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Let $U \subseteq \mathbb{R}^n$ be an open set. Let $f \in W^{1,p}(U)$ and suppose all the weak derivatives of $f$ are continuous. Is $f$ itself continuous?

It is a classic fact that if we assume a-priori $f$ is also continuous, then it is in $C^1$.

Here, however I do not assume $f$ is continuous.

Asaf Shachar
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2 Answers2

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You can show the continuity of $f$ as follows:

If $p = \infty$, then $f$ is Lipschitz continuous and you are done.

In case $p < \infty$, you can consider an open subset $O$ with $\overline O \subset U$. Then, $\overline O$ is compact and the continuous functions $D f$ are bounded on $\overline O$. Now, you can use the Sobolev embedding theorem to get $f \in W^{1,\infty}(O)$. Again, $f$ is Lipschitz on $O$ and (since $O$ was arbitrary), $f$ is continuous on $U$.

Now you know that $f$ is continuous and you can use the technique from the other question to show that $f$ is actually $C^1$.

gerw
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  • Yes, but the fact that if $p=\infty$ then $f$ is Lipschitz isn't quite trivial. – David C. Ullrich May 14 '18 at 13:38
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    Thanks. By the way for the case $p < \infty$ Morrey's inequality suffices to show $f$ is continuous. That is you only need to establish $f \in W^{1,p}_{\text{loc}}$ for some $p>n$. – Asaf Shachar May 14 '18 at 14:15
  • Hi, I came back to this question now, and I think I am suddenly missing something: When $p>n$ the continuity of $f$ follows from Morrey's inequality. However, I don't understand how do you use Sobolev embedding when $p \le n$? (What version of the theorem are you using?) Thanks. – Asaf Shachar Jun 21 '18 at 05:36
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    You have $f \in W^{1,p}(O)$ and $Df \in L^\infty(O)$. Now, the Sobolev embedding theorem implies $f \in L^q(O)$ for a suitable $q > p$. Together with the regularity of the gradient, you get $f \in W^{1,q}(O)$. Now, you can re-iterate to further increase the regularity of $f$. In fact, you need finitely many applications of the embedding theorem (best think in inverses of the exponents: you start with $1/p$ and each applications decreases this number by $1/n$. After finitely many iterations $k$, $1/p - k/n$ is negative and this means the function is continuous). – gerw Jun 21 '18 at 06:56
  • Hmm..., thanks but I think it is a bit more subtle than that. The problem is that you might reach exactly $p=n$, and then you are stuck: it is not clear you are continuous yet, but you cannot proceed another step. Take for example $p= \frac{n}{2}$. Then, after one application of Sobolev's theorem, you get... – Asaf Shachar Jun 21 '18 at 12:35
  • $p^*=n$, so all you have established is $f \in W^{1,n}$, which is not good enough. However, since this question is local, I guess you can always decrease your original exponent $p$ slightly, thus making sure the sequence $1/p - k/n$ ($k=1,2,...$) is never zero. This makes sure you don't "stop accidentally" at the badly-behaved border $p=n$. I think this settles the matter. Thanks again for your time and effort. (I also thought about re-iteration, but somehow I wrongly imagined that you could iterate infinitely many times without surpassing $n$. I now see this is "essentially impossible"). – Asaf Shachar Jun 21 '18 at 12:35
  • $p = n$ is not so bad. In fact this implies $f \in W^{1,q}$ for all $q \in (1,\infty)$. Thus you can do a step to a point $q > n$ and then obtain continuity in the next step. – gerw Jun 21 '18 at 12:49
  • Can you please elaborate? I think that in general there are non-continuous $W^{1,n}$ functions. I do not see how do you "take the next step" from there (using the bounded derivative assumption of course) except by lowering the exponent and then raising it again... (Sobolev inequality is for $p<n$, and Morrey's one is for $p>n$). – Asaf Shachar Jun 21 '18 at 15:22
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    The Sobolev embedding theorem in case $p = n$ implies $f \in L^q(O)$ for all $q \in (1,\infty)$. Together with the gradient, this implies $f \in W^{1,q}(O)$ for all $q \in (1,\infty)$. – gerw Jun 21 '18 at 17:08
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It depends on exactly what we mean by "$f$ is continuous". Taken literally, the answer is no: If $$f(x)=\begin{cases}1,&(x=0), \\0,&(x\ne0)\end{cases}$$then $f$ is not continuous, but the weak derivatives of $f$ equal $0$.

Typically in this sort of context when one says $f$ is continuous one means that $f=g$ almost everywhere, where $g$ is continuous. If that's what we mean then the answer is yes.

Note, inspired by a comment: One has to be careful about jumping to conclusions; the continuity condition above has more or less nothing to do with being continuous almost everywhere! For ease of reference let's label the two:

(i) $f=g$ almost everywhere for some continuous function $g$.

(ii) $f$ is continuous almost everywhere.

On $\Bbb R$, the function $\chi_{(0,\infty)}$ shows that (ii) does not imply (i), while $\chi_{\Bbb Q}$ shows that (i) does not imply (ii).

Exercise investigate the implications, if any, between (i), (ii) and a third version:

(iii) There exists a set $E$ of full measure such that $f_E$ is continuous (wrt to the subspace topology on $E$).

David C. Ullrich
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  • Yeah, I meant continuous almost everywhere (more formally there exist a continuous representative, which is unique if exists). Regarding your comment above on the case $p=\infty$, you can see my added comment about the case $p < \infty$. – Asaf Shachar May 14 '18 at 14:17
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    "more formally"? No, the second condition is not a more formal statement of the first. See edit... – David C. Ullrich May 14 '18 at 14:33