# Is $W_0^{1,p}$ weakly closed?

is $W_0^{1,p}(\Omega)$ weakly closed? $W_0^{1,p}(\Omega)$ is the closure of $C_0^{\infty}(\Omega)$ with respect to the norm of $W_0^{1,p}(\Omega)$, and I've been trying to figure out if it is true that if we have a sequence $u_n\in W_0^{1,p}(\Omega)$ that converges weakly to $u\in W^{1,p}$ then $u\in W_0^{1,p}$. (By weak convergence in $W^{1,p}$ I mean weak convergence in $L^p$ of both $u_n$ and $\nabla u_n$)

The only thing I could think of was to approximate each $u_n$ with a sequence $(u_n^k)_k\in C_0^{\infty}$. Then by the Sobolev embedding theorems we have that weak convergence of $\nabla u_n$ in $L^p$ implies strong convergence of $u_n$ in $L^p$, so it should be possible to approximate $u$ with $C_0^{\infty}$ functions in the $L^p$ norm. But what about the derivatives? I hope I have not been too confusing, thank you.

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Isn't this true trivially? Weak convergence is defined as follows: Given $u_n, u\in W^{1,p}_0(\Omega)$, we say that $u_n\rightharpoonup u$ ($u_n$ converges weakly to $u$), if $\int u_n \varphi \to \int u\varphi$ for every test function $\varphi$... So the limit is an element of your space by definition. Or do you have an alternative definition? – Sam Jan 10 '13 at 0:40

Let $E$ be a Banach space and $C\subset E$ be a convex set.
Theorem: $C$ is weakly closed if and only if $C$ is strong closed (closed in the norm topology).