Let $u\in H^1(B_1^+;\mathbb{R}^3)$ for $B_1^+:=B_1\cap\{x_3>0\} \subset\mathbb{R}^3$. Suppose that $u(x)\in S \subset \mathbb{R}^3$ for $x \in B_1 \cap \{x_3=0\}$ in the sense of trace. The question is why it is possible to choose $r\in(0,1)$ such that $u \in H^1(\partial^+ B^+_r;\mathbb{R}^3)$ where $\partial^+B^+_r:=\partial B_r\cap\{x_3>0\}$. I know that the integrability of $u$ and $\nabla u$ should come from the coarea formula (or Fubini's theorem), but I do not understand why the weak derivative of $u$ exists (satisfying integrating by part) on $\partial^+ B^+_r$ and equals the tangential part of $\nabla u$.
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- 1$\begingroup$ What are the meaning of these notations of the domains? $\endgroup$Liding Yao– Liding Yao2025-03-18 00:23:56 +00:00Commented Mar 18 at 0:23
- $\begingroup$ updated, thanks for letting me know $\endgroup$mnmn1993– mnmn19932025-03-18 00:27:28 +00:00Commented Mar 18 at 0:27
- 1$\begingroup$ I think this is a direct result from Fubini theorem where the total gradient is L^2 for almost every r. $\endgroup$Liding Yao– Liding Yao2025-03-18 00:34:58 +00:00Commented Mar 18 at 0:34
- $\begingroup$ I think Fubini theorem only tells us that $\nabla u$ is $L^2$ on $\partial^+B^+_r$. But why it is the weak derivative of $u$ on $\partial^+B^+_r$ and satisfies the definition of weak derivative on $\partial^+B^+_r$? $\endgroup$mnmn1993– mnmn19932025-03-18 01:04:28 +00:00Commented Mar 18 at 1:04
- 2$\begingroup$ The derivative of $u$ on $\partial^+B_r^+$ is the tangential part of $\nabla u$. Note that $\nabla u\in L^2$ iff $X\cdot\nabla u\in L^2$ for every continuous vector fields $X:B^+\to\mathbb R^3$. $\endgroup$Liding Yao– Liding Yao2025-03-18 01:59:56 +00:00Commented Mar 18 at 1:59
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