If $u \in L^\infty(0, T; H) \cap L^2(0, T; V)$ then $(u \cdot \nabla)u \in L^{4/3}(0,T; L^{6/5})$

Question

Let $H$ be the completion of the space of divergence-free smooth functions with compact support in some bounded set $\Omega \subset \mathbb{R}^3$, and $V = H \cap (H^1_0)^3$. I am following a proof of the following lemma:

If $u \in L^\infty(0, T; H) \cap L^2(0, T; V)$ then $$(u \cdot \nabla)u \in L^{4/3}(0,T; L^{6/5}).$$

The proof states that using the Holder inequality with exponents $5/3$ and $5/2$, $$\|(u \cdot \nabla)u\|_{L^{6/5}} \leq \Big(\int |\nabla u|^{6/5} |u|^{6/5}\Big)^{5/6} \tag{1}$$ $$\leq \Big(\int |\nabla u|^2\Big)^{1/2}\Big(\int |u|^3\Big)^{1/3} \tag{2}$$ $$=\|\nabla u\|_{L^2} \|u\|_{L^3} \tag{3}$$ $$\leq \| \nabla u\|_{L^2} \|u\|_{L^2}^{1/2}\|u\|_{L^6}^{1/2} \tag{4}$$ $$\leq c\|u\|_{H^1}^{3/2}\|u\|_{L^2}^{1/2} \tag{5}.$$ Thus, $$\int_0^T \|(u \cdot \nabla)u\|_{L^{6/5}}^{4/3} \leq c \int_0^T \|u\|_{H^1}^2 \|u\|_{L^2}^{2/3} \tag{6}$$ where the right hand side is finite because $u \in L^\infty(0, T; H) \cap L^2(0, T; H^1)$.

I am new to working with these kind of inequalities. How are (1), (2), (4), and (5) established? In (6), what ensures $\|u\|_{H^1}$ is finite?

Answer 1

$(1)$: We can prove $|(u \cdot \nabla)u| \le |\nabla u| |u|$ by writing everything out explicitly and using Cauchy-Schwartz:

$$|(u \cdot \nabla)u| \le \sqrt {\sum_{i = 1}^n u_i^2} \sqrt {\sum_{i, j = 1}^n (\partial_i u_j)^2} = |u||\nabla u|$$

$(2)$: I'm not sure where the confusion with this bit is. We apply Holder's with exponents $(5/3, 5/2)$ which yields:

$$\lVert (u \cdot \nabla) u\rVert_{L^{6/5}} \le \left(\int |\nabla u|^{(6/5) \times (5/3)}\right)^{(5/6) \times (3/5)} \left(\int |u|^{(6/5) \times (5/2)}\right)^{(5/6) \times (2/5)}$$

$(4)$: This is Lebesgue interpolation, let $\Omega \subseteq \mathbb R^3$ be measurable, $1 \le p \le r \le q \le \infty$ and $u \in L^p(\Omega) \cap L^q(\Omega)$. Then $u \in L^r(\Omega)$ and further:

$$\lVert u\rVert_{L^r} \le \lVert u\rVert_{L^p}^\alpha \lVert u\rVert_{L^q}^{1 - \alpha}$$ where $\alpha$ satisfies: $$\frac 1 r = \frac \alpha p + \frac {1 - \alpha} q$$ Here we are applying Lebesgue interpolation with $(p, q,r) = (2, 6, 3)$, and we find $\alpha = 1/2$.

$(5)$: We see that $\lVert \nabla u\rVert_{L^2} \le \lVert u\rVert_{H^1}$ simply since $\lVert u\rVert_{H^1}^2 = \lVert u\rVert_{L^2}^2 + \lVert \nabla u\rVert_{L^2}^2$.

$(6)$: We are given that $u \in L^2(0, T; V)$. So $t \mapsto \lVert u(t)\rVert_V := \lVert u(t)\rVert_{H^1}$ is $L^2$ on $[0, T]$. Also $u \in L^\infty(0, T; H)$, so $t \mapsto \lVert u(t)\rVert_H := \lVert u(t)\rVert_{L^2}$ is $L^\infty$ on $[0, T]$. The last integral is hence finite.

The book The Three-Dimensional Navier–Stokes Equations by Robinson, Rodrigo and Sadowski also contains this calculation: Lemma 3.4 on page 74. (though not in any more detail than you've written) I'm not sure if you're currently working through it, but statements of Lebesgue interpolation and a few other things are found in the first chapter.