Let $\Omega\subseteq\mathbb R^n$ be open. $u\in\mathcal L^1_\text{loc}(\Omega)$ is called weakly differentiable $:\Leftrightarrow$ $\exists v\in\mathcal L^1_\text{loc}(\Omega;\mathbb R^n)$ with $$\int_\Omega\psi v\;d\lambda^n=-\int_\Omega u\nabla\psi\;\;\;\text{for all }\psi\in C_c^1(\Omega)\;.$$ Now, let $u\in W_0^{1,2}(\Omega)=:H$. I want to show, that $u$ is weakly differentiable. By definition, $H$ is the completion of $C_c^1(\Omega)$ with respect to the norm $$\left\|\psi\right\|_H^2:=\int_\Omega|\nabla\psi|^2+\psi^2\;d\lambda^n\;\;\;\text{for }\psi\in C_c^1(\Omega)\;.$$ So, there is a sequence $\left(\psi_k\right)_{k\in\mathbb N}\subseteq C_c^1(\Omega)$ with $$\psi_k\stackrel H{\to}u\;.\tag{1}$$ Let $\psi\in C_c^1(\Omega)$. Then, $$\left\|\left(u-\psi_k\right)\frac{\partial\psi}{\partial x_i}\right\|_{L^2(\Omega)}\le\left\|u-\psi_k\right\|_{L^2(\Omega)}\left\|\frac{\partial\psi}{\partial x_i}\right\|_{L^2(\Omega)}\stackrel{k\to\infty}{\to}0\;.$$ Hence, $$-\int_\Omega u\frac{\partial\psi}{\partial x_i}\;d\lambda^n=-\lim_{k\to\infty}\int_\Omega\psi_k\frac{\partial\psi}{\partial x_i}\;d\lambda^n\;\;\;\text{in }L^2(\Omega)\;.\tag{2}$$ How do I need to proceed from here? Clearly, by $(1)$ we know that $\nabla\psi_k$ converges in $L^2(\Omega;\mathbb R^n)$ agains some $v$.

  • $\begingroup$ Who told you that every $L^2$ function was weakly differentiable? It's not so. $\endgroup$ Jul 21, 2015 at 19:17
  • $\begingroup$ @DavidC.Ullrich You're right. I was on the wrong track. Actually, I want to show, that each $H:=W_0^{1,2}(\Omega)$-function is weakly differentiable. Many authors define $H$ to be the space of all $L^2(\Omega)$ functions which have a weak derivative $\nabla u\in L^2(\Omega;\mathbb R^n)$. However, I use the (equivalent) definition, where $H$ is the completion of $C_c^1(\Omega)$ with respect to the Sobolev norm and want to verify, that the definitions are equivalent. $\endgroup$
    – 0xbadf00d
    Jul 21, 2015 at 19:29
  • $\begingroup$ You have a sequence $\psi_k$ converging to $u$ in $\lVert\,\cdot\,\rVert_H$. That means not only $(\psi_k)$ is an $L^2$-Cauchy sequence, but also $(\nabla \psi_k)$ is Cauchy in $L^2$. $\endgroup$ Jul 21, 2015 at 19:32
  • $\begingroup$ The sequence $\psi_k$ tends to $u$ in $L^2(\Omega)$, great. That's only using half of the information you have. Since the sequence $\psi_k$ iis Cauchy in $H$ it follows that the sequence $(\nabla \psi_k)$ is a Cauchy sequence in $L^2$. So there is $v$ in $L^2$ such that $\nabla\psi_k\to v$ in $L^2$. Now show that $v$ is locally inetegrable and satisfies the definition of weak differentiability... $\endgroup$ Jul 21, 2015 at 19:35
  • $\begingroup$ Since $v$ is an element of $L^2(\Omega;\mathbb R^n)$ it's trivially locally integrable. But I don't see how I can use $\nabla\psi_k\to v$ in $(2)$. $\endgroup$
    – 0xbadf00d
    Jul 21, 2015 at 20:18

1 Answer 1


Please do not use the same symbol for your function and test functions. You are confusing yourself.

We assume $u\in H$ and $(u_n)\subset C_c^1$ such that $u_n\to u$ in $H$, which in particular means that $u_n\to u$ in $L^2$ and $\partial_i u_n\to v_i$ in $L^2$ for each $i=1,\ldots, N$. Then we observe that, for any $\varphi\in C_c^1$, $$ \int_\Omega u\,\partial_i \varphi\,dx=\lim_{n\to\infty}\int_\Omega u_n\,\partial_i \varphi\,dx = \lim_{n\to \infty}-\int_\Omega\partial_i u_n\,\varphi\,dx = -\int_\Omega v_i\,\varphi\,dx. $$ Both first and last equality are justified by $L^2$ convergence. Finally by definition of weak derivative, you know $v_i$ is $i$-partial weak derivative of $u$.


You must log in to answer this question.