Question

Let $(a_n)_{n=0}^\infty$ be an arbitrary sequence in a real Banach space $X$. Does there exist a smooth function $f: \mathbb {R} \rightarrow X$ such that $f^{(n)}(0)=a_n$ for $n=0,1,2,\ldots $?

Answer 1

According to The Convenient Setting of Global Analysis (Kriegl and Michor), this is due to Wells (1973). The statement given is:

15.4. Borel's Theorem. [Wells, 1973]. Suppose a Banach space $E$ has $C^\infty_b$-bump functions. Then every formal power series with coefficients in $L_{\text{sym}}^n(E;F)$ for another Banach space $F$ is the Taylor-series of a smooth mapping $E \to F$.

In this case, $E = \mathbb{R}$ so you have $C^\infty_b$-bump functions.

References:

• The Convenient Setting of Global Analysis, Kriegl and Michor. MR1471480
• Differentiable functions on Banach spaces with Lipschitz derivatives, Wells. MR370640

Answer 2

The result you are aiming at should follow from the scalar case by using tensor products since $\mathscr C^\infty(\mathbb R,X)= \mathscr C^\infty (\mathbb R) \tilde{\otimes} X$ and $X^{\mathbb N_0} = \mathbb R^{\mathbb N_0} \tilde{\otimes} X$. Because of the nuclearity of $\mathscr C^\infty (\mathbb R)$ and $\mathbb R^{\mathbb N_0}$ the tensor norm does not matter and tensorizing a surjective (hence open) continuous linear operator with the identity leads again to a surjection.