1. ## 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.

2. What does $D_iu$ mean? this seems like an interesting problem and would like to try and help.

3. Originally Posted by putnam120
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.

4. Originally Posted by Jose27
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. 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.