Dual spaces and weak solutions. I have two questions:
(1) $H^{-1}$ space is defined as the dual space of $H_{0}^{1}$, so is the dual space of $H^{1}$ also $H^{-1}$? Or is it correct to act an $T\in H^{-1}$ on a function $u\in H^{1}$?
(2) For the Poisson problem $-\Delta u=f$ with boundary condition $u=g$ for proper conditions on the data and domain. If there exists $w\in H^{1}$ with $w=g$ on the boundary s.t. it holds for all $v\in H_{0}^{1}$ that $\int \nabla w \nabla v=\int fv$, then what can we say about this $w$? Can we get that $w$ is the weak solution of the Poisson problem?

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*$H^1_0$ is contained in $H^1$ and has the same norm, so $H^{-1}$ must be contained in the dual of $H^1$. It turns out that it is a proper subset of the actual dual of $H^1$.

*Sure, this is usually the definition of the weak solution (though we have to interpret "$w=g$ on the boundary" using the trace operator). In particular you are correct to consider only $v \in H^1_0$ rather than $v \in H^1$. I find this easier to understand in the context of the corresponding variational problem: you should be able to add a variation $v$ to a solution candidate $u$ and get another solution candidate. For Dirichlet conditions this means the variations must vanish on the boundary. Then playing with it a bit you find that variations in the variational context are the same as test functions in the distributional context.
