# Are Sobolev spaces $W^{k,p}$ closed subsets of $L^p$?

Let $$u_n\in H^1(\Omega),\ \Omega\subset\mathbb{R}^N, N\geq 1$$ be functions such that $$u_n\to u\in L^2(\Omega)$$ and the convergence takes place in $$L^2(\Omega)$$. Is it true that $$u\in H^1(\Omega)$$?

So, in general, are Sobolev spaces $$W^{k,p}(\Omega)$$ closed subsets in $$L^p(\Omega)$$ with the norm $$\Vert\cdot\Vert_p$$?

No, not in general. Take some $$u \in L^p(\Omega)$$ that is not in $$W^{k, p}(\Omega)$$. I do not know your $$k, p, \Omega$$, but sometimes you can easily construct those with characteristic functions on (half-)lines. E.g. $$\chi_{[0, 1]}(x)$$ has no weak derivative on $$[-1, 1]$$.
It is well known that then there is some sequence $$(u_n)_{n \in \mathbb{N}} \subseteq C_0^\infty(\Omega) \subseteq W^{k, p}(\Omega)$$ that converges to $$u$$ in $$L^p$$. You can research this e.g. in Alt's linear functional analysis.
• In general, I think the characteristic function of a half space, or, more precisely, of $\{x\in\Omega\mid x_1\leq a\}$, has weak partial derivative in $x_1$-direction given by the surface measure of $\{x\in\Omega\mid x_1=a\}$, which is not a function. Of course one has to choose $a$ appropriately to guarantee a "large enough" intersection. Feb 9, 2022 at 15:07
• Yes. That is why I said "sometimes". Sometimes also $\log \log \lvert x \rvert$ works. Depends on your situation Feb 9, 2022 at 17:36