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I assume $u\in H^1(\Omega)$ for some nice domain $\Omega\subset\mathbb R^n$, and you wonder why $Lu\in H^{-1}(\Omega)$. (Correct me if I'm wrong.) For any $v\in H^1_0$, formal integration by parts gives $$ \langle Lu,v\rangle = \int_\Omega -a_{ij}D_juD_iv+b_iD_iub+cuv. $$ This integral makes perfect sense since $u,v,\nabla u,\nabla v\in L^2(\Omega)$. This is actually how one should define $L$ as an operator $L:H^1\to H^{-1}$, and it should not be hard show that it is continuous if the weights $a,b,c$ are good enough. The reason for taking $v\in H^1_0$ instead of $H^1$ is that it's nice to have a formulation without boundary terms that agrees with the classical one for smooth functions.