# How can we glue Sobolev functions?

Let $u:A\cup B\to R$ be a function (where $A$ and $B$ are disjoint connected sets and $A\cup B$ is connected) such that $u$ restrict to $A$ and to $B$ are in $W^{1, p}$. Which result guarantees me that $u$ is $W^{1, p}$?

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Are you assuming $A$ and $B$ are open? –  Stefan Smith Jun 21 '12 at 15:28
If both $A$ and $B$ are either open or closed, then $A\cup B$ won't be connected anymore. However, it may be that $A\cap\overline{B} \ne \emptyset$ or $\overline{A}\cap B \ne \emptyset$. –  Mercy Jun 21 '12 at 16:56
As @bogus noted above, the question needs to be clarified. Sobolev spaces are normally defined on open sets, but the union of disjoint open sets is not connected. –  user31373 Jun 21 '12 at 17:21
Evidently $A$ and $B$ can't be both open sets. $A$ may be closed and $B$ may be an open set such that the union is connected. Suppose that $A$ and $B$ are open such that $\bar{A} \cap \bar{B}$ is a bounday of a smooth manifold, $\bar{A} \cup \bar{B}$ is a connected set and $A \cap B=\emptyset$. A case, for example, is the following one $A=B_1$ and $B=B_2\setminus B_1$. The question is: Let $A$ and $B$ be open sets such that $\bar{A} \cup \bar{B}$ is connected and $u:A\cup B\cup U\to R$ restricted to $A$ and $B$ is in $W^{1,1}(A)$ and $W^{1,1}(B)$, repectivel. When $u$ is in $W^{1,1}$? –  jalmanalves Jun 22 '12 at 12:14
where $U=\bar{A}\cap\bar{B}$ –  jalmanalves Jun 22 '12 at 12:20