I'm working on an old exam problem: Define for $u \in C^2([-1,1])$ the operator $L$ by $[Lu](x) = - \frac{d}{dx} \left( (1-x^2) u'(x) \right)$. Set $\Omega = \{ Lu \mid u \in C^2([-1,1]) \}$. Find the topological closure of $\Omega$ in $L^2([-1,1])$.
I am given the hint that for $f \in \overline{\Omega}$, $Lu=f$ has a solution $u \in L^2([-1,1])$ when the operator $L$ is defined "weakly". But I don't really know what that means . . . The only "weak" definition I know of is the weak convergence of functionals.
Thanks for the help.