Concerning the setting for Navier-Stokes equation, we often see the conditions for various functions involved. To be specific, consider we ought to derive the existence of weak solution of the Navier-Stokes equation with some boundary condition. The equation is as you know $$ \partial \mathbf{v} - \nu \Delta \mathbf{v} + (\mathbf{v} \cdot \nabla) \mathbf{v} + \nabla p = \mathbf{g} $$ with some boundary condition where $\mathbf{v}$ is the unknown vector field and $\mathbf{g}$ stands for the force density. Some authors often requires that $\mathbf{g}$ is in the direct sum $V=L^1(0,T;L_\sigma^2 (\mathbb{R}^3)) + L^2(0,T;W_{0,\sigma}^{-1,2}(\mathbb{R}^3))$ where $L_\sigma^2(\mathbb{R}^3)$ is the completion of $C_{0,\sigma}^\infty ( \mathbb{R}^3)$, the space of smooth divergence-free vector fields, with respect to the norm $\| \cdot \|_2$ and $W_{0,\sigma}^{-1,2}$ is the dual of $W_{0,\sigma}^{1,2}(\mathbb{R})$ where $W_{0,\sigma}^{1,2}(\mathbb{R})$ is the completion of $C_{0,\sigma}^\infty ( \mathbb{R}^3)$ with respect to the norm $\| \cdot \|_{1,2}$. Here I would like to raise two questions.
- Why $V$ is direct sum of the two spaces?
- Why it is natural to require such a seemingly artificial condition?
For the appearence of such a condition, see, for example, https://onlinelibrary.wiley.com/doi/abs/10.1002/mma.1059
1) This is only sum, not direct sum, as the intersection of both spaces is larger than $\{0\}$. However, this does not matter.
2) Because one can prove existence of weak solutions for these right-hand sides with solution satisfying $v\in L^2(0,T, H^1)\cap L^\infty(0,T,L^2)$, which is a common setting for weak formulations of parabolic pdes.