of Proposition \[Norm\], which we recall below:

\[MainBootstrap\] Assume that $U\in X^N_T$ is a solution of the equation $$\label{bo1}
(\partial_t+i\Lambda)U=\frac{i}{4}\sum_{j=1}^2\Lambda R_j\big[|\nabla|\Lambda^{-1}(U+\overline{U})\cdot R_j(U-\overline{U})\big]+\frac{i}{8}\sum_{j=1}^2|\nabla|\big[R_j(U-\overline{U})\cdot R_j(U-\overline{U})\big]$$ with the property that $$\label{bo2}
\|U\|_{X^N_T}\leq\varepsilon_0\leq 1.$$ Then $$\label{bo3}
\sup_{t\in[0,T]}\|e^{it\Lambda}U(t)-U(0)\|_{Y^{N_0}}\lesssim \varepsilon^2_0.$$

We use first the method of normal forms to derive several new formulas describing the solution $U$. Then we use these formulas to prove the desired estimate .

Renormalizations {#renorm}
----------------

Assume that $U\in X^N_T$ satisfies the hypothesis of Proposition \[MainBootstrap\] and let $$U_+:=U,\qquad U_-:=\overline{U}.$$ Let $\widehat{U_+},\widehat{U_-}$ denote the spatial Fourier transforms, $$\widehat{U_\mu}(\xi,t):=\int_{\mathbb{R}^2}U_\mu(x,t)e^{-ix\cdot\xi}\,dx,\qquad \mu\in\{+,-\},\,(\xi,t)\in\mathbb{R}^2\times [0,T].$$ The equation gives $$\label{norm2}
[\partial_t+i\Lambda(\xi)]\widehat{U_+}(\xi,t)=\sum_{(\mu,\nu)\in\{(+,+),(+,-),(-,-)\}}\int_{\mathbb{R}^2}m_{\mu\nu}(\xi,\eta)\widehat{U_\mu}(\xi-\eta,t)\widehat{U_\nu}(\eta,t)\,d\eta,$$ where $$\label{ms1}
m_{++}(\xi,\eta):=c_0i\Big[\frac{-\Lambda(\xi)|\xi-\eta|(\xi\cdot\eta)}{4\Lambda(\xi-\eta)|\xi||\eta|}+\frac{-|\xi|(\eta\cdot (\xi-\eta))}{8|\eta||\xi-\eta|}\Big],$$ $$\label{ms2}
m_{+-}(\xi,\eta):=c_0i\Big[\frac{-\Lambda(\xi)|\eta|(\xi\cdot(\xi-\eta))}{4\Lambda(\eta)|\xi||\xi-\eta|}+\frac{\Lambda(\xi)|\xi-\eta|(\xi\cdot\eta)}{4\Lambda(\xi-\eta)|\xi||\eta|}+\frac{|\xi|(\eta\cdot (\xi-\eta))}{4|\eta||\xi-\eta|}\Big],$$ $$\label{ms3}
m_{--}(\xi,\eta):=c_0i\Big[\frac{\Lambda(\xi)|\xi-\eta|(\xi\cdot\eta)}{4\Lambda(\xi-\eta)|\xi||\eta|}+\frac{-|\xi|(\eta\cdot (\xi-\eta))}{8|\eta||\xi-\eta|}\Big].$$ Letting $$\begin{split}
&V_\mu(t):=e^{\mu it\Lambda}U_\mu(t),\qquad \widehat{U_\mu}(\xi,t)=\widehat{V_\mu}(\xi,t)e^{-\mu it\Lambda(\xi)},
\end{split}$$ for $\mu\in\{+.-\}$, the equation is equivalent to $$\label{norm3}
\frac{d}{dt}[\widehat{V_+}(\xi,t)]=\sum_{(\mu,\nu)\in\{(+,+),(+,-),(-,-)\}}\int_{\mathbb{R}^2}e^{it[\Lambda(\xi)-\mu\Lambda(\xi-\eta)-\nu\Lambda(\eta)]}m_{\mu\nu}(\xi,\eta)\widehat{V_\mu}(\xi-\eta,t)\widehat{V_\nu}(\eta,t)\,d\eta.$$ Therefore, letting $$\label{keyphi} 
\Phi_{\mu\nu}(\xi,\eta):=\Lambda(\xi)-\mu\Lambda(\xi-\eta)-\nu\Lambda(\eta),\qquad (\mu,\nu)\in\{(+,+),(+,-),(-,-)\},$$ it follows that $$\widehat{V_+}(\xi,t)-\widehat{V_+}(\xi,0)=\sum_{(\mu,\nu)\in\{(+,+),(+,-),(-,-)\}}\int_0^t\int_{\mathbb{R}^2}e^{is\Phi_{\mu\nu}(\xi,\eta)}m_{\mu\nu}(\xi,\eta)\widehat{V_\mu}(\xi-\eta,s)\widehat{V_\nu}(\eta,s)\,d\eta ds.$$ It follows from the definition that, for any $(\mu,\nu)\in\{(+,+),(+,-),(-,-)\}$, $$|\Phi_{\mu\nu}(\xi,\eta)|^{-1}\lesssim 1+\min(|\xi|,|\eta|,|\xi-\eta|).$$ Therefore, we can integrate by parts in $s$ to conclude that, for any $t\in[0,T]$, $$\begin{split}
\widehat{V_+}(\xi,t)-\widehat{V_+}(\xi,0)&=\sum_{(\mu,\nu)\in\{(+,+),(+,-),(-,-)\}}-\int_0^t\int_{\mathbb{R}^2}e^{is\Phi_{\mu\nu}(\xi,\eta)}\frac{m_{\mu\nu}(\xi,\eta)}{i\Phi_{\mu\nu}(\xi,\eta)}\frac{d}{ds}\big[\widehat{V_\mu}(\xi-\eta,s)\widehat{V_\nu}(\eta,s)\big]\,d\eta ds\\
&+\sum_{(\mu,\nu)\in\{(+,+),(+,-),(-,-)\}}\int_{\mathbb{R}^2}e^{it\Phi_{\mu\nu}(\xi,\eta)}\frac{m_{\mu\nu}(\xi,\eta)}{i\Phi_{\mu\nu}(\xi,\eta)}\widehat{V_\mu}(\xi-\eta,t)\widehat{V_\nu}(\eta,t)\,d\eta\\
&-\sum_{(\mu,\nu)\in\{(+,+),(+,-),(-,-)\}}\int_{\mathbb{R}^2}\frac{m_{\mu\nu}(\xi,\eta)}{i\Phi_{\mu\nu}(\xi,\eta)}\widehat{V_\mu}(\xi-\eta,0)\widehat{V_\nu}(\eta,0)\,d\eta.
\end{split}$$ For $t\in[0,T]$ let $$\widehat{W}(\xi,t):=\widehat{V_+}(\xi,t)-\sum_{(\mu,\nu)\in\{(+,+),(+,-),(-,-)\}}\int_{\mathbb{R}^2}e^{it\Phi_{\mu\nu}(\xi,\eta)}\frac{m_{\mu\nu}(\xi,\eta)}{i\Phi_{\mu\nu}(\xi,\eta)}\widehat{V_\mu}(\xi-\eta,t)\widehat{V_\nu}(\eta,t)\,d\eta.$$ It follows that $$\label{norm4}
\begin{split}
&\widehat{W}(\xi,t)-\widehat{W}(\xi,0)=\sum_{(\mu,\nu)\in\{(+,+),(+,-),(-,-)\}}i\int_0^tH_{\mu\nu}(\xi,s)\,ds,\\
&H_{\mu\nu}(\xi,s):=\int_{\mathbb{R}^2}e^{is\Phi_{\mu\nu}(\xi,\eta)}\frac{m_{\mu\nu}(\xi,\eta)}{\Phi_{\mu\nu}(\xi,\eta)}\frac{d}{ds}\big[\widehat{V_\mu}(\xi-\eta,s)\widehat{V_\nu}(\eta,s)\big]\,d\eta.
\end{split}$$

To calculate the functions $H_{\mu\nu}$ we use the formulas, see , $$\label{norm14}
\begin{split}
&\frac{d}{ds}[\widehat{V_+}(\xi,s)]=\sum_{(\mu,\nu)\in\{(+,+),(+,-),(-,-)\}}e^{is\Lambda(\xi)}\int_{\mathbb{R}^2}m_{\mu\nu}(\xi,\chi)\widehat{U_\mu}(\xi-\chi,s)\widehat{U_\nu}(\chi,s)\,d\chi,\\
&\frac{d}{ds}[\widehat{V_-}(\xi,s)]=\sum_{(\mu,\nu)\in\{(-,-),(-,+),(+,+)\}}e^{-is\Lambda(\xi)}\int_{\mathbb{R}^2}m'_{\mu\nu}(\xi,\chi)\widehat{U_\mu}(\xi-\chi,s)\widehat{U_\nu}(\chi,s)\,d\chi.
\end{split}$$ where $$m'_{--}(v,w):=\overline{m_{++}(-v,-w)},\qquad m'_{-+}(v,w):=\overline{m_{+-}(-v,-w)},\qquad m'_{++}(v,w):=\overline{m_{--}(-v,-w)}.$$ Therefore $$\begin{split}
&e^{-is\Lambda(\xi)}H_{++}(\xi,s)\\
&=\sum_{(\mu,\nu)\in\{(+,+),(+,-),(-,-)\}}\int_{\mathbb{R}^2\times\mathbb{R}^2}\frac{m_{++}(\xi,\eta)}{\Phi_{++}(\xi,\eta)}\big[\widehat{U_+}(\eta,s)m_{\mu\nu}(\xi-\eta,\chi)\widehat{U_\mu}(\xi-\eta-\chi,s)\widehat{U_\nu}(\chi,s)\big]\,d\eta d\chi\\
&+\sum_{(\mu,\nu)\in\{(+,+),(+,-),(-,-)\}}\int_{\mathbb{R}^2\times\mathbb{R}^2}\frac{m_{++}(\xi,\eta)}{\Phi_{++}(\xi,\eta)}\big[\widehat{U_+}(\xi-\eta,s)m_{\mu\nu}(\eta,\chi)\widehat{U_\mu}(\eta-\chi,s)\widehat{U_\nu}(\chi,s)\big]\,d\eta d\chi,
