Borel's theorem for Banach's space valued functions - MathOverflow

Borel's theorem for Banach's space valued functions

2012-09-03T09:16:15Z

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 by Andrew Stacey for Borel's theorem for Banach's space valued functions

2012-09-03T09:49:20Z

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 by Jochen Wengenroth for Borel's theorem for Banach's space valued functions

2012-09-03T14:24:45Z

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.