Sobolev space - norm $H^1$ and $H^1_0$

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Not all function spaces need their own norm definitions. Some are defined as subspaces of a larger normed space, and inherit the norm automatically.

For example, $\ell_\infty$ is the space of all bounded sequences, $c$ is its subspace that consists of convergent sequences, and $c_0$ consists of the sequences that converge to $0$. We have $\ell_\infty$ norm, but there is no reason to speak of "$c$ norm" or "$c_0$ norm". The norm is already there, inherited from the larger space.

This is how it works with $H^1$ and $H^1_0$. We define a norm on $H^1$. Then we define its subspace $H^1_0$, which already has a norm: it gets it from $H^1$. No need to speak of "$H^1_0$ norm".

That said, on certain domains one can prove that for $u\in H^1_0$, the $H^1$ norm is equivalent to $\|\nabla u\|_{L^2}$ (the homogeneous $H^1$ seminorm), and use $\|\nabla u\|_{L^2}$ as a norm on $H^1_0$.

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Updated on August 01, 2022

Comments

  • jijii
    jijii over 1 year

    When we defined on $H^1_0$ the norm $$||v||_{H^1_0}=||v||_{L^2}+||\nabla v||_{L^2}$$ can we tell that $$||u||_{H^1_0} = ||u||_{H^1}?$$ Thank's

    • Tomás
      Tomás about 10 years
      What norm are you using in $H^1$? or better saying what is the definition of $\|\cdot\|_{H^1}$ for you?
    • Harald Hanche-Olsen
      Harald Hanche-Olsen about 10 years
      What is your question?
    • jijii
      jijii about 10 years
      $$||u||_{H^1}=||u||_{L^2}+||\nabla u||_L^2$$ So, we can telle that $$|u||_{H^1_0}=||u||_{H^1}$$ in the case $$||u||_{H^1_0}=||u||_ {L^2}+||\nabla u||_{L^2}$$??
    • Branimir Ćaćić
      Branimir Ćaćić about 10 years
      Given your definitions of $\|\cdot\|_{H^1}$ and $\|\cdot\|_{H^1_0}$, the equation $$ \forall u \in H_0^1, \quad \|u\|_{H^1_0} = \|u\|_{H^1} $$ is absolutely tautological...
    • Tomás
      Tomás about 10 years
      If this is your definition, then your question is nonsense. If you can't see the obvious, how can we explain it to you?
    • jijii
      jijii about 10 years
      so my question is: we can define on $H^1_0$ the same norme of $H^1$?i.e if $v \in H^1_0$, then $$||v||_{H^1_0}^2=||v||_{L^2}+||\nabla v||_{L^2}$$ is norm on $H^1_0$?
  • Quickbeam2k1
    Quickbeam2k1 about 10 years
    one could add that the space $H^1_0(\Omega)$ is the closure of $C_c^\infty(\Omega)$ functions with respect to the $H^1$-norm. Additionally, i just checked my lecture notes, the poincare inequality for $H^1_0(\Omega)$ does not seem to require regularity of the boundary. Hence you can always prove that the $H^1_0$ norm is equivalent to the $L^2$ norm of the gradient (if the domain is bounded)