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Question about extentions of smooth functions

  1. Apr 9, 2008 #1
    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.

  2. jcsd
  3. Apr 9, 2008 #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.
    Last edited: Apr 9, 2008
  4. Apr 10, 2008 #3
    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

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

    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] ?
  5. Apr 11, 2008 #4


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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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