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The locally convex space of essentially compactly-supported $p$-integrable "functions" $\operatorname{L}_{\mathrm{comp}}^p(\mathbb{R}^d,\mathbb{R})$ is defined as the set $$ \bigcup_{n \in \mathbb{N}} \left\{ f \in L^p(\mathbb{R}^d,\mathbb{R}):\, \operatorname{ess-supp}(f)\subseteq [-n,n]^d \right\}, $$ topologized with the injective limit topology in the category of LCSs with continuous linear maps as morphisms, where the colimit is taken over the injective system $$\left\{L^p(\mathbb{R}^d,\mathbb{R}):\, \operatorname{ess-supp}(f)\subseteq [-n,n]^d\right\}_{n \in \mathbb{N}}.$$

Fix a $k \in \mathbb{N}$, $1\leq p<\infty$ and let $W^{k,p}_{\mathrm{comp}}(\mathbb{R}^d):=W^{k,p}(\mathbb{R}^d) \cap \operatorname{L}_{\mathrm{comp}}^p$. How are the subspace topologies on $W^{k,p}_{\mathrm{comp}}(\mathbb{R}^d)$ comparable? Is the subspace topology induced by restricting the subspace topology on $L^p_{\mathrm{comp}}(\mathbb{R}^d,\mathbb{R})$ to $W^{k,p}_{\mathrm{comp}}(\mathbb{R}^d)$ at-least as fine than the one obtained by restricting the Sobolev topology $W^{k,p}$ to $W^{k,p}_{\mathrm{comp}}(\mathbb{R}^d)$?

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    $\begingroup$ Why should this be true? The $L^p_{comp}$-topology gives nothing for the distributional derivatives! You probably know that because of the stricness of the inductive system the relative topology of $L^p_{comp}$ on the steps is just the $L^p$-topology. $\endgroup$ Mar 25 '20 at 15:05
  • $\begingroup$ But in this post: mathoverflow.net/questions/347318/… don't we find that the topology on $L^p_{comp}$ is strictly finer? $\endgroup$
    – AIM_BLB
    Mar 25 '20 at 15:15
  • $\begingroup$ Indeed, the $L^p_{comp}$ topology is stricly finer than the $L^p$-topology, but on the steps they coincide. There is no contradiction. $\endgroup$ Mar 25 '20 at 15:21
  • $\begingroup$ Oh my brain filtered the word "steps" sorry. I've never seen this terminology, you mean the finite "sub-colimits?". In that case definitely. $\endgroup$
    – AIM_BLB
    Mar 25 '20 at 15:25
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To make my comment an answer: No. To see this, fix a cube $K$ and look at $L^p_K=\{f\in L^p: \text{ ess-supp}(f) \subseteq K\}$. On this space, the $L^p_{comp}$-topology coincides with the $L^p$-topology (this follows from the stricness of the inductive limit) and the $W^{k,p}$-topology is strictly finer on this subspace.

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  • $\begingroup$ Ah, you're right this is very clear. $\endgroup$
    – AIM_BLB
    Mar 25 '20 at 15:26

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