If open and connected , : integrable in for every open is weakly differentiable in and then is constant a.e. in
I don't even know where to begin with this one, the usual proof for when is differentiable does not apply since we don't really have a mean value theorem for this derivatives, so... I'm stuck. Any help is appreciated.
What does mean? this seems like an interesting problem and would like to try and help.
Okay, it's been a while since I posted this but I have an answer now so if anyone's interested here it is:
Originally Posted by Jose27
We're going to prove a generalization of the result, namely: Let (where and satisfies for all and ) and for all then and .
Let where where and let . Define then where and in when , and even more: where the derivative in the right hand side is the weak derivative (this identity follows at once from the identity and the definition of weak derivative).
The argument is inductive, so we start with the case :
Let and be small enough such that then and we have that and so for all which implies that is constant and so in (since in ). Since is connected we get constant a.e. in .
Assume the result for , we want to prove it for :
Assume for a moment that we know that if satisfies the hypothesis with the additional then with . Now let as in the hypothesis and then it's clear that there exists a subsequence such that for all (by the case ), but then and for all and so and , but since we get that and . This gives that and . This proves that is a polynomial in , but in particular it is analytic and is connected, so if we cover with compactly contained sets (countably many), we get that each polynomial representation of is identical in every subset (since polynomials are analytic, intersection of (finitely) open sets is open and the fact that is connected).
Now to prove the first part of the last paragraph it is sufficient to look at where . Let and take as assumed, for fixed let where (at his point it's obvious we're using an inductive argument on the dimension and the case is trivial). Since we get that . The smoothness of implies the smoothness of the coefficients . Now if with then , so by induction are polynomials in variables with . This gives us that is a polynomial in variables with , but the condition for all gives that .
Any comments and suggestions are appreciated.
Edit: It is worth noting that in the thoery of Sobolev spaces there are some useful embeddings for some (like, for example, if we have this last one being the space of Hölder cont. functions with exponent ) with certain properties, and one corollary of these embeddings is that if for all then which makes most of this proof unnecessary.