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I can't see the difference between these two Sobolev spaces:

$H^{1}(\Omega):=\left\{u \in L^{2}(\Omega) | \nabla u \in L^{2}(\Omega)^{d}, u=0 \text { on } \partial \Omega\right\}$

and $H_{0}^{1}(\Omega)=\left\{v \in H^{1}(\Omega) | v_{| \partial \Omega}=0 \text { on } \partial \Omega\right\}$

Where $\gamma_{0}$ is linear from $H^{1}(\Omega) \cap C(\bar{\Omega})$ to $L^{2}(\partial \Omega) \cap C(\overline{\partial \Omega})$ by $\gamma_{0} v:=v_{| \partial \Omega}$

the definition that I usually use is:

$H^{1}(\Omega)=\left\{u \in L^{2}(\Omega) | D_{i} u \in L^{2}(\Omega), \forall i=1, \ldots, N\right\}$

but in the paper I am reading the authors state that:

$H^{1}(\Omega):=\left\{u \in L^{2}(\Omega) | \nabla u \in L^{2}(\Omega)^{d}, u=0 \text { on } \partial \Omega\right\}$

I am a beginner in Sobolev spaces.. someone can give me a hint?

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    The first space of functions need not vanish on the boundary, but the second space of functions are all zero on the boundary.

    I think what the authors of the paper should say is $H_0^1(\Omega)$ for this set, because these functions vanish on the boundary.

    – fGDu94 Jan 28 '20 at 21:23
  • Yes, I agree with you, I was pretty sure about that but I preferred to consult our community, Thank you – Almendrof66 Jan 28 '20 at 21:34

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