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Is it true that $H^1(\Omega_1 \cup \Omega_2 )=H^1(\Omega_1)+H^1(\Omega_2)$?

Below, we already have a counterexample. Let me ask further. If I impose $\Omega_1 \cap \Omega_2=\emptyset$, is there still a counter example?

user33869
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  • Well, if the two underlying sets are disjoint, then yes, trivially so. But see the answer by @PhoemueX. – Harald Hanche-Olsen Sep 14 '14 at 14:55
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    Are $\Omega_1$ and $\Omega_2$ open sets? Then yes, because the existence and integrability of weak derivative can be checked on each connected component separately. –  Sep 14 '14 at 14:58
  • @HaraldHanche-Olsen Well, I think stack exchange is for this kind of questions. Can I ask which definition do you use to verify this trivial fact? – user33869 Sep 14 '14 at 15:09
  • Which definition? We're looking at $L^2$ functions with weak partial derivatives of the first order also in $L^2$. If $\Omega_1$ and $\Omega_2$ are disjoint open sets, then the direct sum of the corresponding $L^2$ spaces is isomorphic to $L^2(\Omega_1\cup\Omega_2)$ in the obvious way, and as @Thursday says, you check the weak derivative on each component separately. I do insist it is trivial; a detailed proof does take some space to write up, though. I am not inclined to do so myself, but perhaps someone else will. – Harald Hanche-Olsen Sep 14 '14 at 15:57
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    Heck, I'll add one tiny bit of detail. Define a map $H^1(\Omega_1 \cup \Omega_2 )\to H^1(\Omega_1)\oplus H^1(\Omega_2)$ by $f\mapsto(f|{\Omega_1},f|{\Omega_2})$. I claim this is an isometry, if $\Omega_1\cap\Omega_2=\emptyset$. Proof omitted due to time constraints and lack of motivation, sorry. – Harald Hanche-Olsen Sep 14 '14 at 16:01
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    @HaraldHanche-Olsen what if $\Omega_1=(0,1)$, $\Omega_2=(1,2)$, then $H^1(\Omega_1\cup \Omega_2) = H^1((0,2))$, which is not equal to $H^1(\Omega_1) + H^1(\Omega_2)$. Functions in the first space are continuous at $1$, while functions in the second not necessarily. – daw Sep 15 '14 at 08:09
  • @daw No, $\Omega_1\cup\Omega_2$ has two components, with $1$ separating the two. There is nothing in the definition requiring continuity at $1$. – Harald Hanche-Olsen Sep 15 '14 at 17:13

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Consider $\Omega_1 = (0,2)$ and $\Omega_2 = (1,3)$.

Then the constant functions $x\mapsto 1$ (one the respective intervals) are in $H^1(\Omega_i)$, but their sum (if extended by 0, otherwise, what do you mean by the sum?) is not continuous (has no continuous representative), hence is not in $H^1(\Omega_1 \cup \Omega_2)$.

PhoemueX
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  • How do you interpret this "sum"? I mean, for example, what is the sum of the functions in your example? – Tomás Sep 15 '14 at 11:04
  • I interpret $H^1(\Omega_i)$ as a subspace of $l^2(\Bbb{R}$ by extending each function by zero on $\Omega_i^c$. Then I add these extensions and restrict to $\Omega_1 \cup \Omega_2$. This may not be the only way to do this, but as the OP did not provide any interpretation, I chose this one. – PhoemueX Sep 15 '14 at 12:06