Smooth Extension of Locally Defined Function on Manifold

[hedipaldi]

how to extend a locally defined function to a smooth function on the whole manifold ,by using a bump function?

 

[Bacle2]

AFAIK, partitions of unity are used to extend. I have not seen bump functions, since I think these are defined on ℝⁿ, and are real-valued. Could you tell us the method you are referring to?

 

[hedipaldi]

Hi,
i quote from the text:If F is a smooth function on a neighbourhood of x,we can multiply it by a bump function to extend it to M ".here M is a differential manifold so there are local coordinates at each point.F is a real function.
i don't see how such an extention is done.
thank's
Hedi

 

[quasar987]

Hi,
i quote from the text:If F is a smooth function on a neighbourhood of x,we can multiply it by a bump function to extend it to M ".here M is a differential manifold so there are local coordinates at each point.F is a real function.
i don't see how such an extention is done.
thank's
Hedi

For instance, say F is defined on a coordinate chart U diffeomorphic to R^n. Let D(1) be the closed subset of U corresponding to the closed unit disk under the coordinate map, and let B(2) be the open subset of U corresponding to the open disk of radius 2 under the coordinate map. Then you know there is a smooth bump function b that is equal to 1 on all of D(1), and has support in B(2), right? So define F', an "extension" of F, by setting

F'(x):=b(x)F(x) if x is in U
F'(x)=0 otherwise

It is obvious that F' is smooth since it is smooth on U and it is smooth "away from U": the only potentially problematic points are at the boundary of U. But at those points, F' is locally identically 0 by construction so all is well.

Now F' is not an extension of F in the strictest sense since we have modified it outside of D(1). But for many purposes, this is just fine.

 

[hedipaldi]

It was very helpfull indeed.Thank"s a lot.
Hedi