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Since my knowledge of functional analysis, $L^p$-, Sobolev- and Hilbert spaces is not very good, I thought I could ask...

Suppose we have a domain $\Omega \subset \mathbb{R}^2$ which is continuously bounded. If we use a triangulation $T^h = \{\tau_1, \tau_2, .., \tau_m \}$ (with $h$ begin the smallest diameter of these triangles) so that $\Omega = \cup_{k=1}^{m} \tau_k$, we can define the space of all piecwise linear functions associated with the triangulation $T^h$ as

\begin{align} V^h_g = \{ v \in C^{\infty}(\overline{\Omega}): \text{$v|_{\tau_k}$ is linear on $\tau_k$ for all $\tau_k \in T^h$ and $v|_{\partial \Omega}$ = g} \} \end{align}

where $g$ denotes the values on the boundary and is in "some good space so we don't get any problems" (maybe $L^2(\partial \Omega)$?).

Now I am not sure about some things and have questions:

Is $V^h_g \subset L^2(\Omega) \subset H^1(\Omega)$ and therefore a Hilber space? Would that mean that if I want to prove convergence in $V^h_g$ I can/should use the $H^1(\Omega)$-norm?

Background: Originally I am intersted in a function $u \in W^{1,p}_g{\Omega)}$, which is the solution of a specific functional. I know that there is a finite element solution $u_h \in V^h_g(\Omega)$ which is close to the real solution as long as $h$ gets small. To compute $u_h$ I want to use a general descent algorithm and show global convergence towards the finite elemet solution $u_h$.

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I don't think this is the definition you want for $V_g^h$.

First of all, an unrelated technical point: the inclusion between $L^2(\Omega)$ and $H^1(\Omega)$ goes the other way: $H^1(\Omega) \subset L^2(\Omega)$. $H^1$ consists of the $L^2$ functions that have an $L^2$ weak derivative, so it is a strictly smaller space than $L^2$.

Second, $V_g^h$ is not even a vector space. The condition $v\big|_{\partial\Omega} = g$ is incompactible with a vector space structure unless $g = 0$.

Third, $V_g^h$ probably doesn't contain the functions you intend for it to contain. There are very few smooth functions that are piecewise linear over a triangulation: the only such functions are in fact globally affine. Smoothness implies that the restrictions of your functions to the triangulations need to match up at the boundaries in a smooth way, and if your functions are linear/affine on the triangulations then this is an extremely restrictive condition. Also, it's not clear that the piecewise linear/affine condition is compatible with the boundary condition: the space could well be empty if $g$ is not itself piecewise affine.

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  • $\begingroup$ First of all, thank you for you answer! Of Course it should be $H^1 \subset L^2$ and clearly I forgot that $V^h_g$ is not even a vector space for $g \not= 0$. Nevertheless $V^h_g$ still contains the functions I Need, because it should contain all possible solution a PDE with boundary condition $g$ in $ \partial \Omega$. $\endgroup$ – superdave99 May 7 at 7:25
  • $\begingroup$ Regarding the boundary condition $g$: If I choose $g \in V^h$ then the space $V^h_g$ is not empty, or am I wrong? $\endgroup$ – superdave99 May 7 at 7:32

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