Suppose we are given a compactly supported test form $\chi\in\Omega_{0}^{m-3}(\mathcal{U})$, then we apply integration by parts once and we have

(3.205)

$$(\phi_{2},\Delta\chi)=\int_{\mathcal{U}}\phi_{2}\wedge(dd^{*}+d^{*}d)\chi=\int_{\mathcal{U}}d\phi_{2}\wedge d^{*}\chi-\int_{\mathcal{U}}d^{*}\phi_{2}\wedge d\chi.$$

Here there is no boundary term because $\phi_{2}=O(r^{-1})$. Notice that (3.133) and (3.190) implies $d\phi_{2}=O^{\prime}(1)$, so

(3.206)

$$\int_{\mathcal{U}}d\phi_{2}\wedge d^{*}\chi=\int_{\mathcal{U}}d^{*}d\phi_{2}\wedge\chi.$$

On the other hand, by (3.139),

(3.207)

$$d^{*}\phi_{2}=-\frac{y_{\alpha}}{2r^{3}}{}dy_{\hat{\alpha}}+\zeta,$$

where $\zeta$ is a $2$-form satisfying $\zeta=O^{\prime}(r^{-1})$. Denote by $S_{\epsilon}^{2}$ the normal geodesic sphere bundle $\{r=\epsilon\}$, then we get that

(3.208)		$\displaystyle-\int_{\mathcal{U}}d^{*}\phi_{2}\wedge d\chi$	$\displaystyle=$	$\displaystyle\int_{\mathcal{U}}dd^{*}\phi_{2}\wedge\chi+\int_{S_{\epsilon}^{2}}d^{*}\phi_{2}\wedge\chi$
			$\displaystyle=$	$\displaystyle\int_{\mathcal{U}}dd^{*}\phi_{2}\wedge\chi+\lim_{\epsilon\rightarrow 0}\frac{1}{2\epsilon^{3}}{}\int_{S_{\epsilon}}(y_{\alpha}dy_{\hat{\alpha}}+\epsilon^{2}\zeta)\wedge\chi.$

By direct calculation of the last term on the right hand side we obtain

(3.209)

$$-\int_{\mathcal{U}}d^{*}\phi_{2}\wedge d\chi=\int_{\mathcal{U}}dd^{*}\phi_{2}\wedge\chi+2\pi{}\int_{P}\chi.$$

This concludes the proof. ∎