I'm studying Sobolev spaces using Evans' PDE book. I can't figure out this simple fact. Let $L$ be an operator in this form: $$Lu= \sum{D_i(a_{ij}D_j(u))+\sum{b_iD_i(u)+cu}}.$$

I can't understand why $L \in H^{-1}$, the dual space of $H_0^1$. I'm struggling because to me it would require that $u$ must have 2 derivatives so $u \in H^2$ to be well defined. Am I missing something? Thanks.

I'm studying Sobolev spaces using Evans' PDE book. I can't figure out this simple fact. Let $L$ be an operator in this form: $$Lu= \sum{D_i(a_{ij}D_j(u))+\sum{b_iD_i(u)+cu}}.$$

I can't understand why $L \in H^{-1}$, the dual space of $H_0^1$. I'm struggling because to me it would require that $u$ must have 2 derivatives so $u \in H^2$ to be well defined. Am I missing something? Thanks.

EDIT: My problem is about this identity $Lu=f$, $f \in L^2$, in $ \Omega$ plus some boundary conditions for example $u \in H_0^1$.I can define a weak solution as $u \in H_0^1$ such that $B[u,v]=(f,v) $ for all $v \in H_0^1$. Ok that's fine, but when i write $Lu=f $ what does it mean?

I'm studying sobolev spaces.(I'm using Evans PDE book) I can't figure out this simple fact. Let $L$ be an operator in this form: $Lu= \sum{D_i(a_{ij}D_j(u))+\sum{b_iD_i(u)+cu}}$

I can't understand why $L \in H^{-1}$, the dual space of $H_0^1$, i'm struggling beacuase to me it would require that $u$ must have 2 derivaties so $u \in H^2$ to be well define. Am I missing something? Thanks.

I'm studying Sobolev spaces using Evans' PDE book. I can't figure out this simple fact. Let $L$ be an operator in this form: $$Lu= \sum{D_i(a_{ij}D_j(u))+\sum{b_iD_i(u)+cu}}.$$

I can't understand why $L \in H^{-1}$, the dual space of $H_0^1$. I'm struggling because to me it would require that $u$ must have 2 derivatives so $u \in H^2$ to be well defined. Am I missing something? Thanks.