. For the second we use the trace regularity result in .

b\) We now show that $$(X^{s,\theta}+ L^2 H^{s+\theta})' \subset X^{-s,-\theta}$$ Let $T$ be a bounded linear functional on $X^{s,\theta} + L^2
H^{s+\theta}$. Due to the second term we can identify $T$ with a function $u \in L^2 H^{-s-\theta}$.

On the other hand, we can apply it to functions $v \in X^{s,\theta}$ of the form $$v =  \sum_{{\lambda}=1}^\infty  \sum_{d=1}^{{\lambda}} S_\lambda v_{\lambda,d}$$ Then we must have the bound $$|Tv|^2  \lesssim \|v\|_{X^{s,\theta}}^2  \lesssim \sum_{{\lambda},d}
\|v_{\lambda,d}\|^2_{X^{s,\theta}_{\lambda,d}} \lesssim
\sum_{{\lambda},d} \left( {\lambda}^{2s}d^{2\theta}\|v_{{\lambda},d}\|^2_{L^2} +
{\lambda}^{2s-2}d^{2\theta-2}\|\tilde
\Box_{g_{<{\sqrt{\lambda}}}}v_{{\lambda},d}\|^2_{L^2}\right)$$ Given the definition of the $X^{s,\theta}_{\lambda,d}$ norms, using succesively the Hahn-Banach theorem and Riesz’s theorem it follows that we can find functions $f_{\lambda,d}$ and $h_{\lambda,d}$ with $$\sum_{{\lambda}=1}^\infty  \sum_{d=1}^{{\lambda}} \lambda^{-2s} d^{-2\theta}
 \|f_{\lambda,d}\|_{L^2}^2
+   \lambda^{2(1-s)} d^{2(1-\theta)} \|h_{\lambda,d}\|_{L^2}^2 = M < \infty
\label{M}$$ so that $$Tv = \sum_{{\lambda}=1}^\infty \sum_{d=1}^{{\lambda}} \int f_{\lambda,d}\,
v_{\lambda,d} + h_{\lambda,d}\cdot \tilde \Box_{g_{<
\sqrt\lambda}}v_{\lambda,d}\, dx dt$$ In particular this must hold for $v$ of the form $v = S_\lambda
v_{\lambda,d}$, $$\int u \, S_\lambda  v_{\lambda,d} dx dt  = \int f_{\lambda,d}\,
v_{\lambda,d} + h_{\lambda,d}\cdot \tilde \Box_{g_{<
\sqrt\lambda}}v_{\lambda,d} \, dx dt$$ For each $\lambda,d$ this yields $$S_\lambda u = f_{\lambda,d} + \tilde \Box_{g_{< \sqrt\lambda}} h_{\lambda,d}$$ Then we can represent $S_\lambda u$ 