# 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 [1] (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 [1].

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 [1].

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.

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

[2] 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