5

I’m sorry if this has already been answered; I couldn’t find a clear answer (maybe there is an indirect one and I failed to make the connection).

In the context of numerical analysis / finite elements methods / non-conforming numerical methods, I often hear non-rigorous statements like “$H^1$ basically means continuous", "$H^1$-conforming functions are in practice continuous"... But why is that? I know this is true in $1$D, but we talk about general dimension, most of the times $2$D or $3$D problems, so I don't see it.

Trying to ask around, I've been told that for $H^1$ functions, jumps are 0 almost everywhere, so, if you also restrict to (piecewise) polynomial spaces, as in finite elements, these two facts together force continuity everywhere. I don’t know how to prove that last implication, but it sounds plausible when I try to visualize it.

However I am confused about the claim itself. (Just to be sure, does “zero jumps almost everywhere” mean that an $H^1$ function can have jumps on sets of measure zero, right?) Why is that true? i.e. why for $H^1$ functions, jumps are 0 almost everywhere?
Also which measure is meant here exactly — Lebesgue measure in $\mathbb{R}^d$? In $2$D, for instance, an interior mesh edge is a curve, hence it has $2$D Lebesgue measure 0. Or maybe the Hausdorff measure?

If this claim is false, misleading, or not the right way to understand the link between $H^1$ and continuity, feel free to address the problem in other ways.

I realize my background is limited; a rigorous and clear explanation would be very much appreciated. Thanks!

alep17
  • 51
  • 1
    Well $H^1$ would mean that all first order partial derivatives are square integrable functions and not deltas, which is what you might need for a 'jump'? But there are unbounded functions of the type $1/|x|^\alpha$ that are in $H^1$ once dimension > 1 – Calvin Khor Sep 29 '25 at 19:07
  • So, it's not even true that an $H^1$ function can have jumps in some points, right? $H^1$ functions must have no jumps at all? They can however be discontinuous because unbounded yes, I see that, thank you. – alep17 Sep 30 '25 at 09:42
  • my previous comment was up to a.e. equivalence (and yes Lebesgue measure). Perhaps also related - https://en.wikipedia.org/wiki/Sobolev_space#Absolutely_continuous_on_lines_(ACL)_characterization_of_Sobolev_functions – Calvin Khor Sep 30 '25 at 13:58

2 Answers2

4

"$H^1$-conforming functions are in practice continuous."

This statement on its own is false when the dimension is $>1$. In the context of finite element methods (or in general any approximation methods for elliptic PDEs), the statement is more formally like

Let $K$ be a simply-connected domain, $\overline{K} = \overline{K_1 \cup K_2} $ with $K_1\cap K_2 = \varnothing$. If $u|_{K_1}\in H^1(K_1)$ and $u|_{K_2}\in H^1(K_2)$, then $u\in H^1(K)$ if $u$ is continuous across $\Gamma := \overline{K_1} \cap \overline{K_2}$.

The proof is pretty straightforward by the definition of the weak derivative using integration by parts separately on each $K_1$ and $K_2$ vs the one on the whole $K$. You can find on any numerical PDE textbook, e.g., Lemma 5.3 in Peter Monk's book. The idea behind the proof is basically that on each element of a mesh, you can define whatever polynomial space (or a finite dimensional approximation space) which is locally $H^1$, now you can "glue" them together to be globally $H^1$ on the mesh.

Shuhao Cao
  • 20,119
  • Thank you for your answer! I’ve just checked the lemma you cited — I didn’t know it before. I guess you are referring to point (1) of the lemma, right? However, I read it as (perhaps) saying that, if $u$ is continuous across $\Gamma$ (and $u\in L^2(K)$) then $u\in H^1(K)$. How would you prove the converse implication? – alep17 Oct 03 '25 at 15:19
  • I report here Lemma 5.3 as it is in Peter Monk's book, for clarity: Suppose $K_1$ and $K_2$ are two non-overlapping Lipschitz domains meeting at a common surface $\Sigma$ (with non-zero measure) so that $K_1 \cap K_2 = \Sigma$.

    (1) Suppose that $p_1 \in H^1(K_1)$ and $p_2 \in H^1(K_2)$, and define $p \in L^2(K_1 \cup K_2 \cup \Sigma)$ by \begin{equation} p = \begin{cases} p_1 & \text{on } K_1, \ p_2 & \text{on } K_2. \end{cases} \end{equation} Then, if $p_1 = p_2$ on $\Sigma$, we have $p \in H^1(K_1 \cup K_2 \cup \Sigma)$.

     – alep17 Oct 03 '25 at 15:34
  • @alep17 The other direction you have to define the continuity weakly in something like $0=\lim_{\epsilon \to 0}\int_{\Sigma} (u(\boldsymbol{x} + \epsilon \boldsymbol{\nu}) - u(\boldsymbol{x} - \epsilon \boldsymbol{\nu})) \partial_{x_i} \phi \nu_i dS(\boldsymbol{x})$, where $\phi\in C^{\infty}_c(K)$ and $\boldsymbol{\nu}$ is the normal vector on $\Sigma$. The test function kinda offsets the potential singularity-type noncontinuity. For simplicity, I changed the statement in the answer in just the "if" direction. – Shuhao Cao Oct 03 '25 at 17:30
3

Caveat Emptor - this answer lacks a considerable amount of rigor.

The set of discontinuity being measure zero is not actually enough - consider the domain $U=\mathbb B^n(0,2)$ , the unit ball of radius two in $\Bbb R^n$. Consider the function(s) $$u_\lambda(x)=\frac{1}{(1-|x|)^\lambda }$$ You can check that there is no combination of values $\lambda <0$, $n\in\mathbb N$, $p\geq 1$ such that $u_\lambda \in W^{1,p}(U)$.

(Note: This is intuitively clear because, if it were in $W^{1,p}(U)$, it must also be in $W^{1,p}(\mathbb B^n(0,1))$. However it cannot be in $W^{1,p}(\mathbb B^n(0,1))$ because it clearly does not have a natural trace on $\partial\mathbb B^n(0,1)$, and all $W^{1,p}$ functions must have a natural trace)

However, we see that the discontinuities of $u_\lambda$ are the set $\partial\mathbb B^n(0,1)$, which are of $\mu^n$-measure zero. You can try coming up with other contrived examples like $u(x_1,x_2)=x_1^{-\lambda}$ but they will all fail because they all fail to meet the ACL criterion mentioned by Calvin Khor.

So, it is not enough for the discontinuity to be measure zero, they must be legitimately "point-like".

K.defaoite
  • 14,333
  • 1
    Ah, thank you! That’s why I was asking which measure we are considering — it didn’t seem right to me to say that discontinuities along a curve are allowed. – alep17 Oct 03 '25 at 15:51