The answer I think must be no. However I can not see what is wrong with the following reasoning.
Assume we have an estimate of the form for the adjoint of an operator $L$ in $L^{2}$ of the form:
$$||\phi||\le||L^{*}\phi||$$
for all $\phi\in C^{\infty}\bigcap H^{k}=V$ and where $||\phi||=\int\phi^{2}$ which is the usual $L^{2}$ norm and $H^{k}$ is higher order Sobolev spaces.
Now $V$ and $H^{k}$ are subspaces of $L^{2}$. Moreover, $V\subset H^{k}$
If we define an inner product space $W$ as the functions $H^{k}$ but with inner product in $L^{2}$ i.e., our Hilbert space is the vector space with elements $f,g\in H^{k}$ but with inner product $(f,g)_{L^{2}}$.
Then we can use Hahn-Banach theorem on the functional
$$k(L^{*}\omega)=\int F\omega$$
to extend the linear functional to $W$, where we are seeing $W$ as the whole space and $V$ as a subspace.
Hence we have a bounded linear functional in $W$. Hence $k\in W^{*}$.
Now using Riez representation theorem there is a $\Psi\in W$ such that
$$(\Psi,L^{*}\omega)_{W}=(\Psi,L^{*}\omega)_{L^{2}}=(F,\omega)_{L^{2}}$$
which is a weak solution, but because $\Psi\in W$ we have that $\Psi\in H^{k}$ ?!
I would appreciate a lot any comments where this went wrong.
The space $W$ as you defined it will typically be an incomplete inner product space, not a Hilbert space. You can't use the Riesz representation theorem there.