Question about extentions of smooth functions

[Aleph-0]

My question is simple :

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

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

Thanks

 

[Pere Callahan]

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.

If you were to replace [0,1] by $\mathbb{R}$, then I'd be with you that the assertion should hold true.

 

[Aleph-0]

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

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

exists and is in $$R$$.


Is there $$g$$ in $$C^\infty(R^n,[-\epsilon, 1+\epsilon])$$ such that $$f=g$$ on $$U$$ ?

 

[mathwonk]

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.