u_\mu\|_{ L^2}$$ This is always stronger than we need. The proof of the lemma is concluded.

We now establish some simple properties of the linear equation $$\Box_g u = f, \qquad u(0) = u_0, \qquad u_t(0) = u_1.
\label{leq}$$ Then

The linear equation is well-posed in $H^s \times
H^{s-1}$ for $0 \leq  s \leq 3$. \[lwp\]

The proof follows easily from energy estimates, see [@MR2153517].

We use this to prove , namely

Assume that $0 \leq  s \leq 3$ and $\theta > 0$. Then the solution $u$ to verifies $$\| u \|_{X^{s,\theta}} \lesssim \|u_0\|_{H^s} +
\|u_1\|_{H^{s-1}} + \|f\|_{L^2 H^{s-1}}$$ \[nhom\]

We decompose the solution $u$ as $$u = \sum_{{\lambda}=1}^\infty S_\lambda  {{\tilde S}}_{\lambda}u$$ and think of this as a part of the sum in which corresponds to $d=1$. Then $$\begin{split}
\|u\|_{X^{s,\theta}}^2 &\lesssim \sum_{{\lambda}=1}^\infty \|{{\tilde S}}_{\lambda}u\|_{X^{s,\theta}_{\lambda,1}}^2 \\
& \approx  \sum_{{\lambda}=1}^\infty {\lambda}^{2s} \| {{\tilde S}}_{\lambda}u\|^2_{L^2} +
\lambda^{2(s-1)} \|\Box_{g_{<{\sqrt{\lambda}}}}  {{\tilde S}}_{\lambda}u\|^2_{L^2}
\\ &\lesssim \|  u\|^2_{L^2 H^s} +  \sum_{{\lambda}=1}^\infty
\lambda^{2(s-1)} \|\Box_{g_{<{\sqrt{\lambda}}}}  {{\tilde S}}_{\lambda}u - {{\tilde S}}_{\lambda}\Box_g
u \|^2_{L^2} + \|f\|^2_{L^2 H^{s-1}}
\end{split}$$ The first term is easily controlled by energy estimates. The second is decomposed as follows: $$\Box_{g_{<{\sqrt{\lambda}}}}  {{\tilde S}}_{\lambda}u - {{\tilde S}}_{\lambda}\Box_g u =
[\Box_{g_{<{\sqrt{\lambda}}}},{{\tilde S}}_{\lambda}] u - {{\tilde S}}_{\lambda}\Box_{g_{>{\sqrt{\lambda}}}} u$$ For the commutator 