What is a good definition of a weak solution? I am interested in any heuristic or formal necessary and sufficient conditions that a good definition of a weak solution of a PDE problem must satisfy.
This question is motivated by trying to show the existence of the wave equation $ \Box_{g} \phi = f $ in a compact subset, $ \Omega $, of a $ 4 $-dimensional manifold.
I can prove that I have an estimate of the form
$$
(\| \omega \|_{K_{1}})^{2} \lesssim (\| \square_{g} \omega \|_{K_{1}})^{2},
$$
where
$$
  \| \phi \|_{K_{1}}
= \left[
  \int^{t}_{0}
  \left( \| \phi \|^{1}_{\Sigma_{\tau}} \right)^{2} ~ \mathrm{d}{\tau}
  \right]^{\frac{1}{2}} \quad \text{and} \quad
  \| \phi \|^{1}_{\Sigma_{\tau}}
= \left[
  \int_{\Sigma_{\tau}}
  \left( \frac{\partial \phi}{\partial \tau} \right)^{2} +
  \sum^{3}_{i = 1} \left( \frac{\partial \phi}{\partial x^{i}} \right)^{2} +
  \phi^{2} \mu_{h}
  \right]^{\frac{1}{2}}.
$$
(I have denoted the volume form on each $ 3 $-surface by $ \mu_{h} $.)
Now, if I define a linear functional $ \displaystyle k(\Box \omega) = \int_{\Omega} f \omega $ with domain $V=\{\Box \omega \text{ s.t. } \omega\in C_{o}^{\infty} \}$.
Then I can prove that this is a bounded linear functional in $ K_{1} = {L^{2}}([0,t],{H^{1}}({\Sigma}_{\tau},\mu_{h})) $ consisting of the functions of $ \tau $ with values in the space $ {H^{1}}(\Sigma_{\tau},\mu_{h}) $.
Extending this linear functional using the Hahn-Banach Theorem to the closure of $V$ with respect the $K_{1}$ norm and the Riesz Representation Theorem, I can guarantee there is a $ \Psi \in K_{1} $ that satisfies:
$$
  (f,\omega)_{L^{2}(\Omega)}
= \int_{\Omega} f \omega
= \langle \Psi,\Box \omega \rangle_{K_{1}}.
$$
for all $\Box\omega\in \overline{V}$.
Is this $ \Psi $ a good weak solution?
I can also prove that there is a $\xi$ 
$$
  \langle f,\omega\rangle_{K_{1}}
= \langle \xi,\Box \omega \rangle_{K_{1}}.
$$
for all $\Box\omega\in \overline{V}$.
Is this $ \xi $ a good weak solution?
EDIT
I will accept the answer from BaronVT as he answered the question before I made this edit. 
However, his answer make me think what if we have an estimate of the form
I can prove that I have an estimate of the form
$$
(\| \omega \|_{K_{1}})^{2} \lesssim (\| \square_{g}^{*} \omega \|_{K_{1}})^{2},
$$
would then the weak solution 
$$
  \langle f,\omega\rangle_{K_{1}}
= \langle \xi,\Box^{*} \omega \rangle_{K_{1}}.
$$
for all $\Box\omega\in \overline{V}$.
Notice that by definition of the adjoint in $K_{1}$ any strong solution is satisfying:
$$
  \langle f,\omega\rangle_{K_{1}}
= \langle \xi,\Box^{*} \omega \rangle_{K_{1}}
=\langle \Box\xi,\omega \rangle_{K_{1}}.
$$
 A: Generally speaking, to define weak solutions, you define a bilinear form corresponding to your PDE that is equivalent to the strong formulation for classical solutions, but also admits weaker solutions (i.e. $H^1$ rather than $C^2$), and is such that any sufficiently regular weak solution is also a strong solution.
For instance for the PDE $-\Delta u = f; x \in \Omega$, the right bilinear form is $a(u,v) = \int_\Omega \nabla u \cdot \nabla v \,dx$. You can check that, for $u,v \in C_0^\infty$, you have
$$
a(u,v) = (-\Delta u, v)_{L^2}
$$
and so if $u$ is a strong solution $a(u,v) = (f,v)_{L^2}$, and this is how you define weak solutions: $u\in H^1_0$ that satisfy $a(u,v) = (f,v)_{L^2}$ for every $v \in H^1_0$. You can prove that $a(u,v)$ is symmetric and positive definite on $H^1_0$, thus you have a Hilbert space (with $a(u,v)$ as the inner product) and so you get existence of solutions (relative to that inner product) by fiat (i.e. Hahn-Banach/ Riesz representation theorem).
Now, the way I have usually seen existence of weak solutions for the wave equation $\square = \partial^2_{tt} - \Delta$ is to define a bilinear form
$$
b(u(t),v) = (u''(t),v)_{L^2} + a(u(t),v)
$$
and look for $u$ so that $b(u(t),v) = (f(t),v)_{L^2}$ for all $t$ and all $v \in H^1$.
The way I have usually seen existence proved is by Galerkin approximation (see also Evans' PDE book), rather than what you're doing (Riesz representation / Lax-Milgram); I'm under the impression that you can't just pull the weak solution from the air like you want, and while I can't give you an entirely satisfactory explanation at the moment, I think essentially the problem is that the bilinear form isn't elliptic/bounded below, i.e. $(H^1_0,b(u,v))$ isn't a Hilbert space.
A few further notes about the weak solutions you've defined so far:
You can calculate
$$
\langle u, v \rangle_{K_1} = (\square u, v)_{L^2} - 2(u'',v)_{L^2} + (u,v)_{L^2}
$$
for $u,v \in C^\infty_0$, so a strong solution would satisfy
$$
\langle u, v \rangle_{K_1} + 2(u'',v)_{L^2} - (u,v)_{L^2} = (f, v)_{L^2}
$$
(the left hand side is how we would define the bilinear form; you can also see it isn't positive definite!) and so in particular, neither
$$
\langle \xi, \square w \rangle_{K_1} = \langle f, \square w \rangle_{K_1}
$$
nor 
$$
\langle \Psi, \square w \rangle_{K_1} = ( f, \square w )_{L^2}
$$
are good weak formulations, because strong solutions won't satisfy the weak equation. 
Regarding your edit, you would need the bilinear form $c(u,v) := \langle u, \square^* v \rangle_{K_1}$ to be an inner product (i.e. symmetric and positive definite). I think symmetry follows ($\square$ is self adjoint), but it is not positive definite (there are nontrivial $u$ solving the homogeneous wave equation).
