A function with a given sequence of its n-th derivative Suppose $h(x)$ is a bump function defined on $\mathbb R.$ Let $\{a_n, n \geqslant 0\}$ be an arbitrary sequence of real numbers and let
$$
f(x)=\sum_{n=0}^\infty \frac{h(\xi_n x)}{n !} a_n x^n, \quad x \in \mathbb{R}
$$
where $\xi_n=n+\sum_{i=0}^n\left|a_i\right|$. Prove that $f$ is a smooth function and
$$
f^{(n)}(0)=a_n, \quad n \geqslant 0
$$
It can be calculated by Libniz's Rule and take termwise calculation that
$$
f^{k}(x)=\sum_{n=0}^{\infty}\frac{a_n}{n!}\sum_{i+j=k}\xi_n^ih^i(\xi_nx)n^{\underline{j}}x^{n-j}
$$
I think it is enough to prove it is uniformly convergent at a neiborhood of $0$,then the termwise derivative is what we want. But I am stuck here and I have no idea why $\xi_n$ is defined so and how to use this condition.
 A: Since $h$ is a bump function it is compactly supported, so you may assume the support is contained in some interval $[-r,r]$. The derivatives of $h$ are also supported in this interval.
First calculate that for $x\in [-\frac r\xi_n, \frac r\xi_n]$ you have that:
$$\left|\frac{a_n}{n!}x^n\right| ≤ \frac{r^n}{n!\cdot \xi_n^{n-1}}≤\frac{r^n}{n!}$$
as we have that $\xi_n≥1$ for $n\neq0$. On the other hand if $x$ is outside of this interval you have that $h(\xi_n x) = 0$ and so the summand is equal to $0$. This gives the following calculation:
$$\left|\sum_{n=0}^N \frac{h(\xi_nx)}{n!}a_n x^n - \sum_{n=0}^\infty \frac{h(\xi_n x)}{n!}a_n x^n\right|≤ \sum_{n=N+1}^\infty\left|\frac{h(\xi_n x^n)}{n!}a_n x^n\right|≤\sum_{n=N+1}^\infty \frac{r^n}{n!}$$
as $N$ grows this converges to $0$ so you have that the series converges uniformly in $x$.
The same calculations - with some variations - can be doen for the derivatives to get that the derivative of the series also converges uniformly. So the defining series and all its derivatives converge uniformly. It follows that the limit is smooth.
