Sobolev Embedding for the weak topology

Question

If $k \in \mathbb{N}$ and $p\geq 1$, the Sobolev embedding theorem states that we have a continuous embedding: $$ W^{k,p}(\mathbb{R}^d) \hookrightarrow L^q(\mathbb{R}^d) $$ if $k<d/p$ where $q \in [p, pd/(d-pk)]$.

In particular, $H^1(\mathbb{R}^3)=W^{1,2}(\mathbb{R}^3)$ is continuously embedded in $L^2(\mathbb{R}^3)$. My question is the following: is this also true when $H^1(\mathbb{R}^3)$ is equipped with the weak topology (or the equivalent weak-star topology)? It feels like the answer should be easier but I'm not used to Sobolev spaces. Any help would be much appreaciated, thank you!

Answer 1

The embedding is not continuous from $H^1(\mathbb{R}^3)$ with the weak topology to $L^2(\mathbb{R}^3)$ with the strong topology. To see this, let $\phi\in C_c^\infty(\mathbb{R}^3)$, $\Vert\phi\Vert_{H^1}=1$, $\mathrm{supp}\,\phi\subseteq B_1(0)$ and define for $n\in\mathbb{N}$: $$\phi_n(x):=\phi(x-(2n,0,0)).$$ Since the sequence $(\phi_n)_n$ is bounded in $H^1$, there is a weakly convergent subsequence $(\phi_{n_k})_k$. However, $\mathrm{supp}\,\phi_n\cap\mathrm{supp}\,\phi_m=\emptyset$ for $m\neq n$ and hence $$\Vert \phi_n-\phi_m\Vert_{L^2}=2\Vert\phi\Vert_{L^2}>0.$$ Thus, $(\phi_n)_n$ cannot have a strongly convergent subsequence in $L^2$.

Remark: The situation is much different if your domain is some open and bounded set $\Omega\subseteq\mathbb{R}^d$ with Lipschitz boundary. In this case the embedding $H^1\hookrightarrow L^2$ would be compact by the Rellich-Kondrachov theorem. Hence, weak convergence in $H^1$ would imply strong convergence in $L^2$ in this case.

Answer 2

Edit: This answer shows that the embedding is continuous from the weak topology to the weak topology. Assume $kp<d$ and $$u_n\rightharpoonup u\quad\text{in }W^{k,p}(\mathbb{R}^d)$$ and let $f\in L^{q'}(\mathbb{R}^d)$ be arbitrary where $q\in\left[p,\frac{dp}{d-kp}\right]$ and $\frac{1}{q}+\frac{1}{q'}=1$. Then, we may define $$T:W^{k,p}(\mathbb{R}^d)\to\mathbb{R}^d,\ v\mapsto\int_{\mathbb{R}^d}vf\,dx.$$ By the Sobolev embedding, $T$ is continuous, hence $$\int_{\mathbb{R}^d}u_nf\,dx=T(u_n)\to T(u)=\int_{\mathbb{R}^d}uf\,dx$$ for all $f\in L^{q'}(\mathbb{R}^d)$. Thus, it follows that $$u_n\rightharpoonup u\quad\text{in }L^q(\mathbb{R}^d).$$