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I have some trouble with proper understanding of $H_0^1(0,1)$ space. Consider the following space $$H_D = \{u\in H^1(0,1): u(0) = u(1) = 0\}.$$ What can we say about the connection between $H_D$ and $H^1_0(0,1)$. Is $H_D$ in $H^1_0(0,1)$? In literature stays, that functions in $H_0^1(0,l)$ are interpreted as $$''u = 0\; \text{on}\; \partial\Omega''.$$

Dina
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2 Answers2

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Do you know anything about the trace operator? There is a theorem that says that if $U$ is a bounded domain and $\partial U$ is $C^1$ and $u \in W^{1,p}$, then $$ u \in W_0^{1,p}(U) \iff Tu=0 \text{ on } \partial U, $$ where $T$ is the trace operator. You want $p = 2$ here.

N.U.
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  • I have heard about the trace operator. I think, that in my case, since $U = [0,1]$ and $u\in H^1(0,1)$, it means that $H_D = H_0^1(0,1)$. Is it correct? – Dina May 14 '13 at 09:07
  • @Dina In your case $U = (0,1)$. Elements in $H^1(0,1)$ don't have a value on the boundary (the points 0 and 1) to begin with, but are given the values that they get from the trace operator. Yes, $H_D = H_0^1(0,1)$. – N.U. May 14 '13 at 10:54
  • N.U., thank you! Now I understand. – Dina May 14 '13 at 11:41
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In one-dimensions, Sobolev functions have a well-defined continuous representative. Therefore, they are well-defined at every point of their domain.

user36236
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