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 $?



Bound of solutions by initial value of Navier Stokes equations
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.


