# Sobolev functions on $\mathbb{R}^N$ cannot be discontinuous on a $(N-1)$-dimensional submanifold

How can one prove (or where can I find a proof) that if $$u \in W^{1,p}(\Omega)$$, where $$\Omega \subset \mathbb{R}^N$$, then $$u$$ cannot have a $$(N-1)$$-manifold of discontinuity points?

• $u\in W^{1,p}$, $p\leq N$ can be discontinuous everywhere. Behavior on $(n-1)$- submanifold is a different story. – Piotr Hajlasz Apr 5 at 23:48
• @PiotrHajlasz Indeed. Could you point out a reference where the result is phrased accurately (and proved)? – Riku Apr 6 at 0:49
• I will write an answer soon. – Piotr Hajlasz Apr 10 at 16:03

Any function in $$W^{1,p}$$, $$p>N$$, has a continuous representative by the Sobolev embedding theorem so there is no issue here. However:

Proposition. There is a function $$f\in W^{1,p}$$, $$p\leq N$$, that is essentially discontinuous everywhere. In fact you can find a function such that the essential supremum on every open set is $$+\infty$$ and the essential infimum on every open set is $$-\infty$$.

Thus the answer to the question the way it is stated is no, the function can be essentially discontinuous everywhere.

There is however, a different point of view which shows that, in fact, a Sobolev function behaves nicely when restricted to an $$(N-1)$$-dimensional manifold and I will present two different approaches to it.

Approach 1. According to Theorem 2 p. 164 in  (I am referring to the first edition) any function $$f\in W^{1,p}$$ has a representative that is absolutely continuous on almost all lines. Here by a representative I mean a Borel function defined everywhere and equal to $$f$$ almost everywhere.

If $$M\subset\Omega$$ is an $$(N-1)$$-dimensional manifold, then almost all lines pass through almost all points on $$M$$ so we can define restriction of $$f$$ to $$M$$ by looking at values of $$f$$ at the points where the lines intersect with $$M$$. Such a restriction is called a trace. Therefore $$f$$ may be discontinuous on $$M$$, but still it behaves nicely on $$M$$.

Approach 2. Any Function in $$W^{1,p}$$, $$p>N$$, has a continuous representative by the Sobolev embedding theorem so there is no issue here.If $$f\in W^{1,N}$$, then $$f\in W^{1,p}_{\rm loc}$$ for ant $$1\leq p so it suffices to discuss the case $$1\leq p only.

The following result is Theorem 1 on p. 160 in .

Theorem 1. There is a representative of $$f\in W^{1,p}(\Omega)$$, $$1\leq p, $$\Omega\subset\mathbb{R}^N$$ that is $$p$$-quasicontinuous. That means for any $$\epsilon>0$$, there is an open set $$V\subset\Omega$$ with $$\operatorname{Cap}_p(V)<\epsilon$$ such that $$f|_{\Omega\setminus V}$$ is continuous.

Here $$\operatorname{Cap}_p$$ stands for the $$p$$-capacity.

Capacity is a certain outer measure. While I will not recall its definition I will explain how it is related to the Hausdorff measure. The next result is Theorem 4 p. 156 and and Theorem 3 p. 193 in .

Theorem 2. If $$1\leq p and $$\operatorname{Cap}_p(A)=0$$, then $$\mathcal{H}^s(A)=0$$ for all $$s>n-p$$. Moreover if $$A$$ is compact, then $$\operatorname{Cap}_1(A)=0$$ if and only if $$\mathcal{H}^{N-1}(A)=0$$.

Now Theorem 1 says that away of a set of arbitrarily small capacity, $$f$$ is continuous so the exceptional set has capacity zero and Theorem 2 says that this exceptional set has vanishing $$(N-1)$$-dimensional measure. Therefore is we consider an $$(N-1)$$-dimensional manifold $$M$$ in $$\Omega$$ this exceptional set has measure zero.

 L. C. Evans, R. F. Gariepy, Measure theory and fine properties of functions. Studies in Advanced Mathematics. CRC Press, Boca Raton, FL, 1992.

 P. Hajłasz, Non-linear elliptic partial differential equations .

• Approach 1 is very nice to have an insight. But can it be turned into a rigorous proof? I don't quite see how to do this because e.g. $f(x,y) = \chi_{\mathbb Q}(x)$ is discontinuous everywhere, but continuous along all vertical lines. – Skeeve Apr 11 at 8:24
• @Skeeve First of all I do not claim that $f$ restricted to $M$ is continuous only that it is well defined a.e. on $M$. Secondly $\chi_{\mathbb{Q}}$ is in fact continuous since it equals zero a.e. so you can find a continuous representative of $\chi_{\mathbb{Q}}$ in the class of functions equal a.e. – Piotr Hajlasz Apr 11 at 14:16
• sure, I agree that $\chi_{\mathbb Q}$ is a bad representative of $0$. But my comment was just to show that even if some function $g$ is continuous along lines then I don't see how to conclude that $g$ cannot be discontinuous on $M$. I mean, how do you proceed after constructing good representative $g$? – Skeeve Apr 11 at 14:30
• @Skeeve As I said, $f$ restricted to $M$ may be discontinuous, because traces are not necessarily continuous. Traces will be continuous on almost all $(N-1)$-dimensional manifolds if $p>N-1$. – Piotr Hajlasz Apr 11 at 14:32
• Excuse me, but the sentence This is how you need to understand that M cannot be the set of discontinuity of f still confuses me. I will try to understand it better, but if you add some details it would be helpful. – Skeeve Apr 11 at 14:42