First of all, note that the $2$–forms $\omega_{\epsilon}^{i}$ have uniformly bounded $C^{0,\alpha}_{\delta-1}$–norm. Indeed, outside the gluing regions $\bm{\underline{\omega}}_{\epsilon}$ is a hyperkähler triple and thus $\omega_{\epsilon}^{i}$ is parallel and bounded. On the gluing regions, $\bm{\underline{\omega}}_{\epsilon}$ differs from the hyperkähler triple $\bm{\underline{\omega}}_{q_{j},\epsilon}$ or $\bm{\underline{\omega}}_{p_{i},\epsilon}$ by terms of order $O(\epsilon\rho^{2}+\epsilon^{3}\rho^{-3})$ (with similar estimates on their derivatives). Finally, $\rho_{\epsilon}^{-\delta+1}$ is bounded above since $\delta<0$.

Now, let $\widetilde{\bm{\underline{\omega}}}$ be an $L^{2}$–orthonormal triple of harmonic self-dual forms with respect to $g_{\epsilon}$. Since

$Q_{\epsilon}$ is close to the identity and $\operatorname{Vol}_{g_{\epsilon}}(M_{\epsilon})=O(\epsilon)$ we can assume that

Finally, observe that for every $u\in C^{0,\alpha}_{\delta-1}$ we have

Indeed, using the definition (6.7) of $\rho_{\epsilon}$ and the construction of $\bm{\underline{\omega}}_{\epsilon}$ it is not difficult to estimate $\|\rho_{\epsilon}^{\delta-1}\|_{L^{2}}\leq C(\epsilon^{\frac{1}{2}}+\epsilon^{\delta+1})$.

Now let $\pi\colon\thinspace C^{0,\alpha}_{\delta-1}(\Lambda^{+}T^{\ast}M_{\epsilon})\rightarrow\mathcal{H}^{+}_{\epsilon}$ be the $L^{2}$–orthogonal projection

and regard $\text{id}-\pi$ as a map $C^{0,\alpha}_{\delta-1}(\Lambda^{+}T^{\ast}M_{\epsilon})\rightarrow C^{0,\alpha}_{\delta-1}(\Lambda^{+}T^{\ast}M_{\epsilon})$. By the remarks above we have

Thus if $\delta\geq-\tfrac{1}{2}$ the projections $\pi$ and $\text{id}-\pi$ are uniformly bounded. Proposition 6.11 and the surjectivity of $L$ then yield the result. ∎