This is an offshoot of another question I asked (If $u$ is a weak solution of the Navier-Stokes equations then $\partial_t u \in L^{4/3}(0,T; V^*)$) which contains all the background information. However here I have tried to omit as many unnecessary details as possible to keep the question more focused.
Let $V \subset H^1$ and $H \subset L^2$ denote function spaces, and let $g$ a linear functional on $V$ defined by: $$g(\varphi) = -\langle \nabla u, \nabla \varphi \rangle - \langle(u \cdot \nabla )u, \varphi \rangle$$ where $u \in L^\infty(0,T; H) \cap L^2(0,T; V)$ is fixed. I would like to show that $$\|g\|_{V^*}^{4/3} \leq c \|(u \cdot \nabla) u\|_{L^{6/5}}^{4/3} + c\|\nabla u \|^{4/3}_{L^2} \tag{1}$$ $$\leq c \|(u \cdot \nabla )u\|_{L^{6/5}}^{4/3} + c\big(1 + \|\nabla u\|_{L^2}^2\big) \tag{2}.$$
I have the following two estimates: $$|\langle \nabla u, \nabla \varphi \rangle| \leq \| \nabla u\|_{L^2} \|\nabla \varphi \|_{L^2} \tag{3}$$ $$| \langle (u \cdot \nabla) u, \varphi \rangle| \leq \|(u \cdot \nabla) u\|_{L^{6/5}} \|\varphi\|_{L^6} \leq c \|(u \cdot \nabla) u\|_{L^{6/5}} \|\nabla \varphi \|_{L^2}. \tag{4}$$
So far what I have is that $$\|g\|_{V^*}^{4/3} = \sup_{\varphi \in V} \Big(\frac{\big|-\langle \nabla u, \nabla \varphi \rangle - \langle(u \cdot \nabla )u, \varphi \rangle\big|}{\|\varphi\|_{L^2}}\Big)^{4/3} \\\leq \sup_{\varphi \in V} \Big(\frac{\big|\langle \nabla u, \nabla \varphi \rangle\big| + \big|\langle(u \cdot \nabla )u, \varphi \rangle\big|}{\|\varphi\|_{L^2}}\Big)^{4/3}$$ $$\leq 2^{1/3}\Bigg[\sup_{\varphi \in V} \Big(\frac{|\langle \nabla u, \nabla \varphi \rangle| }{\|\varphi\|_{L^2}}\Big)^{4/3} + \sup_{\varphi \in V} \Big(\frac{|\langle(u \cdot \nabla )u, \varphi \rangle|}{\|\varphi\|_{L^2}}\Big)^{4/3}\Bigg]$$ which by (3) and (4) continues as $$\leq 2^{1/3}\Bigg[\|\nabla u\|^{4/3}_{L^2}\sup_{\varphi \in V} \Big(\frac{\|\nabla \varphi\|_{L^2} }{\|\varphi\|_{L^2}}\Big)^{4/3} + c \|(u \cdot \nabla) u\|^{4/3}_{L^{6/5}} \sup_{\varphi \in V} \Big(\frac{\|\nabla \varphi \|_{L^2}}{\|\varphi\|_{L^2}}\Big)^{4/3}\Bigg]$$ $$= 2^{1/3}\sup_{\varphi \in V} \Big(\frac{\|\nabla \varphi\|_{L^2} }{\|\varphi\|_{L^2}}\Big)^{4/3}\Bigg[\|\nabla u\|^{4/3}_{L^2} + c \|(u \cdot \nabla) u\|^{4/3}_{L^{6/5}}\Bigg] \tag{5}.$$
This is where I am stuck. How do we know that the supremum in (5) is finite? Even if this term was to be proven finite, both terms in the brackets in (5) will not have the same constant as in (1) so I believe I may be making a mistake somewhere. I am also unsure how to obtain (2). Any help would be much appreciated.
EDIT: I noticed a mistake, when calculating the operator norm of $g$ the norm in the denominator should be the $H^1$ norm. Thus (5) should read $$= 2^{1/3}\sup_{\varphi \in V} \Big(\frac{\|\nabla \varphi\|_{L^2} }{\|\varphi\|_{H^1}}\Big)^{4/3}\Bigg[\|\nabla u\|^{4/3}_{L^2} + c \|(u \cdot \nabla) u\|^{4/3}_{L^{6/5}}\Bigg] \tag{6} $$ Since $\|u\|_{H^1}^2 = \|u\|_{L^2}^2 + \|\nabla u\|_{L^2}^2$ implies $\|\nabla u\|_{L^2} \leq \|u\|_{H^1}$, we have that $$(6) \leq 2^{1/3}\sup_{\varphi \in V} \Big(\frac{\|\varphi\|_{H^1} }{\|\varphi\|_{H^1}}\Big)^{4/3}\Bigg[\|\nabla u\|^{4/3}_{L^2} + c \|(u \cdot \nabla) u\|^{4/3}_{L^{6/5}}\Bigg]\\ = 2^{1/3}\Bigg[\|\nabla u\|^{4/3}_{L^2} + c \|(u \cdot \nabla) u\|^{4/3}_{L^{6/5}}\Bigg] \\\leq C \Big(\|(u \cdot \nabla) u\|_{L^{6/5}}^{4/3} + \|\nabla u \|^{4/3}_{L^2}\Big)$$ after choosing $C$ sufficiently large. I think this should be right.