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My question comes from the topic of Finite Elements (and domain-composition-algorithms).

Consider the following setting: We have some domain $\Omega = (0,1)^2$ which is decomposed into rectangular subdomains. Each subdomain $\Omega_i$ of it with $diam(\Omega) = H$ and $\tau^{h}$ ($h$ the step size between two nodes) a regular triangulation of $\Omega$ by $P_1$-elements (i.e. polynoms $ax + by + c$ on each triangle). Let $\theta_\nu$ be some nodal basis function for $\nu$ a vertex of $\Omega_i$, i.e. $\theta_\nu(x) = 1$ for $x = \nu$ and $\theta_\nu(x) = 0$ for $x$ being any other node of $\Omega$.

Now to the question: Why does $$|\theta_\nu|_{H^1(\Omega_i)} \leq C$$ and $$ ||\theta_\nu||_{L^2(\Omega_i)} \leq C \cdot h^2 $$ independent of $H$ and $h$ hold?

Idea for the first inequality:

Picture of the triangulation

W.l.o.g. triangulate the domain $\Omega$ as done in the picture. On $\Omega_i$ the vertex basis function $\theta_{\nu}$ is per definition only nonzero on the triangle $T$ and there it is a $P_1$ element which is bounded by 1, i.e. ${\theta_{\nu}}_{|T} = ax + by + c \leq 1$ for some reel coefficients $a,b,c \in \mathbb{R}$. In especially we have that $a^2 + b^2 \leq 1$ and thus $$|\theta_\nu|_{H^1(\Omega_i)} = \int_{\Omega_i}|\nabla \Omega_i|^2 d(x,y) = \int_{T}|\nabla(ax + by + c)|^2 d(x,y)$$ $$ = \int_{T} a^2 + b^2 d(x,y) = (a^2 + b^2) \cdot \lambda(T)$$ where $\lambda(T)$ is meant to be the Lebesgue-measure of the triangle T which has finite measure because the triangulation is regular! Are these ideas correct? What do you mean?

Thanks a lot for any approach!!!

Robert

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