For the mollified Navier Stokes equations: $$\partial_t u_{\epsilon} - \Delta u_{\epsilon} + \mathbb P \nabla \cdot((u_{\epsilon} \ast \omega_{\epsilon})\otimes u_{\epsilon})=0 $$ $$\nabla \cdot u_{\epsilon} = 0$$ $$u_{\epsilon} (0, \cdot) = u_0$$ where $\omega_{\epsilon}$ is a standard mollifier.
How can we derive $\|u_{\epsilon}\|_2 \leq \|u_0\|_2 $?
You note first that $$\text{div}(\omega_{\epsilon}\ast u_{\epsilon})= \omega_{\epsilon}\ast\text{div }u_{\epsilon}=0, $$ so that, at least formally, $$ \langle(\omega_{\epsilon}\ast u_{\epsilon})\cdot \nabla u_{\epsilon},u_{\epsilon}\rangle_{L^2}= \langle\partial_j u_{\epsilon}, (\omega_{\epsilon}\ast u_{\epsilon})_ju_{\epsilon}\rangle_{L^2} =-\langle u_{\epsilon},\partial_j (\omega_{\epsilon}\ast u_{\epsilon})_ju_{\epsilon}\rangle_{L^2}, $$ and thus $ \langle(\omega_{\epsilon}\ast u_{\epsilon})\cdot \nabla u_{\epsilon},u_{\epsilon}\rangle_{L^2}=0. $
The sequel is standard, you multiply in $L^2$ the equation by $2u_\epsilon$ and you get by integration in time for $t\ge 0$, $$ \Vert u_\epsilon(t)\Vert_{L^2}^2+2\int_0^t\Vert \nabla u_\epsilon(t')\Vert_{L^2}^2 dt'=\Vert u_\epsilon(0)\Vert_{L^2}^2, $$ yielding the sought result.
