I am working with Sobolev spaces. Let's suppose $\Omega \subset \mathbb{R}^n$ is an open set.

A function $u: \mathbb{R}^n \to \mathbb{R}$ in $L^1(\Omega)$ is said to be weakly differentiable if there exist functions $ g_1,...,g_n $ such that $$\qquad\qquad\qquad\qquad\qquad\int_{\Omega}u\varphi_{x_i}=-\int_{\Omega}g_i \varphi \quad \quad \forall \varphi \in C^{1}_c(\Omega), \forall i=1,...,n. $$

Can every weakly differentiable function be restricted to an open set $\tilde{\Omega}$ such that $u$ is differentiable (classical derivative!) with $m(\Omega - \tilde{\Omega})=0$?

I know this is true for dimension $1$. Is it true for every $\Omega \subset \mathbb{R}^n$?


This is not true for dimension $n>1$. In the book "Some Applications of Functional Analysis in Mathematical Physics" by Sobolev the following example is constructed. Consider function $\varphi(x,y)=f_1(x)+f_2(y)$, where $f_1$ and $f_2$ are continuous on $\mathbb{R}$ nowhere differentiable functions. Then $\varphi$ doesn't have strong (classical) derivatives, but weak derivative $\frac{\partial^2\varphi}{\partial x\partial y}$ exists and equals $0$ on every $\Omega=(a,b)\times(c,d)$. Indeed, $$ \int\limits_\Omega\varphi\frac{\partial^2\psi}{\partial x\partial y}d\mu = \int\limits_\Omega f_1(x)\frac{\partial^2\psi}{\partial x\partial y}d\mu + \int\limits_\Omega f_2(y)\frac{\partial^2\psi}{\partial x\partial y}d\mu. $$ But $\frac{\partial\psi}{\partial x}=\frac{\partial\psi}{\partial y}=0$ on the border $\overline{\Omega}\setminus\Omega$ of $\Omega$, so $$ \int\limits_{\Omega}f_1(x)\frac{\partial^2\psi}{\partial x\partial y}d\mu = \int\limits_a^bf_1(x)\int\limits_c^d \frac{\partial^2\psi}{\partial x\partial y}dy\, dx = \int\limits_a^bf_1(x) \frac{\partial\psi}{\partial x}\Bigg|_{y=c}^{y=d} dx =0 $$ and the same way we get $$ \int\limits_{\Omega}f_2(y)\frac{\partial^2\psi}{\partial x\partial y}d\mu = 0. $$ Therefore $$ \int\limits_{\Omega}\varphi\frac{\partial^2\psi}{\partial x\partial y}d\mu = 0. $$

  • $\begingroup$ Could you explain a little bit why the derivative of $\varphi$ is $0$? I understand intuitivelly that $\varphi$ has no "slope" on the directions $x$ and $y$. $\endgroup$
    – D1X
    Jul 6 '16 at 18:13
  • $\begingroup$ It is not clear if first weak derivative exists for your function. $\endgroup$
    – user99914
    Jul 12 '16 at 19:12
  • $\begingroup$ @ArcticChar, first weak derivatives in this case don't exist, because weak $\frac{\partial\varphi}{\partial x}$ equals weak $\frac{\partial f_1}{\partial x}$ and latter doesn't exist since it's 1-dimension case and $f_1$ is not differentiable a.e. Yeah... I see how my post doesn't answer the question. Will think about it. $\endgroup$
    – Glinka
    Jul 12 '16 at 23:17

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