# Embedding of $\mathcal{D}(\Omega)$ in $W^{k,p}_0(\Omega)$

$$W^{k,p}_0(\Omega)$$ is defined as the closure of the set of all $$C_c^{\infty}(\Omega)$$ under the topology generated by the norm $$W^{k,p}(\Omega)$$. So clearly the identity map from $$\mathcal{D}(\Omega) \rightarrow W^{k,p}_0(\Omega)$$ defines an injection. Is this injection continuous?

If so how to prove it.

P.S:

1. $$\mathcal{D}(\Omega)=(C_c^{\infty}(\Omega),\tau_{LF})$$, i.e., $$C_c^{\infty}(\Omega)$$ endowed with its canonical LF topology.

2. $$W^{k,p}_0(\Omega)=\big(\mathcal{D}(\Omega), \|\cdot\|_{W^{k,p}({\Omega})}\big)$$, i.e., $$C_c^{\infty}(\Omega)$$ endowed with the topology generated by $$W^{k,p}{\Omega}$$ norm.

A clean proof will be greatly appreciated. Thanks in advance.

• The universal property of the locally convex inductive limit topology says that you only need the continuity of the restriction of the embedding to all $\mathcal D(K)=\{f\in C^\infty(\mathbb R^d):$ supp$(f)\subseteq K\}$ endowed with the topology of uniform convergence of all (partial) derivatives are continuous. Commented Oct 20, 2023 at 7:21
• @Jochen Could you please elaborate a bit. I agree with what you mentioned. But it's not clear to me, if it answers my question. Commented Oct 21, 2023 at 22:16
• For each compact set $K$ you can estimate the $W^{k,p}$-norm of $\varphi\in \mathcal D(K)$ by $c\sup\{|\partial^\alpha\varphi(x)|: x\in K, |\alpha|\le k\}$ with some constant $c$. Commented Oct 22, 2023 at 12:04

Enough to prove that if a sequence $$\phi_n \rightarrow 0$$ in $$\mathcal{D}(\Omega)$$ then $$\phi_n \rightarrow 0$$ in $$W^{k,p}$$ norm.
Since $$\phi_n \rightarrow 0$$ in $$\mathcal{D}(\Omega)$$ there exists a compact set $$K \subset \Omega$$ such that $$\operatorname{supp}(\phi_n) \subset K$$ and for every multi index $$\alpha$$ $$\|D^{\alpha}\phi_n\|_{L^{\infty}(K)} \rightarrow 0$$. Consequenlty
\begin{align} \|\phi_n\|_{W^{k,p}(\Omega)} := \sum_{|\alpha| \leq k} \|D^{\alpha}\phi_n\|_{L^p(K)} \leq \mu(K)^{1/q} \sum_{|\alpha| \leq k} \|D^{\alpha}\phi_n\|_{L^{\infty}(K)} \rightarrow 0. \end{align}
• An example of a proof that takes that into account: following the advice of @Jochen, you can equivalently show that the restriction of the identity is continuous from $\mathcal D(K)$ to $W^{k,p}$. This restriction is indeed a map between metric spaces (because $\mathcal D(K)$ is a Fréchet space if I am not wrong), so for that it is enough to show sequential continuity (which is true by the argument in your question, so you are done). Maybe there is an easier remark that allows to go from sequential continuity to actual continuity without involving $\mathcal D(K)$, but I don’t see it. Commented Nov 8, 2023 at 17:53