Question about extentions of smooth functions

  • Thread starter Aleph-0
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  • #1
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My question is simple :

Suppose that [tex]f[/tex] is in [tex]C^\infty(U, [0,1])[/tex] where [tex]U[/tex] is an open of [tex]R^n[/tex] .
Is there [tex]g[/tex] in [tex]C^\infty(R^n,[0,1])[/tex] such that [tex]f=g[/tex] on [tex]U[/tex] ?

I would say yes, but I don't know how to prove it.

Thanks
 

Answers and Replies

  • #2
Did you try it for the special case n=1?

Take for example U=(0,1) and f:U->[0,1] defined by f(x)=x.

This satisfies your conditions, can you extend it to a C^\infty function on R...?

I'd say no but the proof is left to you. :smile:

If you were to replace [0,1] by [itex]\mathbb{R}[/itex], then I'd be with you that the assertion should hold true.
 
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  • #3
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OK, I guess I should add some hypothesis on f and change the conclusion in order to have something that could be true:

Here is the new problem :

Let [tex]\epsilon>0[/tex].
Suppose that [tex]f[/tex] is in [tex]C^\infty(U, [0,1])[/tex] where [tex]U[/tex] is an open of [tex]R^n[/tex] and suppose that for any [tex] x_0 [/tex] in [tex]\partial U[/tex] (the boundary of U), and any n-multi-indice [tex]\alpha[/tex], the limit

[tex]
\lim_{x\in U, x\to x_0} \partial^\alpha f (x)
[/tex]

exists and is in [tex]R[/tex].


Is there [tex]g[/tex] in [tex]C^\infty(R^n,[-\epsilon, 1+\epsilon])[/tex] such that [tex]f=g[/tex] on [tex]U[/tex] ?
 
  • #4
mathwonk
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i would try reflecting f to the other side of the ball U, then using that to extend f a little bit, then chopping f off by a smooth bump function.

at least if U is a nice ball.
 

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