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}
and
\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.