5
$\begingroup$

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 $\partial^\alpha (v'|_{\Omega \backslash Z})$?

$\endgroup$
  • $\begingroup$ @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 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 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 at 13:08
1
$\begingroup$

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 $Z$.

$\endgroup$
  • $\begingroup$ 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 at 11:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.