Don't understand proof of a PDE argument, author uses $w(t,\cdot) \in L^6(\Omega)$ when $w(t,\cdot) \in H^1(\Omega)$. I'm reading this paper.
I attach the details below(you don't have to read it all to understand my question). My question is, $w$ is only assumed to be $H^1$ in space, why then in the manipulations below does the author write expressions like $\lVert w \rVert_{L^6(\Omega)}$? We don't know that $w \in L^6$. Does he use a density argument of some sort? Is it valid to do that? Also, which embedding theorem is it the author uses to get (9)?


 A: The Sobolev embedding theorem implies that in two dimensions, $H^1$ (which is $W^{1,2}$) embeds into $L^q$ for every $1\le q<\infty$.   (By the way, it's not enough for $\Omega$ to be bounded; we need some assumption on $\partial \Omega$, such as smoothness. This is assumed at the beginning of the paper.) 
But in (9) the authors use $H^{3/4}$ (which is $W^{3/4,2}$) , not $H^{1}$. Here the order of smoothness $k$ is $3/4$ and the integrability exponent is still $p=2$. The dimension is $n=2$. The arithmetics of Sobolev embedding is 
$$\frac{1}{q} = \frac{1}{p} - \frac{k}{n} = \frac{1}{2}-\frac{3}{8} = \frac{1}{8}$$
Thus, the $L^8$ norm of $u$ is at most a constant times the $H^{3/4}$ norm of $u$. The authors actually need $L^6$, but want the constant to be small ($\delta$). 
For this we need a simple interpolation-type result: for every $\delta>0$ there is $C_\delta$ such that 
$$\|u\|_{L^6}^6 \le C_\delta \|u\|_{L^2}^6+\delta \|u\|_{L^8}^6 \tag{1}$$
Indeed, by Hölder's inequality 
$$\|u\|_{L^6}^6 = \int |u|^{2/3}|u|^{16/3} \le \left(\int |u|^2 \right)^{1/3}
\left(\int |u|^8 \right)^{2/3} = \|u\|_{L^2}^{2/3}\,\|u\|_{L^8}^{16/3} \tag{2}$$
Recall the Peter-Paul form of Young's inequality: 
$$ab = (\epsilon^{-1}a)(\epsilon b)\le \frac{(\epsilon^{-1}a)^9}{9} + \frac{(\epsilon b)^{9/8}}{9/8} ,\quad a,b\ge 0 \tag{3}$$
Plugging (2) into (3) yields
$$\|u\|_{L^6}^6 \le \frac{1}{9\epsilon^9}\|u\|_{L^2}^{6}+\frac{8\epsilon^{9/8}}
{9}\|u\|_{L^8}^{6}$$
which proves (1).
