$f$ and $g$ are *radial*, and that $K$ is bounded by $1$ and is supported where $$|\xi|\sim 2^{k},\quad|\xi-\eta|\sim 2^{k_{1}},\quad |\eta|\sim 2^{k_{2}};\quad|\Phi(\xi,\eta)|\leq\epsilon,$$ where $\Phi=\Phi_{\sigma\mu\nu}$ is defined in (\[phasedef\]). Then we have$$\label{radimp}\|T(f,g)\|_{L^{2}}\lesssim\min\big(2^{k/2},2^{-\min(k,0)/2}\epsilon^{1/2}\big)2^{-(k_{1}+k_{2})}\|\widehat{f}\|_{L^{1}}\|\widehat{g}\|_{L^{1}}.$$

\(1) This is standard; see [@IP12].

\(2) We may assume that $F=\widehat{f}$ and $G=\widehat{g}$ are nonnegative. Let $\mathcal{F}T(f,g)=H$, by Cauchy-Schwartz we have$$|H(\xi)|^{2}\lesssim\int_{\mathbb{R}^{3}}K(\xi,\eta)F(\xi-\eta)G^{2}(\eta)\,\mathrm{d}\eta\cdot\int_{\mathbb{R}^{3}}K(\xi,\eta)F(\xi-\eta)\,\mathrm{d}\eta.$$ The second factor is bounded by $\|F\|_{L^{2}}$ by Cahchy-Schwartz again, thus we have$$\|H\|_{L^{2}}^{2}\lesssim \|f\|_{L^{2}}\cdot\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}K(\xi,\eta)F(\xi-\eta)G^{2}(\eta)\,\mathrm{d}\eta\mathrm{d}\xi,$$ where the last integral equals$$\int_{\mathbb{R}^{3}}G^{2}(\eta)\,\mathrm{d}\eta\int_{\mathbb{R}^{3}}K(\xi,\eta)F(\xi-\eta)\mathrm{d}\xi\lesssim\int_{\mathbb{R}^{3}}G^{2}(\eta)\cdot \|f\|_{L^{2}}\,\mathrm{d}\eta=\|f\|_{L^{2}}\cdot\|g\|_{L^{2}}^{2},$$ so this proves (2).

\(3) Let $|\widehat{f}(\xi-\eta)|=F(|\xi-\eta|)$ and $|\widehat{g}(\eta)|=G(|\eta|)$, we may assume $\xi=(\lambda,0,0)$ and $\xi-\eta=(x,y,z)$, so that$$|\mathcal{F}T(f,g)(\xi)|\leq\int_{\mathbb{R}^{3}}F(\sqrt{x^{2}+y^{2}+z^{2}})G(\sqrt{(\lambda-x)^{2}+y^{2}+z^{2}})\,\mathrm{d}x\mathrm{d}y\mathrm{d}z.$$ Make the change of variables $$\rho=\sqrt{x^2+y^{2}+z^2},\quad\tau=\sqrt{(\lambda-x)^{2}+y^2+z^2},\quad\theta=\tan^{-1}\frac{z}{y},$$the (inverse) Jacobian being$$J^{-1}:=\left|\begin{array}{ccc}
\rho_{x}&\rho_{y}&\rho_{z}\\
\tau_{x}&\tau_{y}&\tau_{z}\\
\theta_{x}&\theta_{y}&\theta_{z}
\end{array}\right|=\frac{1}{\rho\tau(y^{2}+z^{2})}\left|\begin{array}{ccc}
x&y&z\\
x-\lambda&y&z\\
0&-z&y\end{array}\right|=\frac{\lambda}{\rho\tau},$$ so we have$$\label{radconv}|\mathcal{F}T(f,g)(\xi)|\leq\frac{2\pi}{\lambda}\int_{|\Phi|\leq\epsilon}\rho\tau F(\rho)G(\tau)\,\mathrm{d}\rho\mathrm{d}\tau.$$ Note that $$\|F(\rho)\|_{L_{\rho}^{1}}\sim 2^{-2k_{1}}\|\widehat{f}\|_{L^{1}};\qquad \|G(\tau)\|_{L_{\tau}^{1}}\sim 2^{-2k_{2}}\|\widehat{g}\|_{L^{1}(},$$this implies that$$|\mathcal{F}T(f,g)(\xi)|\lesssim 2^{-k-k_{1}-k_{2}}\|\widehat{f}\|_{L^{1}}\|\widehat{g}\|_{L^{1}},$$ which implies the first part of (\[radimp\]) by Hölder; as for the second part, choose a suitable function $J(\lambda)$ with $\|J\|_{L^{2}}=1$, we have$$\|T(f,g)\|_{L^{2}}\lesssim 2^{k_{1}+k_{2}}\int_{|\Phi|\leq\epsilon}F(\rho)G(\tau)J(\lambda)\,\mathrm{d}\rho\mathrm{d}\tau\mathrm{d}\lambda,$$ then we fix $\rho$ and $\tau$ and notice $|\partial_{\lambda}\Phi|\sim 2^{\min(k,0)}$, so the measure of $\{\lambda:|\Phi|\leq\epsilon\}$ is bounded by $2^{-\min(k,0)}\epsilon$, then use Hölder to conclude.

Proof of Theorem \[mainkg\]: Reduction to $Z$-norm estimate {#energy}
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Let $u$ be a solution to (\[nlkg\]) on $[0,T]$. We define the function $v$ with value in $\mathbb{C}^{2d}$ by $$v_{\nu}=(\partial_{t}-i\Lambda_{\nu})u_{\nu},\qquad\nu\in\mathcal{P},$$ where $u_{-\alpha}=u_{\alpha}$; so we 