Prove $\|F(x,·)\|_{L^{4,1}(E)}^2 \leq C\|F(x,·)\|_{L^{\infty}(E)}\|F(x,·)\|_{L^{2,2}(E)}$ How to derive this inequality? 
$$\|F(x,·)\|_{L^{4,1}(E)}^2 \leq C\|F(x,·)\|_{L^{\infty}(E)}\|F(x,·)\|_{L^{2,2}(E)},$$ 
where $C$ is constant and $$\|F(x,·)\|_{L^{p,n}(E)}=\left(\int_E\displaystyle\sum_{m\leq n}\left|\partial_y^m F(x,y)\right|^pdy\right)^{1/p},$$ $E$ is a bounded domain of $\mathbb{R}^2$.
 A: Ah, I was misled by the fact that $E\subseteq \mathbb{R}^2$. If $E$ has sufficiently smooth boundary, the usual extension and density theorems allows us to consider instead $C^\infty_0(\mathbb{R}^2)$ functions instead. Then the inequality is a simple consequence of integration by parts:
Let $f\in C^\infty_0(\mathbb{R}^d)$, then first we have the interpolation inequality
$$ \int |f|^4 \mathrm{d}x \leq \left(\sup |f|\right)^2 \cdot \int |f|^2 \mathrm{d}x $$
which implies
$$ \|f\|_4^2 \leq \|f\|_\infty \|f\|_2 \tag{*}$$
Next for the first derivative terms, let $\partial_i = \partial_{x^i}$ for some fixed coordinate index $i$, we have the following integration by parts formula
$$ \int |\partial_i f|^4 \mathrm{d}x = \int |\partial_i f|^2 \partial_i f \partial_i f \mathrm{d}x = - \int f \partial_i\left( |\partial_i f|^2 \partial_i f\right) \mathrm{d}x $$
So 
$$ \int |\partial_i f|^4 \mathrm{d}x \leq 3 \|f\|_\infty \int |\partial_i f|^2 \cdot |\partial^2_{ii} f| \mathrm{d}x \leq 3 \|f\|_\infty \| (\partial_i f)^2\|_2 \| \partial^2_{ii} f\|_2  \tag{#} $$
where we used Cauchy-Schwarz in the last step. Noting that
$$ \| (\partial_i f)^2\|_2 = \| \partial_i f\|_4^2 $$ 
and that the left hand side of (#) is $\|\partial_i f\|_4^4$ we have that
$$ \| \partial_i f \|_4^2 \leq 3 \|f\|_\infty \|\partial^2_{ii} f\|_2 \tag{%}$$
Now summing (%) over the index $i$ and combining with (*) gives the desired expression. Notice that the inequality holds in all dimensions, not just in 2 dimensions. (The scaling behaviour of the inequality is compatible with it being dimension independent.)
