Functions series and differentiability

Question

A classical result states that:

Let $\sum_n u_n(x),\quad u_n\in C^m(\mathbb{R})$ be a functions series, such taht:

1. $\sum_n u_n^{(k)}(x),\quad \color{red}{0\leq k\leq m-1}$ are convergent,
2. $\sum_n u_n^{(m)}(x)$ is uniformly convergent.

Then

1. $\sum_n u_n^{(k)}(x),\quad 0\leq k\leq m$ are uniformly convergent,
2. $\sum_n u_n(x)$ is of class $C^m$,
3. $\left(\sum_n u_n(x)\right)^{(k)}=\sum_n u_n^{(k)}(x),\quad 0\leq k\leq m$.

What if we replace the second hypothesis by the following: $$ \sum_n u_n(x) \qquad \text{ is uniformly convergent},$$ the result is still valid? Are there other versions with weak hypotheses?

Answer 1

Counterexample: $u_n(x)=\frac{\cos nx}{n^2}$, $m=1$.

In this case, $\sum_n u_n(x)$ is uniformly convergent and $u_n\in c^1(\mathbb R)$. However, $$\sum_n u_n'(x)=-\sum_n \frac{\sin nx}{n}$$ is convergent but not convergent uniformly on $\mathbb R$.

Actually, we have this result:

Let $a_n\geq 0$ be decreasing, then the series $\sum_{n=1}^\infty a_n\sin nx$ is convergent uniformly on $\mathbb R$ if and only if $$\lim_{n\to\infty} na_n=0.$$