suppose we have the Sobolev space $H^1_0(\Omega)$ over a bounded domain $\Omega \subset \mathbb{R}^2$. With the standard inner product it sure is a Hilbert space. BUT: What if we equip $H^1_0$ with the inner product

\begin{align} \langle \cdot, \cdot \rangle_u = \int_{\Omega} (\epsilon + |\nabla u|)^{p-2} \nabla \cdot \nabla \cdot \end{align}

with $u \in C(\overline{\Omega})$ and $\epsilon > 0$. The induced norm would be

\begin{align} ||v||_u^2 = \int_{ \Omega } ( \epsilon + | \nabla u |^{p-2} ) |v|^2 \end{align}

If I am not wrong we have

\begin{align} \int_{\Omega} ( \epsilon + |\nabla u|)^{p-2} |\nabla v|^2 \leq \sup_{x \in \Omega} (\epsilon + |\nabla u(x)|)^{p-2} \int_{\Omega} |\nabla v|^2 = \sup_{x \in \Omega} (\epsilon + |\nabla u(x)|)^{p-2} ||v||^2 \end{align}


\begin{align} \epsilon ||v||^2=\epsilon \int_{\Omega} |\nabla v|^2 \leq \int_{\Omega} ( \epsilon + |\nabla u|)^{p-2} |\nabla v|^2 = ||v||^2_u \end{align}

Therefore the norms in $H^1_0$ are equivalent. But is this enough to say that $H^1_0 = H^u_0$?


It depends, how the spaces $H^1_0$ and $H^u_0$ are defined:

1) If they are defined as the closure of $C_c^\infty$ with respect to those norms, then they are the same. The norms are equivalent, so Cauchy (or convergent) sequences in one norm are Cauchy (or convergent) in the other norm. A sequence that is convergent with respect to both norms has the same limit in both norms.

2) If they are defined as spaces of $H^1$-functions with zero trace, then these spaces are the same because of the norm equivalence.

  • $\begingroup$ Did you miss a "not" in the second part? Or are they in both cases similiar? I can choose how I define $H^1$, but aren't both ways equal? Though I like the "zero-trace" definition more, because it seems more intuitve for me $\endgroup$ – superdave99 May 11 at 16:29
  • $\begingroup$ The spaces are the same. In order to prove this, one has to use the definition of the spaces. Both ways are the two equivalent standard ways to define $H^1_0$. $\endgroup$ – daw May 11 at 16:31

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