Let $W^{k,p} (\Omega)$ be a Sobolev space, $\Omega \subset \mathbb{R}^N$. Formally, $W^{k,p}(\Omega)$ consists of equivalence classes of functions with finite Sobolev norm. Two functions, $f$ and $g$, are said to be equivalent ('equal') if $\Vert f-g \Vert_{W^{k,p}} =0$.

In many proofs concerning Sobolev functions, say, Poincaré inequality, one first shows the claim for smooth functions and then generalises by using the fact that smooth functions $C_0^{\infty} (\Omega)$ are dense in $W_0^{1,2}(\Omega)$.

However, it is unclear to me what is meant by writing $u \in C_0^{\infty}(\Omega)$. Does it mean that $u$ is equivalent to a smooth function (in the $W^{1,2}$ sense) or that $u$ is itself a smooth function? One proof for Poincaré inequality on $\Omega:=(a, b)\subset \mathbb{R}$ starts by writing the value of $u \in C_0^{\infty}(a, b)$ as $$ u(x)=\int_a^x u dx+u(a), $$ however, if $u$ is considered to be a Sobolev function, one cannot talk about its point wise values, $u(a)$ in particular. Or is it some kind of an agreement that $u(a)$ is the value of the smooth function to which $u$ is equivalent (in $W^{1,2}$)?

Another confusion arises when talking about the trace operator. In the Wiki page, it is said that the trace $T: W^{1,p}(\Omega) \rightarrow L^p (\partial \Omega)$ is such that $Tu = u\mid_{\partial \Omega}$ for any $u \in W^{1,p}(\Omega) \cap C(\overline{\Omega})$. Again, $W^{1,p}$ consist of equivalence classes whereas $C(\overline{\Omega})$ consists of functions. So does the intersection mean functions with finite Sobolev norm which are equivalent to some continuous function? And is $u\mid_{\partial \Omega}$ the boundary function of $u$, or the set of functions equivalent to it?


1 Answer 1


When one speaks of an element of a Sobolev space $u\in W^{k,p}(\Omega)$ being continuous, this is typically meant to mean that there exists an element of the equivalence class $u_c \in u$ such that $u_c$ is a continuous function on $\Omega$ with finite $k,p$ Sobolev norm. Therefore, the set inclusion $u\in C^\infty_0(\Omega)$ reads "the equivalence class $u\in W^{k,p}(\Omega)$ contains an element $u_c$ such that $u_c\in C^\infty_c(\Omega).$"

Since the Sobolev space only cares about function up to a set of measure zero, we could ask questions about whether functions in the space are continuous, strongly differentiable, etc., but those questions are not invariant under modifications on a set of measure zero, so they can only be answered by seeing if there are sufficiently smooth elements of the equivalence class for which these properties apply. Once you have established that such a smooth function $u_c$ exists within an equivalence class $u$, you can then consider pointwise values of $u_c$ and perform classical oeprations on them, like evaluation and strong differentiation.

For the last question about the trace operator, we say that $u\in W^{1,p}(\Omega )\cap C(\overline{\Omega})$ if $u\in W^{1,p}(\Omega)$ contains an element $u_c\in u$ such that $u_c\in C(\overline{\Omega})$. For these functions, $u|_{\partial\Omega}$ is defined to be $u_c|_{\partial\Omega}$ and this defines the trace operator $Tu := u_c|{\partial\Omega}$. However, for a general function $u\in W^{1,p}(\Omega)$ that may not have such a continuous element, $u|_{\partial\Omega}$ is defined to be the limit $\lim_{m\to\infty}Tu_m$, where $u_m\in C^\infty (\overline{\Omega})$ and $\{u_m\}_{m=1}^\infty$ converges to $u$ in $W^{1,p}(\Omega)$. One can then show that $\{Tu_m\}_{m=1}^\infty$ is a Cauchy sequence in $L^p(\partial\Omega)$, so $Tu\in L^p(\partial\Omega)$, which are equivalence classes of functions on the boundary. Here instead of starting with the $L^p$ space and selecting continuous elements, we had to complete the space of continuous functions on the boundary formed by the elements of the Cauchy sequence $\{Tu_m\}_{m=1}^\infty$ so that our function space on the boundary contained their limit point, which necessitated the use of equivalence classes.

  • 2
    $\begingroup$ Perhaps a point to add is that if there is a continuous representative, then it is unique. $\endgroup$
    – LL 3.14
    Apr 16, 2023 at 22:01

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