Let $\Omega \subseteq \mathbb{R}^n$ be open, $p \in [1,\infty]$, $\alpha$ be a multi-index with $n$ entries. If $v,w\in L^p(\Omega)$ we call $w$ the weak-$\alpha$-derivative of $v$ if $$ \forall \varphi \in C_0^\infty (\Omega): \quad \int_\Omega w(x)\varphi(x) dx = (-1)^{|\alpha|} \int_\Omega v(x) \partial^\alpha\varphi(x) dx $$ Now in many cases we have that, if $w$ is the classical derivative $\partial^\alpha v$ almost everywhere, then it also is the weak-$\alpha$-derivative of $v$.

My question is now: does the converse hold? I.e. if $w$ is the weak-$\alpha$-derivative of $v$, can we choose other representatives $v',w'$ which are equivalent to $v,w$ resp. and a Lebesgue-measure-zero set $Z\subseteq \Omega$ such that $w'|_{\Omega\backslash Z}$ is the classical derivative $\partial^\alpha (v'|_{\Omega \backslash Z})$?

  • $\begingroup$ @Surb you're right, but since $1_\mathbb{Q} = 0$ in $L^p$, $0$ is in some sense the classical derivative of $1_\mathbb{Q}|_\mathbb{R}$. I updated the question to make it more precise. $\endgroup$ – Jakob B. Feb 24 '19 at 11:33
  • $\begingroup$ $x\mapsto |x|$ was an example of function s.t. there are no differentiable function $f$ s.t. $f(x)=|x|$. Now, if we try to extend this example, I wouldn't be surprise that an $W^{1,1}$ has just a continuous representative but not a differentiable one... If so, $W^{1,1}$ would be just the set $\{\text{derivable function}\}$. $\endgroup$ – Surb Feb 24 '19 at 13:07
  • $\begingroup$ But I can choose the same map as representative in $L^p(\Omega)$ (with bounded $\Omega$) and $Z=\{0\}$. Then $\DeclareMathOperator{\sgn}{sgn}\sgn|_{\Omega\backslash Z}$ is the classical derivative of that map on $\Omega\backslash Z$. $\endgroup$ – Jakob B. Feb 24 '19 at 13:08

It is true for $n=1$. In this case $W^{1,p}$ is the space of $p$-absolutely continuous functions, and these are differentiable a.e., which is even stronger than your requirement. For higher derivatives you can simply iterate.

It can fail for $n>1$ (of course Sobolev embedding theorems ensure that it still true if the function has enough weak derivatives). Let $\Omega$ be bounded, $(q_k)$ a dense subset of $\Omega$ and $$u(x)=\sum_{k=1}^\infty 2^{-k}\log\log(1+\|x-q_k\|^{-1}).$$ This limit exists in $W^{1,n}(\Omega)$ and is nowhere locally (essentially) bounded. In particular, whichever representative of $u$ and null set $Z$ you choose, there is always a set $\Omega'$ of full measure such that for all $x\in \Omega'$ there exists a sequence $(x_j)$ in $\Omega\setminus Z$ that converges to $x$ and satisfies $|u(x_j)|\to \infty$. Thus $u|_{\Omega\setminus Z}$ is discontinuous for each null set $Z$.

  • $\begingroup$ Thanks! But I still have some questions: (0) why are the summands well-defined? (1) why does the limit exist? (2) why does the limit lie in $W^{1,n}(\Omega)$? $\endgroup$ – Jakob B. Mar 5 '19 at 11:27

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