Is every linear finite element space over a bounded domain a subspace of the sobolev space H^1?

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 $$u_h$$.

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.
• 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 $\partial \Omega$. – superdave99 May 7 at 7:25
• 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? – superdave99 May 7 at 7:32