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Let

  • $d\in\mathbb N$
  • $\Omega\subseteq\mathbb R^d$ be open and $\Lambda\subseteq\mathbb R^d$ be bounded and open with $\overline\Lambda\subseteq\Omega$ (if necessary, we may assume that $\partial\Lambda$ is Lipschitz)
  • $u\in C^1(\Omega)$ with $$\left.u\right|_{\partial\Lambda}=0\tag 1$$

Since $u$ and $\nabla u$ are bounded in $\Lambda$, we obtain that $$\tilde u:=\left.u\right|_{\Lambda}\in H^1(\Lambda)\;.\tag 2$$

Can we even show that $\tilde u\in H_0^1(\Lambda)$?

I know there is a characterization of $H_0^1(\Lambda,\mathbb R^d)$ in terms of the trace operator. However, I don't see that $(1)$ immediately yields $\tilde u\in H_0^1(\Lambda,\mathbb R^d)$.

0xbadf00d
  • 14,208
  • hint : how do you define $H_0^1(\Lambda)$, or the trace operator, or $u|_{\partial \Lambda}$. hint2 : for approximating by $C^\infty_c$ a (uniformly) continuous compactly supported function, use the convolution with a mollifier. – reuns Sep 05 '16 at 14:11
  • Does your $\Omega$ has a nice boundary? (Lipschitz for example) –  Sep 06 '16 at 06:51
  • @JohnMa It's just bounded and open, but if you say that we need a Lipschitz boundary, that would be okay for me too. I guess the Lipschitz boundary is needed anyway for the definition of the trace operator. – 0xbadf00d Sep 06 '16 at 09:00
  • If your boundary is Lipschitz, then you get just accept the answer below. Note $H^1_0$ can be defined for any open set, as completion of $C^\infty_c$ with the $H^1$ norm. So you might add the boundary condition into the question –  Sep 06 '16 at 09:06
  • @JohnMa Even when $\partial\Lambda$ is Lipschitz, I don't see what we have $T\tilde u=\left.\tilde u\right|_{\partial\Lambda}$ from the definition of the trace operator $T$. – 0xbadf00d Sep 06 '16 at 09:12
  • Not from definition, but that can be proved –  Sep 06 '16 at 09:15
  • @JohnMa With that you mean $$H_0^1(\Lambda)=\left{u\in H^1(\Lambda):Tu=T{\rm D}u=0\right};,$$ right? I'm actually aware of that fact, but that doesn't explain why we should have $T\tilde u=\left.\tilde u\right|_{\partial\Lambda}$. – 0xbadf00d Sep 06 '16 at 09:36
  • Then you need to read the construction of trace operator again... it comes from extending $ u\mapsto u|_{\partial \Omega}$ from $C^1(\overline\Omega)$ to $H^1(\Omega)$. –  Sep 06 '16 at 09:55

1 Answers1

1

For $\tilde u \in C(\bar\Lambda)$ you have $T \tilde u = \tilde u|_\Lambda$, where $T$ is the trace operator. Hence, $T \tilde u = 0$ and, consequently, $\tilde u \in H_0^1(\Lambda)$.

gerw
  • 33,373