Borel's lemma

Remarks

Idea of the proof: choose a bump function $\chi$ equal to $1$ on $[-1/2, 1/2]$ and $0$ outside $[-1, 1]$. Set $f(x) = \sum_{n \ge 0} \frac{a_n}{n!} x^n \chi(\lambda_n x)$ with numbers $\lambda_n \ge 1$ that grow fast. The $n$-th term agrees with $\frac{a_n}{n!} x^n$ near $0$. For $k < n$, its $k$-th derivative has supremum at most $2^{-n}$ when $\lambda_n$ is large enough. Then every series of derivatives converges uniformly, so term-by-term differentiation gives $f^{(n)}(0) = a_n$.

The construction also gives $f$ with compact support. Example: for $a_n = (n!)^2$, the Taylor series of $f$ at $0$ is $\sum n! \, x^n$, which converges only at $x = 0$.

Going deeper

For a student

The lemma says that the map sending a smooth function to its Taylor series at $0$, a formal power series, is onto. It is far from one-to-one: the smooth non-analytic function equal to $e^{-1/x}$ for $x > 0$ and to $0$ for $x \le 0$ has zero Taylor series at $0$. Together these facts show how loosely the Taylor series controls a smooth function. The lemma has versions in several variables and with parameters, and it is used in the study of singularities and of partial differential equations.