We proved in the lecture, that the weak solution of the linear hyperbolic equation
\begin{align} u_{tt}+Lu&=f\ \ \ \mathrm{in}\ (0,T)\times\Omega \\ u &= 0\ \ \ \mathrm{on}\ (0,T)\times\partial\Omega \\ u &= g\ \ \ \mathrm{on}\ \{0\}\times\Omega \\ u_t &= h\ \ \ \mathrm{on}\ \{0\}\times\partial\Omega \end{align}
exists and the following bound applies: $$\|u\|_{L^{\infty}(I,W_0^{1,2}(\Omega))}+\|u_t\|_{L^{\infty}(I,L^{2}(\Omega))}+\|u_{tt}\|_{L^{\infty}(I,W_0^{1,2}(\Omega))^*}\leq C(\|f\|_{L^2(I\times\Omega)}+\|g\|_{W_0^{1,2}(\Omega)}+\|h\|_{L^2(\Omega)})$$
Then another theorem follows to prove, that the weak solution is unique - and it is quite complicated. If I understand it correctly - to prove the uniqueness it should suffice to show that $f=g=h=0\Rightarrow u=0$. And we get that automatically from the bound above (or at least so it seems to me - if the right hand side is 0, the norm of $u$ is 0 and therefore $u$=0).
Where am I mistaken? Why do we need a complicated proof (testing the weak formulation with the antiderivative - as in Evans 7.2, Theorem 4)?
I think I get it now. It's as I say in the comments - one can prove by the Galergin approximations that a solution exists and satisfies the bound. However, one does not prove in this manner that any solution to the equation satisfies this bound.
In the case of a parabolic equation, one can prove that these estimates apply to any solution by testing the weak formulation with the solution.
This is not possible in the hyperbolic case, because one would have to test the weak formulation with $u_t$, but $u_t$ is not regular enough for a test function.