In the paper 'On the Existence of the Pressure for Solutions of the Variational Navier-Stokes Equations' by J. Simon he first defines on page 229 that $\partial_t u \in W^{-1,2}\left((0,T);W^{0,2}(\Omega)\right)$ and on page 231 he says that $w=f-\partial_t u + \nu \Delta u - (u \cdot \nabla)u \in W^{-1,\infty}\left((0,T);W^{-1,2}(\Omega)\right)$.
So is the following true?
$$W^{-1,2}\left((0,T);W^{0,2}(\Omega)\right)\subset W^{-1,\infty}\left((0,T);W^{-1,2}(\Omega)\right)$$
where $W^{k,p}=\begin{cases} \{u \in L^p:\partial^\alpha u \in L^p ~\forall |\alpha|\leq k\} &,k \geq 0 \\ (W^{k,p}_0)' &,k<0 \end{cases}$.
My try:
- As $W^{0,2}(\Omega) \subset W^{-1,2}(\Omega)$ we have $W^{-1,2}\left((0,T);W^{0,2}(\Omega)\right) \subset W^{-1,2}\left((0,T);W^{-1,2}(\Omega)\right)$. So IF I could show $$W^{-1,2}\left((0,T);W^{-1,2}(\Omega)\right)\subset W^{-1,\infty}\left((0,T);W^{-1,2}(\Omega)\right),$$ I'd be done. As $W^{-1,2}$ is defined as the dual space of $W^{1,2}_0$ and $W^{-1,\infty}$ as the dual space of $W^{1,1}_0$ I could alternatively show $$W^{1,1}_0\left((0,T);W^{1,2}_0(\Omega)\right) \subset W^{1,2}_0\left((0,T);W^{1,2}_0(\Omega)\right)$$ but unfortunately this is not true. So exploiting $W^{0,2} \subset W^{-1,2}$ at the beginning seems not the way to go.
I am grateful for any hint/comment/help.
Okay, I think I got it. It is true that $\partial_t u \in W^{-1,\infty}(0,T;W^{-1,2}(\Omega))$ in some sense. But this doesn't follow from $W^{-1,2}(0,T;L^2(\Omega)^d)$.
As $u$ is a solution of the Navier-Stokes equations, it is weakly continuous with values in $L^2(\Omega)^d$. Hence $u$ is bounded with values in $L^2(\Omega)^d$ and hence also with values in $H^{-1}(\Omega)^d$. Therefore, $u$ induces a distribution in $W^{-1,\infty}(0,T;H^{-1}(\Omega)^d)$.
And the distributional derivative of $u$ is also in this space.