Is the following series infinitely differentiable?

Question

Let $f:\mathbf{R}\to\mathbf{R},\, f\in C^\infty,\, f(0)=f'(0)=0$ and consdier the series $$\sum_{n=1}^\infty f\left(\frac{x}{n}\right)$$ I want to show that this series defines an infinitely differentiable function $F$ on $\mathbb R.$ I suppose all I need to show is that $\sum f\left(\frac{x}{n}\right)$ converges uniformly, but I don't know how to show that. I know that if $\sum f'\left(\frac{x}{n}\right)$ converges uniformly and $\sum f\left(\frac{x}{n}\right)$ converges pointwise for some $x$, (e.g. 0), then $\sum f\left(\frac{x}{n}\right)$ converges absolutely, but that didn't make it any easier. Is my approach correct? If so, how do I proceed?

Answer 1

A start: For any $x$ Taylor gives

$$\tag 1 f(x) = f(0) + f'(0)x + f''(c)x^2/2= f''(c)x^2/2$$

for some $c$ between $0$ and $x.$ Let $R > 0,$ and let $M_R = \sup_{[-R,R]} |f''|.$ If $|x|\le R,$ then from $(1)$ we see

$$|f(x/n)| \le M_R|x/n|^2/2 \le M_R R^2/n^2.$$

By Weierstrass M, $\sum f(x/n)$ converges uniformly on $[-R,R].$ It follows that $F$ is continuous on $[-R,R].$ Since $R$ is arbitrary, $F$ is continuous on $\mathbb R.$

These ideas should be helpful in showing that $F^{(n)}(x)$ eixsts for all $x\in \mathbb R$ and $n\in \mathbb N.$