# Is a weak derivate of $f$ always a classical derivative of some $g$?

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

• @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. – Jakob B. Feb 24 at 11:33
• $x\mapsto |x|$ 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 $\{\text{derivable function}\}$. – Surb Feb 24 at 13:07
• 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 $\Omega\backslash Z$. – Jakob B. Feb 24 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 $$Z$$.
• 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 $W^{1,n}(\Omega)$? – Jakob B. Mar 5 at 11:27