description/proof of that for \(C^\infty\) function on open neighborhood, there exists \(C^\infty\) function on \(C^\infty\) manifold with boundary that equals function on possibly smaller neighborhood
Topics
About: \(C^\infty\) manifold
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
- 4: Note
Starting Context
- The reader knows a definition of \(C^\infty\) manifold with boundary.
- The reader knows a definition of \(C^k\) map between arbitrary subsets of \(C^\infty\) manifolds with boundary, Where \(k\) Includes \(\infty\).
- The reader admits the proposition that for any \(C^\infty\) manifold with boundary, 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 with boundary, there exists a \(C^\infty\) function on the whole manifold with boundary 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: Structured Description
Here is the rules of Structured Description.
Entities:
\(M\): \(\in \{\text{ the } C^\infty \text{ manifolds with boundary }\}\)
\(p_0\): \(\in M\)
\(U_{p_0}\): \(\in \{\text{ the open neighborhoods of } p_0 \text{ on } M\}\)
\(f\): \(: U_{p_0} \to \mathbb{R}\), \(\in \{\text{ the } C^\infty \text{ maps }\}\)
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Statements:
\(\exists \widetilde{f}: M \to \mathbb{R}, \exists N_{p_0} \in \{\text{ the neighborhoods of } p_0 \text{ on } M\} \text{ such that } N_{p_0} \subseteq U_{p_0} (f \vert_{N_{p_0}} = \widetilde{f} \vert_{N_{p_0}})\)
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2: Natural Language Description
For any \(C^\infty\) manifold with boundary, \(M\), any point, \(p_0 \in M\), any open neighborhood, \(U_{p_0} \subseteq M\), of \(p_0\), and any \(C^\infty\) map, \(f: U_{p_0} \to \mathbb{R}\), there is a \(C^\infty\) function, \(\widetilde{f}: M \to \mathbb{R}\), that equals the original function, \(f\), on a possibly smaller neighborhood, \(N_{p_0} \subseteq M\), of the point, \(p_0\).
3: Proof
Whole Strategy: Step 1: take 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, \(N_{p_0} \subseteq U_{p_0}\); Step 2: define \(\widetilde{f} (p) := b (p) f (p) \text{ for } p \in U_{p_0}; 0 \text{ for } p \notin U_{p_0}\); Step 3: see that \(\widetilde{f}\) satisfies the requirements.
Step 1:
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 with boundary, 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: take \(\{p_0\}\) as the closed subset and \(U_{p_0}\) as the open neighborhood.
Step 2:
Let us define \(\widetilde{f} (p) := b (p) f (p) \text{ for } p \in U_{p_0}; 0 \text{ for } p \notin U_{p_0}\).
Step 3:
\(\widetilde{f}\) is \(C^\infty\), because for each \(p \in U_{p_0}\), it is a product of \(C^\infty\) functions, \(C^\infty\); for each \(p \notin U_{p_0}\), \(p \notin supp \text{ } b\) 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, and so, there is a neighborhood of \(p\) contained in \(M \setminus supp \text{ } b\) in which \(b\) is \(0\), and so, \(\widetilde{f}\) is \(0\) there, \(C^\infty\); as \(\widetilde{f}\) is \(C^\infty\) on an open neighborhood of each point on \(M\), \(\widetilde{f}\) is \(C^\infty\).
On \(N_{p_0}\), \(\widetilde{f} = f\) because \(b = 1\) there.
4: Note
An expression like "\(f\) can be extended to \(\widetilde{f}\)" tends to be used, but \(\widetilde{f}\) is not really any extension of \(f\), because \(\widetilde{f}\) is not guaranteed to equal \(f\) on the whole \(U_{p_0}\).
