$\displaystyle\qquad\times\left.\left(\frac{1}{\zeta^{\lambda_{j}}-z^{\lambda_{j}}}-\frac{1}{\zeta^{\lambda_{0}}-z^{\lambda_{0}}}\right)\,dA(\zeta^{\lambda_{0}})\right|_{L^{q}(\Omega)}.$

By estimate (5.1) applied to $\gamma=q$ and $r=1$ and Hölder’s inequality, we
have

$\displaystyle D_{1}$ $\displaystyle\leq
C_{1,0}C(q,\Omega)\left(|(\delta^{\lambda_{j},\lambda_{0}}A)\widehat{X_{1}}(\cdot,t,\lambda)|_{L^{1}(\Omega)}+|A(\cdot,\lambda_{0})(D^{\lambda_{j}}\widehat{X_{1}}(\cdot,t))|_{L^{1}(\Omega)}\right)$
$\displaystyle\leq
C_{1,0}C(q,\Omega)\left(|\delta^{\lambda_{j},\lambda_{0}}A|_{L^{p}(\Omega)}|\widehat{X_{1}}(\cdot,t,\lambda)|_{L^{q}(\Omega)}+|A(\zeta,\lambda_{0})|_{L^{p}(\Omega)}|D^{\lambda_{j}}\widehat{X_{1}}|_{L^{q}(\Omega)}\right),$

where $1<p^{\prime}\leq q<2$. By assumption, this expression goes to $0$.
Similarly we can show for $D_{2}$ converges to $0$, and also

$D_{3}\leq
C_{1,0}C(q,\Omega)\left|\Gamma^{\lambda_{j}}-\Gamma^{\lambda_{0}}\right|_{1}\left\\{\left|A(\cdot,\lambda_{0})\right|_{L^{p}(\Omega)}+\left|B(\cdot,\lambda_{0})\right|_{L^{p}(\Omega)}\right\\}|\widehat{X_{1}}(\cdot,t,\lambda_{0})|_{L^{q}(\Omega)}.$
$\displaystyle D_{4}$
$\displaystyle\leq\left|\iint_{\Omega}\left(A(\zeta,\lambda_{0})\widehat{X_{1}}(\zeta,t,\lambda_{0})+B(\zeta,\lambda_{0})\overline{\widehat{X_{1}}(\zeta,t,\lambda_{0})}\right)\frac{\delta_{\zeta,z}(\Gamma^{\lambda_{0}}-\Gamma^{\lambda_{j}})}{\zeta^{\lambda_{j}}-z^{\lambda_{j}}}\right.$
$\displaystyle\qquad\times\left.\frac{1}{\zeta^{\lambda_{0}}-z^{\lambda_{0}}}\,dA(\zeta^{\lambda_{0}})\right|_{L^{q}_{z}(\Omega)}.$
$\displaystyle\leq
C_{1,0}C(q,\Omega)\left|\Gamma^{\lambda_{0}}-\Gamma^{\lambda_{j}}\right|_{1}\left\\{\left|A(\cdot,\lambda_{0})\right|_{L^{p}(\Omega)}+\left|B(\cdot,\lambda_{0})\right|_{L^{p}(\Omega)}\right\\}|\widehat{X_{1}}(\cdot,t,\lambda_{0})|_{L^{q}(\Omega)}.$

Hence $D_{3},D_{4}\to 0$ as $|\lambda_{j}-\lambda_{0}|\to 0$. In view of
(5.12)
$|P^{\lambda_{j}}\widehat{X_{1}}(\cdot,t,\lambda_{j})-P^{\lambda_{0}}\widehat{X_{1}}(\cdot,t)|_{L^{q}(\Omega)}\to
0$ as $|\lambda_{j}-\lambda_{0}|\to 0$.

Letting $j\to\infty$ in equation (5.11), we have

$\displaystyle 0$
$\displaystyle=\widehat{X_{1}}(z,t)-T_{\mathbb{C}}[A(\cdot,\lambda_{0})\widehat{X_{1}}(\cdot,t)+B(\cdot,\lambda_{0})\overline{\widehat{X_{1}}(\cdot,t)}](z,t)$

holds in $L^{q}(\Omega)$. By Proposition 7.21, the equation has only the
trivial solution, so $\widehat{X_{1}}\equiv 0$. But
$|\widehat{X_{1}}(\cdot,t)|_{L^{q}(\Omega)}=\lim_{j\to\infty}|\widehat{X_{1}}(\cdot,t,\lambda_{j})|_{L^{q}(\Omega)}=1$,
contradiction. Thus $|X_{1}(\cdot,t,\lambda)|_{L^{q}(\Omega)}$ is bounded, for
any $q$ with $1<p^{\prime}\leq q<2$.

Next, we show that $X_{1}(\cdot,t,\lambda)$ is continuous in the
$L^{q}(\Omega)$-norm, for any $q$ with $1<p^{\prime}\leq q<2$. Otherwise there
exists some sequence $\lambda_{j}\to\lambda_{0}$, such that
$|X_{1}(\cdot,t,\lambda_{j})-X_{1}(\cdot,t,\lambda_{0})|_{L^{q}(\Omega)}\geq\delta>0$.
Since we have already shown that $|X_{1}(\cdot,t,\lambda)|_{L^{q}(\Omega)}$ is
bounded uniformly in $\lambda$, by the computation above,
$|P^{\lambda_{j}}X_{1}(\cdot,t,\lambda_{j})|_{L^{q}_{\alpha}(\Omega)}$ is
bounded uniformly in $\lambda$. Passing to a subsequence if necessary,
$P^{\lambda_{j}}X_{1}(\cdot,t,\lambda_{j})$ converges to some function in
$L^{q}(\Omega)$. The right-hand side of the integral equation (5.8) converges
to $\frac{1}{t^{\lambda_{0}}-z^{\lambda_{0}}}$ in $L^{q}({\Omega})$. Indeed,
we have

$\left(\iint_{\Omega}\left|\frac{1}{t^{\lambda_{j}}-z^{\lambda_{j}}}-\frac{1}{t^{\lambda_{0}}-z^{\lambda_{0}}}\right|^{q}\,dA(z)\right)^{\frac{1}{q}}\leq
C_{1,0}|\Gamma^{\lambda_{j}}-\Gamma^{\lambda_{0}}|_{1}\left(\iint_{\Omega}\frac{1}{|t-z|^{q}}\,dA(z)\right)^{\frac{1}{q}},$

which converges to $0$ as $|\lambda_{j}-\lambda_{0}|\to 0$, since $\Gamma\in
C^{1,0}(\overline{\Omega})$. It follows from equation (5.8) that
$X_{1}(z,t,\lambda_{j})$ converges to some $X^{\ast}_{1}(z,t)$ in
$L^{q}(\Omega)$. As before we can show that
$|P^{\lambda_{j}}X_{1}(\cdot,t,\lambda_{j})-P^{\lambda_{0}}X_{1}^{\ast}(\cdot,t)|_{L^{q}(\Omega)}\to
0$. Taking $j\to\infty$ in equation (5.8) we obtain

$X_{1}^{\ast}(z,t)-P^{\lambda_{0}}X_{1}^{\ast}(z,t)=\frac{1}{2(t^{\lambda_{0}}-z^{\lambda_{0}})},\quad\quad\text{in
$L^{q}(\Omega)$}.$

On the other hand, since $X_{1}(z,t,\lambda_{0})$ satisfies the same equation.
By uniqueness of solution we have
$X_{1}(\cdot,t,\lambda_{0})=X_{1}^{\ast}(\cdot,t)$ in $L^{q}(\Omega)$. But
$|X_{1}^{\ast}(\cdot,t)-X_{1}(\cdot,t,\lambda_{0})|_{L^{q}(\overline{\Omega})}\geq\delta>0$,
which is a contradiction. This shows that of $X_{1}(\cdot,t,\lambda)$ in
continuous in $\lambda$ in the $L^{q}(\Omega)$\- norm.

Finally to obtain estimate (5.10) we write $X_{1}$ in the form (3.26):

(5.13)
$X_{1}(z,t,\lambda)=\frac{e^{\omega_{1}(z,t,\lambda)}}{t^{\lambda}-z^{\lambda}}.$

By estimate (5.3), we have

$|\omega_{j}(\cdot,t,\lambda)|_{C^{\alpha_{p}}(\overline{\Omega})}\leq
C_{1,0}C(p,\Omega)\left(|A(\cdot,\lambda)|_{L^{p}(\Omega)}+|B(\cdot,\lambda)|_{L^{p}(\Omega)}\right),$

where $\alpha_{p}=\frac{p-2}{p}$, and the constants are independent of
$\lambda$ and $t\in K$. Estimate (5.10) then readily follows. ∎

(ii) As in part (i) we only need to prove the statements for $X_{1}$. Let
$D^{\lambda_{1},\lambda_{2}}X_{1}=X_{1}(z,t,\lambda_{1})-X_{1}(z,t,\lambda_{2})$.
Then in view of equation (5.3), we see that $D^{\lambda_{1},\lambda_{2}}X_{1}$
satisfies the following integral equation

$D^{\lambda_{1},\lambda_{2}}X_{1}(z,t)+P^{\lambda_{2}}(D^{\lambda_{1},\lambda_{2}}X_{1})(z,t)=\sum_{i=1}^{4}R_{i}(z,t,\lambda_{1}),$

where

$\displaystyle
P^{\lambda_{2}}f(z,t)=T_{\mathbb{C}}\left[A(\cdot,\lambda_{2})f(\cdot,t)+B(\cdot,\lambda_{2})\overline{f(\cdot,t)}\right]$
$\displaystyle
R_{1}(z,t,\lambda_{1})=\iint_{\Omega}\frac{\left(A(\cdot,\lambda_{1})-A(\cdot,\lambda_{2})\right)X_{1}(\cdot,t,\lambda_{1})}{\zeta^{\lambda_{1}}-z^{\lambda_{1}}}\,dA(\zeta^{\lambda_{1}}).$
$\displaystyle
R_{2}(z,t,\lambda_{1})=\iint_{\Omega}\frac{A(\cdot,\lambda_{2})X(\cdot,t,\lambda_{1})(a^{\lambda_{1}}-a^{\lambda_{2}})}{\zeta^{\lambda_{1}}-z^{\lambda_{1}}}\,dA(\zeta).$
$\displaystyle
R_{3}(z,t,\lambda_{1})=\iint_{\Omega}A(\cdot,\lambda_{2})X(\cdot,t,\lambda_{1})\left(\frac{1}{\zeta^{\lambda_{1}}-z^{\lambda_{1}}}-\frac{1}{\zeta^{\lambda_{2}}-z^{\lambda_{2}}}\right)\,dA(\zeta^{\lambda_{2}}).$
$\displaystyle
R_{4}(z,t,\lambda_{1})=\frac{1}{t^{\lambda_{1}}-z^{\lambda_{1}}}-\frac{1}{t^{\lambda_{2}}-z^{\lambda_{2}}}.$

Since $|X_{1}(\cdot,t,\lambda)|_{L^{q}(\Omega)}$ is bounded, we can show by
the same proof as in (i) that $\sup_{t\in K}$
$|R_{i}(z,t,\lambda_{1})|_{L^{q}(\Omega)}$ converges to $0$ as
$|\lambda_{1}-\lambda_{2}|\to 0$, $1\leq i\leq 4$. Let
$(I+P^{\lambda_{2}})^{-1}:L^{q}(\Omega)\to L^{q}(\Omega)$ be the inverse
operator of $I+P^{\lambda_{2}}$. By the open mapping theorem,
$(I+P^{\lambda_{2}})^{-1}$ is continuous. Denoting its operator norm by
$M_{\lambda_{2}}$, we have

$\sup_{t\in K}|D^{\lambda_{1},\lambda_{2}}X_{1}(\cdot,t)|_{L^{q}(\Omega)}\leq
M_{\lambda_{2}}\sum_{i=1}^{4}\sup_{t\in
K}\left|R_{i}(z,t,\lambda_{1})\right|_{L^{q}(\Omega)},$

which converges to $0$ as $|\lambda_{1}-\lambda_{2}|\to 0$. ∎

(iii) We write $X_{1}$ in the form (3.26):

(5.14)
$X_{1}(z,t,\lambda)=\frac{e^{\omega_{1}(z,t,\lambda)}}{t^{\lambda}-z^{\lambda}}.$

where $\omega_{1}$ is given by expression (5.2). By Lemma 5.2, if
$|A(\cdot,\lambda)|_{L^{p}(\Omega)}$, $|B(\cdot,\lambda)|_{L^{p}(\Omega)}$ are
bounded uniformly in $\lambda$, then
$|\omega_{1}(\cdot,t,\lambda)|_{C^{\alpha}(\overline{\Omega})}$ is bounded
uniformly in $\lambda\in[0,1]$ and $t\in K$, for $\alpha=\frac{p-2}{p}$. By
(5.14), $|X_{1}(\cdot,t,\lambda)|_{C^{\alpha}(E_{\varepsilon})}$ is bounded
uniformly in $\lambda$ and $t$. By part (ii), $\sup_{t\in
K}|X_{1}(\cdot,t,\lambda_{1})-X_{1}(\cdot,t,\lambda_{2})|_{L^{1}(\Omega)}$
converges to $0$ as $|\lambda_{1}-\lambda_{2}|\to 0$. Applying Proposition
7.23 we get

$\sup_{t\in
K}|X_{1}(\cdot,t,\lambda_{1})-X_{1}(\cdot,t,\lambda_{2})|_{C^{0}(\Omega\cap
E_{\varepsilon})}\longrightarrow 0.\qquad\qed$

######  Lemma 5.4.

Let $2<p<\infty$. Suppose $\lambda\mapsto A(\cdot,\lambda),B(\cdot,\lambda)$
are continuous maps from $[0,1]$ to $L^{p}(\mathbb{C})$, where for each
$\lambda\in[0,1]$, $A(\cdot,\lambda),B(\cdot,\lambda)\equiv 0$ outside
$\Omega$. For $j=1,2$, let

$\omega_{j}(z,t,\lambda)=\frac{t^{\lambda}-z^{\lambda}}{\pi}\iint_{\Omega^{\lambda}}\frac{A(\zeta,\lambda)X_{j}(\zeta,t,\lambda)+B(\zeta,\lambda)\overline{X_{j}(\zeta,t,\lambda)}}{(\zeta^{\lambda}-z^{\lambda})(t^{\lambda}-\zeta^{\lambda})X_{j}(\zeta,t,\lambda)}\,dA(\zeta^{\lambda}).$

Then for each $t$, $\lambda\mapsto\omega_{j}(\cdot,t,\lambda)\in
C^{0}(\overline{\Omega})$ is a continuous map from $[0,1]$ to
$C^{0}(\overline{\Omega})$. Moreover for any compact subset $K$ in
$\mathbb{C}$, we have

(5.15) $\sup_{t\in
K}|\omega_{j}(\cdot,t,\lambda_{1})-\omega_{j}(\cdot,t,\lambda_{2})|_{C^{0}(\overline{\Omega})}\to
0\quad\text{as $|\lambda_{1}-\lambda_{2}|\to 0$.}$

Proof. Rewrite $\omega_{j}$ as

$\omega_{j}(z,t,\lambda)=\frac{1}{\pi}\iint_{\Omega}\left[\frac{1}{\zeta^{\lambda}-z^{\lambda}}-\frac{1}{\zeta^{\lambda}-t^{\lambda}}\right]\\\
\left[A(\zeta,\lambda)+B(\zeta,\lambda)\frac{\overline{X_{j}(\zeta,t,\lambda)}}{X_{j}(\zeta,t,\lambda)}\right]a(\zeta,\lambda)\,dA(\zeta),$

where $a(\zeta,\lambda)=|(\det\Gamma^{\lambda})(\zeta)|$. We have

(5.16) $|\omega_{j}(z,t,\lambda_{1})-\omega_{j}(z,t,\lambda_{2})|\leq
D_{1}+D_{2}+D_{3},$

where

$\displaystyle D_{1}$
$\displaystyle=\iint_{\mathbb{C}}\left\\{\left|\frac{1}{\zeta^{\lambda_{1}}-z^{\lambda_{1}}}-\frac{1}{\zeta^{\lambda_{2}}-z^{\lambda_{2}}}\right|+\left|\frac{1}{\zeta^{\lambda_{1}}-t^{\lambda_{1}}}-\frac{1}{\zeta^{\lambda_{2}}-t^{\lambda_{2}}}\right|\right\\}$
$\displaystyle\qquad\times\left|A(\zeta,\lambda_{1})+B(\zeta,\lambda_{1})\frac{\overline{X_{j}(\zeta,t,\lambda_{1})}}{X_{j}(\zeta,t,\lambda_{1})}\right|\,dA(\zeta^{\lambda_{1}}).$
$\displaystyle
D_{2}=\iint_{\mathbb{C}}\left|\frac{1}{\zeta^{\lambda_{2}}-z^{\lambda_{2}}}+\frac{1}{t^{\lambda_{2}}-\zeta^{\lambda_{2}}}\right||(Aa)(\zeta,\lambda_{1})-(Aa)(\zeta,\lambda_{2})|\,dA(\zeta).$
$\displaystyle D_{3}$
$\displaystyle=\iint_{\mathbb{C}}\left|\frac{1}{\zeta^{\lambda_{2}}-z^{\lambda_{2}}}+\frac{1}{t^{\lambda_{2}}-\zeta^{\lambda_{2}}}\right|$
$\displaystyle\qquad\times\left|(Ba)(\zeta,\lambda_{1})\frac{\overline{X_{j}(\zeta,t,\lambda_{1})}}{X_{j}(\zeta,t,\lambda_{1})}-(Ba)(\zeta,\lambda_{2})\frac{\overline{X_{j}(\zeta,t,\lambda_{2})}}{X_{j}(\zeta,t,\lambda_{2})}\right|\,dA(\zeta).$

Let $p^{\prime}$ be the conjugate of $p$, with $1<p^{\prime}<2$. By Hölder’s
inequality we have

$\displaystyle D_{1}$ $\displaystyle\leq
C_{1,0}|\Gamma^{\lambda_{1}}-\Gamma^{\lambda_{2}}|_{1}\int_{\Omega}\left(\frac{1}{|\zeta-z|}+\frac{1}{|\zeta-t|}\right)\left|A(\zeta,\lambda_{1})+B(\zeta,\lambda_{1})\frac{\overline{X_{j}(\zeta,t,\lambda_{1})}}{X_{j}(\zeta,t,\lambda_{1})}\right|\,dA(\zeta^{\lambda_{1}})$
$\displaystyle\leq
C_{1,0}|\Gamma^{\lambda_{1}}-\Gamma^{\lambda_{2}}|_{1}\left(\|(\zeta-z)^{-1}\|_{L^{p^{\prime}}(\Omega)}+\|(\zeta-t)^{-1}\|_{L^{p^{\prime}}(\Omega)}\right)\left(|A(\cdot,\lambda)|_{L^{p}(\Omega)}+|B(\cdot,\lambda)|_{L^{p}(\Omega)}\right)$
$\displaystyle\leq
C_{1,0}C_{p}|\Gamma^{\lambda_{1}}-\Gamma^{\lambda_{2}}|_{1}\left(|A(\cdot,\lambda)|_{L^{p}(\Omega)}+|B(\cdot,\lambda)|_{L^{p}(\Omega)}\right),$

and similarly for $D_{2}$:

$D_{2}\leq
C_{1,0}C_{p}\left(|\delta^{\lambda_{1}\lambda_{2}}A|_{L^{p}(\Omega)}+|A(\cdot,\lambda)|_{L^{p}(\Omega)}|\delta^{\lambda_{1}\lambda_{2}}\Gamma|_{1}\right).$

Hence by our assumption, $D_{1}$ and $D_{2}$ converge to $0$ and the
convergence is uniform in $t\in K$. For $D_{3}$ we have

$\displaystyle D_{3}\leq D_{31}+D_{32},$

where

$D_{31}=\int_{\mathbb{C}}\left|\frac{1}{\zeta^{\lambda_{2}}-z^{\lambda_{2}}}+\frac{1}{t^{\lambda_{2}}-\zeta^{\lambda_{2}}}\right|\left|\delta^{\lambda_{1},\lambda_{2}}(Ba)\right|\left|\frac{\overline{X_{j}(\zeta,t,\lambda_{1})}}{X_{j}(\zeta,t,\lambda_{1})}\right|\,dA(\zeta),$

and

$D_{32}=\int_{\mathbb{C}}\left|\frac{(Ba)(\zeta,\lambda_{2})}{\zeta^{\lambda_{2}}-z^{\lambda_{2}}}+\frac{(Ba)(\zeta,\lambda_{2})}{t^{\lambda_{2}}-\zeta^{\lambda_{2}}}\right|\left|\frac{\overline{X_{j}(\zeta,t,\lambda_{1})}}{X_{j}(\zeta,t,\lambda_{1})}-\frac{\overline{X_{j}(\zeta,t,\lambda_{2})}}{X_{j}(\zeta,t,\lambda_{2})}\right|\,dA(\zeta).$

By Hölder’s inequality,

$D_{31}\leq
C_{1,0}C_{p}|(Ba)(\cdot,\lambda_{1})-(Ba)(\cdot,\lambda_{2})|_{L^{p}(\Omega)}$

which by assumption goes to $0$ as $|\lambda_{1}-\lambda_{2}|\to 0$. On the
other hand,

(5.17) $\displaystyle D_{32}\leq
2\int_{\mathbb{C}}\left|\frac{(Ba)(\zeta,\lambda_{2})}{\zeta^{\lambda_{2}}-z^{\lambda_{2}}}+\frac{(Ba)(\zeta,\lambda_{2})}{t^{\lambda_{2}}-\zeta^{\lambda_{2}}}\right|\left|\frac{X_{j}(\zeta,t,\lambda_{1})-X_{j}(\zeta,t,\lambda_{2})}{X_{j}(\zeta,t,\lambda_{1})}\right|\,dA(\zeta).$

By (5.5), $|\omega_{j}(\zeta,t,\lambda)|$ is bounded uniformly for any
$\zeta,t\in\mathbb{C}$ and $\lambda\in[0,1]$, and

$X_{1}(\zeta,t,\lambda)=\frac{e^{\omega_{1}(\zeta,t,\lambda)}}{2(t^{\lambda}-\zeta^{\lambda})},\quad
X_{2}(\zeta,t,\lambda)=\frac{e^{\omega_{2}(\zeta,t,\lambda)}}{2i(t^{\lambda}-\zeta^{\lambda})}.$

Hence

$\left|X_{j}(\zeta,t,\lambda)\right|\geq\frac{C_{1,0}}{|\zeta-t|}.$

In what follows we write $L^{1}_{\zeta}(\Omega)$ to indicate the integral is
with respect to $\zeta$. Using the above estimate in (5.17) we have

$\displaystyle D_{32}$ $\displaystyle\leq
C_{1,0}\left(\left|\frac{B(\zeta,\lambda_{2})}{\zeta^{\lambda_{2}}-z^{\lambda_{2}}}\left[\delta^{\lambda_{1}\lambda_{2}}X_{j}(\zeta,t)\right]\right|_{L^{1}_{\zeta}(\Omega)}+\left|B(\zeta,\lambda_{2})\left[\delta^{\lambda_{1}\lambda_{2}}X_{j}(\zeta,t)\right]\right|_{L^{1}_{\zeta}(\Omega)}\right)$
$\displaystyle\leq
C_{1,0}\left|B(\cdot,\lambda_{2})\right|_{L^{p}(\overline{\Omega})}\left(\left|\frac{\delta^{\lambda_{1}\lambda_{2}}X_{j}(\zeta,t)}{\zeta^{\lambda_{2}}-z^{\lambda_{2}}}\right|_{L^{p^{\prime}}_{\zeta}(\overline{\Omega})}+\left|\delta^{\lambda_{1}\lambda_{2}}X_{j}(\zeta,t)\right|_{L^{p^{\prime}}_{\zeta}(\overline{\Omega})}\right)$
$\displaystyle\leq
C_{1,0}\left|B(\cdot,\lambda_{2})\right|_{L^{p}(\overline{\Omega})}\left\\{\left(\int_{\\{\zeta:|\zeta-z|<\delta^{\prime}\\}}\left|\frac{\delta^{\lambda_{1}\lambda_{2}}X_{j}(\zeta,t)}{\zeta-z}\right|^{p^{\prime}}\,dA(\zeta)\right)^{\frac{1}{p^{\prime}}}\right.$
$\displaystyle\quad+\left.\left(\int_{\\{\zeta:|\zeta-z|\geq\delta^{\prime}\\}}\left|\frac{\delta^{\lambda_{1}\lambda_{2}}X_{j}(\zeta,t)}{\zeta-z}\right|^{p^{\prime}}\,dA(\zeta)\right)^{\frac{1}{p^{\prime}}}+\left|\delta^{\lambda_{1}\lambda_{2}}X_{j}(\zeta,t)\right|_{L^{p^{\prime}}_{\zeta}(\overline{\Omega})}\right\\}$

where $\delta^{\prime}>0$ is to be determined. By Lemma 5.3 (ii), $\sup_{t\in
K}|\delta^{\lambda_{1}\lambda_{2}}X_{j}(\cdot,t)|_{L^{p^{\prime}}(\Omega)}\to
0$ as $|\lambda_{1}-\lambda_{2}|\to 0$. So for fixed $\delta^{\prime}$ the
last two terms in the above bracket converge to $0$. By estimate (5.6),

$\displaystyle|\omega_{j}(z,t,\lambda_{1})-\omega_{j}(z,t,\lambda_{2})|$
$\displaystyle\leq|\omega_{j}(z,t,\lambda_{1})|+|\omega_{j}(z,t,\lambda_{2})|$
$\displaystyle\leq
C_{1,0}C_{p}\left(|A(\cdot,\lambda)|_{L^{p}(\overline{\Omega})}+|B(\cdot,\lambda)|_{L^{p}(\overline{\Omega})}\right)|z-t|^{\alpha}.$

For any $\varepsilon>0$, we can pick $r>0$ such that for any
$z\in\overline{\Omega}$ and $t\in K$ with $|z-t|<r$, one has
$|\omega_{j}(z,t,\lambda_{1})-\omega_{j}(z,t,\lambda_{2})|<\varepsilon$. Then
it suffices to consider the points $z$ for which $|z-t|\geq r$.

Now for this fixed $r$, we can let $\delta^{\prime}=\frac{r}{2}$ be such that
for any $z$ with $|z-t|\geq r$ and $|\zeta-z|<\delta^{\prime}$, we have
$|\zeta-t|\geq|t-z|-|z-\zeta|\geq\delta^{\prime}=\frac{r}{2}$, and so

$\\{\zeta:|\zeta-z|<\delta^{\prime}\\}\subset\\{\zeta:|\zeta-t|\geq\delta^{\prime}\\}.$

Hence

$\displaystyle\left(\int_{\\{\zeta:|\zeta-z|<\delta^{\prime}\\}}\left|\frac{\delta^{\lambda_{1}\lambda_{2}}X_{j}(\zeta,t)}{\zeta-z}\right|^{p^{\prime}}\,dA(\zeta)\right)^{\frac{1}{p^{\prime}}}$
$\displaystyle\leq\left(\int_{\\{\zeta:|\zeta-t|\geq\delta^{\prime}\\}}\left|\frac{\delta^{\lambda_{1}\lambda_{2}}X_{j}(\zeta,t)}{\zeta-z}\right|^{p^{\prime}}\,dA(\zeta)\right)^{\frac{1}{p^{\prime}}}$
$\displaystyle\leq C_{p^{\prime}}\sup_{t\in
K}|\delta^{\lambda_{1}\lambda_{2}}X_{j}(\zeta,t)|_{C^{0}(\\{\zeta:|\zeta-t|\geq\delta^{\prime}\\})}.$

The last expression goes to $0$ as $|\lambda_{1}-\lambda_{2}|\to 0$, by Lemma
5.3 (iii). This shows that $D_{32}$ converges to $0$ as
$|\lambda_{1}-\lambda_{2}|\to 0$ and the convergence is uniform in $t\in K$.
The proof of (5.15) is now complete. ∎

##  6\. Families of multiply-connected domains

In this section we prove Theorem 1.1 for families of bounded multiply-
connected domains. We will first follow Vekua to reduce the problem to a
singular integral equation. We then obtain regularities in all variables by
transforming it into a equivalent Fredholm equation, so that compactness
argument can be used.

In view of Proposition 2.10, it suffices to consider the following problem:

(6.1) $\displaystyle\Delta
U^{\lambda}+a^{\lambda}(x,y)U_{x}^{\lambda}+b^{\lambda}(x,y)U_{y}^{\lambda}=f^{\lambda}(x,y),\quad\text{in
$\Omega^{\lambda}$; }$ (6.2)
$\displaystyle\alpha^{\lambda}U^{\lambda}_{x}-\beta^{\lambda}U^{\lambda}_{y}=\gamma^{\lambda},\quad\text{on
$b\Omega^{\lambda}$, }$

where
$a^{\lambda},b^{\lambda},f^{\lambda}\in\mathcal{C}^{k+\mu,j}(\overline{\Omega^{\lambda}})$
and
$\alpha^{\lambda},\beta^{\lambda},\gamma^{\lambda}\in\mathcal{C}^{k+1+\mu,j}(b\Omega^{\lambda})$.
As remarked at the end of Section 3, if we set
$w^{\lambda}=U^{\lambda}_{x}-iU^{\lambda}_{y}$ on $\Omega^{\lambda}$ for each
fixed $\lambda\in[0,1]$, then $w^{\lambda}$ is a solution to the following
Problem A on $\Omega^{\lambda}$:

(6.3)
$\displaystyle\partial{}_{\overline{z}}w^{\lambda}+A^{\lambda}w^{\lambda}+B^{\lambda}\overline{w^{\lambda}}=F^{\lambda},\quad\text{in
$\Omega^{\lambda}$.}$ (6.4) $\displaystyle
Re[\overline{l^{\lambda}}w^{\lambda}]=\gamma^{\lambda},\quad\text{on
$b\Omega^{\lambda}$,}$

where
$A^{\lambda},B^{\lambda},F^{\lambda}\in\mathcal{C}^{k+\mu,j}(\overline{\Omega^{\lambda}})$
and
$l^{\lambda},\gamma^{\lambda}\in\mathcal{C}^{k+1+\mu,j}(b\Omega^{\lambda})$.
In addition $U^{\lambda}$ and $w^{\lambda}$ are related by the formula

$U^{\lambda}(z^{\lambda})=c_{0}(\lambda)+Re\int_{z_{0}^{\lambda}}^{z^{\lambda}}w^{\lambda}(\zeta)\,d\zeta^{\lambda},\quad
z^{\lambda}=x^{\lambda}+iy^{\lambda}\in\Omega^{\lambda},$

where $z_{0}^{\lambda}=x_{0}^{\lambda}+iy_{0}^{\lambda}$ is some fixed point
in $\Omega^{\lambda}$, and the integral can be taken over any path going from
$z_{0}^{\lambda}$ to $z^{\lambda}$ and lying in $\Omega^{\lambda}$.

######  Theorem 6.1 (Problem A on multiply-connected domains).

Let $k$ and $j$ be non-negative integers satisfying $k\geq j$, and $0<\mu<1$.
Let $\Omega$ be a bounded domain in $\mathbb{C}$ whose boundary is $C^{1+\mu}$
and which has $m+1$ connected components. ($\Omega$ has $m$ holes inside.) Let
$\Gamma^{\lambda}:\Omega\to\Omega^{\lambda}$ be a family of embeddings of
class $\mathcal{C}^{1,0}(\Omega)$. Fix a normally distributed set (Definition
3.4) $\\{\\{z_{r}^{\lambda}\\}_{r=1}^{N_{0}}$,
$\\{z_{s}^{\lambda}\\}_{s=1}^{N_{1}}\\}$ in $\Omega$ so that
$\\{z_{r}^{\lambda}=\Gamma^{\lambda}(z_{r}),z_{a}^{\lambda}=\Gamma^{\lambda}(z_{s})\\}$
forms a normally distributed set in $\Omega^{\lambda}$. For each $\lambda$,
let $w^{\lambda}$ be the unique solution to Problem A on $\Omega^{\lambda}$:

(6.5)
$\begin{cases}\partial{}_{\overline{z}}w^{\lambda}(z)+A^{\lambda}(z)w^{\lambda}(z)+B^{\lambda}(z)\overline{w^{\lambda}(z)}=F^{\lambda}(z)&\text{in
$\Omega^{\lambda}$,}\\\
Re[\overline{l^{\lambda}(\zeta)}w^{\lambda}(\zeta)]=\gamma^{\lambda}(\zeta)&\text{on
$b\Omega^{\lambda}$,}\end{cases}$

where the index $n$ (Definition 3.1) satisfies $n>m-1$ and $w^{\lambda}$
satisfies the additional conditions

(6.6) $\displaystyle
w^{\lambda}(z_{r}^{\lambda})=a_{r}(\lambda)+ib_{r}(\lambda),\quad\quad(r=1,...,N_{0}),$
(6.7) $\displaystyle
w^{\lambda}((z_{s}^{\prime})^{\lambda})=l_{j}(z_{s}^{\prime},\lambda)(\gamma_{j}(z_{s}^{\prime},\lambda)+ic_{s}(\lambda)),\quad\quad(s=1,...,N_{1}).$

(i) Suppose for $2<p<\infty$, $\lambda\to A^{\lambda},B^{\lambda},F^{\lambda}$
are continuous maps from $[0,1]$ to $L^{p}(\Omega)$, and that
$l^{\lambda},\gamma^{\lambda}\in\mathcal{C}^{\mu,0}(b\Omega^{\lambda})$,
$a_{r},b_{r},c_{s}\in C^{0}([0,1])$. Then
$w^{\lambda}\in\mathcal{C}^{\nu,0}(\overline{\Omega^{\lambda}})$, where
$\nu=min(\alpha_{p},\mu)$ and $\alpha_{p}=\frac{p-2}{p}$.
(ii) If $b\Omega\in C^{k+1+\mu}$,
$\Gamma^{\lambda}\in\mathcal{C}^{k+1+\mu,0}(\overline{\Omega})$,
$A^{\lambda},B^{\lambda},F^{\lambda}\in\mathcal{C}^{k+\mu,0}(\overline{\Omega^{\lambda}})$,
$l^{\lambda},\gamma^{\lambda}\in\mathcal{C}^{k+1+\mu,0}(b\Omega^{\lambda})$,
then $w^{\lambda}\in\mathcal{C}^{k+1+\mu,0}(\overline{\Omega^{\lambda}})$, and
there exists some constant $C$ independent of $\lambda$ such that

$\displaystyle\|w^{\lambda}\|_{0,0}\leq
C\left(\|F^{\lambda}\|_{0,0}+\|\gamma^{\lambda}\|_{\mu,0}+\sum_{r}|a_{r}|_{0}+|b_{r}|_{0}+\sum_{s}|c_{s}|_{0}\right),$
$\displaystyle\|w^{\lambda}\|_{k+1+\mu,0}\leq
C\left(\|F^{\lambda}\|_{k+\mu,0}+\|\gamma^{\lambda}\|_{k+1+\mu,0}+\sum_{r}|a_{r}|_{0}+|b_{r}|_{0}+\sum_{s}|c_{s}|_{0}\right).$

(iii) If $b\Omega\in C^{k+1+\mu}$,
$\Gamma^{\lambda}\in\mathcal{C}^{k+1+\mu,j}(\overline{\Omega})$,
$A^{\lambda},B^{\lambda},F^{\lambda}\in\mathcal{C}^{k+\mu,j}(\overline{\Omega^{\lambda}})$,
$l^{\lambda},\gamma^{\lambda}\in\mathcal{C}^{k+1+\mu,j}(b\Omega^{\lambda})$,
then $w^{\lambda}\in\mathcal{C}^{k+1+\mu,j}(\overline{\Omega^{\lambda}})$, and
there exists some constant $C$ independent of $\lambda$ such that

$\|w^{\lambda}\|_{k+1+\mu,j}\leq
C\left(\|F^{\lambda}\|_{k+\mu,j}+\|\gamma^{\lambda}\|_{k+1+\mu,j}+\sum_{r=1}^{N_{0}}|a_{r}|_{j}+|b_{r}|_{j}+\sum_{s=1}^{N_{1}}|c_{s}|_{j}\right).$

As a consequence of Theorem 6.1, we can now prove Theorem 1.1 for the case
$\varkappa>m-1$, where $\varkappa$ is the index for the oblique derivative
boundary value problem. (See Definition 4.2).

######  Theorem 6.2 (Multiply-connected domains, $\varkappa>m-1$ ).

Let $k$ and $j$ be non-negative integers satisfying $k\geq j$, and $0<\mu<1$.
Let $\Omega$ be a bounded domain in $\mathbb{C}$ with $\mathcal{C}^{k+2+\mu}$
boundary. Let $\Gamma^{\lambda}$ be a family of maps that embed $\Omega$ to
$\Omega^{\lambda}$, with
$\Gamma^{\lambda}\in\mathcal{C}^{k+2+\mu,j}(\overline{\Omega})$. Consider the
following oblique derivative problem:

(6.8) $\displaystyle
a^{\lambda}(x,y)u^{\lambda}_{xx}+2b^{\lambda}(x,y)u^{\lambda}_{xy}+c^{\lambda}(x,y)u^{\lambda}_{yy}+d^{\lambda}(x,y)u^{\lambda}_{x}+e^{\lambda}(x,y)u^{\lambda}_{y}=f^{\lambda},\quad\text{in
$\Omega^{\lambda}$, }$ (6.9)
$\displaystyle\frac{du^{\lambda}}{d\Upsilon^{\lambda}}=\gamma^{\lambda}\quad\text{on
$b\Omega^{\lambda}$, }$

with index $\varkappa>m-1$. Here
$\Upsilon^{\lambda}=\xi^{\lambda}+i\eta^{\lambda}$ is a non-vanishing vector
field along $b\Omega^{\lambda}$. Suppose
$a^{\lambda},b^{\lambda},c^{\lambda},d^{\lambda},e^{\lambda},f^{\lambda}$ are
functions in $\Omega^{\lambda}$ such that
$a^{\lambda},b^{\lambda},c^{\lambda}\in\mathcal{C}^{k+1+\mu,j}(\overline{\Omega^{\lambda}})$,
and
$d^{\lambda},e^{\lambda},f^{\lambda}\in\mathcal{C}^{k+\mu,j}(\overline{\Omega^{\lambda}})$.
Suppose further that $\Upsilon^{\lambda},\gamma^{\lambda}$ are functions in
$b\Omega^{\lambda}$ such that
$\Upsilon^{\lambda},\gamma^{\lambda}\in\mathcal{C}^{k+1+\mu,j}(b\Omega^{\lambda})$.
For each $\lambda\in[0,1]$, let $\\{\\{z_{r}^{\lambda}\\}_{r=1}^{N_{0}}$,
$\\{z_{s}^{\lambda}\\}_{s=1}^{N_{1}}\\}$ be a normally distributed set for
$\Omega^{\lambda}$, and let $u^{\lambda}$ be the unique solution to
(6.8)-(6.9) on $\Omega^{\lambda}$ satisfying the conditions (see [VA62]*p. 287
Theorem 4.13)

(6.10) $\displaystyle
u^{\lambda}_{x}(z_{r}^{\lambda})-iu^{\lambda}_{y}(z_{r}^{\lambda})=a_{r}(\lambda)+ib_{r}(\lambda)\quad\quad(r=1,...,N_{0}),$
(6.11) $\displaystyle
u^{\lambda}_{x}((z_{s}^{\prime})^{\lambda})-iu^{\lambda}_{y}((z_{s}^{\prime})^{\lambda})=l^{\lambda}((z_{s}^{\prime})^{\lambda})(\gamma^{\lambda}((z_{s}^{\prime})^{\lambda})+ic_{s}(\lambda)),\quad\quad(s=1,...,N_{1}),$
(6.12) $\displaystyle u^{\lambda}(z_{0}^{\lambda})=e(\lambda),\quad
z_{0}^{\lambda}\in\Omega^{\lambda},$

where $a_{r},b_{r},c_{s},e\in C^{j}([0,1])$. Then
$u^{\lambda}\in\mathcal{C}^{k+2+\mu,j}(\overline{\Omega^{\lambda}})$.

###### Proof.

Let $\varphi^{\lambda}$ be the maps defined as in Proposition 2.10. Then
$\mathcal{U}^{\lambda}:=u^{\lambda}\circ(\varphi^{\lambda})^{-1}$ is a
solution to the following problem

$\displaystyle\Delta\,\mathcal{U}^{\lambda}+p^{\lambda}(\xi,\eta)\,\mathcal{U}^{\lambda}_{\xi}+q^{\lambda}(\xi,\eta)\,\mathcal{U}^{\lambda}_{\eta}=h^{\lambda}(\xi,\eta),\quad\text{in
$D^{\lambda}$.}$
$\displaystyle\nu_{1}^{\lambda}\,\mathcal{U}^{\lambda}_{\xi}+\nu_{2}^{\lambda}\,\mathcal{U}^{\lambda}_{\eta}=g^{\lambda}(\xi,\eta),\quad\text{on
$bD^{\lambda}$,}$

where
$p^{\lambda},q^{\lambda},h^{\lambda}\in\mathcal{C}^{k+\mu,j}(\overline{D^{\lambda}})$
and
$\nu_{1}^{\lambda},\nu_{2}^{\lambda},g^{\lambda}\in\mathcal{C}^{k+1+\mu,j}(\overline{D^{\lambda}})$.
Furthermore $\mathcal{U}^{\lambda}$ satisfies the conditions:

$\displaystyle\mathcal{U}^{\lambda}_{\xi}(\zeta_{r}^{\lambda})-i\,\mathcal{U}^{\lambda}_{\eta}(\zeta_{r}^{\lambda})=\widetilde{a_{r}}(\lambda)+i\widetilde{b_{r}}(\lambda)\quad\quad(r=1,...,N_{0}),$
$\displaystyle\mathcal{U}^{\lambda}_{\xi}((\zeta_{s}^{\prime})^{\lambda})-i\,\mathcal{U}^{\lambda}_{\eta}((\zeta_{s}^{\prime})^{\lambda})=\widetilde{c}_{s}(\lambda)+i\widetilde{d}_{s}(\lambda),\quad\quad(s=1,...,N_{1}),$
$\displaystyle\mathcal{U}^{\lambda}(\zeta_{0}^{\lambda})=\widetilde{e}(\lambda),\quad
z_{0}^{\lambda}\in\Omega^{\lambda},$

where $\\{\zeta_{r}^{\lambda},(\zeta^{\prime}_{s})^{\lambda}\\}$ is the image
of $\\{z^{\lambda}_{r},(z^{\prime}_{s})^{\lambda}\\}$ under the map
$\varphi^{\lambda}$, and
$\widetilde{a_{r}},\widetilde{b_{r}},\widetilde{c_{s}},\widetilde{d_{s}}\widetilde{e}\in
C^{j}([0,1])$.

Let $w^{\lambda}=\mathcal{U}^{\lambda}_{\xi}-i\,\mathcal{U}_{\eta}^{\lambda}$
be the unique solution in $\mathcal{C}^{k+1+\mu,j}(\overline{\Omega})$ to
problem (6.5) as given in Theorem 6.1, where we take

$\displaystyle
A^{\lambda}=\frac{1}{4}\left(p^{\lambda}+iq^{\lambda}\right),\quad
B^{\lambda}=\frac{1}{4}\left(p^{\lambda}-iq^{\lambda}\right),\quad
F^{\lambda}=\frac{1}{2}h^{\lambda},$ $\displaystyle
l^{\lambda}=\nu_{1}^{\lambda}-i\nu_{2}^{\lambda},\quad\gamma^{\lambda}=g^{\lambda}.$

and $w^{\lambda}$ satisfies

$\displaystyle
w^{\lambda}(\zeta_{r}^{\lambda})=\widetilde{a_{r}}(\lambda)+i\widetilde{b_{r}}(\lambda),\quad\quad(r=1,\dots,h_{0}),$
$\displaystyle
w^{\lambda}((\zeta^{\prime}_{s})^{\lambda})=\widetilde{c}_{s}(\lambda)+i\widetilde{d}_{s}(\lambda),\quad\quad(s=1,\dots,h_{1}).$

Set

$\displaystyle\mathcal{U}^{\lambda}(\zeta^{\lambda})=c_{0}(\lambda)+Re\int_{\zeta_{0}^{\lambda}}^{z^{\lambda}}w^{\lambda}(\zeta)\,d\zeta^{\lambda},\quad
z^{\lambda}=x^{\lambda}+iy^{\lambda}\in\Omega^{\lambda}.$ $\displaystyle
u^{\lambda}(z^{\lambda})=\mathcal{U}^{\lambda}(\varphi^{\lambda}(z^{\lambda}))=c_{0}(\lambda)+Re\int_{\zeta_{0}^{\lambda}}^{\varphi^{\lambda}(z^{\lambda})}w^{\lambda}(\zeta)\,d\zeta^{\lambda}.$

In view of (6.12) and $e\in C^{j}([0,1])$, we have $c_{0}\in C^{j}([0,1])$.
Hence $u^{\lambda}\in\mathcal{C}^{k+2+\mu,j}(\overline{\Omega^{\lambda}})$. ∎

For the rest of the section we prove Theorem 6.1. As the proof is quite
technical we will divide it into several steps. In §6.1 we reduce problem A
(6.3)-(6.4) to a singular integral equation with parameter. We then transform
the singular integral equation into a Fredholm equation and derive all the
relevant estimates. In the end we apply standard compactness argument to show
that the solution
$w^{\lambda}\in\mathcal{C}^{\nu^{\prime},0}(b\Omega^{\lambda})$, for any
$0<\nu^{\prime}<\nu$.

In §6.2 we use the $\mathcal{C}^{\nu-,0}(b\Omega^{\lambda})$ estimate to prove
the $\mathcal{C}^{\nu,0}(b\Omega^{\lambda})$ estimate. For our proof we apply
a trick of Vekua to reduce the problem to the case for families of simply-
connected domains. In §6.3, we derive the
$\mathcal{C}^{k+1+\mu,0}(b\Omega^{\lambda})$ estimate for $w^{\lambda}$. In
§6.4, we take the derivatives in $\lambda$ and derive the
$\mathcal{C}^{k+1+\mu,j}(b\Omega^{\lambda})$ estimate for $w^{\lambda}$. This
completes the proof.

###  6.1. The $\mathcal{C}^{\nu^{-},0}$ estimate

Recall the notation: $f(z,\lambda):=f^{\lambda}(z^{\lambda})$, for
$(z,\lambda)\in\overline{\Omega}\times[0,1]$.

Step 1: Setup of the singular integral equation.

For each $\lambda$, let $w^{\lambda}$ be the unique solution to the boundary
value problem (6.5) satisfying conditions (6.6) and (6.7). Since
$Re[\overline{l(t,\lambda)}w(t,\lambda)]=\gamma(t,\lambda)$ on $b\Omega$,
there exists a real-valued function $\eta$ such that

(6.13)
$w(t,\lambda)=l(t,\lambda)\gamma(t,\lambda)+il(t,\lambda)\eta(t,\lambda),\quad(t,\lambda)\in
b\Omega\times[0,1].$

Then the solution $w^{\lambda}$ can be represented by formula (3.35):

(6.14)
$w(z,\lambda)=w_{\gamma}(z,\lambda)+w_{\eta}(z,\lambda)+w_{F}(z,\lambda),\quad
z\in\Omega,$

where

(6.15) $\displaystyle w_{\gamma}(z,\lambda)$ $\displaystyle:=\frac{1}{2\pi
i}\int_{b\Omega}G_{1}(z,t,\lambda)l(t,\lambda)\gamma(t,\lambda)\,dt^{\lambda}-\frac{1}{2\pi
i}\int_{b\Omega}G_{2}(z,t,\lambda)\overline{l(t,\lambda)}\gamma(t,\lambda)\,\overline{dt^{\lambda}},$
(6.16) $\displaystyle w_{\eta}(z,\lambda)$
$\displaystyle:=\frac{1}{2\pi}\int_{b\Omega}G_{1}(z,t,\lambda)l(t,\lambda)\eta(t,\lambda)\,dt^{\lambda}-\frac{1}{2\pi}\int_{b\Omega}G_{2}(z,t,\lambda)\overline{l(t,\lambda)}\eta(t,\lambda)\,\overline{dt^{\lambda}},$
(6.17) $\displaystyle w_{F}(z,\lambda)$
$\displaystyle:=-\frac{1}{\pi}\iint_{\Omega}G_{1}(z,\zeta,\lambda)F(\zeta,\lambda)\,dA(\zeta^{\lambda})-\frac{1}{\pi}\iint_{\Omega}G_{2}(z,\zeta,\lambda)\overline{F(\zeta,\lambda)}\,dA(\zeta^{\lambda}).$

Here $G_{i}(z,t,\lambda)$ is the fundamental kernel of the equation:

$\partial{}_{\overline{z}}w+A(z,\lambda)w+B(z,\lambda)\overline{w}=0,\quad
z\in\mathbb{C},$

where for $z\notin\Omega$, we take $A(z,\lambda)\equiv B(z,\lambda)\equiv 0$.

According to (3.31), $G_{i}(z,t,\lambda)$ has the representations:

$G_{1}(z,t,\lambda)=\frac{e^{\omega_{1}(z,t,\lambda)}+e^{\omega_{2}(z,t,\lambda)}}{t^{\lambda}-z^{\lambda}},\quad
G_{2}(z,t,\lambda)=\frac{e^{\omega_{1}(z,t,\lambda)}-e^{\omega_{2}(z,t,\lambda)}}{t^{\lambda}-z^{\lambda}},$

where $\omega_{i}(z,t,\lambda)^{\prime}s$ are given by formula (5.2).

In view of (3.33), It can be shown easily that for fixed $\lambda$ and
$F(\cdot,\lambda)\in L^{p}(\Omega)$, $w_{F}(\cdot,\lambda)$ is continuous in
the whole complex plane.

Referring to [VA62]*p.234, $\eta$ satisfies the singular integral equation

(6.18)
$\mathcal{K}^{\lambda}\eta(\zeta,\lambda)=\int_{b\Omega}K(\zeta,t,\lambda)\eta(t,\lambda)\,ds=\gamma_{0}(\zeta,\lambda),$

where

(6.19)
$\gamma_{0}(\zeta,\lambda)=\gamma(\zeta,\lambda)-Re[\overline{l(\zeta,\lambda)}w_{\gamma}^{+}(\zeta,\lambda)]-Re[\overline{l(\zeta,\lambda)}w_{F}(\zeta,\lambda)],$

and

$K(\zeta,t,\lambda)=-\frac{1}{2\pi}Re\left[l(t,\lambda)(\rho^{\lambda})^{\prime}(s)\left(G_{1}(\zeta,t,\lambda)\overline{l(\zeta,\lambda)}+\overline{G_{2}(\zeta,t,\lambda)}l(\zeta,\lambda)\right)\right].$

In addition, $\eta$ satisfies the following conditions:

(6.20) $\displaystyle
w_{\eta}(z_{i},\lambda)=a_{i}(\lambda)+ib_{i}(\lambda)-w_{\gamma}(z_{i},\lambda)-w_{F}(z_{i},\lambda)$
(6.21) $\displaystyle
w_{\eta}^{+}(z_{i}^{\prime},\lambda)=l(z_{i}^{\prime},\lambda)(\gamma(z_{i}^{\prime},\lambda)+ic_{i}(\lambda))-w_{\gamma}^{+}(z_{i}^{\prime},\lambda)-w_{F}(z_{i}^{\prime},\lambda)$

where $\\{z_{i},z_{i}^{\prime}\\}$ for $i=1,\dots,k$ and
$i^{\prime}=1,\dots,k^{\prime}$ is a normally distributed set on $\Omega$.
Here we use $w_{\gamma}^{+}$ (resp.$w_{\eta}^{+}$) to denote the limiting
value of $w_{\gamma}(z,\lambda)$ (resp.$w_{\eta}^{+}(z,\lambda)$) as $z$
approaches to $z_{j}^{\prime}\in b\Omega$ from the interior of $\Omega$. For
fixed $\lambda$, $\eta(\cdot,\lambda)\in C^{\nu}(\overline{\Omega})$ by
Proposition 3.6 (i), and thus according to Proposition 7.14, $w_{\eta}^{+}$ is
given by formula (7.8).

Referring to [VA62]*p. 235, the homogeneous integral equation

(6.22) $\mathcal{K}^{\lambda}\eta(\zeta,\lambda)=0$

has $2n+1$ linearly independent solutions. We apply the theory of one
dimensional singular integrals. The reader can refer to the classic book of
Muskhelishvili [MI92].

Step 2: Estimates for $w_{\gamma}^{\lambda}$ and $w_{F}^{\lambda}$.
We first show that the right-hand side of (6.18) is in
$C^{\nu,0}(\overline{\Omega^{\lambda}})$.

######  Lemma 6.3.

Let $\gamma^{\lambda}\in\mathcal{C}^{\mu,0}(b\Omega^{\lambda})$, and let
$w_{\gamma}$ be defined by formula (6.15), then
$w_{\gamma}^{\lambda}\in\mathcal{C}^{\mu,0}(b\Omega^{\lambda})$.

Proof. By equations (3.32) $w_{\gamma}^{\lambda}$ satisfies the equation

$\overline{\partial}_{z^{\lambda}}w_{\gamma}+A(z,\lambda)w_{\gamma}(z,\lambda)+B(z,\lambda)\overline{w_{\gamma}(z,\lambda)}=0.$

By Proposition 7.15, $w_{\gamma}^{\lambda}$ is the unique solution to the
integral equation:

(6.23)
$w_{\gamma}(z,\lambda)+P^{\lambda}w_{\gamma}(z,\lambda)=\Phi(z,\lambda),\quad
z\in\Omega,$

where

(6.24) $\displaystyle
P^{\lambda}w_{\gamma}(z,\lambda):=T_{\Omega^{\lambda}}(A^{\lambda}w^{\lambda}_{\gamma}+B^{\lambda}\overline{w^{\lambda}_{\gamma}})(z^{\lambda}),$
(6.25) $\displaystyle\Phi(z,\lambda):=\frac{1}{2\pi
i}\int_{b\Omega}\frac{(l\gamma)(t,\lambda)}{t^{\lambda}-z^{\lambda}}\,dt^{\lambda}.$

By Proposition 7.13 we know for fixed $\lambda$,
$w_{\gamma}(\cdot,\lambda)\in\mathcal{C}^{\nu}(\overline{\Omega})$, where
$\nu=\min(\frac{p-2}{p},\mu)$.

We make a few observations:
1) By Proposition 7.4, we have

(6.26)
$\displaystyle\left|P^{\lambda}w_{\gamma}(\cdot,\lambda)\right|_{C^{\alpha_{p}}(\overline{\Omega})}$
$\displaystyle\leq
C_{1,0}C_{p}(|A(\cdot,\lambda)|_{L^{p}(\overline{\Omega})}+|B(\cdot,\lambda)|_{L^{p}(\overline{\Omega})})|w_{\gamma}(\cdot,\lambda)|_{C^{0}(\overline{\Omega})}.$

In other words, the operator $P^{\lambda}$ is a compact operator for each
fixed $\lambda$.

2) By standard estimates for Cauchy integral (or see [B-G14]*p.176), we have

(6.27) $\|\Phi\|_{0,0}\leq
C_{1,0}\|l\gamma\|_{\mu,0},\quad\|\Phi\|_{\mu,0}\leq
C_{1+\mu,0}\|l\gamma\|_{\mu,0}.$

3) The homogeneous integral equation

$w_{\gamma}(z,\lambda)+T_{\Omega^{\lambda}}(A^{\lambda}w^{\lambda}_{\gamma}+B^{\lambda}\overline{w^{\lambda}_{\gamma}})(z^{\lambda})=0,\quad
z\in\Omega$

has only the trivial solution for each $\lambda$. For the existence and
uniqueness, see the argument in the proof of Proposition 7.15.

First, we show that $|w_{\gamma}(\cdot,\lambda)|_{0}$ is bounded uniformly in
$\lambda$. Seeking contradiction, suppose that for some sequence
$\\{\lambda_{j}\\}$ converging to $\lambda_{0}$, one has
$N_{j}:=|\gamma(\cdot,\lambda_{j})|_{0}\to\infty$. Dividing both sides of
(6.23) by $N_{j}$, we have

(6.28)
$\widehat{w_{\gamma}}(z,\lambda_{j})+P^{\lambda_{j}}\widehat{w_{\gamma}}(z,\lambda_{j})=\frac{1}{N_{j}}\Phi(z,\lambda_{j}),\quad
z\in\Omega.$

where
$\widehat{w_{\gamma}}(z,\lambda_{j})=\frac{1}{N_{j}}\widetilde{w_{\gamma}}(z,\lambda_{j})$.
By estimate (6.26),
$|P^{\lambda}\widehat{w_{\gamma}}(\cdot,\lambda)|_{\alpha}$ is bounded by a
constant uniform in $\lambda$. By Arzela-Ascoli theorem and passing to a
subsequence if necessary, $P^{\lambda_{j}}\widehat{w_{\gamma}}$ converges
uniformly to some function. The right-hand side of the above equation
converges to $0$ by (6.27), so in view of (6.28) $\widehat{w_{\gamma}}$
converges uniformly to some function $\widetilde{w}$. One can check that
$P^{\lambda_{j}}\widehat{w_{\gamma}}$ converges uniformly to
$P^{\lambda_{0}}\widetilde{w}$, thus passing to the limit as $j\to\infty$ we
get

$\widetilde{w}+P^{\lambda_{0}}\widetilde{w}=0.$

By above observation 3), $\widetilde{w}\equiv 0$ which contradicts with the
fact that
$|\widetilde{w}|_{0}=\lim_{j\to\infty}|\widehat{w_{\gamma}}(\cdot,\lambda)|_{0}=1$.
Hence $|w_{\gamma}(\cdot,\lambda)|_{0}$ is bounded uniformly in $\lambda$.
Applying estimate (6.26) we get

$\displaystyle\|w_{\gamma}^{\lambda}\|_{\nu,0}$
$\displaystyle\leq\|-P^{\lambda}w_{\gamma}^{\lambda}\|_{\alpha_{p},0}+\|\Phi^{\lambda}\|_{\mu,0}$
$\displaystyle\leq
C(\|w_{\gamma}^{\lambda}\|_{0,0}+\|l^{\lambda}\gamma^{\lambda}\|_{\mu,0}).$

where $C$ is some constant independent of $\lambda$.

Next we show $w_{\gamma}(\cdot,\lambda)$ is continuous in $\lambda$ in the
$C^{0}(b\Omega)$-norm. Suppose not. Then there exists some $\lambda_{0}$ and a
sequence $\lambda_{j}\to\lambda_{0}$, such that
$|w_{\gamma}(\cdot,\lambda_{j})-w_{\gamma}(\cdot,\lambda_{0})|_{0}\geq\varepsilon>0$.
Since we have shown that $|w_{\gamma}^{\lambda}|_{\nu}$ is bounded uniformly
in $\lambda$. Passing to a subsequence if necessary, we have that
$w_{\gamma}^{\lambda_{j}}$ converges to some $w_{\ast}$ in the
$C^{\nu^{\prime}}(b\Omega)$-norm, for $0<\nu^{\prime}<\nu$. As before we can
show that
$|P^{\lambda_{j}}w_{\gamma}^{\lambda_{j}}-P^{\lambda_{0}}w_{\ast}|_{0}\to 0$.
Letting $|\lambda_{j}-\lambda_{0}|\to 0$ in (6.23) we get

$w_{\ast}(z)+P^{\lambda_{0}}w_{\ast}(z,\lambda)=\Phi(z,\lambda_{0}),\quad
z\in\Omega.$

Since we know that $w^{\lambda_{0}}_{\gamma}$ satisfies the same equation,
again by above observation 3) we must have

$w_{\ast}=w_{\gamma}(\cdot,\lambda_{0}).$

But this contradicts with
$|w_{\gamma}(\cdot,\lambda_{j})-w_{\gamma}(\cdot,\lambda_{0})|\geq\varepsilon>0$.
Hence $w_{\gamma}(\cdot,\lambda)$ is continuous in $\lambda$ in the
$C^{0}(b\Omega)$ norm. By Proposition 2.3, the results imply that
$w_{\gamma}^{\lambda}\in\mathcal{C}^{\nu,0}(b\Omega^{\lambda})$. ∎

Next we prove that
$w_{F}^{\lambda}\in\mathcal{C}^{\alpha_{p},0}(b\Omega^{\lambda})$.

######  Lemma 6.4.

Let $2<p<\infty$ and $\alpha_{p}=\frac{p-2}{p}$. Suppose $\lambda\mapsto
A(\cdot,\lambda),B(\cdot,\lambda),F(\cdot,\lambda)$ are continuous maps from
$[0,1]$ to $L^{p}(\Omega)$. For $z\in\Omega$, let $w^{\lambda}_{F}$ be given
by formula (6.17). Then for each $\lambda$, $w_{F}^{\lambda}$ can be extended
to a function in $C^{\alpha_{p}}(\overline{\Omega})$, and
$w^{\lambda}_{F}\in\mathcal{C}^{\alpha_{p},0}(\overline{\Omega^{\lambda}})$.
Moreover we have the following estimate

$\|w_{F}^{\lambda}\|_{\alpha_{p},0}\leq
C(p,\Omega)\left(\sup_{\lambda\in[0,1]}|A(\cdot,\lambda)|_{L^{p}(\Omega)}+\sup_{\lambda\in[0,1]}|B(\cdot,\lambda)|_{L^{p}(\Omega)}\right),$

where the constant depends only on $p$ and $\Omega$ and is independent of
$\lambda$.

Proof. By Hölder’s inequality we have

(6.29) $\displaystyle|w_{F}(z,\lambda)|$ $\displaystyle\leq
C\left(|G_{1}(\zeta,\cdot,\lambda)|_{L^{p^{\prime}}(\Omega)}+|G_{2}(\zeta,\cdot,\lambda)|_{L^{p^{\prime}}(\Omega)}\right)|F(\cdot,\lambda)|_{L^{p}(\Omega)}$
$\displaystyle\leq
C_{1,0}C_{p}\left(\Sigma_{i=1,2}\|\omega_{i}\|_{0,0}\right)|F(\cdot,\lambda)|_{L^{p}(\Omega)},\quad
2<p<\infty,$

where
$\|\omega_{i}\|_{0,0}=\sup_{\zeta\in\overline{\Omega},\lambda\in[0,1]}|\omega_{i}(\cdot,t,\lambda)|_{0}$.
We have

(6.30) $\displaystyle|w_{F}(z,\lambda)-w_{F}(z^{\prime},\lambda)|$
$\displaystyle\quad\leq
C\sum_{i=1,2}\iint_{\Omega}\left|G_{i}(z,\zeta,\lambda)-G_{i}(z^{\prime},\zeta,\lambda)\right||F(\zeta,\lambda)|\,dA(\zeta^{\lambda})$
$\displaystyle\quad\leq
C\iint_{\Omega}\frac{\Sigma_{i=1,2}\left|e^{\omega_{i}(z,\zeta,\lambda)}-e^{\omega_{i}(z^{\prime},\zeta,\lambda)}\right|}{|\zeta^{\lambda}-z^{\lambda}|}|F(\zeta,\lambda)|\,dA(\zeta^{\lambda})$
$\displaystyle\qquad+C\iint_{\Omega}\left(\Sigma_{i=1,2}\left|e^{\omega_{i}(z^{\prime},\zeta,\lambda)}\right|\right)\left|\frac{1}{\zeta^{\lambda}-z^{\lambda}}-\frac{1}{\zeta^{\lambda}-(z^{\prime})^{\lambda}}\right||F(t,\lambda)|\,dA(\zeta^{\lambda}).$

Denote the first integral above by $I$ and the second integral by $II$.
Applying Hölder’s inequality for $p>2$ we have

$\displaystyle I$ $\displaystyle\leq
C_{1,0}(\Sigma_{i=1,2}\|\omega_{i}\|_{\alpha,0})|z-z^{\prime}|^{\alpha_{p}}\iint_{\Omega}\frac{|F(\zeta,\lambda)|}{|z-\zeta|}\,dA(\zeta)$
$\displaystyle\leq
C_{1,0}C_{p}(\Sigma_{i=1,2}\|\omega_{i}\|_{\alpha,0})|F(\cdot,\lambda)|_{L^{p}(\Omega)}|z-z^{\prime}|^{\alpha_{p}},$

where
$\|\omega_{i}\|_{\alpha_{p},0}:=\sup_{\zeta\in\overline{\Omega},\lambda\in[0,1]}|\omega_{i}(\cdot,t,\lambda)|_{\alpha_{p}}$.

$\displaystyle II$
$\displaystyle=\left\\{\iint_{|z-\zeta|<2|z-z^{\prime}|}+\iint_{|z-\zeta|\geq|z-z^{\prime}|}\right\\}\left(\Sigma_{i=1,2}\left|e^{\omega_{i}(z^{\prime},\zeta,\lambda)}\right|\right)$
$\displaystyle\qquad\times\left|\frac{1}{z^{\lambda}-\zeta^{\lambda}}-\frac{1}{(z^{\prime})^{\lambda}-\zeta^{\lambda}}\right||F(\zeta,\lambda)|\,dA(\zeta^{\lambda})$
$\displaystyle\leq
C_{1,0}(\Sigma_{i=1,2}\|\omega_{i}\|_{0,0})\left\\{\iint_{|z-\zeta|<2|z-z^{\prime}|}\frac{|F(\zeta,\lambda)|}{|z-\zeta|}\,dA(\zeta)+\iint_{|z^{\prime}-\zeta|<3|z-z^{\prime}|}\frac{|F(\zeta,\lambda)|}{|z^{\prime}-\zeta|}\,dA(\zeta)\right.$
$\displaystyle\qquad\left.+\iint_{|z-\zeta|\geq
2|z-z^{\prime}|}\left|\frac{1}{z^{\lambda}-\zeta^{\lambda}}-\frac{1}{(z^{\prime})^{\lambda}-\zeta^{\lambda}}\right||F(\zeta,\lambda)|\,dA(\zeta)\right\\}$
$\displaystyle\leq
C_{1,0}(\Sigma_{i=1,2}\|\omega_{i}\|_{0,0})\left\\{|F(\cdot,\lambda)|_{L^{p}(\overline{\Omega})}|z-z^{\prime}|^{\alpha}+|z-z^{\prime}|\iint_{|z-\zeta|\geq
2|z-z^{\prime}|}\frac{|F(\zeta,\lambda)|}{|z-\zeta||z^{\prime}-\zeta|}\,dA(\zeta)\right\\}$
$\displaystyle\leq
C_{1,0}C_{p}(\Sigma_{i=1,2}\|\omega_{i}\|_{0,0})|F(\cdot,\lambda)|_{L^{p}(\overline{\Omega})}\left\\{|z-z^{\prime}|^{\alpha}+|z-z^{\prime}|(C+|z-z^{\prime}|^{-\frac{2}{p}})\right\\}$
$\displaystyle\leq
C_{1,0}C_{p}(\Sigma_{i=1,2}\|\omega_{i}\|_{0,0})|F(\cdot,\lambda)|_{L^{p}(\overline{\Omega})}|z-z^{\prime}|^{\alpha}.$

Thus we obtain

(6.31) $\displaystyle|w_{F}(\cdot,\lambda)|_{\alpha_{p}}$ $\displaystyle\leq
C_{1,0}C_{p}(\Sigma_{i=1,2}\|\omega_{i}\|_{\alpha,0})|F(\cdot,\lambda)|_{L^{p}(\Omega)}$
(6.32) $\displaystyle\leq
C\left(|A(\cdot,\lambda)|_{L^{p}(\Omega)}+|B(\cdot,\lambda)|_{L^{p}(\Omega)}\right),$

where we used (5.3), and the constant $C$ depends only on $p$ and $\Omega$ and
is independent of $\lambda$.

Next we show $w_{F}^{\lambda}$ is continuous in $C^{0}$-norm. For simplicity
we omit the constant $-\frac{1}{\pi}$. Set

$dA(\zeta^{\lambda})=a(\zeta,\lambda)dA(\zeta),\quad
a(\zeta,\lambda)=|det(D\Gamma^{\lambda})|.$

We have

$\displaystyle|w_{F}(z,\lambda_{1})-w_{F}(z,\lambda_{2})|\leq\sum_{i=1,2}\left|\iint_{\Omega}\delta^{\lambda_{1},\lambda_{2}}[G_{i}(z,\zeta)F(t)a(\zeta)]\,dA(\zeta)\right|$
$\displaystyle\quad\leq\sum_{i=1,2}\iint_{\Omega}[\delta^{\lambda_{1}\lambda_{2}}G_{i}(z,t)](Fa)(\zeta,\lambda_{1})\,dA(\zeta)+\iint_{\Omega}G_{i}(z,\zeta,\lambda_{2})[\delta^{\lambda_{1}\lambda_{2}}(Fa)(t)]\,dA(\zeta)$
$\displaystyle\quad\leq\iint_{\Omega}\frac{\Sigma_{i=1,2}|e^{\omega_{i}(z,\zeta,\lambda_{1})}-e^{\omega_{i}(z,\zeta,\lambda_{2})}|}{|\zeta^{\lambda_{1}}-z^{\lambda_{1}}|}|(Fa)(\zeta,\lambda_{1})|+\iint_{\Omega}\Sigma_{i}|e^{\omega_{i}(z,\zeta,\lambda_{2})}||(Fa)(\zeta,\lambda_{1})|$
$\displaystyle\qquad\left|\frac{1}{\zeta^{\lambda_{1}}-z^{\lambda_{1}}}-\frac{1}{\zeta^{\lambda_{2}}-z^{\lambda_{2}}}\right|\,dA(\zeta)+\iint_{\Omega}\frac{\Sigma_{i}|e^{\omega_{i}(z,\zeta,\lambda_{2})}|}{|\zeta^{\lambda_{2}}-z^{\lambda_{2}}|}|\delta^{\lambda_{1}\lambda_{2}}(Fa)(t)|\,dA(\zeta).$

By Hölder’s inequality, this is bounded by

$C_{1,0}C_{p}\left\\{\Sigma_{i}\sup_{\zeta\in\overline{\Omega}}|\delta^{\lambda_{1},\lambda_{2}}\omega_{i}(\cdot,\zeta)|_{0}|F|_{L^{p}(\Omega)}+(\Sigma_{i}\|\omega_{i}\|_{0,0})\left(|\Gamma^{\lambda_{1}}-\Gamma^{\lambda_{2}}|_{1}|F|_{L^{p}(\Omega)}+|\delta^{\lambda_{1}\lambda_{2}}F|_{L^{p}(\Omega)}\right)\right\\},$

where the constants are independent of $\lambda$. By Lemma 5.4 (iii),
$\sup_{\zeta\in\overline{\Omega}}|\omega_{i}(z,\zeta,\lambda_{1})-\omega_{i}(z,\zeta,\lambda_{2})|_{0}\to
0$ , so the first term above goes to $0$. Since
$\Gamma^{\lambda}\in\mathcal{C}^{1,0}$, and $F(\cdot,\lambda)$ is continuous
in $\lambda$ in the $L^{p}(\Omega)$-norm, the second and third terms above
converge to $0$ as $|\lambda_{1}-\lambda_{2}|\to 0$. ∎

Step 3: Reduction of the singular integral operator.

To estimate $w_{\eta}^{\lambda}$ ((6.16)), we first need to estimate
$\eta(\cdot,\lambda)$. Write

$\mathcal{K}^{\lambda}\eta(\zeta,\lambda)=\mathcal{K}_{1}^{\lambda}\eta(\zeta,\lambda)+\mathcal{N}_{1}^{\lambda}\eta(\zeta,\lambda),$

where

$\displaystyle\mathcal{K}_{1}^{\lambda}\eta(\zeta,\lambda):=-\frac{1}{2\pi}\int_{b\Omega}Re\left[l(t,\lambda)\overline{l(\zeta,\lambda)}(\rho^{\lambda})^{\prime}(s)G_{1}(\zeta,t,\lambda)\right]\eta(t,\lambda)\,ds,$
(6.33)
$\displaystyle\mathcal{N}_{1}^{\lambda}\eta_{1}(\zeta,\lambda):=-\frac{1}{2\pi}\int_{b\Omega}Re\left[l(t,\lambda)l(\zeta,\lambda)(\rho^{\lambda})^{\prime}(s)\overline{G_{2}(\zeta,t,\lambda)}\right]\eta(t,\lambda)\,ds.$

The operator $\mathcal{N}_{1}^{\lambda}$ is completely continuous, since by
estimate (5.6),

$\displaystyle|G_{2}(\zeta,t,\lambda)|$
$\displaystyle=\left|\frac{e^{\omega_{1}(\zeta,t,\lambda)}-e^{\omega_{2}(\zeta,t,\lambda)}}{2(t^{\lambda}-\zeta^{\lambda})}\right|\leq
C|t-\zeta|^{-\frac{2}{p}}$ $\displaystyle\leq
C\frac{|\omega_{1}(\zeta,t,\lambda)-\omega_{2}(\zeta,t,\lambda)|}{|t^{\lambda}-\zeta^{\lambda}|}$
$\displaystyle\leq
C\frac{|\omega_{1}(\zeta,t,\lambda)|+|\omega_{2}(\zeta,t,\lambda)|}{t-\zeta}$
$\displaystyle\leq
C_{1,0}C_{p}\left(|A(\cdot,\lambda)|_{L^{p}(\Omega)}+|B(\cdot,\lambda)|_{L^{p}(\Omega)}\right)|t-\zeta|^{-\frac{2}{p}},$

where $2<p<\infty$. We shall call $\mathcal{N}_{1}^{\lambda}$ a _Fredholm
operator of the first kind_ , using the same terminologies as in [MI92].

On the other hand, using that $w_{j}(t,t,\lambda)=0$ for all $t$ and estimate
(5.7), we have

$\displaystyle G_{1}(\zeta,t,\lambda)-\frac{1}{t^{\lambda}-\zeta^{\lambda}}$
$\displaystyle=\frac{e^{\omega_{1}(\zeta,t,\lambda)}+e^{\omega_{2}(\zeta,t,\lambda)}}{2(t^{\lambda}-\zeta^{\lambda})}-\frac{2}{2(t^{\lambda}-\zeta^{\lambda})}$
$\displaystyle=\frac{e^{\omega_{1}(\zeta,t,\lambda)}-e^{\omega_{1}(t,t,\lambda)}}{2(t^{\lambda}-\zeta^{\lambda})}+\frac{e^{\omega_{2}(\zeta,t,\lambda)}-e^{\omega_{2}(t,t,\lambda)}}{2(t^{\lambda}-\zeta^{\lambda})}$
$\displaystyle\leq
C_{1,0}C_{p}\left(|A(\cdot,\lambda)|_{L^{p}(\Omega)}+|B(\cdot,\lambda)|_{L^{p}(\Omega)}\right)|\zeta-t|^{-\frac{2}{p}}.$

It follows that

$\mathcal{K}^{\lambda}_{1}\eta(\zeta,\lambda)=\mathcal{K}_{2}^{\lambda}\eta(\zeta,\lambda)+\mathcal{N}_{2}^{\lambda}\eta(\zeta,\lambda).$

where

$\mathcal{K}^{\lambda}_{2}\eta(\zeta,\lambda):=-\frac{1}{2\pi}\int_{b\Omega}Re\left[\frac{l(t,\lambda)(\rho^{\lambda})^{\prime}(s)\overline{l(\zeta,\lambda)}}{t^{\lambda}-\zeta^{\lambda}}\right]\eta(t,\lambda)\,ds$

and

(6.34)
$\mathcal{N}_{2}^{\lambda}\eta(\zeta,\lambda):=-\frac{1}{2\pi}\int_{b\Omega}Re\left[l(t,\lambda)\overline{l(\zeta,\lambda)}(\rho^{\lambda})^{\prime}(s)\sum_{i=1,2}\frac{e^{\omega_{i}(\zeta,t,\lambda)}-e^{\omega_{i}(t,t,\lambda)}}{2(t^{\lambda}-\zeta^{\lambda})}\right]\eta(t,\lambda)\,ds.$

The operator $\mathcal{N}_{2}^{\lambda}$ is completely continuous, and

$\mathcal{K}_{2}^{\lambda}\eta(\zeta,\lambda)=-\frac{1}{4\pi}\int_{b\Omega}\left[\frac{l(t,\lambda)(\rho^{\lambda})^{\prime}(s)\overline{l(\zeta,\lambda)}}{t^{\lambda}-\zeta^{\lambda}}+\frac{\overline{l(t,\lambda)}\,\overline{(\rho^{\lambda})^{\prime}(s)}l(\zeta,\lambda)}{\overline{t^{\lambda}-\zeta^{\lambda}}}\right]\eta(t,\lambda)\,ds.$

Since

$\displaystyle\frac{ds}{\overline{t^{\lambda}-\zeta^{\lambda}}}=\frac{(\rho^{\lambda})^{\prime}(s)\,d\overline{t^{\lambda}}}{\overline{t^{\lambda}-\zeta^{\lambda}}}$
$\displaystyle=(\rho^{\lambda})^{\prime}(s)\,d\log(\overline{t}-\overline{\zeta^{\lambda}})$
$\displaystyle=(\rho^{\lambda})^{\prime}(s)\,d\log(t^{\lambda}-\zeta^{\lambda})+(\rho^{\lambda})^{\prime}(s)\,d\log\frac{\overline{t^{\lambda}-\zeta^{\lambda}}}{t^{\lambda}-\zeta^{\lambda}}$
$\displaystyle=\frac{(\rho^{\lambda})^{\prime}(s)\,dt^{\lambda}}{t^{\lambda}-\zeta^{\lambda}}+(\rho^{\lambda})^{\prime}(s)\,d\log\frac{\overline{t^{\lambda}}-\overline{\zeta^{\lambda}}}{t^{\lambda}-\zeta^{\lambda}},$

we can rewrite $\mathcal{K}_{2}^{\lambda}\eta$ as

$\mathcal{K}_{2}^{\lambda}\eta(\zeta,\lambda)=\mathcal{K}^{\lambda}_{3}\eta(\zeta,\lambda)+\mathcal{N}_{3}^{\lambda}\eta(\zeta,\lambda),$

where

$\displaystyle\mathcal{K}^{\lambda}_{3}\eta(\zeta,\lambda):=-\frac{1}{4\pi}\int_{b\Omega}\frac{\left(l(t,\lambda)\overline{l(\zeta,\lambda)}+\overline{l(t,\lambda)}l(\zeta,\lambda)\right)\eta(t,\lambda)\,dt^{\lambda}}{t^{\lambda}-\zeta^{\lambda}}$
(6.35)
$\displaystyle\mathcal{N}_{3}^{\lambda}\eta(\zeta,\lambda):=-\frac{1}{4\pi}\int_{b\Omega}\overline{l(t,\lambda)}l(\zeta,\lambda)\eta(t,\lambda)\,d\,log\frac{\overline{t^{\lambda}-\zeta^{\lambda}}}{t^{\lambda}-\zeta^{\lambda}}.$

Now, $\mathcal{N}_{3}^{\lambda}$ is a compact operator, and we can write
$\mathcal{K}_{3}^{\lambda}\eta^{\lambda}$ in the form

(6.36)
$\mathcal{K}_{3}^{\lambda}\eta(\zeta,\lambda)=A(\zeta,\lambda)\eta(\zeta,\lambda)+\frac{1}{\pi
i}\int_{b\Omega}\frac{S(\zeta,t,\lambda)\eta(t,\lambda)}{t^{\lambda}-\zeta^{\lambda}}\,dt^{\lambda},$

where $A(\zeta,\lambda)\equiv 0$, and

(6.37)
$S(\zeta,t,\lambda):=-\frac{i}{4}\,\left(l(t,\lambda)\overline{l(\zeta,\lambda)}+\overline{l(t,\lambda)}l(\zeta,\lambda)\right).$

In particular we have $S(t,t,\lambda)=-\frac{i}{2}$. Now, the integral
equation (6.18) can be written as

(6.38)
$\mathcal{K}^{\lambda}\eta(\zeta,\lambda)=\mathcal{K}_{3}^{\lambda}\eta(\zeta,\lambda)+\mathcal{N}_{1}^{\lambda}\eta(\zeta,\lambda)+\mathcal{N}_{2}^{\lambda}\eta(\zeta,\lambda)+\mathcal{N}_{3}^{\lambda}\eta(\zeta,\lambda)=\gamma_{0}(\zeta,\lambda),$

where $\mathcal{K}^{\lambda}_{3}$ is a singular integral operator and
$N^{\lambda}_{j}$-s are compact operators. Following [MI92] (P. 118, (45.4)),
the index of the singular integral operator $\mathcal{K}^{\lambda}_{3}$ is
defined to be

$\frac{1}{2\pi}\left[\arg\frac{A(t,\lambda)-S(t,t,\lambda)}{A(t,\lambda)+S(t,t,\lambda)}\right]_{b\Omega}=\frac{1}{2\pi}[\arg(-1)]=0.$

One can find a “ reducing operator” $\mathcal{M}^{\lambda}$ so that the
composition
$\mathcal{F}^{\lambda}=\mathcal{M}^{\lambda}\circ\mathcal{K}^{\lambda}$ is a
Fredholm operator. Moreover by Vekua’s equivalence theorem [MI92]*p. 149, case
1, if the index of $\mathcal{K}_{0}^{\lambda}$ is nonnegative, one can choose
$\mathcal{M}^{\lambda}$ so that the equation

$\mathcal{M}^{\lambda}\varphi=0$

has only the trivial solution $\varphi\equiv 0$. Therefore $\eta^{\lambda}$ is
a solution to the Fredholm equation
$\mathcal{F}^{\lambda}\eta^{\lambda}=\mathcal{M}^{\lambda}\gamma_{0}$ if and
only if $\eta^{\lambda}$ is a solution to the singular integral equation
$\mathcal{K}^{\lambda}\eta^{\lambda}=\gamma_{0}$, in other words it suffices
to consider the equation

$\mathcal{F}^{\lambda}\eta^{\lambda}=\mathcal{M}^{\lambda}\gamma_{0}^{\lambda}.$

There are many choices for such $\mathcal{M}^{\lambda}$, for example, one can
take $\mathcal{M}^{\lambda}$ to be the dominant part of the adjoint operator
of $\mathcal{K}_{3}^{\lambda}$ (See the remark after Definition 7.19. ) For an
operator $\mathcal{T}$, we use the notation

$\mathcal{T}\eta(\zeta,\lambda):=\mathcal{T}^{\lambda}\eta^{\lambda}(\zeta^{\lambda}).$

We have

(6.39) $\mathcal{M}\eta(\zeta,\lambda):=-\frac{S(\zeta,\zeta,\lambda)}{\pi
i}P.V.\int_{b\Omega}\frac{\eta(t,\lambda)\,dt^{\lambda}}{t^{\lambda}-\zeta^{\lambda}}=\frac{1}{2\pi}P.V.\int_{b\Omega}\frac{\eta(t,\lambda)\,dt^{\lambda}}{t^{\lambda}-\zeta^{\lambda}}.$

Since $S$ is Hölder continuous in both $\zeta$ and $t$, we can apply formula
(7.18) to get

$\displaystyle\mathcal{F}^{\lambda}\eta^{\lambda}(\zeta^{\lambda})=\mathcal{M}\mathcal{K}_{3}\eta(\zeta,\lambda)$
$\displaystyle\quad=-[S(\zeta,\zeta,\lambda)]^{2}\eta(\zeta,\lambda)-\frac{S(\zeta,\zeta,\lambda)}{(\pi
i)^{2}}\int_{b\Omega}\left[\int_{b\Omega}\frac{S(t_{1},t,\lambda)}{(t_{1}^{\lambda}-\zeta^{\lambda})(t^{\lambda}-t_{1}^{\lambda})}\,dt_{1}^{\lambda}\right]\eta(t,\lambda)\,dt^{\lambda}$
$\displaystyle\quad=\frac{1}{4}\eta(\zeta,\lambda)-\frac{i}{2\pi^{2}}\int_{b\Omega}\left[\int_{b\Omega}\frac{l(t,\lambda)\overline{l(t_{1},\lambda)}+\overline{l(t,\lambda)}l(t_{1},\lambda)}{(t_{1}^{\lambda}-\zeta^{\lambda})(t^{\lambda}-t_{1}^{\lambda})}\,dt_{1}^{\lambda}\right]\eta(t,\lambda)\,dt^{\lambda},$

where we used that $S(t,t,\lambda)=-\frac{i}{2}$. We have

(6.40)
$\displaystyle\int_{b\Omega}\frac{l(t,\lambda)\overline{l(t_{1},\lambda)}}{(t_{1}^{\lambda}-\zeta^{\lambda})(t^{\lambda}-t_{1}^{\lambda})}=\frac{l(t,\lambda)}{t^{\lambda}-\zeta^{\lambda}}[v_{0}(\zeta,{\lambda})-v_{0}(t,\lambda)],$

where we set

(6.41) $\displaystyle v_{0}(\zeta,\lambda)$
$\displaystyle=P.V.\int_{b\Omega}\frac{\overline{l(\tau,\lambda)}}{\tau^{\lambda}-\zeta^{\lambda}}\,d\tau^{\lambda}$
$\displaystyle=\int_{b\Omega}\frac{\overline{l(\tau,\lambda)}-\overline{l(\zeta,\lambda)}}{\tau^{\lambda}-\zeta^{\lambda}}\,d\tau^{\lambda}+(\pi
i)\overline{l(\zeta,\lambda)}.$

By Lemma 7.10, $v_{0}\in C^{\mu,0}(b\Omega^{\lambda})$ and

(6.42) $\|v_{0}\|_{\mu,0}\leq C_{1,0}\|l\|_{\mu,0}.$

In view of expression (6.40), we have

(6.43)
$\mathcal{F}\eta(\zeta,\lambda)=\frac{1}{4}\eta(\zeta,\lambda)+\int_{b\Omega}\frac{k_{0}(\zeta,t,\lambda)}{\zeta^{\lambda}-t^{\lambda}}\eta(t,\lambda)\,dt^{\lambda},$

where $k_{0}(\zeta,t,\lambda)$ is given by

(6.44)
$k_{0}(\zeta,t,\lambda)=l(t,\lambda)\left(v_{0}(\zeta,\lambda)-v_{0}(t,\lambda)\right).$

In view of (6.42) we have

(6.45) $|k_{0}(\zeta,t,\lambda)|\leq A_{0}|\zeta-t|^{\mu},$

where $A_{0}$ is some constant independent of $\lambda\in[0,1]$ and $t\in
b\Omega$.

Next, we compute the composition operators
$\mathcal{M^{\lambda}}\mathcal{N}_{1}^{\lambda}\eta^{\lambda}$,
$\mathcal{M^{\lambda}}\mathcal{N}_{2}^{\lambda}\eta^{\lambda}$,
$\mathcal{M^{\lambda}}\mathcal{N}_{3}^{\lambda}\eta^{\lambda}$. All of them
turn out to be Fredholm operators of the first kind. For simplicity, we will
omit all the non-zero constants in the following computations. In view of
(6.33), we have

$\displaystyle\mathcal{M}\mathcal{N}_{1}\eta(\zeta,\lambda)$
$\displaystyle\quad=\int_{b\Omega}\frac{1}{t^{\lambda}-\zeta^{\lambda}}\left[\int_{b\Omega}Re\left[l(t_{1},\lambda)l(t,\lambda)\overline{G_{2}(t,t_{1},\lambda)}(\rho^{\lambda})^{\prime}(s_{1})\right]\eta(t_{1},\lambda)\,ds_{1}\right]\,dt^{\lambda}$
$\displaystyle\quad=\int_{b\Omega}\left[\int_{b\Omega}\frac{1}{t^{\lambda}-\zeta^{\lambda}}Re\left(l(t_{1},\lambda)l(t,\lambda)\overline{G_{2}(t,t_{1},\lambda)}(\rho^{\lambda})^{\prime}(s_{1})\right)\,dt^{\lambda}\right]\eta(t_{1},\lambda)\,ds_{1}$
$\displaystyle\quad=\int_{b\Omega}\left[\int_{b\Omega}\frac{1}{t^{\lambda}-\zeta^{\lambda}}\left(l(t_{1},\lambda)l(t,\lambda)\overline{G_{2}(t,t_{1},\lambda)}(\rho^{\lambda})^{\prime}(s_{1})\right.\right.$
$\displaystyle\qquad+\overline{l(t_{1},\lambda)}\,\overline{l(t,\lambda)}\,\overline{(\rho^{\lambda})^{\prime}(s_{1})}G_{2}(t,t_{1},\lambda)\left.\right)\,dt^{\lambda}\left.\right]\eta(t_{1},\lambda)\,ds_{1},$

where we can easily justify the interchange of integrals. The inner integral
can be expanded as

(6.46) $\displaystyle
l(t_{1},\lambda)(t_{1}^{\lambda})^{\prime}(s_{1})\int_{b\Omega}\frac{l(t,\lambda)\left(\overline{e^{\omega_{1}(t,t_{1},\lambda)}}-\overline{e^{\omega_{2}(t,t_{1},\lambda)}}\right)}{(t^{\lambda}-\zeta^{\lambda})(\overline{t_{1}^{\lambda}-t^{\lambda}})}\,dt^{\lambda}$
$\displaystyle+\,\overline{l(t_{1},\lambda)}\,\overline{(t_{1}^{\lambda})^{\prime}(s_{1})}\int_{b\Omega}\frac{\overline{l(t,\lambda)}\left(e^{\omega_{1}(t,t_{1},\lambda)}-e^{\omega_{2}(t,t_{1},\lambda)}\right)}{(t^{\lambda}-\zeta^{\lambda})(t_{1}^{\lambda}-t^{\lambda})}\,dt^{\lambda}.$

Denoting the first and second integrals in the above expression by $I_{1}$ and
$I_{2}$ respectively, we have

(6.47) $\displaystyle
I_{1}(\zeta,t_{1},\lambda)=\int_{b\Omega}\frac{l(t,\lambda)e^{2iarg(t_{1}^{\lambda}-t^{\lambda})}}{(t^{\lambda}-\zeta^{\lambda})(t_{1}^{\lambda}-t^{\lambda})}\left[\overline{e^{\omega_{1}(t,t_{1},\lambda)}}-\overline{e^{\omega_{2}(t,t_{1},\lambda)}}\right]\,dt^{\lambda}$
$\displaystyle\quad=\frac{1}{t_{1}^{\lambda}-\zeta^{\lambda}}\left\\{P.V.\int_{b\Omega}\frac{F(t,t_{1},\lambda)}{t^{\lambda}-\zeta^{\lambda}}dt^{\lambda}\right.-P.V.\left.\int_{b\Omega}\frac{F(t,t_{1},\lambda)}{t^{\lambda}-t_{1}^{\lambda}}\,dt^{\lambda}\right\\}.$
$\displaystyle\quad=\frac{1}{t_{1}^{\lambda}-\zeta^{\lambda}}[v^{\prime}_{1}(\zeta,t_{1},\lambda)-v^{\prime}_{1}(t_{1},t_{1},\lambda)],$

where we set

$\displaystyle
v^{\prime}_{1}(\zeta,t_{1},\lambda):=\int_{b\Omega}\frac{F(t,t_{1},\lambda)-F(\zeta,t_{1},\lambda)}{t^{\lambda}-\zeta^{\lambda}}\,dt^{\lambda}+\pi
iF(\zeta,t_{1},\lambda),$ $\displaystyle
v^{\prime}_{1}(t_{1},t_{1},\lambda):=\int_{b\Omega}\frac{F(t,t_{1},\lambda)-F(t_{1},t_{1},\lambda)}{t^{\lambda}-t_{1}^{\lambda}}\,dt^{\lambda}+\pi
iF(t_{1},t_{1},\lambda),$

and

(6.48)
$F(t,t_{1},\lambda):=l(t,\lambda)e^{2iarg(t_{1}^{\lambda}-t^{\lambda})}\left[\overline{e^{\omega_{1}(t,t_{1},\lambda)}}-\overline{e^{\omega_{2}(t,t_{1},\lambda)}}\right].$

Note that since $\omega_{1}(t_{1},t_{1})=\omega_{2}(t_{1},t_{1})=0$, we have
$F(t_{1},t_{1},\lambda)\equiv 0$ on $b\Omega$. We show that
$v_{1}^{\prime}(\cdot,t_{1},\lambda)\in C^{\nu,0}(b\Omega^{\lambda})$. By
Lemma 7.10, it suffices to see that $F(\cdot,t,\lambda)$ is $\nu$-Hölder
continuous in the first argument, and this follows from Lemma 7.24, and
estimate (5.3). In fact we have

$|F(\cdot,t_{1},\lambda)|_{\nu}\leq
C_{1+\mu,0}|l(\cdot,\lambda)|_{\mu}(\Sigma_{i=1,2}|\omega_{i}(\cdot,t_{1},\lambda)|_{\alpha_{p}}),\quad\nu=min(\alpha_{p},\mu),$

where the constant is uniform in $\lambda$ and $t_{1}\in b\Omega$. Hence by
Lemma 7.10,

(6.49) $\|v^{\prime}_{1}(\cdot,t_{1},\lambda)\|_{\nu,0}\leq
C_{1,0}\|F(\cdot,t_{1},\lambda)\|_{\nu,0}\leq
C_{1+\mu,0}\|l\|_{\mu,0}(\Sigma_{i=1,2}\|\omega_{i}\|_{\alpha_{p},0}).$

In a similar way, we can show that

(6.50)
$I_{2}(\zeta,t_{1},\lambda)=\frac{1}{t_{1}^{\lambda}-\zeta^{\lambda}}[v^{\prime\prime}_{1}(t_{1},\zeta,\lambda)-v^{\prime\prime}_{1}(t_{1},t_{1},\lambda)],$

where $v^{\prime\prime}_{1}$ satisfies

$\|v^{\prime\prime}_{1}(\cdot,t_{1},\lambda)\|_{\nu,0}\leq
C_{1+\mu,0}\|l\|_{\mu,0}(\Sigma_{i=1,2}\|\omega_{i}\|_{\alpha_{p},0}).$

Putting (6.47) and (6.50) in (6.46), we have

(6.51)
$\mathcal{M}\mathcal{N}_{1}\eta(\zeta,\lambda)=\int_{b\Omega}\frac{k_{1}(\zeta,t_{1},\lambda)}{t_{1}^{\lambda}-\zeta^{\lambda}}\eta(t_{1},\lambda)\,ds_{1},\quad\quad\nu=min(\alpha_{p},\mu)$

where $k_{1}(\zeta,t_{1},\lambda)$ is given by

(6.52) $\displaystyle k_{1}(\zeta,t_{1},\lambda)$
$\displaystyle:=l(t_{1},\lambda)(\rho^{\lambda})^{\prime}(s_{1})[v_{1}^{\prime}(\zeta,t_{1},\lambda)-v^{\prime}(t_{1},t_{1},\lambda)]$
(6.53)
$\displaystyle\quad+\overline{l(t_{1},\lambda)}\overline{(\rho^{\lambda})^{\prime}(s_{1})}[v_{1}^{\prime\prime}(\zeta,t_{1},\lambda)-v^{\prime\prime}(t_{1},t_{1},\lambda)],$

and $k_{1}$ satisfies

(6.54) $|k_{1}(\zeta,t_{1},\lambda)|\leq A_{1}|\zeta-t_{1}|^{\nu}.$

Here $A_{1}$ is some constant independent of $\lambda\in[0,1]$ and $t_{1}\in
b\Omega$. By similar procedures applied to (6.34), we can show that the
following holds (up to a nonzero constant)

(6.55)
$\mathcal{M}\mathcal{N}_{2}\eta(\zeta,\lambda)=\int_{b\Omega}\frac{k_{2}(\zeta,t_{1},\lambda)}{t_{1}^{\lambda}-\zeta^{\lambda}}\eta(t_{1},\lambda)\,ds_{1},\quad\quad\nu=min(\alpha,\mu)$

where $k_{2}(\zeta,t_{1},\lambda)$ is given by

(6.56) $\displaystyle k_{2}(\zeta,t_{1},\lambda)$
$\displaystyle:=l(t_{1},\lambda)(\rho^{\lambda})^{\prime}(s_{1})[v_{2}^{\prime}(\zeta,t_{1},\lambda)-v^{\prime}_{2}(t_{1},t_{1},\lambda)]$
$\displaystyle\quad+\overline{l(t_{1},\lambda)}\overline{(\rho^{\lambda})^{\prime}(s_{1})}[v_{2}^{\prime\prime}(\zeta,t_{1},\lambda)-v^{\prime\prime}_{2}(t_{1},t_{1},\lambda)]$

where $v_{2}^{\prime}$ and $v_{2}^{\prime\prime}$ satisfy the estimates

$\displaystyle\|v^{\prime}_{2}(\cdot,t_{1},\lambda)\|_{\nu,0}\leq
C_{1+\mu,0}\|l\|_{\mu,0}(\Sigma_{i=1,2}\|\omega_{i}\|_{\alpha,0}),$
$\displaystyle\|v^{\prime\prime}_{2}(\cdot,t_{1},\lambda)\|_{\nu,0}\leq
C_{1+\mu,0}\|l\|_{\mu,0}(\Sigma_{i=1,2}\|\omega_{i}\|_{\alpha,0}).$

Hence

(6.57) $|k_{2}(\zeta,t_{1},\lambda)|\leq A_{2}|\zeta-t_{1}|^{\nu},$

where $A_{2}$ is some constant independent of $\lambda\in[0,1]$ and $t_{1}\in
b\Omega$. It remains to compute
$\mathcal{M}^{\lambda}\mathcal{N}_{3}^{\lambda}\eta^{\lambda}$. From (6.35),
we have

$\displaystyle\mathcal{N}_{3}\eta(\zeta,\lambda)$
$\displaystyle=-\frac{1}{4\pi}\int_{b\Omega}\overline{l(t,\lambda)}l(\zeta,\lambda)\eta(t,\lambda)\,d\,log\frac{\overline{t^{\lambda}-\zeta^{\lambda}}}{t^{\lambda}-\zeta^{\lambda}}$
$\displaystyle=-\frac{(-2i)}{4\pi}\int_{b\Omega}\overline{l(t,\lambda)}l(\zeta,\lambda)\eta(t,\lambda)\,d_{t^{\lambda}}\arg(t^{\lambda}-\zeta^{\lambda})$
$\displaystyle=\frac{i}{2\pi}\int_{b\Omega}\overline{l(t,\lambda)}l(\zeta,\lambda)\eta(t,\lambda)\partial{}_{\tau^{\lambda}(s)}\arg(t^{\lambda}-\zeta^{\lambda})\,ds,$

where $\tau^{\lambda}$ is the unit tangent vector of $b\Omega^{\lambda}$ at
$t^{\lambda}$. Omitting the constant in the front, we have

$\displaystyle\mathcal{M}\mathcal{N}_{3}\eta(\zeta,\lambda)$
$\displaystyle=\int_{b\Omega}\frac{1}{t^{\lambda}-\zeta^{\lambda}}\left[\int_{b\Omega}\overline{l(t_{1},\lambda)}l(t,\lambda)[\partial{}_{\tau^{\lambda}(s_{1})}\arg(t_{1}^{\lambda}-t^{\lambda})]\eta(t_{1},\lambda)\,ds_{1}\right]\,dt^{\lambda}$
$\displaystyle=\int_{b\Omega}\overline{l(t_{1},\lambda)}\left[\int_{b\Omega}\frac{l(t,\lambda)[\partial{}_{\tau^{\lambda}(s_{1})}\arg(t_{1}^{\lambda}-t^{\lambda})](t_{1}^{\lambda}-t^{\lambda})}{(t^{\lambda}-\zeta^{\lambda})(t_{1}^{\lambda}-t^{\lambda})}\,dt^{\lambda}\right]\eta(t_{1},\lambda)\,ds_{1}.$

Denote

$F(t,t_{1},\lambda)=l(t,\lambda)(t_{1}^{\lambda}-t^{\lambda})[\partial{}_{\tau^{\lambda}(s_{1})}\arg(t_{1}^{\lambda}-t^{\lambda})].$

As before we show that $|F(\cdot,t_{1},\lambda)|_{\mu}$ is bounded uniformly
in $\lambda\in[0,1]$ and $t_{1}\in b\Omega$. Since
$l^{\lambda}\in\mathcal{C}^{\mu,0}(b\Omega^{\lambda})$, we only need to show
the function

$[\partial{}_{\tau^{\lambda}(s_{1})}\arg(t_{1}^{\lambda}-t^{\lambda})](t_{1}^{\lambda}-t^{\lambda})$

is Hölder $\mu$ continuous in $t$, and the Hölder norm is uniformly bounded in
$\lambda$ and $t_{1}$.

Let $\delta=|t-\widetilde{t}|$. If $|t-t_{1}|<2|t-\widetilde{t}|$, then
$|\widetilde{t}-t_{1}|\leq|\widetilde{t}-t|+|t-t_{1}|\leq 3|t-\widetilde{t}|$,
and we have

$\displaystyle\left|[\partial{}_{\tau^{\lambda}(s_{1})}\arg(t_{1}^{\lambda}-t^{\lambda})](t_{1}^{\lambda}-t^{\lambda})-\partial{}_{\tau^{\lambda}(s_{1})}\arg(t_{1}^{\lambda}-\widetilde{t}^{\lambda})](t_{1}^{\lambda}-\widetilde{t}^{\lambda})\right|$
$\displaystyle\quad=\left|\partial{}_{s_{1}}\arg(\rho^{\lambda}(s_{1})-t^{\lambda})(t_{1}^{\lambda}-t^{\lambda})-[\partial{}_{s_{1}}\arg(\rho^{\lambda}(s_{1})-\widetilde{t}^{\lambda})](t_{1}^{\lambda}-\widetilde{t}^{\lambda})\right|$
$\displaystyle\quad\leq\left|\partial{}_{s_{1}}\arg(\rho^{\lambda}(s_{1})-t^{\lambda})\right||t_{1}^{\lambda}-t^{\lambda}|+\left|\partial{}_{s_{1}}\arg(\rho^{\lambda}(s_{1})-\widetilde{t}^{\lambda})\right||t_{1}^{\lambda}-\widetilde{t}^{\lambda}|.$

By Lemma 7.24 (i), the above is bounded by

$\displaystyle
C_{1+\mu,0}\left(|t_{1}-t|^{\mu-1}|t_{1}-t|+|t_{1}-\widetilde{t}|^{\mu-1}|t_{1}-\widetilde{t}|\right)$
$\displaystyle\quad\leq
C_{1+\mu,0}(|t_{1}-t|^{\mu}+|t_{1}-\widetilde{t}|^{\mu})=C_{1+\mu,0}^{\prime}|t-\widetilde{t}|^{\mu}.$

On the other hand if $|t-t_{1}|\geq 2|t-\widetilde{t}|$, we have

$\displaystyle\left|[\partial{}_{\tau^{\lambda}(s_{1})}\arg(t_{1}^{\lambda}-t^{\lambda})](t_{1}^{\lambda}-t^{\lambda})-\partial{}_{\tau^{\lambda}(s_{1})}\arg(t_{1}^{\lambda}-\widetilde{t}^{\lambda})](t_{1}^{\lambda}-\widetilde{t}^{\lambda})\right|$
$\displaystyle\quad\leq\left|\left\\{\partial{}_{\tau^{\lambda}(s_{1})}\arg(t_{1}^{\lambda}-t^{\lambda})-\partial{}_{\tau^{\lambda}(s_{1})}\arg(t_{1}^{\lambda}-\widetilde{t}^{\lambda})\right\\}(t_{1}^{\lambda}-t^{\lambda})\right|$
$\displaystyle\qquad+\left|\partial{}_{\tau^{\lambda}(s_{1})}\arg(t_{1}^{\lambda}-\widetilde{t}^{\lambda})](t^{\lambda}-\widetilde{t}^{\lambda})\right|.$

By Lemma 7.24 (ii), since $|t-t_{1}|\geq 2|t-\widetilde{t}|$, the above is
bounded by

$\displaystyle
C_{1+\mu,0}\left(\frac{|t-\widetilde{t}|}{|t-t_{1}|^{2-\mu}}|t_{1}-t|+|t_{1}-t|^{\mu-1}|t-\widetilde{t}|\right)\leq
C_{1+\mu,0}^{\prime}|t-\widetilde{t}|^{\mu}.$

Now, by the same proof as that for
$\mathcal{M}^{\lambda}\mathcal{N}_{1}^{\lambda}$ and
$\mathcal{M}^{\lambda}\mathcal{N}_{2}^{\lambda}$, we can show that

(6.58)
$\mathcal{M}\mathcal{N}_{3}\eta(\zeta,\lambda)=\int_{b\Omega}\frac{k_{3}(\zeta,t_{1},\lambda)}{t_{1}^{\lambda}-\zeta^{\lambda}}\eta(t_{1},\lambda)\,ds_{1},$

where $k_{3}(\zeta,t_{1},\lambda)$ is of the form

(6.59)
$k_{3}(\zeta,t_{1},\lambda):=\overline{l(t_{1},\lambda)}\left(v_{3}(\zeta,t_{1},\lambda)-v_{3}(t_{1},t_{1},\lambda)\right),$

and $v_{3}$ satisfies the estimate

$\|v_{3}(\cdot,t_{1},\lambda)\|_{\mu,0}\leq C_{1+\mu,0}\|l\|_{\mu,0}.$

Hence

(6.60) $|k_{3}(\zeta,t_{1},\lambda)|\leq A_{3}|\zeta-t_{1}|^{\mu},$

where $A_{3}$ is some constant independent of $\lambda\in[0,1]$ and $t_{1}\in
b\Omega$.

We now apply $\mathcal{M}$ to both sides of integral equation (6.38). By
(6.43), (6.51), (6.55), (6.58) we have

(6.61)
$(I+\mathcal{N})\eta(\zeta,\lambda):=\eta(\zeta,\lambda)+\int_{b\Omega}\frac{k(\zeta,t,\lambda)}{\zeta^{\lambda}-t^{\lambda}}\eta(t,\lambda)\,dt^{\lambda}=\mathcal{M}\gamma_{0}(\zeta,\lambda),$

where
$k(\zeta,t,\lambda):=k_{0}(\zeta,t,\lambda)+k_{1}(\zeta,t,\lambda)+k_{2}(\zeta,t,\lambda)+k_{3}(\zeta,t,\lambda)$.
In view of estimates (6.45), (6.54), (6.57) and (6.60), we have

(6.62) $|k(\zeta,t,\lambda)|\leq
A|\zeta-t|^{\nu},\qquad\nu=\min\left(\mu,\>\alpha_{p}\right),$

for some constant $A$ independent of $\lambda\in[0,1]$ and $t\in b\Omega$.

Step 4: Estimates for the right-hand side of the reduced equation.

We now prove
$\mathcal{M}^{\lambda}\gamma_{0}^{\lambda}\in\mathcal{C}^{\nu^{\prime},0}(\overline{\Omega^{\lambda}})$,
for any $0<\nu^{\prime}<\nu$ and $\nu=\min(\mu,\alpha_{p})$. Recall that
$\gamma_{0}$ is given by (6.19):

(6.63)
$\gamma_{0}(\zeta,\lambda)=\gamma(\zeta,\lambda)-Re[\overline{l(\zeta,\lambda)}w_{\gamma}^{+}(\zeta,\lambda)]-Re[\overline{l(\zeta,\lambda)}w_{F}(\zeta,\lambda)].$

Since for each $\lambda$, $l(\cdot,\lambda)$ and $\gamma(\cdot,\lambda)$ are
Hölder continuous, we can apply the jump formula for generalized Cauchy
integrals (Proposition 7.14) to get

(6.64) $\displaystyle w_{\gamma}^{+}(\zeta,\lambda)$
$\displaystyle=\frac{1}{2}\,l(\zeta,\lambda)\gamma(\zeta,\lambda)+\frac{1}{2\pi
i}P.V.\int_{b\Omega}G_{1}(\zeta,t,\lambda)l(t,\lambda)\gamma(t,\lambda)\,dt^{\lambda}$
$\displaystyle\quad-\frac{1}{2\pi
i}\int_{b\Omega}G_{2}(\zeta,t,\lambda)\overline{l(t,\lambda)}\gamma(t,\lambda)\,\overline{dt^{\lambda}}$
$\displaystyle=P_{1}(\zeta,\lambda)+P_{2}(\zeta,\lambda)+P_{3}(\zeta,\lambda)+P_{4}(\zeta,\lambda),$

where we set

(6.65) $\displaystyle
P_{1}(\zeta,\lambda):=\frac{1}{2}\,l(\zeta,\lambda)\gamma(\zeta,\lambda),\quad\quad
P_{2}(\zeta,\lambda):=\pi i[l(\zeta,\lambda)\gamma(\zeta,\lambda)],$ (6.66)
$\displaystyle
P_{3}(\zeta,\lambda):=\int_{b\Omega}\frac{1}{2(t^{\lambda}-\zeta^{\lambda})}\times$
$\displaystyle\quad\left\\{\left(\sum_{i=1}^{2}e^{\omega_{i}(\zeta,t,\lambda)}\right)[l(t,\lambda)\gamma(t,\lambda)]-\left(\sum_{i=1}^{2}e^{\omega_{i}(t,t,\lambda)}\right)[l(\zeta,\lambda)\gamma(\zeta,\lambda)]\right\\}\,dt^{\lambda},$
(6.67) $\displaystyle P_{4}(\zeta,\lambda):=-\frac{1}{2\pi
i}\int_{b\Omega}\frac{e^{\omega_{1}(\zeta,t,\lambda)}-e^{\omega_{2}(\zeta,t,\lambda)}}{2(t^{\lambda}-\zeta^{\lambda})}\overline{l(t,\lambda)}\gamma(t,\lambda)\,\overline{dt^{\lambda}}.$

By assumption, $P_{1}^{\lambda},P_{2}^{\lambda}\in
C^{\mu,0}(b\Omega^{\lambda})$. Using (6.39), we have

$\displaystyle\mathcal{M}(\overline{l}P_{1})(\zeta,\lambda)$
$\displaystyle=\frac{1}{2\pi}P.V.\int_{b\Omega}\frac{\overline{l}(l\gamma)(t,\lambda)}{t^{\lambda}-\zeta^{\lambda}}\,dt^{\lambda}$
$\displaystyle=\frac{1}{2\pi}P.V.\int_{b\Omega}\frac{\gamma(t,\lambda)}{t^{\lambda}-\zeta^{\lambda}}\,dt^{\lambda}$
$\displaystyle=\frac{1}{2\pi}\left\\{\int_{b\Omega}\frac{\gamma(t,\lambda)-\gamma(\zeta,\lambda)}{t^{\lambda}-\zeta^{\lambda}}\,dt^{\lambda}+\pi
i\gamma(\zeta,\lambda)\right\\}.$

By Lemma 7.10, we have

(6.68) $\|\mathcal{M}(\overline{l}P_{1})\|_{\mu,0}\leq
C_{1,0}\|\gamma\|_{\mu},\quad\|\mathcal{M}(\overline{l}P_{2})\|_{\mu,0}\leq
C_{1,0}\|\gamma\|_{\mu}.$

For $P_{3}$, since $\omega_{i}(t,t,\lambda)=0$, we have

(6.69) $\displaystyle P_{3}(\zeta,\lambda)$
$\displaystyle=\sum_{i=1,2}\int_{b\Omega}\frac{l(t,\lambda)\gamma(t,\lambda)}{2(t^{\lambda}-\zeta^{\lambda})}\left[e^{\omega_{i}(\zeta,t,\lambda)}-e^{\omega_{i}(t,t,\lambda)}\right]\,dt^{\lambda}$
$\displaystyle\quad+\int_{b\Omega}\frac{1}{t^{\lambda}-\zeta^{\lambda}}[l(t,\lambda)\gamma(t,\lambda)-l(\zeta,\lambda)\gamma(\zeta,\lambda)]\,dt^{\lambda},$

and

$\displaystyle\mathcal{M}(\overline{l}P_{3})(\zeta,\lambda)=\sum_{i=1,2}\int_{b\Omega}\frac{\overline{l(t,\lambda)}}{t^{\lambda}-\zeta^{\lambda}}\left[\int_{b\Omega}\frac{l(t_{1},\lambda)\gamma(t_{1},\lambda)}{2(t_{1}^{\lambda}-t^{\lambda})}\left[e^{\omega_{i}(t,t_{1},\lambda)}-e^{\omega_{i}(t_{1},t_{1},\lambda)}\right]\,dt_{1}^{\lambda}\right]\,dt^{\lambda}$
$\displaystyle\qquad+\int_{b\Omega}\frac{\overline{l(t,\lambda)}}{t^{\lambda}-\zeta^{\lambda}}\left[\int_{b\Omega}\frac{1}{t_{1}^{\lambda}-t^{\lambda}}[l(t_{1},\lambda)\gamma(t_{1},\lambda)-l(t,\lambda)\gamma(t,\lambda)]\,dt_{1}^{\lambda}\right]\,dt^{\lambda}$
$\displaystyle\quad=\sum_{i=1,2}\int_{b\Omega}l(t_{1},\lambda)\gamma(t_{1},\lambda)\left[\int_{b\Omega}\frac{f_{i}(t_{1},t,\lambda)}{(t^{\lambda}-\zeta^{\lambda})(t_{1}^{\lambda}-t^{\lambda})}\,dt^{\lambda}\right]\,dt_{1}^{\lambda}$
$\displaystyle\qquad+\int_{b\Omega}\left[\int_{b\Omega}\frac{g(t_{1},t,\lambda)}{(t^{\lambda}-\zeta^{\lambda})(t_{1}^{\lambda}-t^{\lambda})}\,dt^{\lambda}\right]dt_{1}^{\lambda}$
$\displaystyle\quad=\sum_{i=1,2}\int_{b\Omega}\frac{l(t_{1},\lambda)\gamma(t_{1},\lambda)}{t_{1}^{\lambda}-\zeta^{\lambda}}\left[P.V.\int_{b\Omega}\frac{f_{i}(t_{1},t,\lambda)}{t^{\lambda}-\zeta^{\lambda}}\,dt^{\lambda}-\int_{b\Omega}\frac{f_{i}(t_{1},t,\lambda)}{t^{\lambda}-t_{1}^{\lambda}}\,dt^{\lambda}\right]\,dt_{1}^{\lambda}$
$\displaystyle\qquad+\int_{b\Omega}\frac{1}{t_{1}^{\lambda}-\zeta^{\lambda}}\left[P.V.\int_{b\Omega}\frac{g(t_{1},t,\lambda)}{t^{\lambda}-\zeta^{\lambda}}\,dt^{\lambda}-\int_{b\Omega}\frac{g(t_{1},t,\lambda)}{t^{\lambda}-t_{1}^{\lambda}}\,dt^{\lambda}\right]\,dt^{\lambda}_{1},$

where we denote

$\displaystyle
f_{i}(t_{1},t,\lambda):=\overline{l(t,\lambda)}\left(e^{\omega_{i}(t,t_{1},\lambda)}-e^{\omega_{i}(t_{1},t_{1},\lambda)}\right),$
$\displaystyle
g(t_{1},t,\lambda):=\overline{l(t,\lambda)}l(t_{1},\lambda)\gamma(t_{1},\lambda)-\gamma(t,\lambda).$

We can write $\mathcal{M}(\overline{l}P_{3})$ as

(6.70)
$\mathcal{M}(\overline{l}P_{3})(\zeta,\lambda)=\int_{b\Omega}\frac{k(\zeta,t_{1},\lambda)}{t_{1}^{\lambda}-\zeta^{\lambda}}\,dt_{1}^{\lambda},$

where

$\displaystyle k(\zeta,t_{1},\lambda)$
$\displaystyle:=l(t_{1},\lambda)\gamma(t_{1},\lambda)\sum_{i=1}^{2}\left(v_{i}(t_{1},\zeta,\lambda)-v_{i}(t_{1},t_{1},\lambda)\right)$
$\displaystyle\quad+\left(w(t_{1},\zeta,\lambda)-w(t_{1},t_{1},\lambda)\right),$

and

$\displaystyle v_{i}(t_{1},\zeta,\lambda):$
$\displaystyle=P.V.\int_{b\Omega}\frac{f_{i}(t_{1},t,\lambda)}{t^{\lambda}-\zeta^{\lambda}}\,dt^{\lambda}$
$\displaystyle=\int_{b\Omega}\frac{f_{i}(t_{1},t,\lambda)-f_{i}(t_{1},\zeta,\lambda)}{t^{\lambda}-\zeta^{\lambda}}\,dt^{\lambda}+\pi
if_{i}(t_{1},\zeta,\lambda);$ $\displaystyle w(t_{1},\zeta,\lambda)$
$\displaystyle:=P.V.\int_{b\Omega}\frac{g(t_{1},t,\lambda)}{t^{\lambda}-\zeta^{\lambda}}\,dt^{\lambda}$
$\displaystyle=\int_{b\Omega}\frac{g(t_{1},t,\lambda)-g(t_{1},\zeta,\lambda)}{t^{\lambda}-\zeta^{\lambda}}\,dt^{\lambda}+\pi
ig(t_{1},\zeta,\lambda).$

Let $\|\omega_{i}\|_{\alpha,0}:=\sup_{t_{1}\in
b\Omega,\lambda\in[0,1]}|\omega_{i}(\cdot,t_{1},\lambda)|_{\alpha_{p}}$. By
Lemma 7.10, for $\nu=\min(\alpha_{p},\mu)$ we have

$\displaystyle\|v_{i}(t_{1},\cdot,\lambda)\|_{\nu,0}\leq
C_{1,0}\|f_{i}(t_{1},\cdot,\lambda)\|_{\nu,0}\leq
C_{1,0}\|\omega_{i}\|_{\alpha_{p},0}|l(\cdot,\lambda)|_{\mu}.$
$\displaystyle\|w(t_{1},\cdot,\lambda)\|_{\mu,0}\leq
C_{1,0}|g(t_{1},\cdot,\lambda)|_{\mu}\leq
C_{1,0}\left(|\gamma(\cdot,\lambda)|_{0}|l(\cdot,\lambda)|_{\mu}+|\gamma(\cdot,\lambda)|_{\mu}\right),$

where the constants are uniform in $t_{1}\in b\Omega$. It follows from (6.70)
and Lemma 7.11 that given any $0<\nu^{\prime}<\nu$, one has

(6.71) $\displaystyle\|\mathcal{M}(\overline{l}P_{3})\|_{\nu^{\prime},0}$
$\displaystyle\leq M_{0},$

for some $M_{0}$ independent of $\lambda$. Similarly we can show

(6.72) $\displaystyle\|\mathcal{M}(\overline{l}P_{4})\|_{\nu^{\prime},0}\leq
M_{1},$

for some $M_{1}$ independent of $\lambda$. Combining (6.68), (6.71) and
(6.72), we obtain

(6.73)
$\|\mathcal{M}(Re[\overline{l^{\lambda}}(w_{\gamma}^{\lambda})^{+}])\|_{\nu^{\prime},0}\leq
M,\quad 0<\nu^{\prime}<\nu,$

where $M$ depends only on $\nu^{\prime}$, $\|\omega_{i}\|_{\alpha_{p},0}$,
$\|l\|_{\mu,0}$ and $\|\gamma\|_{\nu,0}$.

Next, we show
$\lambda\to\left|\mathcal{M}\left(Re[\overline{l^{\lambda}}(w_{\gamma}^{\lambda})^{+}]\right)\right|$
is a continuous map from $[0,1]$ to $C^{0}(b\Omega)$. For simplicity, we omit
the constant $\frac{1}{2\pi}$ in the operator $\mathcal{M}$.

$\displaystyle\left|\mathcal{M}\left(Re\left[\overline{l^{\lambda_{1}}}(w_{\gamma}^{\lambda_{1}})^{+}\right]\right)-\mathcal{M}\left(Re\left[\overline{l^{\lambda_{2}}}(w_{\gamma}^{\lambda_{2}})^{+}\right]\right)\right|$
$\displaystyle\quad=\left|P.V.\int_{b\Omega}\frac{Re[(\overline{l}w_{\gamma}^{+})(t,\lambda_{1})]}{t^{\lambda_{1}}-\zeta^{\lambda_{1}}}\,dt^{\lambda_{1}}-P.V.\int_{b\Omega}\frac{Re[(\overline{l}w_{\gamma}^{+})(t,\lambda_{2})]}{t^{\lambda_{2}}-\zeta^{\lambda_{2}}}\,dt^{\lambda_{2}}\right|.$

We parametrize $b\Omega^{\lambda}$ by $\rho^{\lambda}(s)$ and set
$\zeta^{\lambda}=\rho^{\lambda}(s_{0})$. The above is bounded by

(6.74) $\displaystyle\left|(\pi
i)\left\\{Re[\overline{l}w_{\gamma}^{+}](\zeta,\lambda_{1})-Re[\overline{l}w_{\gamma}^{+}](\zeta,\lambda_{2})\right\\}\right|$
$\displaystyle+\left|\int_{b\Omega}\frac{Re[(\overline{l}w_{\gamma}^{+})(t,\lambda_{1})]-Re[(\overline{l}w_{\gamma}^{+})(\zeta,\lambda_{1})]}{t^{\lambda_{1}}-\zeta^{\lambda_{1}}}dt^{\lambda_{1}}\right.$
$\displaystyle\quad-\left.\int_{b\Omega}\frac{Re[(\overline{l}w_{\gamma}^{+})(t,\lambda_{2})]-Re[(\overline{l}w_{\gamma}^{+})(\zeta,\lambda_{2})]}{t^{\lambda_{2}}-\zeta^{\lambda_{2}}}dt^{\lambda_{2}}\right|.$

We can rewrite the above as

(6.75)
$\displaystyle\pi\left|\left\\{Re[\overline{l}w_{\gamma}^{+}](\zeta,\lambda_{1})-Re[\overline{l}w_{\gamma}^{+}](\zeta,\lambda_{2})\right\\}\right|+\left|\int_{b\Omega}\frac{F^{\lambda_{1},\lambda_{2}}(s)-F^{\lambda_{1},\lambda_{2}}(s_{0})}{t^{\lambda_{1}}-\zeta^{\lambda_{1}}}\,ds\right|$
$\displaystyle\quad+\left.\int_{b\Omega}\left\\{Re[(\overline{l}w_{\gamma}^{+})(t,\lambda_{2})]-Re[(\overline{l}w_{\gamma}^{+})(\zeta,\lambda_{2})]\right\\}\left(\frac{1}{t^{\lambda_{1}}-\zeta^{\lambda_{1}}}-\frac{1}{t^{\lambda_{2}}-\zeta^{\lambda_{2}}}\right)dt^{\lambda_{2}}\right|$

where

(6.76)
$F^{\lambda_{1},\lambda_{2}}(s)=Re[\overline{l}w_{\gamma}^{+}](s,\lambda_{1})(\rho^{\lambda_{1}})^{\prime}(s)-Re[\overline{l}w_{\gamma}^{+}](s,\lambda_{2})(\rho^{\lambda_{2}})^{\prime}(s),$

and we write

$Re[\overline{l}w_{\gamma}^{+}](s,\lambda):=Re[\overline{l^{\lambda}}(w_{\gamma}^{\lambda})^{+}](t^{\lambda})=Re[\overline{l^{\lambda}}(w_{\gamma}^{\lambda})^{+}](\rho^{\lambda}(s)).$

We claim that:

  1. (1)

$w_{\gamma}^{+}\in\mathcal{C}^{\nu,0}(b\Omega^{\lambda})$, that is,

    1. (a)

$|w_{\gamma}^{+}(\cdot,\lambda)|_{\nu}$ is uniformly bounded in $\lambda$.

    2. (b)

$|w_{\gamma}^{+}(\cdot,\lambda_{1})-w_{\gamma}^{+}(\cdot,\lambda_{2})|_{0}\to
0$, as $|\lambda_{1}-\lambda_{2}|\to 0$.

  2. (2)

For any $0<\nu^{\prime}<\nu$, we have
$|F^{\lambda_{1},\lambda_{2}}|_{\nu^{\prime}}\to 0$ as
$|\lambda_{1}-\lambda_{2}|\to 0$.

Assuming the above for now, the expression (6.75) is then bounded by

$\pi\left|\delta^{\lambda_{1},\lambda_{2}}Re[\overline{l}w^{+}_{\gamma}]\right|_{0}+C_{1,0}|F^{\lambda_{1},\lambda_{2}}|_{\nu^{\prime}}\int_{b\Omega}\frac{ds}{|t-\zeta|^{1-\nu^{\prime}}}+C_{1,0}\|lw_{\gamma}\|_{\nu,0}|\Gamma^{\lambda_{1}}-\Gamma^{\lambda_{2}}|_{1}\int_{b\Omega}\frac{dt}{|t-\zeta|^{1-\nu}}$

which converges to $0$ uniformly in $\zeta$ as $|\lambda_{1}-\lambda_{2}|\to
0$.

We now verify the claims. In view of (6.76) and our assumptions on
$l^{\lambda}$ and $\rho^{\lambda}$, to prove (2) it suffices to show that
$|w_{\gamma}^{+}|_{\nu^{\prime}}\to 0$.

By Proposition 2.3, it suffices to verify claim (1). By (6.64) we have

$w_{\gamma}^{+}(\zeta,\lambda):=P_{1}(\zeta,\lambda)+P_{2}(\zeta,\lambda)+P_{3}(\zeta,\lambda)+P_{4}(\zeta,\lambda).$

By assumption $P_{1},P_{2}\in C^{\nu,0}(b\Omega^{\lambda})$, and

(6.77)
$\|P_{1}\|_{\mu,0}\leq\|l\gamma\|_{\mu,0},\quad\|P_{2}\|_{\mu,0}\leq\|l\gamma\|_{\mu,0}.$

Recall expression (6.69) for $P_{3}$:

$\displaystyle P_{3}(\zeta,\lambda)$
$\displaystyle=\sum_{i=1,2}\int_{b\Omega}\frac{l(t,\lambda)\gamma(t,\lambda)}{2(t^{\lambda}-\zeta^{\lambda})}\left[e^{\omega_{i}(\zeta,t,\lambda)}-e^{\omega_{i}(t,t,\lambda)}\right]\,dt^{\lambda}$
$\displaystyle\quad+\int_{b\Omega}\frac{1}{t^{\lambda}-\zeta^{\lambda}}[l(t,\lambda)\gamma(t,\lambda)-l(\zeta,\lambda)\gamma(\zeta,\lambda)]\,dt^{\lambda}.$

By Lemma 7.11, for any $0<\nu^{\prime}<\nu$ we have

(6.78) $\|P_{3}(\cdot,\lambda)\|_{\nu^{\prime},0}\leq
C_{\nu^{\prime}}C_{1,0}\|l\gamma\|_{\mu,0},$

where the constants are independent of $\lambda$. In the same way we have

(6.79) $\|P_{4}\|_{\alpha^{\prime},0}\leq
C_{\alpha^{\prime}}C_{1,0}\|l\gamma\|_{0,0}$

for any $0<\alpha^{\prime}<\alpha$. Putting together (6.77), (6.78) and
(6.79), we obtain

$\|w_{\gamma}^{+}\|_{\nu^{\prime},0}\leq
A_{\nu^{\prime}}C_{1,0}\|l\gamma\|_{\mu,0}$

for any $0<\nu^{\prime}<\nu<1$. This proves claim (1). Start with

Next we show the continuity of $w_{\gamma}^{+}$ in the $C^{0}$-norm.

(6.80)
$\left|w_{\gamma}^{+}(\zeta,\lambda_{1})-w_{\gamma}^{+}(\zeta,\lambda_{2})\right|\leq\sum_{i=1}^{4}\left|P_{i}(\zeta,\lambda_{1})-P_{i}(\zeta,\lambda_{2})\right|.$

In view of (6.65), for $i=1,2$,

$|P_{i}(\zeta,\lambda_{1})-P_{i}(\zeta,\lambda_{2})|\leq
C|l(\zeta,\lambda_{1})\gamma(\zeta,\lambda_{1})-l(\zeta,\lambda_{2})\gamma(\zeta,\lambda_{2})|,$

which converges to $0$ as $|\lambda_{1}-\lambda_{2}|\to 0$, since $l,\gamma\in
C^{\mu,0}(b\Omega^{\lambda})$. In view of expression (6.69) for $P_{3}$, we
have

$\displaystyle|P_{3}(\zeta,\lambda_{1})-P_{3}(\zeta,\lambda_{2})|$
$\displaystyle\leq\sum_{i=1,2}\int_{b\Omega}\frac{(l\gamma)(t,\lambda_{1})}{2(t^{\lambda_{1}}-\zeta^{\lambda_{1}})}\left(e^{\omega_{i}(\zeta,t,\lambda_{1})}-e^{\omega_{i}(t,t,\lambda_{1})}\right)(\rho^{\lambda_{1}})^{\prime}(s)\,ds$
$\displaystyle\qquad-\int_{b\Omega}\frac{(l\gamma)(t,\lambda_{2})}{2(t^{\lambda_{2}}-\zeta^{\lambda_{2}})}\left(e^{\omega_{i}(\zeta,t,\lambda_{2})}-e^{\omega_{i}(t,t,\lambda_{2})}\right)(\rho^{\lambda_{2}})^{\prime}(s)\,ds$
$\displaystyle\qquad+\int_{b\Omega}\frac{1}{t^{\lambda_{1}}-\zeta^{\lambda_{1}}}\left\\{(l\gamma)(t,\lambda_{1})-(l\gamma)(\zeta,\lambda_{1})\right\\}(\rho^{\lambda_{1}})^{\prime}(s)\,ds$
$\displaystyle\qquad-\int_{b\Omega}\frac{1}{t^{\lambda_{2}}-\zeta^{\lambda_{2}}}\left\\{(l\gamma)(t,\lambda_{2})-(l\gamma)(\zeta,\lambda_{2})\right\\}(\rho^{\lambda_{2}})^{\prime}(s)\,ds.$

The above expression can be bounded by a sum of the following terms:

(6.81)
$\sum_{i=1,2}\int_{b\Omega}\left\\{(l\gamma\rho^{\prime})(s,\lambda_{1})-(l\gamma\rho^{\prime})(s,\lambda_{2})\right\\}\frac{e^{\omega_{i}(\zeta,t,\lambda_{1})}-e^{\omega_{i}(t,t,\lambda_{1})}}{t^{\lambda_{1}}-\zeta^{\lambda_{1}}}\,ds,$
(6.82)
$\sum_{i=1,2}\int_{b\Omega}(l\gamma)(s,\lambda_{2})\frac{F_{i}^{\lambda_{1},\lambda_{2}}(\zeta)-F_{i}^{\lambda_{1},\lambda_{2}}(t)}{t^{\lambda_{1}}-\zeta^{\lambda_{1}}}\,dt^{\lambda_{2}},$
(6.83)
$\sum_{i=1,2}\int_{b\Omega}(l\gamma)(s,\lambda_{2})\left(e^{\omega_{i}(\zeta,t,\lambda_{2})}-e^{\omega_{i}(t,t,\lambda_{2})}\right)\left(\frac{1}{t^{\lambda_{1}}-\zeta^{\lambda_{1}}}-\frac{1}{t^{\lambda_{2}}-\zeta^{\lambda_{2}}}\right)\,dt^{\lambda_{2}},$
(6.84)
$\int_{b\Omega}\frac{\widetilde{F}^{\lambda_{1},\lambda_{2}}(t)-\widetilde{F}^{\lambda_{1},\lambda_{2}}(\zeta)}{t^{\lambda_{1}}-\zeta^{\lambda_{1}}}\,ds$
(6.85)
$\int_{b\Omega}\left\\{(l\gamma)(t,\lambda_{2})-(l\gamma)(\zeta,\lambda_{2})\right\\}\left(\frac{1}{t^{\lambda_{1}}-\zeta^{\lambda_{1}}}-\frac{1}{t^{\lambda_{2}}-\zeta^{\lambda_{2}}}\right)\,dt^{\lambda_{2}},$

where we set

$\displaystyle
F_{i}^{\lambda_{1},\lambda_{2}}(\zeta)=e^{\omega_{i}(\zeta,t,\lambda_{1})}-e^{\omega_{i}(\zeta,t,\lambda_{2})},$
$\displaystyle\widetilde{F}^{\lambda_{1},\lambda_{2}}(\zeta)=(l\gamma\rho^{\prime})(\zeta,\lambda_{1})-(l\gamma\rho^{\prime})(\zeta,\lambda_{2}).$

As before we let $\|\omega_{i}\|_{\alpha_{p},0}=\sup_{\lambda\in[0,1],t\in
b\Omega}|\omega_{i}(\cdot,t,\lambda)|_{\alpha_{p}}$. Then expression (6.81) is
bounded by

$C_{1,0}(\Sigma_{i=1,2}\|\omega_{i}\|_{\alpha_{p},0})|\delta^{\lambda_{1},\lambda_{2}}(l\gamma\rho^{\prime})|_{0}\int_{b\Omega}\frac{1}{|\zeta-t|^{1-\alpha_{p}}}\,ds,\quad\alpha_{p}=\frac{p-2}{p},$

which converges to $0$ as $|\lambda_{1}-\lambda_{2}|\to 0$ by assumption. Also
expression (6.83) is bounded by

$C_{1,0}(\Sigma_{i=1,2}\|\omega_{i}\|_{\alpha_{p},0})|\Gamma^{\lambda_{1}}-\Gamma^{\lambda_{2}}|_{1}\int_{b\Omega}\frac{1}{|t-\zeta|^{1-\alpha_{p}}}\,dt,$

which converges to $0$ as $|\lambda_{1}-\lambda_{2}|\to 0$, and expression
(6.85) is bounded by

$C_{1,0}\|l\gamma\|_{\mu,0}|\Gamma^{\lambda_{1}}-\Gamma^{\lambda_{2}}|_{1}\int_{b\Omega}\frac{1}{|t-\zeta|^{1-\mu}}\,dt$

which converges to $0$ as $|\lambda_{1}-\lambda_{2}|\to 0$. Finally in order
to show (6.82) and (6.84) converges to $0$ it suffices to see that some Hölder
norms of $F_{i}^{\lambda_{1},\lambda_{2}}$ and
$\widetilde{F}^{\lambda_{1},\lambda_{2}}$ converge to $0$. This follows from
Lemma 2.3 and the following facts:

  1. (1)

$|F_{i}^{\lambda_{1},\lambda_{2}}|_{\alpha_{p}}\leq
C\|\omega_{i}\|_{\alpha_{p},0}$;

  2. (2)

$|F_{i}^{\lambda_{1},\lambda_{2}}|_{0}\to 0$ as $|\lambda_{1}-\lambda_{2}|\to
0$, by Lemma 5.4;

  3. (3)

$\Gamma^{\lambda}\in\mathcal{C}^{1+\mu,0}(\overline{\Omega})$,
$l^{\lambda},\gamma^{\lambda}\in C^{\mu,0}(b\Omega^{\lambda})$.

Hence we have shown that $\lambda\mapsto P_{3}(\cdot,\lambda)$ is a continuous
map from $[0,1]$ to $C^{0}(b\Omega)$. Same can be said for
$P_{4}(\cdot,\lambda)$, of which the proof is similar and we leave it to the
reader.

Now claim (1) (b) follows from (6.80). By earlier remarks this proves that the
map
$\lambda\to\left|\mathcal{M}^{\lambda}\left(Re[\overline{l^{\lambda}}(w_{\gamma}^{\lambda})^{+}]\right)\right|$
is continuous from $[0,1]$ to $C^{0}(b\Omega)$. Combining this with (6.73), we
have shown the following

$\mathcal{M}\left(Re[\overline{l^{\lambda}}(w_{\gamma}^{\lambda})^{+}]\right)\in
C^{\nu^{\prime},0}(b\Omega^{\lambda}),$

for any $0<\nu^{\prime}<\nu$.

Next, we show that $\mathcal{M}\left(Re[\overline{l}w_{F}]\right)\in
C^{\nu,0}(b\Omega^{\lambda})$. We have

$\displaystyle\mathcal{M}(Re[\overline{l}w_{F}])(\zeta,\lambda)=\frac{1}{2\pi}P.V.\int_{b\Omega}\frac{Re[\overline{l}w_{F}]}{t^{\lambda}-\zeta^{\lambda}}\,dt^{\lambda}$
$\displaystyle\qquad=\frac{1}{2\pi}\left\\{\int_{b\Omega}\frac{Re[\overline{l}w_{F}](t,\lambda)-Re[\overline{l}w_{F}](\zeta,\lambda)}{t^{\lambda}-\zeta^{\lambda}}\,dt^{\lambda}+\pi
iRe[\overline{l}w_{F}](\zeta,\lambda)\right\\}.$

By Lemma 7.10, we get

(6.86) $\displaystyle\|\mathcal{M}(Re[\overline{l}w_{F}])\|_{\nu,0}$
$\displaystyle\leq C_{1,0}\|lw_{F}\|_{\nu,0}$ $\displaystyle\leq
C_{1,0}\|l\|_{\mu,0}\|w_{F}\|_{\alpha_{p},0},$

Using estimate (6.31) in (6.86) we get

(6.87) $\|\mathcal{M}(Re[\overline{l}w_{F}])\|_{\nu,0}\leq
C_{1,0}C_{p}\|l\|_{\mu,0}(\Sigma_{i=1,2}\|\omega_{i}\|_{\alpha_{p},0})\sup_{\lambda\in[0,1]}|F(\cdot,\lambda)|_{L^{p}(\Omega)}.$

On the other hand, as we have seen, in order to show that
$\mathcal{M}(Re[\overline{l}w_{F}])$ is continuous in $\lambda$ in the sup
norm, it is suffcient to show
$w_{F}^{\lambda}\in\mathcal{C}^{\alpha_{p},0}(b\Omega^{\lambda})$, which is
given by Lemma 6.4. Thus we have proved that
$\mathcal{M}\left(Re[\overline{l}w_{F}]\right)\in\mathcal{C}^{\nu,0}(b\Omega^{\lambda})$,
and it satisfies estimate (6.87). Combining with estimates (6.73) we obtain
that for any $0<\nu^{\prime}<\nu$,

(6.88)
$\displaystyle\|M^{\lambda}\gamma_{0}(\cdot,\lambda)\|_{\nu^{\prime},0}\leq
M_{1}$

where $M_{1}$ is some positive constant depending on $\nu^{\prime}$,
$\|\omega_{i}\|_{\alpha_{p},0}$, $\|l\|_{\mu,0}$, $\|\gamma\|_{\mu,0}$, and
$\sup_{\lambda\in[0,1]}|F(\cdot,\lambda)|_{L^{p}(\Omega)}$.

Step 5: Estimates for $\eta^{\lambda}$

We are now ready to show that
$\eta^{\lambda}\in\mathcal{C}^{\nu,0}(b\Omega^{\lambda})$. Recall from (6.61)
that $\eta(\cdot,\lambda)$ satisfies the reduced Fredholm integral equation:

(6.89)
$(I+\mathcal{N})\eta(\zeta,\lambda):=\eta(\zeta,\lambda)+\int_{b\Omega}\frac{k(\zeta,t,\lambda)}{t^{\lambda}-\zeta^{\lambda}}\,\eta(t,\lambda)\,dt^{\lambda}=\mathcal{M}\gamma_{0}(\zeta,\lambda),$

where $k=\Sigma_{i=1}^{4}k_{i}$, and the $k_{i}$-s are given by expressions
(6.44), (6.52), (6.56) and (6.59).

Let $w=w_{\eta}(z,\lambda)+w_{\gamma}(z,\lambda)+w_{F}(z,\lambda)$ where
$w_{\eta}^{\lambda}$, $w_{\gamma}^{\lambda}$ and $w_{F}^{\lambda}$ are given
by formulas (6.15) (6.16) and (6.17) respectively. For $(\zeta,\lambda)\in
b\Omega\times[0,1]$, we have
$w(\zeta,\lambda)=l(\zeta,\lambda)\gamma(\zeta,\lambda)+il(\zeta,\lambda)\eta(\zeta,\lambda)$.
Hence we have the relation

(6.90)
$w_{\eta}(\zeta,\lambda)=l(\zeta,\lambda)\gamma(\zeta,\lambda)+il(\zeta,\lambda)\eta(\zeta,\lambda)-w_{\gamma}(\zeta,\lambda)-w_{F}(\zeta,\lambda),\quad(\zeta,\lambda)\in
b\Omega\times[0,1].$

Now, for each fixed $\lambda$ we know by Proposition 3.6 that
$\eta^{\lambda}\in C^{\nu}(b\Omega^{\lambda})$, and we know that
$w_{\eta}(\cdot,\lambda)$ is the solution to the following Problem A:

$\displaystyle\partial{}_{\overline{z}}w_{\eta}^{\lambda}(z)+A^{\lambda}(z)w_{\eta}(z)+B^{\lambda}(z)\overline{w_{\eta}^{\lambda}(z)}=0\qquad\text{in
$\Omega^{\lambda}$.}$ (6.91) $\displaystyle
Re[\overline{l^{\lambda}(\zeta)}w_{\eta}^{\lambda}(\zeta)]=\gamma^{\lambda}(\zeta)-Re[\overline{l^{\lambda}}w^{\lambda}_{\gamma}](\zeta)-Re[\overline{l^{\lambda}}w_{F}^{\lambda}](\zeta)\qquad\text{on
$b\Omega^{\lambda}$}.$

In addition $w_{\eta}^{\lambda}$ satisfies conditions (6.20) and (6.21):

(6.92)
$w_{\eta}(z_{i},\lambda)=a_{i}(\lambda)+ib_{i}(\lambda)-w_{\gamma}(z_{i},\lambda)-w_{F}(z_{i},\lambda),\qquad
i=1,...,k,$ (6.93)
$w_{\eta}^{+}(z_{i}^{\prime},\lambda)=l(z_{i}^{\prime},\lambda)(\gamma(z_{i}^{\prime},\lambda)+ic_{i}(\lambda))-w_{\gamma}^{+}(z_{i}^{\prime},\lambda)-w_{F}(z_{i}^{\prime},\lambda),\qquad
i=1,...,k^{\prime},$

on some normally distributed set $\\{z_{i},z_{i}^{\prime}\\}$ in
$\overline{\Omega}$.

We now prove that $\eta^{\lambda}\in\mathcal{C}^{\nu,0}(b\Omega^{\lambda})$.
First, we show that $|\eta|_{0}$ is bounded in $\lambda$. Seeking
contradiction, suppose that for some sequence $\\{\lambda_{j}\\}$ converging
to $\lambda_{0}$, such that $N_{j}:=|\eta(\cdot,\lambda_{j})|_{0}\to\infty$.
Let $\widehat{\eta}(\cdot,\lambda_{j})=\frac{1}{N_{j}}\eta(\cdot,\lambda)$.
Then $|\widehat{\eta}(\cdot,\lambda)|_{0}=1$. Dividing both sides of (6.89) by
$N_{j}$, we have

(6.94)
$(I+\mathcal{N})\widehat{\eta}(\zeta,\lambda_{j})=\frac{1}{N_{j}}\mathcal{M}\gamma_{0}.$

The corresponding $w_{\widehat{\eta}}(\cdot,\lambda_{j})$ is the solution to
the following boundary value problem:

$\displaystyle\partial{}_{\overline{z}}w_{\widehat{\eta}}^{\lambda_{j}}(z)+A(z,\lambda_{j})w_{\widehat{\eta}}(z,\lambda_{j})+B(z,\lambda_{j})\overline{w_{\widehat{\eta}}(z,\lambda_{j})}=0\quad\text{in
$\Omega$, }$ (6.95) $\displaystyle
Re[\overline{l(\zeta,\lambda_{j})}w_{\widehat{\eta}}(\zeta,\lambda_{j})]=\frac{1}{N_{j}}\left\\{\gamma(\zeta,\lambda_{j})-Re[\overline{l}w_{\gamma}](\zeta,\lambda_{j})-Re[\overline{l}w_{F}](\zeta,\lambda_{j})\right\\}\quad\text{on
$b\Omega$},$

and in addition $w_{\widehat{\eta}}^{\lambda_{j}}$ satisfies the condition

(6.96)
$w_{\widehat{\eta}}(z_{i},\lambda_{j})=\frac{1}{N_{j}}\left\\{a_{i}(\lambda_{j})+ib_{i}(\lambda_{j})-w_{\gamma}(z_{i},\lambda_{j})-w_{F}(z_{i},\lambda_{j})\right\\},\qquad
i=1,...,k.$ (6.97)
$w_{\eta}^{+}(z_{i}^{\prime},\lambda_{j})=\frac{1}{N_{j}}\left\\{l(z_{i}^{\prime},\lambda_{j})(\gamma(z_{i}^{\prime},\lambda_{j})+ic_{i}(\lambda_{j}))-w_{\gamma}^{+}(z_{i}^{\prime},\lambda_{j})-w_{F}(z_{i}^{\prime},\lambda_{j})\right\\},\qquad
i=1,...,k^{\prime}.$

By Lemma 7.7 and our earlier estimate for $k$, we have for any
$0<\nu^{\prime}<\nu$,
$|\mathcal{N}\widehat{\eta}(\cdot,\lambda_{j})|_{\nu^{\prime}}$ is uniformly
bounded in $\lambda$. By Arzela-Ascoli theorem, and passing to subsequences if
necessary, $\mathcal{N}\widehat{\eta}(\cdot,\lambda_{j})$ converges to some
function in the $C^{\tau}(b\Omega)$ norm, for any $0<\tau<\nu$. It follows
from equation (6.94) that $\widehat{\eta}(\cdot,\lambda_{j})$ converges to
some function $\widetilde{\eta}$ in $C^{\tau}(b\Omega)$.

Let

$\displaystyle w_{\widehat{\eta}}(z,\lambda_{j})$
$\displaystyle=\frac{1}{2\pi}\int_{b\Omega}G_{1}(z,t,\lambda_{j})l(t,\lambda_{j})\widehat{\eta}(t,\lambda_{j})\,dt^{\lambda_{j}}-\frac{1}{2\pi}\int_{b\Omega}G_{2}(z,t,\lambda_{j})\overline{l(t,\lambda_{j})}\widehat{\eta}(t,\lambda_{j})\,\overline{dt^{\lambda_{j}}},$

and

(6.98) $\displaystyle w_{\widetilde{\eta}}^{\lambda_{0}}(z)$
$\displaystyle=\frac{1}{2\pi}\int_{b\Omega}G_{1}(z,t,\lambda_{0})l(t,\lambda_{0})\widetilde{\eta}(t)\,dt^{\lambda_{0}}-\frac{1}{2\pi}\int_{b\Omega}G_{2}(z,t,\lambda_{0})\overline{l(t,\lambda_{0})}\widetilde{\eta}(t)\,\overline{dt^{\lambda_{0}}}.$

From (6.94) we have

$\widehat{\eta}(\zeta,\lambda_{j})=-\mathcal{N}\widehat{\eta}(\zeta,\lambda_{j})+\frac{1}{N_{j}}\mathcal{M}\gamma_{0}.$

By Lemma 7.7, we have for any $0<\nu^{\prime}<\nu$,

(6.99) $|\widehat{\eta}(\cdot,\lambda_{j})|_{\nu^{\prime}}\leq
C_{\nu^{\prime}}C_{1,0}|\widehat{\eta}(\cdot,\lambda_{j})|_{0}+\frac{1}{N_{j}}\|\mathcal{M}\gamma_{0}\|_{\nu^{\prime}}=C,$

where the constant is independent of $\lambda$.

Let $\\{\\{z_{i}\\}_{i=1}^{N_{1}},\\{z_{i^{\prime}}\\}_{i=1}^{N_{2}}\\}$ be a
normally distributed set in $\Omega$. Then it is easy to see that
$w_{\widehat{\eta}}(z_{i},\lambda_{i})$ converges to
$w_{\widetilde{\eta}}^{\lambda_{0}}(z_{i})$, since for $z_{i}\in\Omega$, the
kernels $G_{1}$ and $G_{2}$ are uniformly bounded in absolute value by a
constant.

On the other hand, for $z_{i}^{\prime}\in b\Omega$, we apply the jump formula
(7.8) for generalized Cauchy integral to get

$\displaystyle w_{\widehat{\eta}}^{+}(z_{i}^{\prime},\lambda)$
$\displaystyle=\frac{1}{2}\,l(z_{i}^{\prime},\lambda)\widehat{\eta}(z_{i}^{\prime})+P.V.\frac{1}{2\pi
i}\int_{b\Omega}G_{1}(z_{i}^{\prime},t,\lambda)l(t,\lambda)\widehat{\eta}(t,\lambda)\,dt^{\lambda}$
$\displaystyle\quad-\frac{1}{2\pi
i}\int_{b\Omega}G_{2}(z_{i}^{\prime},t,\lambda)l(t,\lambda)\widehat{\eta}(t,\lambda)\,dt^{\lambda}.$

In the same way as we did earlier for $w^{+}_{\gamma}$, together with (6.99),
one can prove that

$|w_{\widehat{\eta}}^{+}(z_{i}^{\prime},\lambda_{j})-(w_{\widetilde{\eta}}^{\lambda_{0}})^{+}(z_{i}^{\prime})|\longrightarrow
0,\quad i=1,\dots,N_{2}$

as $|\lambda_{j}-\lambda_{0}|\to 0$. In fact same proof shows that

$|w_{\widehat{\eta}}(\cdot,\lambda_{j})-w_{\widetilde{\eta}}^{\lambda_{0}}(\cdot)|_{C^{0}(\overline{\Omega})}\longrightarrow
0.$

Gathering the results and letting $|\lambda_{j}-\lambda_{0}|\to 0$ in (6.95),
(6.96) and (6.97), we see that $w_{\widetilde{\eta}}$ is the solution to the
boundary value problem

$\begin{cases}\partial{}_{\overline{z}}w_{\widetilde{\eta}}(z)+A^{\lambda_{0}}(z)w_{\widetilde{\eta}}(z)+B^{\lambda_{0}}(z)\overline{w_{\widetilde{\eta}}(z)}=0&\text{in
$\Omega^{\lambda_{0}}$.}\\\
Re[\overline{l^{\lambda_{0}}(\zeta)}w_{\widetilde{\eta}}(\zeta)]=0&\text{on
$b\Omega^{\lambda_{0}}$},\end{cases}$

satisfying the conditions

$w_{\widetilde{\eta}}^{\lambda_{0}}(z_{i})=0,\quad(w_{\widetilde{\eta}}^{\lambda_{0}})^{+}(z_{i}^{\prime})=0,$

for $i=1,...,N_{1}$ and $i^{\prime}=1,...,N_{2}$. By Proposition 3.5, we have
$w_{\widetilde{\eta}}^{\lambda_{0}}\equiv 0$ in $\Omega^{\lambda_{0}}$. By
(6.90), we have

$w_{\eta}(t,\lambda_{j})+w_{\gamma}(t,\lambda_{j})+w_{F}(t,\lambda_{j})=l(t,\lambda_{j})\gamma(t,\lambda_{j})+il(t,\lambda_{j})\eta(t,\lambda_{j}),\quad
t\in b\Omega.$

Dividing on both sides by $N_{j}$ we get

$w_{\widehat{\eta}}(t,\lambda_{j})+\frac{1}{N_{j}}w_{\gamma}(t,\lambda_{j})+\frac{1}{N_{j}}w_{F}(t,\lambda_{j})=\frac{1}{N_{j}}l(t,\lambda_{j})\gamma(t,\lambda_{j})+il(t,\lambda_{j})\widehat{\eta}(t,\lambda_{j}),\quad
t\in b\Omega.$

Letting $|\lambda_{j}-\lambda_{0}|\to 0$, the left-hand side converges
uniformly to $w_{\widetilde{\eta}}^{\lambda_{0}}\equiv 0$. Hence the right-
hand side converges uniformly to $0$. Since $|l(t,\lambda_{j})|$ is bounded
below by a positive constant uniform in $t$ and $\lambda$, and
$\widehat{\eta}(t,\lambda_{j})$ converges uniformly to $\widetilde{\eta}$, we
conclude that $\widetilde{\eta}\equiv 0$.

On the other hand we have
$|\widetilde{\eta}|_{0}=\lim_{j\to\infty}|\widehat{\eta}(\cdot,\lambda_{j})|_{0}=1$,
which is a contradiction. This shows that $|\eta(\cdot,\lambda)|_{0}$ is
bounded in $\lambda$. It then follows by (6.89), which can be written as
$\eta^{\lambda}=-\mathcal{N}^{\lambda}\eta^{\lambda}+\mathcal{M}^{\lambda}\gamma_{0}^{\lambda}$,
and Lemma 7.11 that

$\displaystyle\|\eta\|_{\nu^{\prime},0}$
$\displaystyle=\|-\mathcal{N}\eta\|_{\nu^{\prime},0}+\|\mathcal{M}\gamma_{0}\|_{\nu,0}$
$\displaystyle\leq C_{\nu^{\prime}}C_{1,0}\|\eta\|_{0,0}.$

for any $0<\nu^{\prime}<\nu$.

Next we show $\eta(\cdot,\lambda)$ is continuous in $\lambda$ in the
$C^{0}(b\Omega)$-norm. Suppose not, then there exists some $\lambda_{0}$ and a
sequence $\lambda_{j}\to\lambda_{0}$, such that
$|\eta(\cdot,\lambda_{j})-\eta(\cdot,\lambda_{0})|_{0}\geq\varepsilon>0$.
Since we have shown that $|\eta(\cdot,\lambda)|_{\nu^{\prime}}$ is bounded for
any $0<\nu^{\prime}<\nu$, passing to a subsequence if necessary, we have that
$\eta(\cdot,\lambda_{j})$ converges to some $\eta^{\ast}$ in the
$C^{\nu^{\prime}}(b\Omega)$-norm. As before we can show that
$w_{\eta}(\zeta,\lambda_{0})$ converges to $w_{\eta_{\ast}}(\zeta)$ uniformly
on $b\Omega$. Letting $|\lambda_{j}-\lambda_{0}|\to 0$ in (6.91), (6.92) and
(6.93) we see that $w_{\eta_{\ast}}$ is a solution to the following Problem A

(6.100)
$\begin{cases}\partial{}_{\overline{z}}w_{\eta_{\ast}}(z)+A^{\lambda_{0}}(z)w_{\eta_{\ast}}(z)+B^{\lambda_{0}}(z)\overline{w_{\eta_{\ast}}(z)}=0&\text{in
$\Omega$.}\\\
Re[\overline{l^{\lambda_{0}}(\zeta)}w_{\eta_{\ast}}(\zeta)]=\gamma(\zeta,\lambda_{0})-Re[\overline{l^{\lambda_{0}}(\zeta)}w^{\lambda_{0}}_{\gamma}(\zeta)]-Re[\overline{l^{\lambda_{0}}(\zeta)}w^{\lambda_{0}}_{F}(\zeta)]&\text{on
$b\Omega$},\end{cases}$

and in addition $w_{\eta_{\ast}}$ satisfies the condition:

$\displaystyle
w_{\eta_{\ast}}(z_{s}^{\lambda_{0}})=a_{s}(\lambda_{0})+ib_{s}(\lambda)-w_{\gamma}(z_{s},\lambda_{0})-w_{F}(z_{s},\lambda_{0}),\quad\quad
s=1,...,N_{1},$ $\displaystyle
w_{\eta_{\ast}}^{+}(z_{s}^{\lambda_{0}})=l(z_{s}^{\prime},\lambda_{0})(\gamma(z_{s}^{\prime},\lambda_{0})+ic_{s}(\lambda_{0}))-w_{\gamma}^{+}(z_{s}^{\prime},\lambda_{0})-w_{F}(z_{s}^{\prime},\lambda_{0}),\qquad
s=1,...,N_{2}.$

On the other hand, we had assumed that $\eta(\cdot,\lambda_{0})$ is also a
solution to (6.100) satisfying the above conditions. Hence by Proposition 3.5
we have $w_{\eta}(\cdot,\lambda_{0})=w_{\eta_{\ast}}(\cdot)$. Now by (6.90),
we have

(6.101)
$w_{\eta}(\zeta,\lambda_{0})=l(\zeta,\lambda_{0})\gamma(\zeta,\lambda_{0})+il(\zeta,\lambda_{0})\eta(\zeta,\lambda_{0})-w_{\gamma}(\zeta,\lambda_{0})-w_{F}(\zeta,\lambda_{0}),\quad\zeta\in
b\Omega.$

and

(6.102)
$w_{\eta}(\zeta,\lambda_{j})=l(\zeta,\lambda_{j})\gamma(\zeta,\lambda_{j})+il(\zeta,\lambda_{j})\eta(\zeta,\lambda_{j})-w_{\gamma}(\zeta,\lambda_{j})-w_{F}(\zeta,\lambda_{j}),\quad\zeta\in
b\Omega.$

By using the fact that $|\eta^{\lambda}|_{\nu^{\prime}}$ ($\nu^{\prime}>0$) is
uniformly bounded in $\lambda$, we see by the same arguement as before that
$w_{\eta}(\cdot,\lambda_{j})$ converges uniformly to $w_{\eta^{\ast}}$ on
$b\Omega$, as $|\lambda_{j}-\lambda_{0}|\to 0$. Passing the limit in equation
(6.102) we get

$w_{\eta}(\zeta,\lambda_{0})=w_{\eta_{\ast}}(\zeta)=l(\zeta,\lambda_{0})\gamma(\zeta,\lambda_{0})+il(\zeta,\lambda_{0})\eta_{\ast}(\zeta)-w_{\gamma}(\zeta,\lambda_{0})-w_{F}(\zeta,\lambda_{0}),\quad\zeta\in
b\Omega.$

Comparing this with equation (6.101), we get
$\eta_{\ast}(\zeta)=\eta(\zeta,\lambda_{0})$. But we also have
$|\eta^{\ast}(\cdot)-\eta(\cdot,\lambda_{0})|_{0}\geq\varepsilon>0$, which is
impossible. Hence we conclude that $\eta(\cdot,\lambda)$ is continuous in
$\lambda$ in the $C^{0}(b\Omega)$ norm, and
$\eta^{\lambda}\in\mathcal{C}^{\nu^{\prime},0}(b\Omega^{\lambda})$ for any
$0<\nu^{\prime}<\nu$.

Now by the proof of Lemma 6.3 in Step 2 one can show that
$w^{\lambda}_{\eta}\in\mathcal{C}^{\nu^{\prime},0}(b\Omega^{\lambda})$.

###  6.2. The $\mathcal{C}^{\nu,0}$ estimate

In this section we prove part (ii) of Theorem 6.1, namely we show that the
solution $w^{\lambda}\in\mathcal{C}^{\nu-,0}(\overline{\Omega^{\lambda}})$
found in part (i) is in fact in the class
$\mathcal{C}^{\nu,0}(\overline{\Omega^{\lambda}})$.

First we make a reduction to the case $F^{\lambda}\equiv 0$. Recall that
$A^{\lambda}$, $B^{\lambda}$ and $F^{\lambda}$ are taken to be zero outside
$\Omega^{\lambda}$. So we can write

(6.103)
$w_{F}(z,\lambda)=-\frac{1}{\pi}\iint_{\mathbb{C}}G_{1}(z,\zeta,\lambda)F(\zeta,\lambda)\,dA(\zeta^{\lambda})-\frac{1}{\pi}\iint_{\mathbb{C}}G_{2}(z,\zeta,\lambda)\overline{F}(\zeta,\lambda)\,dA(\zeta^{\lambda}).$

By Lemma 6.4, $w_{F}^{\lambda}\in\mathcal{C}^{\alpha_{p},0}(\mathbb{C})$.
Furthermore, $w_{F}^{\lambda}$ satisfies the equation

$\partial{}_{\overline{z^{\lambda}}}w_{F}^{\lambda}(z^{\lambda})+A^{\lambda}(z^{\lambda})w_{F}^{\lambda}(z^{\lambda})+B^{\lambda}(z^{\lambda})\overline{w_{F}^{\lambda}(z^{\lambda})}=F^{\lambda}(z^{\lambda}),\quad
z\in\Omega.$

Now let $w_{r}^{\lambda}$ be the solution to the following boundary value
problem

$\begin{cases}\partial{}_{\overline{z^{\lambda}}}w_{r}^{\lambda}(z^{\lambda})+A^{\lambda}(z^{\lambda})w_{r}^{\lambda}(z^{\lambda})+B^{\lambda}(z^{\lambda})\overline{w_{r}^{\lambda}(z^{\lambda})}=0&\quad
z\in\Omega,\\\
Re[\overline{l^{\lambda}}w_{r}^{\lambda}](\zeta^{\lambda})=\gamma^{\lambda}(\zeta^{\lambda})-Re[\overline{l^{\lambda}}w^{\lambda}_{F}](\zeta^{\lambda})&\quad\zeta\in
b\Omega,\end{cases}$

and in addition $w_{r}^{\lambda}$ satisfies the conditions:

$\displaystyle
w_{r}(z_{i},\lambda)=a_{i}(\lambda)+ib_{i}(\lambda)-w_{F}(z_{i},\lambda)$
$\displaystyle
w_{r}^{+}(z_{i}^{\prime},\lambda)=l(z_{i}^{\prime},\lambda)(\gamma(z_{i}^{\prime},\lambda)+ic_{i}(\lambda))-w_{F}(z_{i}^{\prime},\lambda)$

on the normally distributed set
$\\{\\{z_{i}\\}_{i=1}^{N_{0}},\\{z_{i}^{\prime}\\}_{i=1}^{N_{1}}\\}$. Then any
solution to the original boundary value problem (6.5) satisfying condition
(6.6) and (6.7) is the sum of $w_{r}^{\lambda}$ and $w_{F}^{\lambda}$. Since
$w^{\lambda}_{F}\in\mathcal{C}^{\alpha_{p}}(\overline{\Omega^{\lambda}})$, it
suffices to consider the following boundary value problem for
$w_{r}^{\lambda}$:

(6.104)
$\begin{cases}\partial{}_{\overline{z^{\lambda}}}w_{r}^{\lambda}(z^{\lambda})+A(z,\lambda)w_{r}^{\lambda}(z^{\lambda})+B(z,\lambda)\overline{w_{r}^{\lambda}(z^{\lambda})}=0&\quad
z\in\Omega,\\\
Re[\overline{l^{\lambda}}w_{r}^{\lambda}](\zeta^{\lambda})=\gamma^{\lambda}(\zeta^{\lambda})&\quad\zeta\in
b\Omega,\end{cases}$

where $\gamma^{\lambda}\in\mathcal{C}^{\nu,0}(b\Omega^{\lambda})$, for
$\nu=\min(\alpha_{p},\mu)$.

We now use an idea of Vekua to reduce problem (6.104) to one on simply-
connected domain. First we introduce some notations. For a bounded domain
$\Omega$ whose boundary consists of $m+1$ connected component, we denote by
$b\Omega_{0}$ the outer boundary of $\Omega$, and we let $\Omega_{0}$ be the
bounded simply-connected domain interior to the contour $b\Omega_{0}$. We
denote the boundary of the holes in the interior of $\Omega_{0}$ by
$b\Omega_{j}$, for $1\leq j\leq m$, and we let $\Omega_{j}$ be the unbounded
domain exterior to the contour $b\Omega_{j}$.

In view of formula (3.35), we can write the solution of problem (6.104) in the
form

(6.105) $w_{r}(z,\lambda)=w_{0}(z,\lambda)+\cdots+w_{m}(z,\lambda),\quad\quad
z\in\Omega,$

where for $0\leq j\leq m$, we set

(6.106) $w_{j}(z,\lambda)=\frac{1}{2\pi
i}\int_{b\Omega_{j}}G_{1}(z,t,\lambda)w_{r}(t,\lambda)\,dt^{\lambda}+G_{2}(z,t,\lambda)\overline{w_{r}(t,\lambda)}\,\overline{dt^{\lambda}}.$

Fix $j_{0}$ with $0\leq j_{0}\leq m$, we claim that for all $j\neq j_{0}$,
$w_{j}^{\lambda}\in\mathcal{C}^{\nu,0}(b\Omega_{j_{0}}^{\lambda})$. Assuming
this for now, then $w_{j_{0}}^{\lambda}$ is a solution to the boundary value
problem

(6.107)
$\displaystyle\partial{}_{\overline{z^{\lambda}}}w^{\lambda}_{j_{0}}(z^{\lambda})+A(z,\lambda)w^{\lambda}_{j_{0}}(z^{\lambda})+B(z,\lambda)\overline{w_{j_{0}}^{\lambda}(z^{\lambda})}=0\quad\text{in
$\Omega_{j_{0}}^{\lambda}$.}$ (6.108) $\displaystyle
Re[\overline{l(z,\lambda)}w_{j_{0}}^{\lambda}(z^{\lambda})]=\gamma_{j_{0}}(z,\lambda)\quad\text{on
$b\Omega_{j_{0}}^{\lambda}$}$

where

$\gamma_{j_{0}}(z,\lambda):=\gamma(z,\lambda)-\sum_{j\neq
j_{0}}Re[\overline{l(z,\lambda)}w_{j}^{\lambda}(z^{\lambda})],$

and $\gamma_{j_{0}}^{\lambda}\in\mathcal{C}^{\nu,0}(b\Omega_{j}^{\lambda})$.

If $j_{0}=0$, the above problem reduces to earlier case on bounded simply-
connected domain. For $j_{0}\geq 1$, we need to invert the unbounded domain
$\Omega_{j_{0}}$. We fix some point $a_{j_{0}}$ in
$\mathbb{C}\setminus\Omega_{j_{0}}$ such that
$\operatorname{dist}(a_{j_{0}},b\Omega^{\lambda}_{j_{0}})\geq c>0$ for all
$\lambda$. By a substitution
$z^{\lambda}=\varphi(\zeta^{\lambda})=\frac{1}{\zeta^{\lambda}-a_{j_{0}}}$, we
have for each fixed $\lambda\in[0,1]$,

$\displaystyle\partial{}_{\overline{\zeta^{\lambda}}}w_{j_{0}}^{\lambda}\left(\varphi(\zeta^{\lambda}))\right)$
$\displaystyle=\partial{}_{\overline{z^{\lambda}}}w^{\lambda}_{j_{0}}(z^{\lambda})\cdot\partial{}_{\overline{\zeta^{\lambda}}}\overline{\varphi}$
$\displaystyle=-\overline{\partial{}_{\zeta^{\lambda}}\varphi}A^{\lambda}(z^{\lambda})w^{\lambda}_{j_{0}}(z^{\lambda})-\overline{\partial{}_{\zeta^{\lambda}}\varphi}B^{\lambda}(z^{\lambda})\overline{w^{\lambda}_{j_{0}}(z^{\lambda})}.$

Set
$w^{\ast}_{j_{0}}(\zeta,\lambda)=w^{\lambda}_{j_{0}}(\varphi(\zeta^{\lambda}))$,
$A^{\ast}(\zeta,\lambda)=(\overline{\partial{}_{\zeta^{\lambda}}\varphi})A^{\lambda}(\varphi^{-1}(\zeta^{\lambda}))$,
$B^{\ast}(\zeta,\lambda)=$
$(\overline{\partial{}_{\zeta^{\lambda}}\varphi})B^{\lambda}(\varphi^{-1}(\zeta^{\lambda})$,
$l^{\ast}(\zeta,\lambda)=l^{\lambda}(\varphi(\zeta^{\lambda}))$ and
$\gamma^{\ast}_{j}(\zeta,\lambda)=\gamma^{\lambda}_{j}(\varphi(\zeta^{\lambda}))$,
where $\Gamma^{\lambda}\in\mathcal{C}^{1+\mu,0}(\mathbb{C})$ is the extension
of $\Gamma^{\lambda}$. Equations (6.107)-(6.108) then become

$\displaystyle\overline{\partial}_{\zeta^{\lambda}}w^{\ast}_{j_{0}}(\zeta,\lambda)+A^{\ast}(\zeta,\lambda)w^{\ast}_{j_{0}}(\zeta,\lambda)+B^{\ast}(\zeta,\lambda)\overline{w^{\ast}_{j_{0}}(\zeta,\lambda)}=0,\quad\text{in
$D^{\lambda}_{j_{0}}$}.$ $\displaystyle
Re\left[\overline{l^{\ast}(\zeta,\lambda)}w^{\ast}_{j_{0}}(\zeta)\right]=\gamma_{j_{0}}^{\ast}(\zeta,\lambda),\quad\text{on
$bD^{\lambda}_{j_{0}}$, }$

where $D^{\lambda}_{j_{0}}$ (resp. $bD^{\lambda}_{j_{0}}$) is the image of
$\Omega^{\lambda}_{j_{0}}$ (resp. $b\Omega^{\lambda}_{j_{0}}$), under the map
$\varphi^{-1}:z^{\lambda}\mapsto\zeta^{\lambda}=\frac{1}{z^{\lambda}}+a$. The
family $D^{\lambda}_{j_{0}}$ is now bounded simply-connected in the class
$\mathcal{C}^{1+\mu,0}$. Note that since $A^{\lambda},B^{\lambda}$ vanish
outside $\Omega^{\lambda}$ and in particular near infinity, we have
$A_{\ast}^{\lambda}\equiv B_{\ast}^{\lambda}\equiv 0$ when $\zeta^{\lambda}$
is near $a_{j_{0}}$, and $A_{\ast}(\cdot,\lambda),B_{\ast}(\cdot,\lambda)$
satisfy the same assumptions as $A(\cdot,\lambda)$ and $B(\cdot,\lambda)$. It
is clear that
$l^{\ast},\gamma^{\ast}_{j_{0}}\in\mathcal{C}^{\nu,0}(bD_{j_{0}}^{\lambda})$.

We can now apply Theorem 4.6 to the family of bounded simply-connected domains
$D_{j_{0}}^{\lambda}$, and we get
$w_{j_{0}}^{\ast}\in\mathcal{C}^{\nu,0}(\overline{D_{j_{0}}^{\lambda}})$, for
each $0\leq j_{0}\leq m$. (Note that the conditions on the normally
distributed set are trivially satisfied since we already know from part (i)
that $w_{r}^{\lambda}\in\mathcal{C}^{\nu^{-},0}(\overline{\Omega^{\lambda}})$,
and hence by Proposition 7.13 one has
$w_{j_{0}}^{\lambda}\in\mathcal{C}^{\nu^{-},0}(\overline{\Omega_{j_{0}}^{\lambda}})$.)
Since

$w^{\lambda}_{j_{0}}(z^{\lambda})=w^{\ast}_{j_{0}}(\varphi^{-1}(z^{\lambda})),$

we have
$w_{j_{0}}^{\lambda}\in\mathcal{C}^{\nu,0}(\overline{\Omega_{j}^{\lambda}})$,
for each $0\leq j_{0}\leq m$. In view of (6.105), we get
$w_{r}^{\lambda}\in\mathcal{C}^{\nu,0}(\overline{\Omega^{\lambda}})$.

It remains to prove for each $j_{0}$ and $j\neq j_{0}$, we have
$w_{j}^{\lambda}\in\mathcal{C}^{\nu,0}(b\Omega_{j_{0}}^{\lambda})$. First we
show that $w_{j}^{\lambda}$ is continuous in the
$C^{0}(b\Omega_{j_{0}})$-norm. Let
$d_{j}=\operatorname{dist}(b\Omega_{j},b\Omega_{j_{0}})$ and let
$dt^{\lambda}=a(s,\lambda)ds$. In view of (6.106), we have for $z\in
b\Omega_{j_{0}}$,

$\displaystyle|w_{j}(z,\lambda_{1})-w_{j}(z,\lambda_{2})|$
$\displaystyle\quad\leq
C\sum_{i=1,2}\int_{b\Omega_{j}}|\delta^{\lambda_{1}\lambda_{2}}G_{i}(z,\cdot)||w_{r}(t,\lambda_{1})|\,dt^{\lambda_{1}}+|G_{i}(z,t,\lambda_{2})||\delta^{\lambda_{1}\lambda_{2}}(w_{r}a)(t)|\,ds$
$\displaystyle\quad\leq\|w_{r}\|_{0,0}\int_{b\Omega_{j}}\frac{\Sigma_{i=1,2}|e^{\omega_{i}(z,t,\lambda_{1})}-e^{\omega_{i}(z,t,\lambda_{2})}|}{|t^{\lambda_{1}}-z^{\lambda_{1}}|}\,dt^{\lambda_{1}}$
$\displaystyle\qquad+\|w_{r}\|_{0,0}\int_{b\Omega_{j}}\sum_{i=1}^{2}\left|e^{\omega_{i}(z,t,\lambda_{2})}\right|\left|\frac{1}{|t^{\lambda_{1}}-z^{\lambda_{1}}|}-\frac{1}{|t^{\lambda_{2}}-z^{\lambda_{2}}|}\right|\,dt^{\lambda_{1}}$
$\displaystyle\qquad+\int_{b\Omega_{j}}\frac{\sum_{i=1}^{2}\left|e^{\omega_{i}(z,t,\lambda_{2})}\right|}{|t^{\lambda_{2}}-z^{\lambda_{2}}|}|\delta^{\lambda_{1},\lambda_{2}}(w_{r}a)(s)|\,ds$

Since $z\in b\Omega_{j_{0}}$ and $t\in b\Omega_{j}^{\prime}$, we have
$|t^{\lambda}-z^{\lambda}|$ is bounded below by some constant $c_{j}$ uniform
in $\lambda$. Hence the first term is bounded by

$C_{1,0}C_{j}\|w_{r}\|_{0,0}\sum_{i=1,2}\sup_{t\in
b\Omega_{j}}|\omega_{i}(\cdot,t,\lambda_{1})-\omega_{i}(\cdot,t,\lambda_{2})|_{C^{0}(b\Omega_{j_{0}})}.$

Similarly the second term is bounded by

$C_{1,0}C_{j}\|w_{r}\|_{0,0}\left(\Sigma_{i=1,2}\|\omega_{i}\|_{0,0}\right)|\Gamma^{\lambda_{1}}-\Gamma^{\lambda_{2}}|_{1},$

and the third term is bounded by

$C_{1,0}C_{j}\left(\Sigma_{i=1,2}\|\omega_{i}\|_{0,0}\right)|\delta^{\lambda_{1},\lambda_{2}}(w_{r}a)|_{C^{0}(\overline{\Omega})},$

where $\|w_{r}\|_{0,0}=\sup_{t\in
b\Omega_{j},\lambda\in[0,1]}|w_{r}(t,\lambda)|$ and
$\|\omega_{i}\|_{0,0}=\sup_{z\in b\Omega_{j_{0}},t\in
b\Omega_{j}\lambda\in[0,1]}|\omega_{i}(z,t,\lambda)|$.

By part (i), we have
$w_{r}^{\lambda}\in\mathcal{C}^{\nu^{-},0}(\overline{\Omega^{\lambda}})$ so in
particular $|w_{r}^{\lambda_{1}}-w_{r}^{\lambda_{2}}|_{0}\to 0$. Combined with
(5.4), we readily see that the above expressions converge to $0$ as
$|\lambda_{1}-\lambda_{2}|\to 0$. This shows that for any $j\neq j_{0}$,
$w_{j}(\cdot,\lambda)$ is continuous in $\lambda$ in the
$C^{0}(b\Omega_{j_{0}})$\- norm.

Next, we show that $|w_{j}(\cdot,\lambda)|_{C^{\nu}(b\Omega_{j_{0}})}$ is
bounded uniformly in $\lambda$. In fact we will show the boundedness of
$|w_{j}(\cdot,\lambda)|_{C^{\alpha_{p}}(b\Omega_{j_{0}})}$. Recall that
$w_{j}^{\lambda}$ is given by formula (6.106):

(6.109) $w_{j}(z,\lambda)=\frac{1}{2\pi
i}\int_{b\Omega_{j}}G_{1}(z,t,\lambda)w_{r}(t,\lambda)\,dt^{\lambda}+G_{2}(z,t,\lambda)\overline{w_{r}(t,\lambda)}\,\overline{dt^{\lambda}},\quad
z\in\Omega_{j},$

where $G_{1}$ and $G_{2}$ are given by

$\displaystyle
G_{1}(z,t,\lambda)=\frac{e^{\omega_{1}(z,t,\lambda)}+e^{\omega_{1}(z,t,\lambda)}}{t^{\lambda}-z^{\lambda}},\qquad
G_{2}(z,t,\lambda)=\frac{e^{\omega_{1}(z,t,\lambda)}-e^{\omega_{1}(z,t,\lambda)}}{t^{\lambda}-z^{\lambda}},$

and $\omega_{i}$ are given by formula (5.2), with $A$ and $B$ replaced by $A$
and $B$. By the proof of Lemma 5.2, we can show that

$\sup_{t\in b\Omega_{j}}|\omega_{j}(\cdot,t,\lambda)|_{\alpha_{p}}\leq
C_{1,0}C(p,\Omega)\left(|A(\cdot,\lambda)|_{L^{p}(\Omega)}+|B(\cdot,\lambda)|_{L^{p}(\Omega)}\right),\quad
2<p<\infty.$

Now for $z\in b\Omega_{j_{0}}$ and $t\in b\Omega_{j}$ where $j\neq j_{0}$, one
has $|t^{\lambda}-z^{\lambda}|$ is bounded below by a positive constant
$c_{j}$ uniform in $\lambda$. Hence it follows by the above formula that
$|G_{i}(\cdot,\lambda,\lambda)|_{C^{\alpha_{p}}(b\Omega_{0})}$ is bounded
uniformly in $t\in b\Omega_{j}$ and $\lambda\in[0,1]$. Since we know that
$w_{r}^{\lambda}\in\mathcal{C}^{\nu-,0}(\overline{\Omega^{\lambda}})$, in
particular $|w_{r}(t,\lambda)|$ is bounded above uniformly in $t\in
b\Omega_{j}$ and $\lambda\in[0,1]$. (6.109) then implies that
$|w_{j}(\cdot,\lambda)|_{C^{\nu}(b\Omega_{j_{0}})}$ is bounded uniformly in
$\lambda$. This proves the earlier claim and the proof is now complete.

###  6.3. Differentiability in space variable: The $\mathcal{C}^{k+1+\mu,0}$
estimate

In this subsection we prove part (ii) of Theorem 6.1, namely the following:

######  Proposition 6.5.

Let $\Omega$ be a bounded domain in $\mathbb{C}$ such that $b\Omega$ has $m+1$
connected components. Let $k$ be a non-negative integer and $0<\mu<1$. Let
$\Gamma^{\lambda}:\Omega\to\Omega^{\lambda}$ be a $\mathcal{C}^{k+1+\mu,0}$
embedding. For each $\lambda$, let $w^{\lambda}$ be the unique solution to
problem A on $\Omega^{\lambda}$ with index $n>m-1$,

$\begin{cases}\partial{}_{\overline{z}}w^{\lambda}(z^{\lambda})+A^{\lambda}(z^{\lambda})w^{\lambda}(z^{\lambda})+B^{\lambda}(z^{\lambda})\overline{w^{\lambda}(z^{\lambda})}=F^{\lambda}(z^{\lambda})&\text{in
$\Omega^{\lambda}$.}\\\
Re[\overline{l^{\lambda}(z^{\lambda})}w^{\lambda}(z^{\lambda})]=\gamma^{\lambda}(z^{\lambda})&\text{on
$b\Omega^{\lambda}$, }\end{cases}$

satisfying

$\displaystyle
w^{\lambda}(\zeta_{r}^{\lambda})=a_{r}(\lambda)+ib_{r}(\lambda),\quad\quad(r=1,...,N_{0}),$
$\displaystyle
w^{\lambda}((\zeta_{s}^{\prime})^{\lambda})=l^{\lambda}((\zeta_{s}^{\prime})^{\lambda})(\gamma^{\lambda}((\zeta_{s}^{\prime})^{\lambda})+ic_{s}(\lambda)),\quad\quad(s=1,...,N_{1}),$

on a normally distributed set $\\{\zeta_{r}^{\lambda},\zeta_{s}^{\prime}\\}$
in $\Omega$. Suppose
$A^{\lambda},B^{\lambda},F^{\lambda}\in\mathcal{C}^{k+\mu,0}(\overline{\Omega^{\lambda}})$,
$l^{\lambda},\gamma^{\lambda}\in\mathcal{C}^{k+1+\mu,0}(b\Omega^{\lambda})$,
and $a_{r},b_{r},c_{s}\in C^{0}([0,1])$. Then
$w^{\lambda}\in\mathcal{C}^{k+1+\mu,0}(\overline{\Omega^{\lambda}})$.
Furthermore, there exists some constant $C$ independent of $\lambda$ such that

(6.110) $\displaystyle\|w^{\lambda}\|_{0,0}\leq
C\left(\|F^{\lambda}\|_{0,0}+\|\gamma^{\lambda}\|_{\mu,0}+\sum_{r=1}^{N_{0}}|a_{r}|_{0}+|b_{r}|_{0}+\sum_{s=1}^{N_{1}}|c_{s}|_{0}\right),$
$\displaystyle\|w^{\lambda}\|_{k+1+\mu,0}\leq
C\left(\|F^{\lambda}\|_{k+\mu,0}+\|\gamma^{\lambda}\|_{k+1+\mu,0}+\sum_{r=1}^{N_{0}}|a_{r}|_{0}+|b_{r}|_{0}+\sum_{s=1}^{N_{1}}|c_{s}|_{0}\right).$

Proof. As before we write $A(z,\lambda):=A^{\lambda}(z^{\lambda})$,
$B(z,\lambda):=B^{\lambda}(z^{\lambda})$ and
$F(z,\lambda):=F^{\lambda}(z^{\lambda})$. Note for fixed $\lambda$,
$w^{\lambda}\in C^{k+1+\mu}(\overline{\Omega^{\lambda}})$ by Proposition 3.6.
By Proposition 2.3, it suffices to show that $w(\cdot,\lambda)$ is continuous
in $\lambda$ in the $C^{0}(\overline{\Omega})$ norm and bounded uniformly in
$\lambda$ in the $C^{k+1+\mu}(\overline{\Omega})$-norm.

Like before we can make a reduction to the case $F^{\lambda}\equiv 0$. Define
$w_{F^{\lambda}}$ by formula (6.103). For the reduction to work we need to
show that
$w_{F}^{\lambda}\in\mathcal{C}^{k+1+\mu,0}(\overline{\Omega^{\lambda}})$. Now
$w_{F}^{\lambda}$ satisfies the equation:

(6.111)
$\partial{}_{\overline{z^{\lambda}}}w_{F^{\lambda}}+A^{\lambda}w_{F^{\lambda}}+B^{\lambda}\overline{w_{F^{\lambda}}}=F^{\lambda}\quad\text{
in $\Omega^{\lambda}$. }$

Since by assumption $F^{\lambda}$ is at least Hölder continuous, Lemma 6.4
shows that
$w_{F^{\lambda}}\in\mathcal{C}^{1^{-},0}(\overline{\Omega^{\lambda}})$. It
then follows from (6.111) that
$\partial{}_{\overline{z^{\lambda}}}w_{F}^{\lambda}\in\mathcal{C}^{\mu,0}(\overline{\Omega^{\lambda}})$.
For each fixed $\lambda$, we apply Cauchy-Green formula to $w^{\lambda}_{F}$
on $\Omega^{\lambda}$ to get

$w_{F}(z,\lambda)=T_{\Omega^{\lambda}}(\partial{}_{\overline{z^{\lambda}}}w_{F^{\lambda}})+\frac{1}{2\pi
i}\int_{b\Omega}\frac{w_{F}(t,\lambda)}{t^{\lambda}-z^{\lambda}}dt^{\lambda},\quad
z\in\Omega$

where we write $w_{F}(z,\lambda)=w_{F^{\lambda}}(z^{\lambda})$. Taking
$\partial{}_{z^{\lambda}}$ on both sides, we have

(6.112)
$\partial{}_{z^{\lambda}}w_{F^{\lambda}}(z^{\lambda})=\Pi_{\Omega^{\lambda}}(\partial{}_{\overline{z^{\lambda}}}w_{F^{\lambda}})+\frac{1}{2\pi
i}\int_{b\Omega}\frac{w_{F}(t,\lambda)}{(t^{\lambda}-z^{\lambda})^{2}}\,dt^{\lambda},$

where $\Pi_{\Omega^{\lambda}}:=\partial{}_{z}T_{\Omega^{\lambda}}$. Since
$w_{F^{\lambda}}\in\mathcal{C}^{1^{-},0}(\overline{\Omega^{\lambda}})$, the
second term is in $\mathcal{C}^{1-,0}(\overline{\Omega^{\lambda}})$. Now
applying Lemma 7.9 to
$\partial{}_{\overline{z^{\lambda}}}w_{F^{\lambda}}\in\mathcal{C}^{\mu,0}(\overline{\Omega^{\lambda}})$,
we have
$\Pi_{\Omega^{\lambda}}(\partial{}_{\overline{z^{\lambda}}}w_{F^{\lambda}})\in\mathcal{C}^{\mu,0}(\overline{\Omega^{\lambda}})$.
It then follows from (6.112) that
$\partial{}_{z^{\lambda}}w_{F^{\lambda}}(z^{\lambda})\in\mathcal{C}^{\mu,0}(\overline{\Omega^{\lambda}})$.
This shows that
$w_{F^{\lambda}}\in\mathcal{C}^{1+\mu,0}(\overline{\Omega^{\lambda}})$. We can
iterate this argument by using (6.111) and (6.112) so that eventually we get
$w_{F^{\lambda}}\in\mathcal{C}^{k+1+\mu,0}(\overline{\Omega^{\lambda}})$. This
proves the reduction.

The problem now is reduced to show that if $w_{r}^{\lambda}$ solves (6.104),
where $\gamma^{\lambda}\in\mathcal{C}^{k+1+\mu,0}(b\Omega^{\lambda})$, then
$w_{r}^{\lambda}\in\mathcal{C}^{k+1+\mu,0}(\overline{\Omega^{\lambda}})$. As
in the proof of part (ii), we reduce the problem to one on simply-connected
domain, and we use the same notation $w_{j}$ and $b\Omega^{j}$ as before.

Note also that by the assumptions and the result of part (ii),
$w_{r}^{\lambda}\in\mathcal{C}^{\frac{p-2}{p},0}(\overline{\Omega^{\lambda}})$
for any $2<p<\infty$. Hence
$w_{r}^{\lambda}\in\mathcal{C}^{1-,0}(\overline{\Omega^{\lambda}})$.

In view of formula (3.35) with $F\equiv 0$, we can write the solution of
problem (6.104) in the form

$w_{r}(z,\lambda)=w_{0}(z,\lambda)+\cdots+w_{m}(z,\lambda),\quad\quad
z\in\Omega,$

where we set

$w_{j}(z,\lambda)=\frac{1}{2\pi
i}\int_{b\Omega_{j}}G_{1}(z,t,\lambda)w_{r}(t,\lambda)\,dt^{\lambda}+G_{2}(z,t,\lambda)\overline{w_{r}(t,\lambda)}\,\overline{dt^{\lambda}},$

for $0\leq j\leq m$.

Fix $j_{0}$ with $0\leq j_{0}\leq m$, we need to show that for all $j\neq
j_{0}$, we have
$w_{j}^{\lambda}\in\mathcal{C}^{k+1+\mu,0}(b\Omega_{j_{0}}^{\lambda})$. The
rest of the argument follows the same as in part (ii).

That $w_{j}^{\lambda}$ is continuous in the $C^{0}(b\Omega_{j_{0}})$-norm is
proved in part (ii), so we only need to show that
$|w_{j}(\cdot,\lambda)|_{C^{k+1+\mu}(b\Omega_{j_{0}})}$ is bounded in
$\lambda$. Recall that $w_{j}^{\lambda}$ satisfies the equation:

$\partial{}_{\overline{z^{\lambda}}}w^{\lambda}_{j}(z^{\lambda})+A^{\lambda}(z^{\lambda})w^{\lambda}_{j}(z^{\lambda})+B^{\lambda}(z^{\lambda})\overline{w_{j}^{\lambda}(z^{\lambda})}=0,\quad
z^{\lambda}\in\Omega_{j}^{\lambda}.$

By part (i), we have
$w_{r}^{\lambda}\in\mathcal{C}^{1^{-},0}(\overline{\Omega^{\lambda}})$. In
view of Proposition 7.13 and (6.106),
$w_{j}^{\lambda}\in\mathcal{C}^{1^{-},0}(\overline{\Omega_{j}^{\lambda}})$.
Fix an open set $V_{j}$ so that
$V_{j}\subset\subset(\Omega_{j}^{\lambda}\cap\Omega^{\lambda})$ and
$\operatorname{dist}(b\Omega_{j_{0}}^{\lambda},V_{j})\geq\delta_{0}$ for all
$\lambda$. Certainly $w^{\lambda}_{j}$ satisfy the same equation on $V_{j}$:

(6.113)
$\partial{}_{\overline{z}}w^{\lambda}_{j}(z)+A^{\lambda}(z)w^{\lambda}_{j}(z)+B^{\lambda}(z)\overline{w_{j}^{\lambda}(z)}=0,\quad
z\in V_{j}.$

Here
$A^{\lambda},B^{\lambda}\in\mathcal{C}^{k+\mu,0}(\overline{V_{j}})\subset\mathcal{C}^{\mu,0}(\overline{V_{j}})$
and $w^{\lambda}_{j}\in\mathcal{C}^{1^{-},0}(\overline{V_{j}})$, thus equation
(6.113) implies that
$\partial{}_{\overline{z}}w_{j}^{\lambda}\in\mathcal{C}^{\mu,0}(\overline{V_{j}})$.

Applying Cauchy-Green formula to $V_{j}$, we have for each fixed $\lambda$,

(6.114)
$w_{j}^{\lambda}(z)=T_{V_{j}}(\partial{}_{\overline{z}}w_{j})(z)+\frac{1}{2\pi
i}\int_{bV_{j}}\frac{w_{j}^{\lambda}(\zeta)}{\zeta-z}\,d\zeta,\quad z\in
V_{j}.$

Taking $\partial{}_{z}$ on both sides we get

(6.115)
$\partial{}_{z}w_{j}^{\lambda}(z)=\partial{}_{z}T_{V_{j}}(\partial{}_{\overline{z}}w_{j}^{\lambda})(z)+\frac{1}{2\pi
i}\int_{bV_{j}}\frac{w_{j}^{\lambda}(\zeta)}{(\zeta-z)^{2}}\,\,d\zeta,\quad
z\in V_{j}.$

Now, for $\zeta\in bV_{j}$ and $z\in b\Omega_{j_{0}}^{\lambda}$, $|\zeta-z|$
is bounded below by a positive constant $\delta_{0}$ independent of $\lambda$,
so the second term above is bounded by $C|w_{j}^{\lambda}|_{C^{0}(bV_{j})}$
for some $C$ independent of $\lambda$. On the other hand by Lemma 7.9,

$|\partial{}_{z}T_{V_{j}}(\partial{}_{\overline{z}}w_{j}^{\lambda})|_{C^{\mu}(\overline{V_{j}})}\leq
C|\partial{}_{\overline{z}}w_{j}^{\lambda}|_{C^{\mu}(\overline{V_{j}})}.$

Thus

$|\partial{}_{z}w_{j}^{\lambda}|_{C^{\mu}(\overline{V_{j}})}\leq
C\left(|\partial{}_{\overline{z}}w_{j^{\prime}}^{\lambda}|_{C^{\mu}(\overline{V_{j}})}+|w_{j}^{\lambda}|_{C^{0}(\overline{V_{j}})}\right),$

where $C$ is some constant independent of $\lambda$. Hence
$|w_{j}^{\lambda}|_{C^{1+\mu}(V_{j})}$ is uniformly bounded in $\lambda$. By
repeating the argument using (6.113) and (6.115), we can show that
$|w_{j}^{\lambda}|_{C^{k+1+\mu}(V_{j})}$ is uniformly bounded in $\lambda$.
Since $\Gamma^{\lambda}\in\mathcal{C}^{k+1+\mu,0}(\overline{\Omega})$ and
$b\Omega_{j_{0}}\subset V_{j}$,
$|w_{j}^{\lambda}(z^{\lambda})|_{C^{k+1+\mu}(b\Omega_{j_{0}})}$ is bounded
uniformly in $\lambda$. This completes the proof of the claim and thus the
theorem. ∎

###  6.4. Differentiability in parameter: the $\mathcal{C}^{k+1+\mu,j}$
estimate

In this subsection we prove part (iii) of Theorem 6.1, namely the following:

######  Theorem 6.6.

Let $\Omega$ be a bounded domain in $\mathbb{C}$ such that $b\Omega$ has $m+1$
connected component. Let $k$ and $j$ be non-negative integers satisfying
$k\geq j\geq 1$ and let $0<\mu<1$. Suppose
$\Gamma^{\lambda}:\Omega\to\Omega^{\lambda}$ is a $\mathcal{C}^{k+1+\mu,j}$
embedding. For each $\lambda$, let $w^{\lambda}$ be the unique solution to
Problem A on $\Omega^{\lambda}$ with index $n>m-1$,

(6.116)
$\begin{cases}\partial{}_{\overline{z}}w^{\lambda}(z^{\lambda})+A^{\lambda}(z^{\lambda})w^{\lambda}(z^{\lambda})+B^{\lambda}(z^{\lambda})\overline{w^{\lambda}(z^{\lambda})}=F^{\lambda}(z^{\lambda})&\text{in
$\Omega^{\lambda}$.}\\\
Re[\overline{l^{\lambda}(z^{\lambda})}w^{\lambda}(z^{\lambda})]=\gamma^{\lambda}(z^{\lambda})&\text{on
$b\Omega^{\lambda}$, }\end{cases}$

satisfying

(6.117) $\displaystyle
w^{\lambda}(\zeta_{r}^{\lambda})=a_{r}(\lambda)+ib_{r}(\lambda),\quad\quad(r=1,...,N_{0}),$
(6.118) $\displaystyle
w^{\lambda}((\zeta_{s}^{\prime})^{\lambda})=l^{\lambda}((\zeta_{s}^{\prime})^{\lambda})(\gamma^{\lambda}((\zeta_{s}^{\prime})^{\lambda})+ic_{s}(\lambda)),\quad\quad(s=1,...,N_{1}),$

on a normally distributed set $\\{\zeta_{r}^{\lambda},\zeta_{s}^{\prime}\\}$
in $\Omega$. Suppose
$A^{\lambda},B^{\lambda},F^{\lambda}\in\mathcal{C}^{k+\mu,j}(\overline{\Omega^{\lambda}})$,
$l^{\lambda},\gamma^{\lambda}\in\mathcal{C}^{k+1+\mu,j}(b\Omega^{\lambda})$,
and $a_{r},b_{r},c_{s}\in C^{j}([0,1])$. Then
$w^{\lambda}\in\mathcal{C}^{k+1+\mu,j}(\overline{\Omega^{\lambda}})$.
Furthermore, there exists some constant $C$ independent of $\lambda$ such that

(6.119) $\|w^{\lambda}\|_{k+1+\mu,j}\leq
C\left(\|F^{\lambda}\|_{k+\mu,j}+\|\gamma^{\lambda}\|_{k+1+\mu,j}+\sum_{r=1}^{N_{0}}|a_{r}|_{j}+|b_{r}|_{j}+\sum_{s=1}^{N_{1}}|c_{s}|_{j}\right).$

###### Proof.

We do induction on $j$. The $j=0$ case is proved in Theorem (6.1) part (ii).
Suppose the statement has been proved for $j-1$. In particular we know that
$w^{\lambda}\in\mathcal{C}^{k+1+\mu,j-1}(\overline{\Omega^{\lambda}})$ for
$k\geq j\geq 1$.

Let $\Psi^{\lambda}=(\Gamma^{\lambda})^{-1}$ and
$v^{\lambda}(\zeta)=w^{\lambda}\circ\Gamma^{\lambda}(\zeta)$. Then
$w^{\lambda}=v^{\lambda}\circ\Psi^{\lambda}$,
$\Psi^{\lambda}(z^{\lambda})=\zeta$. For fixed $\lambda$, we have

$\displaystyle\partial{}_{\overline{z^{\lambda}}}w^{\lambda}$
$\displaystyle=\frac{\partial\Psi^{\lambda}}{\partial\overline{z^{\lambda}}}(\Gamma^{\lambda}(\zeta))\partial{}_{\zeta}v^{\lambda}+\frac{\partial\overline{\Psi^{\lambda}}}{\partial\overline{z^{\lambda}}}(\Gamma^{\lambda}(\zeta))\partial{}_{\overline{\zeta}}v^{\lambda}$
$\displaystyle=h^{\lambda}(\zeta)\left(\partial{}_{\overline{\zeta}}v^{\lambda}+e^{\lambda}(\zeta)\partial{}_{\zeta}v^{\lambda}\right),$

where we denote

$e^{\lambda}(\zeta):=\frac{\partial\Psi^{\lambda}}{\partial\overline{z^{\lambda}}}(\Gamma^{\lambda}(\zeta))\left(\frac{\partial\overline{\Psi^{\lambda}}}{\partial\overline{z^{\lambda}}}(\Gamma^{\lambda}(\zeta))\right)^{-1},\quad
h^{\lambda}(\zeta):=\frac{\partial\overline{\Psi^{\lambda}}}{\partial\overline{z^{\lambda}}}(\Gamma^{\lambda}(\zeta)).$

Note that since $\Psi^{0}=\Gamma^{0}=id$, we have $e^{0}\equiv 0$ and
$h^{0}\equiv 1$ on $\Omega$. Thus for $|\lambda|$ small, $|e^{\lambda}|<1$.
Now $v^{\lambda}$ is the solution to

(6.120)
$\displaystyle\begin{cases}\partial{}_{\overline{\zeta}}v^{\lambda}+e^{\lambda}(\zeta)\partial{}_{\zeta}v^{\lambda}+A_{1}^{\lambda}(\zeta)v^{\lambda}(\zeta)+B_{1}^{\lambda}(\zeta)\overline{v^{\lambda}(\zeta)}=F_{1}^{\lambda}(\zeta)&\text{in
$\Omega$.}\\\
Re[\overline{l_{1}^{\lambda}(\zeta)}v^{\lambda}(\zeta)]=\gamma_{1}^{\lambda}(\zeta)&\text{on
$b\Omega$},\end{cases}$

where

$\displaystyle
A_{1}^{\lambda}(\zeta)=(h^{\lambda})^{-1}A^{\lambda}(\Gamma^{\lambda}(\zeta)),\quad
B_{1}^{\lambda}(\zeta)=(h^{\lambda})^{-1}B^{\lambda}(\Gamma^{\lambda}(\zeta)),$
$\displaystyle
F_{1}^{\lambda}(\zeta)=(h^{\lambda})^{-1}F^{\lambda}(\Gamma^{\lambda}(\zeta)),\quad
l_{1}^{\lambda}(\zeta)=l^{\lambda}(\Gamma^{\lambda}(\zeta)),\quad\gamma_{1}^{\lambda}(\zeta)=\gamma^{\lambda}(\Gamma^{\lambda}(\zeta)).$

By assumption,
$e^{\lambda},A^{\lambda}_{1},B^{\lambda}_{1},F^{\lambda}_{1}\in\mathcal{C}^{k+\mu,j}(\Omega^{\lambda})$,
and
$l_{1}^{\lambda},\gamma_{1}^{\lambda}\in\mathcal{C}^{k+1+\mu,j}(b\Omega^{\lambda})$.
Subtracting equation (6.120) by the same equation with $\lambda$ replaced by
$\lambda_{0}$ and dividing the resulting difference by $\lambda-\lambda_{0}$
we get

(6.121)
$\displaystyle\begin{cases}\partial{}_{\overline{\zeta}}(D^{\lambda,\lambda_{0}}v)+e^{\lambda_{0}}\partial{}_{\zeta}(D^{\lambda,\lambda_{0}}v^{\lambda})+A_{1}^{\lambda_{0}}D^{\lambda,\lambda_{0}}v+B_{1}^{\lambda_{0}}\overline{D^{\lambda,\lambda_{0}}v}=F_{2}^{\lambda,\lambda_{0}}&\text{in
$\Omega$.}\\\
Re[\overline{l_{1}^{\lambda_{0}}}D^{\lambda,\lambda_{0}}v]=\gamma_{2}^{\lambda,\lambda_{0}}&\text{on
$b\Omega$},\end{cases}$

where we used the notation
$D^{\lambda,\lambda_{0}}f:=(f^{\lambda}-f^{\lambda_{0}})/(\lambda-\lambda_{0})$,
and

(6.122) $\displaystyle
F_{2}^{\lambda,\lambda_{0}}:=D^{\lambda,\lambda_{0}}F_{1}-(D^{\lambda,\lambda_{0}}e)\partial{}_{\zeta}v^{\lambda}-(D^{\lambda,\lambda_{0}}A_{1})v^{\lambda}-(D^{\lambda,\lambda_{0}}B_{1})v^{\lambda},$
(6.123)
$\displaystyle\gamma_{2}^{\lambda,\lambda_{0}}:=D^{\lambda,\lambda_{0}}\gamma_{1}-Re[\overline{(D^{\lambda,\lambda_{0}}l_{1})}v^{\lambda}].$

In addition, $D^{\lambda,\lambda_{0}}v$ satisfies the condition

(6.124)
$\begin{gathered}D^{\lambda,\lambda_{0}}v(\zeta_{r})=D^{\lambda,\lambda_{0}}a_{r}+iD^{\lambda,\lambda_{0}}b_{r},\quad
r=1,\dots N_{0}.\\\
D^{\lambda,\lambda_{0}}v(\zeta_{s}^{\prime})=D^{\lambda,\lambda_{0}}\left\\{l(\zeta_{s}^{\prime},\cdot)[\gamma(\zeta_{s}^{\prime},\cdot)+ic_{s}(\cdot)]\right\\},\quad\quad
s=1,\dots N_{1}.\end{gathered}$

Consider the problem

(6.125)
$\displaystyle\begin{cases}\overline{\partial}_{\zeta}u^{\lambda}+e^{\lambda}\partial{}_{\zeta}u^{\lambda}+A_{1}^{\lambda}u^{\lambda}+B_{1}^{\lambda}\overline{u^{\lambda}}=F_{2}^{\lambda}&\text{in
$\Omega$.}\\\
Re[\overline{l_{1}^{\lambda}}u^{\lambda}]=\gamma_{2}^{\lambda}&\text{on
$b\Omega$},\end{cases}$

where

(6.126) $\displaystyle
F_{2}^{\lambda}:=(\partial{}_{\lambda}F_{1}^{\lambda})-(\partial{}_{\lambda}e^{\lambda})\partial{}_{\zeta}v^{\lambda}-(\partial{}_{\lambda}A_{1}^{\lambda})v^{\lambda}-(\partial{}_{\lambda}B_{1}^{\lambda})v^{\lambda},$
(6.127)
$\displaystyle\gamma_{2}^{\lambda}:=(\partial{}_{\lambda}\gamma_{1}^{\lambda})-Re[\overline{(\partial{}_{\lambda}l_{1}^{\lambda})}v^{\lambda}].$

In addition, we require that $u^{\lambda}$ satisfies the condition:

(6.128)
$\begin{gathered}u^{\lambda}(\zeta_{r})=a_{r}^{\prime}(\lambda)+i\partial{}b_{r}^{\prime}(\lambda),\quad
r=1,\dots N_{0}.\\\
u^{\lambda}(\zeta_{s}^{\prime})=\partial{}_{\lambda}\left\\{l(\zeta_{s}^{\prime},\lambda)\left[\gamma(\zeta^{\prime}_{s},\lambda)+ic_{s}(\lambda)\right]\right\\},\quad
s=1,\dots,N_{1}.\end{gathered}$

By assumption,
$\partial{}_{\lambda}e^{\lambda},\partial{}_{\lambda}A_{1}^{\lambda},\partial{}_{\lambda}B_{1}^{\lambda},\partial{}_{\lambda}F_{1}^{\lambda}\in\mathcal{C}^{k-1+\mu,j-1}(\overline{\Omega^{\lambda}})$,
and
$\partial{}_{\lambda}\gamma_{1}^{\lambda},\partial{}_{\lambda}l_{1}^{\lambda}\in\mathcal{C}^{k+\mu,j-1}(b\Omega^{\lambda})$.
By the induction hypothesis,
$v^{\lambda}=w^{\lambda}\circ\Gamma^{\lambda}\in\mathcal{C}^{k+1+\mu,j-1}(\overline{\Omega^{\lambda}})$,
and there exists some constant $C$ independent of $\lambda$ such that

(6.129) $\|v^{\lambda}\|_{\mathcal{C}^{k+1+\mu,j-1}(\Omega)}\leq
C\left(\|F^{\lambda}\|_{\mathcal{C}^{k+\mu,j-1}(\Omega^{\lambda})}+\|\gamma^{\lambda}\|_{\mathcal{C}^{k+1+\mu,j-1}(b\Omega^{\lambda})}\right).$

Thus in view of (6.126) and (6.127) we have
$F_{2}^{\lambda}\in\mathcal{C}^{k-1+\mu,j-1}(\overline{\Omega^{\lambda}})$,
and $\gamma_{2}^{\lambda}\in\mathcal{C}^{k+\mu,j-1}(b\Omega^{\lambda})$. To