A description/proof of that for \(C^\infty\) function on open neighborhood, there exists \(C^\infty\) function on manifold that equals function on possibly smaller neighborhood
Topics
About: \(C^\infty\) manifold
The table of contents of this article
Starting Context
- The reader knows a definition of \(C^\infty\) manifold.
- The reader knows a definition of \(C^\infty\) function on \(C^\infty\) manifold.
- The reader admits the proposition that for any \(C^\infty\) manifold, any closed subset, and any open neighborhood of the subset, there is a \(C^\infty\) bump function that is supported in the open neighborhood and is \(1\) on a neighborhood of the closed subset.
Target Context
- The reader will have a description and a proof of the proposition that for any \(C^\infty\) function on any point open neighborhood of any \(C^\infty\) manifold, there exists a \(C^\infty\) function on the whole manifold that equals the original function on a possibly smaller neighborhood of the point.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Description
For any \(C^\infty\) manifold, \(M\), any point, \(p_0 \in M\), any open neighborhood, \(U_{p_0}\), of \(p_0\), and any \(C^\infty\) function, \(f: U_{p_0} \to \mathbb{R}\), there is a \(C^\infty\) function, \(\tilde{f}: M \to \mathbb{R}\), that equals the original function, \(f\), on a possibly smaller neighborhood, \(N_{p_0}\), of the point, \(p_0\).
2: Proof
There is a \(C^\infty\) bump function, \(b: M \to \mathbb{R}\), that is supported in \(U_{p_0}\) and equals \(1\) on a possibly smaller neighborhood, \(N_{p_0} \subseteq U_{p_0}\), by the proposition that for any \(C^\infty\) manifold, any closed subset, and any open neighborhood of the subset, there is a \(C^\infty\) bump function that is supported in the open neighborhood and is \(1\) on a neighborhood of the closed subset. \(\tilde{f} (p) := b (p) f (p) \text{ for } p \in U_{p_0}; 0 \text{ for } p \notin U_{p_0}\), a \(C^\infty\) function, because for any \(p \in U_{p_0}\), it is a product of \(C^\infty\) functions, and for any \(p \notin U_{p_0}\), \(p \notin supp \text{ } b\) because \(supp \text{ } b \subseteq U_{p_0}\), and as \(supp \text{ } b\) is closed, \(M \setminus supp \text{ } b\) is open where \(p\) belongs, so, there is a neighborhood of \(p\) contained in \(M \setminus supp \text{ } b\) in which \(b (p) = 0\) because \(p\) is outside the support of \(b\), so \(\tilde{f} (p) = 0\). On \(N_{p_0}\), \(\tilde{f} = f\) because \(b = 1\) there.
3: Note
An expression like "\(f\) can be extended to \(\tilde{f}\)" tends to be used, but \(\tilde{f}\) is not really any extension of \(f\), because \(\tilde{f}\) is not guaranteed to equal \(f\) on the whole \(U_{p_0}\).
