Doubt obtaining the inequality (3) via (1) I am reading a proof from the 2009-paper Travelling waves for the Gross-Pitevskii equation II (Béthuel, Gravejat, Saut) and I am really stuck in one step. Please, help me with this:
In the paper, they prove that $$\Vert \Delta u_\lambda\Vert_{L^2+L^{4/3}}\leq 10\alpha\qquad (1)$$ and, by means of $$ \frac{\lambda}{2}\int \vert u - u_\lambda\vert \ dx + \int e(u_\lambda) \ dx \leq \int e(u)\ dx ,\qquad (2)$$
they end up with the inequality $$\Vert \nabla u_\lambda\Vert_{H^1+W^{1,4/3}}\leq K\alpha\qquad (3)$$ where $K$ is a universal constant. Here $e(v) = \frac{1}{2} \vert \nabla v\vert ^2 + \frac{1}{4}(1-|v|^2)^2$ and, given $u\in H^1$, $u_\lambda\in H^1$ is the solution to the minimization problem ($\lambda > 1$)
$$ F_\lambda(u_\lambda) =  \inf \lbrace F_\lambda(v)=\frac{\lambda}{2}\int \vert u - v\vert \ dx + \int e(v) \ dx\ | \ v\in H^1\rbrace$$  The integrals and functional spaces are understood over the 2-dimensional torus. My question is how can I show (3) from (1) using (2)? Thanks in advance. Any hint or idea is welcome.
 A: Note that using $(1)$ you only need to bound $$
\Vert \nabla u_\lambda\Vert_{L^2+L^{4/3}}\leq K\alpha.
$$
Now, note that we can bound the left-hand side of $(2)$ from below by: $$
\dfrac{\lambda}{2}\int\vert u-u_\lambda\vert +\dfrac{1}{2}\int \vert  \nabla u_\lambda\vert^2+\dfrac{1}{4}\int(1-u_\lambda^2)^2 \geq \dfrac{1}{2}\int\vert \nabla u_\lambda\vert^2\geq \dfrac{1}{2}\Vert \nabla u_\lambda\Vert_{L^2+L^{4/3}}^2
$$
Thus, using $(2)$ combined with the previous inequality and taking square root we obtain $$
\Vert \nabla u_\lambda\Vert_{L^2+L^{4/3}}\leq K\left(\int e(u)\right)^{1/2}.
$$
Now, comparing your notation with the one from the paper I notice that the $\alpha$ coming from $(1)$ is given by* $$
\alpha=\left(\int e(u)\right)^{1/2},
$$
which is exactly the same term as in the previous inequality. Thus, combining $(1)$ with the previous inequality we obtain $$
\Vert \nabla u_\lambda \Vert_{H^1+W^{1,4/3}}\leq \tilde K\alpha,
$$
where $\tilde K$ is an universal constant.
Edit: Some details of *: Actually their definition of $\alpha$ consider a  factor $\delta\lambda+1$, which is certainly bigger than $1$, so it's even better, you only need to bound the inequality for $\Vert \nabla u_\lambda\Vert_{L^2+L^{4/3}}$ from above by the same term times a constant bigger than $1$. Since this is in the definition of $\alpha$ does not affect the universality of $\tilde K$.
