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The approximation theorem states that if $U$ is bounded, $u \in W^{1,p}(U)$ for some $1 \leq p < \infty$ then there are functions $u_m \in C^\infty(U) \cap W^{k,p}(U)$, such that $u_m \rightarrow u$ in $W^{k,p}(U)$.

This is probably very simple, but I'm trying to follow along with an example in a book I am reading. I'm looking at the set $\Omega = (-1,0) \cup (0,1)$ and the function $u : \Omega \rightarrow \mathbb{R}$ defined by

$u(x) = \left\{ \begin{array}{ll} 1, & x > 0 \\ 0, & x < 0 \end{array} \right.$.

It's clear to me that this function is in $W^{1,p}(\Omega)$ for $1 \leq p \leq \Omega$; however, I'm having trouble proving that it cannot be approximated by functions in $C^\infty(\overline{\Omega})$. It seems that any sequence approaching $u$ would approach the heaviside function which I know is not in $L^p$ and therefore not in $W^{1,p}$; however, actually proving this rigorously has eluded me for a while now.

I also had a question about a global form of the approximation theorem. If we also add that $\partial U$ is $C^1$ to the hypothesis, then we have functions $u_m \in C^\infty(\overline{U})$ such that $u_m \rightarrow u$ in $W^{k,p}(U)$. I know that $W^{k,p}$ is Banach so I was wondering if in this case we are able to assume that $\lim\limits_{m \rightarrow \infty} u_m$ exists in $W^{1,p}$. I'm just not sure if the $C^\infty$ functions intersected with the $W^{k,p}$ functions are complete. Anyway, thanks a lot for looking. I appreciate any help you guys can provide.

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BTW, the Heaviside function is in $L^p$ (on bounded sets). The problem does not come from the fact that the limit function is not integrable but from the missing regularity of the limit function on the closure of the set. – Dirk Oct 1 '12 at 6:23

For the first question: in one dimension $W^{1,1}(U)$ embeds into $C_b(U)$ (bounded continuous functions with supremum norm). Therefore convergence in $W^{1,1}$ implies uniform convergence, which is clearly impossible. Also, on bounded domains the larger values of $p$ correspond to smaller spaces, so the case $p=1$ is in fact the general case.

I don't really understand the second question. Convergence in $W^{k,p }$ for $k\ge1$ certainly implies convergence in $W^{1,p}$. The intersection of $C^{\infty }$ with any Sobolev space is never complete; in most cases (namely finite $p$) it is dense.

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