Borel's theorem for Banach's space valued functions - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T06:54:42Z http://mathoverflow.net/feeds/question/106228 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106228/borels-theorem-for-banachs-space-valued-functions Borel's theorem for Banach's space valued functions Jan 2012-09-03T09:16:15Z 2012-09-03T14:24:45Z <p>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 $n=0,1,2,\ldots$?</p> http://mathoverflow.net/questions/106228/borels-theorem-for-banachs-space-valued-functions/106229#106229 Answer by Andrew Stacey for Borel's theorem for Banach's space valued functions Andrew Stacey 2012-09-03T09:49:20Z 2012-09-03T10:31:40Z <p>According to <em>The Convenient Setting of Global Analysis</em> (Kriegl and Michor), this is due to Wells (1973). The statement given is:</p> <blockquote> <p><strong>15.4. Borel's Theorem.</strong> [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 $E \to F$.</p> </blockquote> <p>In this case, $E = \mathbb{R}$ so you have $C^\infty_b$-bump functions.</p> <h3>References:</h3> <ul> <li><em>The Convenient Setting of Global Analysis</em>, Kriegl and Michor. <a href="http://www.ams.org/mathscinet-getitem?mr=1471480" rel="nofollow">MR1471480</a></li> <li><em>Differentiable functions on Banach spaces with Lipschitz derivatives</em>, Wells. <a href="http://www.ams.org/mathscinet-getitem?mr=370640" rel="nofollow">MR370640</a></li> </ul> http://mathoverflow.net/questions/106228/borels-theorem-for-banachs-space-valued-functions/106248#106248 Answer by Jochen Wengenroth for Borel's theorem for Banach's space valued functions Jochen Wengenroth 2012-09-03T14:24:45Z 2012-09-03T14:24:45Z <p>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.</p>