Every real sequence is the derivative sequence of some function

I am looking for the proof of the following theorem:

Let $(a_n)$ be a sequence of real numbers. Then there exists a function $f$ which is infinitely differentiable at 0, and $$ \frac{d^nf}{dx^n}(0) = a_n, \ \ \text{for all } n.$$

I would appreciate either a sketch of the proof or an online reference to it. A general case is listed as Borel's lemma in Wikipedia, without proof.

The hard part is when the power series $\sum_n \frac{a_n}{n!}x^n$ has a zero radius of convergence.

Edit: Thanks for the answers!

Solution 1:

As you mentioned, this is a result of Borel. Proofs can be found in several sources, see for example the book on "Complex variables" by Berenstein and Gay (it may just be an exercise there, but at least there is a "hint"). The idea is to try a power series. Of course, this may not converge, so you use the kind of functions that come in constructions of smooth partitions of unity to help the convergence; the point is that you can ensure these functions decay to zero sufficiently fast.

Edit: There is a proof in Wikipedia, actually. See here.

And there is another question where proofs are sketched. In particular, in my answer I sketch a proof of Peano that is different from the standard proof I reference above.

Solution 2:

This is a famous theorem of Borel. If you have Hormander's "The analysis of partial differential operators I" it's Theorem 1.2.6 there. He proves the result in any dimension. The basic idea is to write a sum of functions $\sum_n {a_n \over n!}\phi(m_n x)x^n$ where $\phi(x)$ is $1$ near $x = 0$ and is equal to zero outside of a small set containing $0$. If the $m_n$ are chosen carefully then the sum will have the desired properties.