While reviewing some lecture notes, I stumbled upon the following proposition. $\newcommand{\vertiii}[1]{{\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert #1 \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}}$

Let $\Omega \subset \mathbb{R}^d$ be a bounded Lipschitz domain and consider the Sobolev space $H^1 (\Omega) := W^{1,2}(\Omega)$. Then the following norms are equivalent on $H^1 (\Omega) $: \begin{align} \Vert v \Vert_1^2 &:= \int_\Omega v^2 dx + \int_{\Omega} \vert \nabla v \vert^2 dx =: \Vert v \Vert_0^2 + \vert v\vert_1^2 \\ \vertiii{v}_1^2 &:= \int_{\partial \Omega} v^2 d\sigma + \vert v \vert_1^2 \end{align}

Showing this proposition consists of applying a theorem, that was stated without proof:

Let $\lbrace f_i \rbrace_{i=1}^l$ be a system with the properties

  1. $f_i : H^1(\Omega) \rightarrow \mathbb{R}_0^+$ is a semi norm
  2. $\exists C_i >0$ s.t. $0 \leq f_i(v) \leq C_i \Vert v \Vert_1 \quad \forall v \in H^1 (\Omega)$
  3. $f_i$ is a norm on the polynomials of degree $0$

then the $\Vert \cdot \Vert_1$-norm and \begin{equation} \vertiii{v}^2 := \sum\limits_{i=1}^l f_i^2(v) + \vert v \vert_1^2 \end{equation} are equivalent.

Proving that $f_1(v) := \int_{\partial \Omega} v^2 d\sigma$ indeed possesses the properties of the theorem is straight forward (continuity of the trace operator for boundedness, ...). However, I am interested in the proof of the theorem itself.

I would be grateful for references in the literature, hints for doing the proof myself or comments.


The main reason is the Poincaré inequality, the proof of which you can find, for example, here: $$ \int_\Omega u^2 \le C\int_\Omega |\nabla u|^2 \quad \text{ provided } \int_\Omega u = 0 $$ The space $H^1(\Omega)$ is the direct sum of the one-dimensional space of constant functions, called $L$, and its orthogonal complement $M$, which consists of the functions with zero integral. By the Poincaré inequality, the norms are comparable on $M$. But the assumption (3), they are also comparable on $L$. This implies they are comparable on the entire space (all choices of norms on the direct sum of two normed spaces are equivalent.)

  • $\begingroup$ Thank you! I assume one can generalize these notions to Sobolev spaces $W^{k,p} (\Omega)$ using similar arguments (decomposition into the space of polynomials of degree k-1 and the orthogonal complement)!? $\endgroup$
    – user404633
    Jan 24 '17 at 14:00
  • $\begingroup$ Yes. The $W^{k,p}$ norm is equivalent to $L^p$ of the $k$th order derivatives + whatever it takes to prevent the polynomials of degree $\le k-1$ from having zero norm., $\endgroup$
    – user357151
    Jan 24 '17 at 14:37

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