= Solution
In three dimensions, the <Sobolev inequality> gives $\lVert v\rVert_{L^6(U)}\leq C_U\lVert v\rVert_{H_0^1(U)}$. Therefore
$$
\lVert w^3\rVert_{L^2(U_T)}^2
=\int_0^T\lVert w(t)\rVert_6^6dt
\leq C_U^6T\lVert w\rVert_{L_t^\infty H_x^1}^6,
$$
and hence
$$
\boxed{\lVert w^3\rVert_{L^2(U_T)}
\leq C_U^3T^{1/2}\lVert w\rVert_{L_t^\infty H_x^1}^3.}
$$
Using $w^3-\widetilde w^3=(w-\widetilde w)(w^2+w\widetilde w+\widetilde w^2)$, the <Holder inequality> and the same Sobolev embedding give at each time
$$
\lVert w^3-\widetilde w^3\rVert_2
\leq C_U^3\lVert w-\widetilde w\rVert_{H^1}
\big(\lVert w\rVert_{H^1}^2+\lVert\widetilde w\rVert_{H^1}^2\big).
$$
Taking the $L^2$ norm in time supplies the required estimate with the factor $T^{1/2}$.
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