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)?
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).
