# Why is $T: C^\infty(\mathbb{R}) \ni f \mapsto (a_n) = (f^{(n)}(0)) \in \mathbb{R}^\mathbb{N}$ surjective?

I am reading "Linear Algebra" by Takeshi Saito.

Let $$f \in C^\infty(\mathbb{R})$$.
Let $$T$$ be a mapping such that $$T: C^\infty(\mathbb{R}) \ni f \mapsto (a_n) = (f^{(n)}(0)) \in \mathbb{R}^\mathbb{N}$$.

$$C^\infty(\mathbb{R})$$ and $$\mathbb{R}^\mathbb{N}$$ are linear spaces.
$$T$$ is a linear mapping.

In this book, the author says that $$T$$ is surjective without a proof.

Why is $$T$$ surjective?

• I don't know if there's an easy way to see this. Borel's lemma should help in either case. – MSDG May 12 '19 at 10:09

## 1 Answer

The approach of @Monadologie works if you cut off the polynomials early enough. This is the proof presented on Wikipedia for Borel's lemma from a bird's view