- #1

- 210

- 0

- Thread starter hedipaldi
- Start date

- #1

- 210

- 0

- #2

Bacle2

Science Advisor

- 1,089

- 10

- #3

- 210

- 0

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

- #4

quasar987

Science Advisor

Homework Helper

Gold Member

- 4,783

- 18

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

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

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.

- #5

- 210

- 0

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

Hedi

Hedi