The following is from the book "The Three-Dimensional Navier-Stokes Equations" by Robinson, Rodrigo, and Sadowski.
For some background, let $H$ be the completion (in the $L^2$ norm) of the space of divergence-free smooth functions with compact support in some bounded set $\Omega \subset \mathbb{R}^3$, and $V = H \cap (H^1_0)^3$. Let $C_{c, \sigma}^{\infty}(\Omega)$ be the set of all divergence free smooth vector fields with compact support. Any norm that has no subscript denotes the $L^2$ norm, $\| \cdot\| = \| \cdot\|_{L^2}$.
They define $u$ to be a weak solution of the Navier-Stokes equations satisfying the initial condition $u_0 \in H$ if
- $u \in L^\infty(0,T; H) \cap L^2(0,T; V)$ for all $T > 0$ and
- $u$ satisfies the equation $$\int_0^s -\langle u, \partial_t \varphi\rangle + \int_0^s \langle \nabla u, \nabla \varphi \rangle + \int_0^s \langle(u \cdot \nabla)u, \varphi \rangle = \langle u_0, \varphi(0)\rangle - \langle u(s), \varphi(s)\rangle \tag{3.3} $$ for all test functions $\varphi \in C^\infty_{c, \sigma}$.
In their proof they reference the following:
Corollary 1.32 Suppose that $u, g \in L^1(0, T; X)$ and that for every $f \in X^*$ $$\langle f, u(t_2)\rangle - \langle f, u(t_1)\rangle = \int_{t_1}^{t_2} \langle f, g(s) \rangle ds$$ for almost every pair $(t_1, t_2)$ with $0 \leq t_1 \leq t_2 \leq T$. Then $\partial_t u = g$ in the sense of Definition 1.30 (which gives the definition for a weak time derivative, which I have omitted in this post).
Lemma 3.4 If $u \in L^\infty(0,T; H) \cap L^2(0, T; V)$ then $$(u \cdot \nabla)u \in L^{4/3}(0,T; L^{6/5}).$$
Lemma 3.6 If $u$ is a weak solution of the Navier-Stokes equations then $$\int_{t_1}^{t_2} \langle \nabla u, \nabla \phi\rangle + \int_{t_1}^{t_2} \langle(u \cdot \nabla )u, \phi \rangle = \langle u(t_1), \phi \rangle - \langle u(t_2), \phi \rangle \tag{3.4}$$ for $\phi \in C_{c, \sigma}^\infty(\Omega)$ and almost every $t_2 \geq t_1$, for almost every $t_1 \geq 0$ including $t=0$.
I am having some trouble following their proof of the following lemma:
Lemma 3.7 If $u$ is a weak solution of the Navier-Stokes equations then $$\partial_t u \in L^{4/3}(0,T; V^*).$$
Their proof:
From Lemma 3.6 we have, for almost all $t > 0$ and for every choice of $\varphi \in C_{c, \sigma}^\infty(\Omega)$, $$\langle u(t), \varphi \rangle - \langle u_0, \varphi \rangle = \int_0^t \langle g, \varphi\rangle_{V^* \times V},$$ where $g$ is given by $$\langle g, \varphi\rangle_{V^* \times V} = -\langle \nabla u, \nabla \varphi \rangle - \langle(u \cdot \nabla )u, \varphi \rangle.$$ From our assumption on the integrability of $u$, it follows that $g$ is a bounded linear functional on $V$ for almost all $s$, because $$|\langle \nabla u, \nabla \varphi \rangle| \leq \| \nabla u\| \|\nabla \varphi \|$$ and $$| \langle (u \cdot \nabla) u, \varphi \rangle| \leq \|(u \cdot \nabla) u\|_{L^{6/5}} \|\varphi\|_{L^6} \leq c \|(u \cdot \nabla) u\|_{L^{6/5}} \|\nabla \varphi \|,$$ where we used the estimate in the proof of Lemma 3.4 and the Sobolev embedding $H^1 \subset L^6$.
I am able to follow everything until this point. From here the proof continues:
Hence, Lemma 3.4 yields $$\|g\|_{V^*}^{4/3} \leq c \|(u \cdot \nabla) u\|_{L^{6/5}}^{4/3} + c\|\nabla u \|^{4/3} \\\leq c \|(u \cdot \nabla )u\|_{L^{6/5}}^{4/3} + c(1 + \|\nabla u\|^2) \in L^1(0,T).$$ It follows that $g \in L^{4/3}(0, T; V^*)$ and using Corollary 1.32 we obtain $$\partial_t u \in L^{4/3}(0, T; V^*).$$
I do not understand much of this part. How do they obtain these bounds for $g$ in the $V^*$ norm, and in particular how does this all follow from Lemma 3.4? How are the terms on the right hand side in $L^1$, and how does that imply $g \in L^{4/3}(0, T; V^*)$? How does Corollary 1.32 lead to the result?