Why function $f(x)=\sum_{n=1}^{\infty} \frac{\sin(nx)}{n^3}$is differentiable?

Question

My lecturer said me that $f(x)=\sum_{n=1}^{\infty} \frac{\sin(nx)}{n^3}$ is differentiable on $\mathbb{R}$. I don't know why. Can you explain it me ?

The formal derivative is an absolutely convergent series since $|\cos(nx)|\leq 1$ and $\sum_{n\geq 1}\frac{1}{n^2}<+\infty$. — Jack D'Aurizio, Sep 16 '16 at 15:45
Some useful theorems: http://math.stackexchange.com/questions/1754033/integration-and-differentiation-of-fourier-series/1770532#1770532 — Winther, Sep 16 '16 at 15:55

score 1 · Accepted Answer · answered Sep 16 '16 at 16:25

1

The series $f(x)$ converges pointwise for at least one $x$ (actually for any $x$ since it is normally convergent)
The series $\displaystyle\sum_{n\ge1}\frac{n\cos(nx)}{n^3}=\sum_{n\ge1}\frac{\cos(nx)}{n^2}$ is uniformly convergent, since it is normally convergent – bounded above by $\displaystyle\sum_{n\ge1}\frac1{n^2}$.

These two conditions ensure the series $f(x) is differentiable,and its derivative is obtained differentiating term by term.

answered Sep 16 '16 at 16:25

Bernard

So normal covergence implies pointwise and uniform covergence ? – Happy man Sep 16 '16 at 18:02
Why, yes, since you can find an $N$ such that &c., which doesn't depend on $x$. – Bernard Sep 16 '16 at 18:49

1 Answers1