actually $C^1$) on $\mathbb{R}$ with derivative $$\sum_{k\geq 0}ka_k\cos(kx)-kb_k\sin(kx).$$

Note that the only estimates needed to get all this are $|\cos y|\leq 1$ and $|\sin y|\leq 1$ for every $y\in\mathbb{R}$.

I'll conclude by a proof of the fact I've used.

Claim: let $f(x)=\sum_{n\geq 0}f_n(x)$ be a series of differentiable (resp. $C^1$) functions on $\mathbb{R}$. If $f$ converges pointwise on $\mathbb{R}$ (actually, it suffices that it converges at a single point give the following assumption) and if $\sum_{n\geq 0}f_n'(x)$ converges normally on $\mathbb{R}$, then $f$ is differentiable (resp. $C^1$) on $\mathbb{R}$ and $f'(x)=\sum_{n\geq 0}f_n'(x)$.

Proof: Fix $x_0$. By the mean value theorem $f_n(x)-f_n(x_0)=f'_n(c_x)(x-x_0)$ hence $\left|\frac{f_n(x)-f_n(x_0)}{x-x_0}\right|\leq \sup |f'_n|$ for every $x\neq x_0$ and every $n\geq 0$. Therefore $$ \frac{f(x)-f(x_0)}{x-x_0}=\sum_{n\geq 0}\frac{f_n(x)-f_n(x_0)}{x-x_0} $$ is normally convergent. By dominated convergence, we can swap limit at $x_0$ and sum, which yields the differentiability at $x_0$ and the formula for the derivative. In the $C^1$ case, normal convergence of the series of derivatives impies its continuity. QED.

Answer 2

Hint: Try the comparison test.