# Prove Borel's Lemma (Pugh's book #35) [duplicate]

Given any sequence whatsoever of real numbers (a_r), there is a smooth function $f: \mathbb{R} \to \mathbb{R}$ such that $f^{(r)}(0) = a_r$. Pugh's hint says to try $f=\sum \beta_k(x)a_kx^k/k!$, where $\beta_k$ is a well-chosen bump function.

I'm working with a few other people and we're trying to define a bump function $\beta_k(x)$ = bump $*b_k*x$ where from 1/2 to 1, $(exp(\frac{1}{|x|-1}))(1-exp(\frac{-1}{|x|-1/2}) + C_k$ but have no idea how to define

## marked as duplicate by Andrés E. Caicedo, user63181, Claude Leibovici, Stefan Hansen, Yiorgos S. SmyrlisFeb 7 '14 at 9:00

In Every power series is the Taylor series of some $C^{\infty}$ function, Andrew Stacey pointed to Kriegl and Michor's book A Convenient Setting for Global Analysis, section 15.4. I present a version of the proof from this book below.
Let $\rho(x)$ be a bump function which is equal to $1$ in the neighborhood of $0$ and has support contained in $(-1,1)$. For any number $t_k>1$, to be chosen later, the function $f_k(x) = \dfrac{a_k}{k!} x^k \rho(t_k x)$ has the desired $k$th derivative at $0$, with all other derivatives vanishing at $x=0$. It remains to arrange $t_k$ so that the series $\sum f_k$ converges to a $C^\infty$ function; that is, the series of $n$th derivatives $f_k^{(n)}$ converges uniformly, for every $n$.
It suffices to consider $k\ge 2n$. We have $$f_k^{(n)}(x) = a_k \sum_{m=0}^n \binom{n}{m}\dfrac{1}{(k-m)!} x^{k-m} t_k^{n-m}\rho^{(n-m)}(t_k x)$$ Let $B_n=\max_{m\le n}\sup |\rho^{(m)}|$. Since $|\rho^{(n-m)}(t_k x)|\le B_{k}\chi_{\{|x|\le t_k^{-1}\}}$, it follows that the supremum of $|f_k^{(n)}|$ is at most $$a_k B_k \sum_{m=0}^n \binom{n}{m}\dfrac{1}{(k-m)!} t_k^{m-k } t_k^{n-m} \le a_k B_k t_k^{-k/2}\sum_{m=0}^n \binom{n}{m}\dfrac{1}{(k-m)!}$$ Let $M_k$ denote the maximum of $\sum_{m=0}^n \binom{n}{m}\frac{1}{(k-m)!}$ over $n\le k/2$. Whatever $a_k$, $B_k$ and $M_k$ are, we can choose $t_k$ large enough so that $$a_k B_k M_k t_k^{-k/2} \le 2^{-k}$$ and uniform convergence follows.