Considering the simplest boundary value problem. $$\Delta u=f,\text{in}\,\Omega.$$ $$u=0,\text{in}\,\partial\Omega.$$
In evans we always assume $f\in L^{2}(\Omega)$ and use integral by parts to $v\Delta u=fv$ where $v$ is a test function.
My question is,how to derive the weak solution formula when $f\in H^{-1}$?
I know the right hand side should be understood as $\langle f,v\rangle=\int f^{0}v+\sum f^{i}v_{x^{i}}.$
But how can I understand the left hand side $\langle\Delta u,v\rangle$?In order to integrate by part it must just be the integral $\int \Delta u v$.But I cannot see why.
The notation $f \in H^{-1}$ means that $f$ is in the dual space of $H_0^1$, i.e., $f$ is a linear mapping of $v \in H_0^1$ to (in this context most likely) $\mathbb R$. Thus, it might make more sense to write in this case $f(v)$.
So the space from which $v$ stems is in this case already fixed to $H^1_0$. While you are in theory still free to choose the space $U$ for $u \in U$, it makes most sense to pick the same as for $v$, as then the standard theory (Lax-Milgram, error-estimates, ...) applies. Otherwise, you have to start figuring out stuff on your known, based on how the spaces $H^1_0$ and $U$ are related to one another.
For $u \in H^1_0$, you then obtain the standard LHS
$$ \int_\Omega \Delta u v \mathrm d x = \int_\Omega \nabla \cdot (v \nabla u) - \nabla v \cdot \nabla u \mathrm d x \overset{\text{Divergence Theorem & B.C.}}{=} -\int_\Omega \nabla v \cdot \nabla u \mathrm d x $$