Do we need to identify dual spaces in PDEs? In PDEs we often use the fact that we can identify dual spaces eg. $L^2(0,T;V)^* = L^2(0,T;V^*)$ in the sense that
$$u_t + Au = f$$
where $u_t$, $f \in L^2(0,T;V^*)$ and $A:L^2(0,T;V) \to [L^2(0,T;V)]^*$ (eg. $A=\Delta$). Because we identify the dual spaces, $Au$ and $f$ both lie in the same space hence the equality makes sense.
There are more complicated examples with $L^p(0,T;V)^*$ and $L^q(0,T;V^*)$ where $p$ and $q$ are conjugate.
My question is, is this identification always necessary when dealing with PDE problems? Every paper I read seems to use this identification. Suppose I have a very weird space and I cannot show that I can identify the dual spaces (eg. if $V$ is not Banach/reflexive). Can one pose the problem differently? 
 A: I would like to give an opinion. First I think that the need of such properties, for example, the properties of a space being Hilbert, or the propertie of a space being reflexive, motivated the discovery of such spaces. When solving, for example, an equation like you have proposed, if such space can be identified, then we can solve it with some ease, however this is not the only method to solve such equation.
We can look for example to the Perron method (page 51). The technique is different (note that the spaces here are not reflexive) and there is no need to idenfity dual spaces.
Another example is the proposition 4.2.1 of Pucci-Serrin book. They work in a non-reflexive space, however they do not use a variational method, they just use a topological method, in fact they show that to find a solution to the proposed problem, is equivalently to find a fixed point for some funtion.
My conclusion is: the identification is not necessary. Some problems can be solved by different ways and the choice of such way depends on who will solve the problem. 
A: I would like to add that the approach using pairs of dual spaces is most useful for parabolic and elliptic problem classes that have linear versions as special cases (so this includes elliptic quasilinear problems in divergence form). The reason is that such problems can sometimes be thought of as "study a map $\Phi : V \to V^\ast$" where $\Phi = F^\ast \circ \phi \circ F$, and $F: V \to W$ is linear (a differentiation operator), $\phi: W \to W^\ast$ is nonlinear (e.g. a substitution operator), and $F^\ast$ is something like an adjoint to $F$. 
For fully nonlinear elliptic problems as well as for genuinely nonlinear hyperbolic ones it is known that this does not work. The examples given by @Tomas show this very clearly.
This is not just an "argument from ignorance". For example it is known that hyperbolic problems that are well-posed in spaces related to $L^2$ must essentially be linear problems (or perturbations of such problems).  
