2022-05-08

71: For C Function on Open Neighborhood, There Exists C Function on C Manifold with Boundary That Equals Function on Possibly Smaller Neighborhood

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description/proof of that for C function on open neighborhood, there exists C function on C manifold with boundary that equals function on possibly smaller neighborhood

Topics


About: C manifold

The table of contents of this article


Starting Context



Target Context


  • The reader will have a description and a proof of the proposition that for any C function on any point open neighborhood of any C manifold with boundary, there exists a C 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: { the C manifolds with boundary }
p0: M
Up0: { the open neighborhoods of p0 on M}
f: :Up0R, { the C maps }
//

Statements:
f~:MR,Np0{ the neighborhoods of p0 on M} such that Np0Up0(f|Np0=f~|Np0)
//


2: Natural Language Description


For any C manifold with boundary, M, any point, p0M, any open neighborhood, Up0M, of p0, and any C map, f:Up0R, there is a C function, f~:MR, that equals the original function, f, on a possibly smaller neighborhood, Np0M, of the point, p0.


3: Proof


Whole Strategy: Step 1: take a C bump function, b:MR, that is supported in Up0 and equals 1 on a possibly smaller neighborhood, Np0Up0; Step 2: define f~(p):=b(p)f(p) for pUp0;0 for pUp0; Step 3: see that f~ satisfies the requirements.

Step 1:

There is a C bump function, b:MR, that is supported in Up0 and equals 1 on a possibly smaller neighborhood, Np0Up0, by the proposition that for any C manifold with boundary, any closed subset, and any open neighborhood of the subset, there is a C bump function that is supported in the open neighborhood and is 1 on a neighborhood of the closed subset: take {p0} as the closed subset and Up0 as the open neighborhood.

Step 2:

Let us define f~(p):=b(p)f(p) for pUp0;0 for pUp0.

Step 3:

f~ is C, because for each pUp0, it is a product of C functions, C; for each pUp0, psupp b because supp bUp0, and as supp b is closed, Msupp b is open where p belongs, and so, there is a neighborhood of p contained in Msupp b in which b is 0, and so, f~ is 0 there, C; as f~ is C on an open neighborhood of each point on M, f~ is C.

On Np0, f~=f because b=1 there.


4: Note


An expression like "f can be extended to f~" tends to be used, but f~ is not really any extension of f, because f~ is not guaranteed to equal f on the whole Up0.


References


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