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$W^{k,p}_0(\Omega)$ 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.

Daniele Tampieri
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Veronica
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  • 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. – Jochen Oct 20 '23 at 07:21
  • @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. – Veronica Oct 21 '23 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 $c$. – Jochen Oct 22 '23 at 12:04

1 Answers1

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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}

Daniele Tampieri
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Veronica
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  • The statement is true, no question about it, but your proof is not entirely convincing written like that. Two strictly different topologies on the same space might have the same converging sequences when not induced by a metric (https://math.stackexchange.com/questions/76691/example-of-different-topologies-with-same-convergent-sequences ), which implies that in general functions between topological spaces that are sequentially continuous are not necessarily continuous. It is probably easily fixable, but I still feel something is missing. – Lorenzo Pompili Nov 08 '23 at 17:42
  • 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. – Lorenzo Pompili Nov 08 '23 at 17:53