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We want to find $u \in H^{1}_{0}(\Omega)$, such that $$\int_\Omega \lambda \nabla u\nabla v dx =\int_\Omega fvdx \text{ } \forall v \in H^{1}_{0}(\Omega) $$ We then define the bilinear form $a(u,v) :=\int_\Omega \lambda \nabla u\nabla v dx $, which is bounded: $$ |a(u,v)| = |\lambda| |\int_\Omega \nabla u\nabla v dx| \leq ||u||_{H^{1}_{0}(\Omega)} ||v||_{H^{1}_{0}(\Omega)}$$

We can now say, that the functional $v \rightarrow a(u,v)$ lives in the dualspace $H^{1}_{0}(\Omega)^*$. By our first equation, we get $$\int_\Omega fvdx = a(u,v)\text{ }\forall v \in H^{1}_{0}(\Omega)$$ Since $H^{1}_{0}(\Omega)$ is a Hilbert Space, we can identify it with its dual, so we can now say: $$f(v) = a(u,v)$$ Is that correct? I One thing that confuses me, is that my book does the derivation of the weak formulation of the stationary heat conduction in almost the same way, but uses a testfunction $v \in C^{\infty}_0$ instead of $H^{1}_{0}(\Omega)$. What makes the difference in that? Bonusquestion: I always assumed that the 0 in the subscript of $H^{1}_{0}(\Omega)$ claims compact support. But does it really??

Thanks!

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The $0$ subscript does not mean compact support. The definition is

$$H^1_0(\Omega) = \text{Closure}(C^\infty_0(\Omega)),$$

where closure is under the $H^1$ norm. Thus, $C^\infty_0(\Omega)$ is dense in $H^1_0(\Omega)$, which means any $u\in H^1_0(\Omega)$ can be approximated arbitrarily well (in the $H^1$ norm) by a smooth function with compact support, even though $u$ may not itself have compact support.

Since $C^\infty_0(\Omega)$ is dense in $H^1_0(\Omega)$, it does not matter whether your test functions belong to $C^\infty_0(\Omega)$ or $H^1_0(\Omega)$, the notion of weak solution is the same. Indeed, by density you can show that if the weak formulation held for just $C^\infty_0$ test functions, then it would hold for $H^1_0$ functions as well (by approximation).

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