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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.

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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.)

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  • $\begingroup$ 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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