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Math Help - Weak Derivatives

  1. #1
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    Weak Derivatives

    If \Omega \subset \mathbb{R} ^n open and connected , u \in L_{loc}^1 ( \Omega) = \{ u: \Omega \longrightarrow \mathbb{R} : u integrable in \Omega _1 for every open \Omega _1 \hspace{2 mm} such \hspace{2 mm} that \hspace{2 mm} \overline{ \Omega_1} \subset \Omega \hspace{2 mm} is \hspace{2 mm} compact \hspace{2 mm} in \hspace{2 mm} \Omega \} is weakly differentiable in \Omega and D_{i} u =0 \forall i=1, ... , n then u is constant a.e. in \Omega

    I don't even know where to begin with this one, the usual proof for when u 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.
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  2. #2
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    What does D_iu mean? this seems like an interesting problem and would like to try and help.
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  3. #3
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    Quote Originally Posted by putnam120 View Post
    What does D_iu mean? this seems like an interesting problem and would like to try and help.
    D_{i} u is the i-eth (spell?) weak derivative of u ie. it is a function v_i \in L_{loc} ^1 ( \Omega) such that \int_{ \Omega} u \frac{ \partial \phi}{ \partial x_i } = - \int_{ \Omega} v_{i} \phi for all \phi \in C_{c} ^{\infty} ( \Omega) (the space of all infinitely differentiable function to \mathbb{R} with compact ( in \Omega ) support).

    The weak derivatives satisfy the common propierties of the partial derivatives, ie, it's unique(modulo the relation of being equal a.e.), the weak derivative of a sum is the sum of the derivatives, etc.
    Last edited by Jose27; June 8th 2009 at 07:33 PM.
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  4. #4
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    Quote Originally Posted by Jose27 View Post
    If \Omega \subset \mathbb{R} ^n open and connected , u \in L_{loc}^1 ( \Omega) = \{ u: \Omega  \longrightarrow \mathbb{R} : u integrable in \Omega _1 for every open \Omega _1 \hspace{2 mm} such  \hspace{2 mm} that \hspace{2 mm} \overline{ \Omega_1} \subset \Omega  \hspace{2 mm} is \hspace{2 mm} compact \hspace{2 mm} in \hspace{2 mm}  \Omega \} is weakly differentiable in \Omega and D_{i} u =0 \forall i=1, ... , n then u is constant a.e. in \Omega

    I don't even know where to begin with this one, the usual proof for when u 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.
    Okay, it's been a while since I posted this but I have an answer now so if anyone's interested here it is:

    We're going to prove a generalization of the result, namely: Let u\in W_{loc} ^{k,p} (\Omega ) := \{ u\in L_{loc}^p(\Omega ) :  D^{\alpha } u \in L_{loc}^p (\Omega) \ \mbox{for all} \ |\alpha| \leq k  \} (where \alpha =(\alpha_1,...,\alpha _n) \in \mathbb{N}  ^n and D^{\alpha } u satisfies \int _V uD^{\alpha }\phi  =(-1)^{| \alpha |}\int_V D^{\alpha }u\phi for all \phi \in  C_0^{\infty } (V) and V \subset \subset \Omega ) and D^{\alpha } u =0 for all |\alpha | =k then u \in \mathbb{R} [x_1,...,x_n] and \deg (u) \leq  k-1.

    Let \eta : \mathbb{R} ^n \rightarrow  \mathbb{R} where \eta (x) = ce^{\frac{1}{\| x \| ^2  -1}} where c = \int_{\mathbb{R} ^n} \eta and let \eta _{\varepsilon } (x)= \varepsilon ^{-n} \eta \left(  \frac{x}{\varepsilon } \right). Define u_{\varepsilon } = u  \ast \eta _{\varepsilon } then u_{\varepsilon } \in  C^{\infty } (\Omega _{\varepsilon } ) where \Omega  _{\varepsilon } = \{ x\in \Omega : d(x, \partial \Omega ) > \epsilon  \} and u_{\varepsilon } \rightarrow u in L_{loc} ^p ( \Omega ) when \varepsilon \rightarrow  0, and even more: D^{\alpha } (u_{\varepsilon }) =  (D^{\alpha } u)_{\varepsilon } where the derivative in the right hand side is the weak derivative (this identity follows at once from the identity D_x^{\alpha } \eta_{\varepsilon } (x-y) = (-1)^{|\alpha  |}D_y^{\alpha } \eta _{\varepsilon } (x-y) and the definition of weak derivative).

    The argument is inductive, so we start with the case k=1:

    Let V \subset \subset  \Omega and \varepsilon >0 be small enough such that \overline{V} \subset \Omega _{\varepsilon } then u_{\varepsilon } \in W^{1,p} (V) and we have that (D^iu)_{\varepsilon } = D^i(u_{ \varepsilon }) and so D^i(u_{\varepsilon}) =0 for all i=1,...,n which implies that u_{\varepsilon } is constant and so u=c in V (since u_{\varepsilon  }  \rightarrow u in L^p (V)). Since \Omega is connected we get u constant a.e. in \Omega.

    Assume the result for m\leq k, we want to prove it for k+1:

    Assume for a moment that we know that if f: \Omega \rightarrow \mathbb{R} satisfies the hypothesis with the additional f\in C^{\infty }  (\Omega ) then f\in \mathbb{R} [x_1,...,x_n] with \deg (f) \leq k-1. Now let u \in W_{loc}^{k+1,p}  (\Omega ) as in the hypothesis and V \subset \subset  \Omega then it's clear that there exists a subsequence \varepsilon _l \rightarrow  0 such that D^{\beta}u_{\varepsilon _l}=\mbox{constant} = D^{\beta }u for all |\beta |= k (by the case k=1), but then u-u_{\varepsilon _l} \in W^{k,p}(V) and D^{\beta  }(u-u_{\varepsilon _l})=0 for all |\beta |=k and so u-u_{\varepsilon _l} \in \mathbb{R} [x_1,...,x_n] and \deg (u-u_{\varepsilon _l}) \leq k-1, but since D^{\alpha } (u_{\varepsilon } )= (D^{\alpha }u)_{\varepsilon  } we get that u_{\varepsilon _l} \in \mathbb{R}  [x_1,...,x_n] and \deg (u_{\varepsilon _l}) \leq k. This gives that u\in \mathbb{R} [x_1,...,x_n] and \deg (u) \leq k. This proves that u is a polynomial in V, but in particular it is analytic and \Omega is connected, so if we cover \Omega with compactly contained sets (countably many), we get that each polynomial representation of u is identical in every subset (since polynomials are analytic, intersection of (finitely) open sets is open and the fact that \Omega is connected).

    Now to prove the first part of the last paragraph it is sufficient to look at V\subset \subset \Omega where V=\{ (a_1,b_1)  \times ...\times (a_n,b_n) : a_i,b_i \in \mathbb{R}, \ a_i<b_i  \}. Let x=(x_1,x_2)\in \mathbb{R} ^{n-1} \times  \mathbb{R} and take f as assumed, for fixed x_1 let f_{x_1}(x_2) = f(x_1,x_2) where f_{x_1} : (a_n,b_n) \rightarrow \mathbb{R} (at his point it's obvious we're using an inductive argument on the dimension and the case n=1 is trivial). Since f_{x_1}^{(k)} \equiv  0 we get that f_{x_1} (x_2)= \sum_{j=1}^{k-1}  c_j(x_1)x_2^{j}. The smoothness of f implies the smoothness of the coefficients c_j(x_1). Now if |\beta | =k with \beta \in \mathbb{N} ^{n-1} then \sum_{j=1}^{k-1}x_2^jD^{\beta }c_j(x_1)= D^{\beta } f  =0, so by induction c_j(x_1) are polynomials in n-1 variables with \deg (c_j) \leq k-1. This gives us that f is a polynomial in n variables with \deg (f) \leq 2(k-1), but the condition D^{\alpha } f=0 for all |\alpha | =k gives that \deg (f) \leq k-1.

    Any comments and suggestions are appreciated.

    Edit: It is worth noting that in the thoery of Sobolev spaces W^{k,p}(\Omega) there are some useful embeddings for some \Omega (like, for example, if n>p we have W^{k,p}(\mathbb{R} ^n) \hookrightarrow C^{0,\gamma }(\mathbb{R} ^n) this last one being the space of Hölder cont. functions with exponent \gamma =1-\frac{n}{p}) with certain properties, and one corollary of these embeddings is that if u\in W^{k,p}(\Omega ) for all k\in \mathbb{N} then u\in C^{\infty }(\Omega ) which makes most of this proof unnecessary.
    Last edited by Jose27; April 25th 2010 at 07:11 PM.
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