If $\boldsymbol{v} \in [W^{1, 2}_{loc}(\overline{\Omega})]^3$, then $\boldsymbol{v}\cdot \nabla v \in [L^{3/2}_{loc}(\overline{\Omega})]^3$.

Question

I am reading An Introduction to the Mathematical Theory of the Navier-Stokes Equations of G.P.Galdi and there some point in a proof I do not understand.

Let $\Omega \subset \mathbb R^3$ be an open set (not necessarily bounded). We have $\boldsymbol{v} \in [W^{1, 2}_{loc}(\overline{\Omega})]^3$, where $u \in W^{1, 2}_{loc}(\overline{\Omega})$ if $$u \in W^{1, 2}(\Omega') \quad \text{for all } \Omega' \text{ bounded with } \Omega' \subset \Omega.$$ Now, at some point in a proof, the autor use the fact that as $\boldsymbol{v} \in [W^{1, 2}_{loc}(\overline{\Omega})]^3$, then $$\boldsymbol{v}\cdot \nabla v \in [L^{3/2}_{loc}(\overline{\Omega})]^3.$$ I do not really see why this should be true. Obviously he is using some Sobolev embedding into $L^{n/n-1}_{loc}$ but as $\boldsymbol{v}\cdot \nabla v$ in only in $L^1_{loc}$, not in any Sobolev space, I am a bit confuse. Could anyone help me with that?

Answer 1

Look at individual components. If $v_i \in W^{1,2}(\Omega')$ then $\displaystyle \frac{\partial v_i}{\partial x_j} \in L^2(\Omega')$ and by the embedding theorem $v_i \in L^6(\Omega')$. According to Holder's inequality the product of a function in $L^2$ and a function in $L^6$ lies in $L^{3/2}$.