Edit: This question has been substantially modified on January 12th, 2015.
I have been studying Michael Struwe's paper "On Partial Results for the Navier-Stokes Equations", Comm. Pure Appl. Math 41 (1988), no. 4, 437-458, and I have a question regarding the rigorous justification of the formal calculations in the proof of the Serrin condition in Section 3 of the paper. I have also studied the original proof by Serrin.
Consider the weak solutions of the Navier-Stokes system. Let's assume $n=3$ for simplicity. A weak solution $u \in L^{2, \infty}(Q; \mathbb R^3)$, with $|\nabla u| \in L^2(Q)$, is defined by requiring that $\nabla \cdot u=0$ and $$ \int_Q -u \cdot \partial_t \varphi + \nabla u \cdot \nabla \varphi - [(u \cdot \nabla)\varphi \cdot u] \, dx \, dt =0 $$ for every $\varphi \in C_0^\infty(Q)$ with $\nabla \cdot \varphi=0$.
By choosing a test function $\varphi = \nabla \times \psi$ with a compactly supported smooth $\psi=(\psi_1, \psi_2, \psi_3)$ it is possible to show that the vorticity $\omega = \text{curl}\ u$ satisfies the vorticity equation $$\tag 2 -\sum_{i=1}^3 \int_Q \omega_i\partial_t\psi_i+\omega_i\Delta \psi_i+ \sum_{j=1}^3 (u_i\omega_j-\omega_iu_j)\frac{\partial \psi_i}{\partial x_j} \, dx \, dt= 0. $$
Then, in the proof by Struwe, one formally chooses a test function $\psi=\omega|\omega|^{2s-2}\zeta^2$, where $\zeta$ is a standard cut-off function. In order to rigorously justify this formal choice of the test function, one considers an approximated vorticity equation (2) with $u_i$ replaced by $u_i^k$, where $u_i^k$ is any smooth approximation of $u_i$ in $L^2$-norm. Then one obtains an approximative solution $\omega_i^k$, which has enough regularity for the formal calculations to be rigorously justified.
The final step of the proof is to show that $\omega_i^k \to \omega_i$ weakly in $L^2$. By some analysis, it is possible to show that $\omega_i^k$, indeed, converge to a function $\tilde \omega_i$, and also that for each $k$, the function $\omega_i^k$ is a unique solution of the approximative equation.
Now my question is that how do we know that $\omega_i = \tilde\omega_i$?
I believe that it is possible to choose the approximative sequence $u_i^k$ to be smooth solutions of the Navier-Stokes equation. In this case, it is clear that $\omega_i^k = \nabla \times u_i^k$ and, as discussed below in the comments section, then we may use the continuity of differential operators in the distributional topology to conclude that $\omega_i^k = \nabla \times u_i^k \to \nabla \times u = \omega_i$ as $k \to \infty$. However, it seems like a non-trivial result that $u_i^k$ can be chosen to be smooth solutions of the Navier-Stokes equation, and Struwe does not state or prove such result.
I am mostly interested in the methodology rather than the result itself and, thus, I would like to understand if it is possible to avoid using the fact that $u_i^k$ solves the equation. In particular, if $u_i^k$ is any smooth approximation of $u_i$, I do not see why $\omega_i^k$ needs to satisfy $\omega_i^k = \nabla \times u_i^k$. Perhaps this is something which follows from the fact that $\omega_i^k$ is a unique solution of the approximative equation? My first guess would be to try showing by Biot-Savart law that $\nabla \times u_i^k$ is always a solution and then to rely on uniqueness, but I would prefer not to reinvent the wheel if someone knows how to do this.
