I want to take the weak gradient operator

$$ \begin{aligned} \nabla: L^2(\Omega) &\to W^{-1,2}(\Omega,\mathbb{R}^d) \\ \langle \nabla u, \phi \rangle_{W^{-1,2},W_0^{1,2}}&:=(u,\text{div}\phi)_{L^2}=\int_\Omega u(x) \,\text{div}\phi(x) ~\text{d}x \end{aligned} $$ for all $\phi \in W_0^{1,2}(\Omega,\mathbb{R}^d)$.

Assume $\nabla u=0$. Show that $u$ is constant.

I know that if I interpret $u$ as a tempered distribution and take the gradient as a distributional derivative it should be possible to conclude $u$ is constant. But I'd like to take the definition of the weak gradient operator above.

We have $$ \int_\Omega u(x) \, \text{div}\phi(x) ~dx=0 \text{ for all } \phi \in W_0^{1,2}(\Omega,\mathbb{R}^d).$$

If $d=1$ I could conclude $u=\text{const}$ by the fundamental lemma of calculus of variations. But in general $$ \int_\Omega u(x) \, (\partial_{1}\phi_1(x)+...+\partial_d \phi_d(x)) ~d(x_1,...,x_d)=0 \text{ for all } \phi \in W_0^{1,2}(\Omega,\mathbb{R}^d)$$ and I don't know if there is a similar theorem.

  • $\begingroup$ How would you do it for $d=1$? I don't see why this case should be special. The proof I know considers smooth approximations by convolution. $\endgroup$ – Michał Miśkiewicz Jul 8 '17 at 14:42
  • $\begingroup$ @MichałMiśkiewicz Thank you for your comment. When I look for the 'Fundamental Lemma of Calculus of Variations' I only find the version if $\int_a^b u(x) \phi'(x) \, dx=0$ for all $\phi \in C_0^\infty(a,b)$ then $u=\text{constant}$. Do you say that its proof can easily be adapted to higher dimensions? I have a look into this. $\endgroup$ – Fritz Jul 8 '17 at 15:18
  • $\begingroup$ (The basic version of) the fundamental lemma holds in all dimensions: if $u \in L^1(\Omega)$ and $\int_\Omega u \varphi = 0$ for all $\varphi \in C_c^\infty(\Omega)$, then $u = 0$ a.e. in $\Omega$. $\endgroup$ – Michał Miśkiewicz Jul 8 '17 at 17:28
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    $\begingroup$ You can find an answer here. $\endgroup$ – Michał Miśkiewicz Jul 8 '17 at 17:35

As posted by Miskiewicz, one can prove it with a mollification method. That means choosing a ball $B$ such that $\overline{B} \subset \Omega$ and a Dirac sequence $\{ \phi_\epsilon \}_{\epsilon>0}$. Then from $\nabla(q*\phi_\epsilon)=0$ in $B$ we can deduce that $q*\phi_\epsilon$ is constant in $B$. And $q * \phi_\epsilon \to q$ in $L^2(B)$ yields the statement.

Another proof can be done as follows:

Suppose we have a $u\in L^2(\Omega)$ such that $\nabla u=0$ in $W^{-1,2}(\Omega,\mathbb{R}^d)$ hence by definition of the gradient operator \begin{equation*} \int_{\Omega} u \text{div} \zeta \text{ d}x=0 \text{ for all } \zeta \in W_0^{1,2}(\Omega,\mathbb{R}^d) \end{equation*} and for all $\zeta \in W_0^{1,2}(\Omega,\mathbb{R}^d)$ \begin{equation*} \begin{aligned} 0=\sum_{i=1}^d \int_\Omega u_i \partial_i \zeta_i \text{ d}x &= \sum_{i=1}^d \langle [u_i], \partial_i \zeta_i \rangle_{W_0^{1,2}(\Omega)} \\ &= - \sum_{i=1}^d \langle \partial_i [u_i],\zeta_i \rangle_{W_0^{1,2}(\Omega)}. \end{aligned} \end{equation*}
By choosing successively test functions $\zeta \in C_c^\infty(\Omega,\mathbb{R}^d)$ in which $d-1$ components vanish, we have \begin{equation*} \partial_i [u_i] = 0 \text{ in } \mathcal{D}'(\Omega) \text{ for all } i \in \{1,2,...,d\} \end{equation*} hence $u_i=constant$ almost everywhere in $\Omega$ as $\Omega$ is connected.


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