we use the fixed time bound along with square summability in $\lambda$. The second part is controlled by .

The result in the next Lemma implies the estimate for the spaces $X,Y$:

Assume that $0 \leq s \leq 3$ and $\frac12 < \theta < 1$. Then the operator $\Box_g^{-1}$ has the mapping property $$\Box_g^{-1}: X^{s-1,\theta-1} \to X^{s,\theta}$$

Let $f \in X^{s-1,\theta-1}$. We use the representation in , $$f = f_0 + \sum_{{\lambda}=1}^\infty \sum_{d=1}^{{\lambda}}\Box_{g_{<{\sqrt{\lambda}}}}
S_{{\lambda}} f_{{\lambda},d}$$ By Definition  the function $$u = \sum_{{\lambda}=1}^\infty \sum_{d=1}^{{\lambda}} S_{{\lambda}} f_{{\lambda},d}$$ belongs to $X^{s,\theta}$. The difference $v = u - \Box_g^{-1} f$ solves $$\Box_g v = \Box_g u - f, \qquad v(0) = u(0), \qquad v_t(0) = u_t(0)$$ To estimate it we use Lemma \[nhom\]. The initial data is controlled due to Corollary \[trace\], so it remains to bound the inhomogeneous term in $L^2 H^{s-1}$. Thus we need to show that $$\| \sum_{{\lambda}=1}^\infty  \sum_{d=1}^{{\lambda}} \Box_{g_{>{\sqrt{\lambda}}}}
S_{{\lambda}} f_{{\lambda},d} \|_{L^2H^{s-1}}^2 \lesssim \sum_{{\lambda},d}
\|f_{{\lambda},d}\|_{X^{s,\theta}_{{\lambda},d}}^2$$ Considering the trace regularity result in Corollary \[trace\] this would follow from $$\| \sum_{{\lambda}=1}^\infty \Box_{g_{>{\sqrt{\lambda}}}} S_{{\lambda}} f_{{\lambda}}
\|_{L^2H^{s-1}}^2 \lesssim \sum_{{\lambda}}  \|\nabla f_{{\lambda}}\|_{L^\infty
H^{s-1}}^2, \qquad f_{\lambda}= \sum_{d=1}^{\lambda}f_{{\lambda},d}$$ which in turn is a consequence of the fixed time bound .

We finish this section by proving a 