The classical Schwarz kernel for the upper half-plane \(\mathbb{C}^{+}\) is
$$\frac{z-\overline{z}}{|x-z|^{2}}, \quad (x,z)\in(-\infty ,+\infty)\times\mathbb{C}^{+} $$
and satisfies (see [14])
$$\begin{aligned}& \lim_{z\in\mathbb{C}^{+}, z\rightarrow t} \frac {1}{2\pi i} \int_{c}^{d}\rho(x) \biggl[ \frac{1}{x-z}- \frac {1}{x-\overline{z}} \biggr]\,\mathrm{d}x \\& \quad = \rho(t), \quad t\in(c,d)\subset(-\infty,+ \infty), \end{aligned}$$
(23)
with \(\rho\in C([c,d],\mathbb{C})\), \(c< d\).
Lemma 3.1
For
\(\rho\in C(\partial\Omega,\mathbb{C})\), \(z\in\Omega\), we obtain
$$\begin{aligned}& \lim_{z\rightarrow t} \frac{1}{2\pi i} \int_{[a,\mu ]}\rho(\xi) \biggl[ \frac{1}{\xi-z}- \frac{1}{\xi+\overline {z}-2a} \biggr]\,\mathrm{d}\xi= \rho(t), \quad t\in(a,\mu), \\& \lim_{z\rightarrow t} \frac{1}{2\pi i} \int_{[\mu,\nu ]}\rho(\xi) \biggl[ \frac{1}{\xi-z}- \frac{1}{\xi-\overline {z}-2bi} \biggr]\,\mathrm{d}\xi= \rho(t), \quad t\in(\mu,\nu), \\& \lim_{z\rightarrow t} \frac{1}{2\pi i} \int_{[\nu ,0]}\rho(\xi) \biggl[ \frac{1}{\xi-z}- \frac{1}{\xi+\overline {z}} \biggr]\,\mathrm{d}\xi= \rho(t), \quad t\in(\nu,0). \end{aligned}$$
Proof
When \(\xi\in[a,\mu]\) and letting \(\xi=a+iy\), then \(y\in[0,b]\) and
$$\begin{aligned}& \lim_{z\rightarrow t, t\in(a,\mu )} \frac{1}{2\pi i} \int_{[a,\mu]}\rho(\xi) \biggl[ \frac{1}{\xi-z}- \frac{1}{\xi+\overline{z}-2a} \biggr]\,\mathrm{d}\xi \\& \quad =\lim_{z\rightarrow t, t\in(a,\mu )} \frac{1}{2\pi i} \int_{0}^{b}\rho(a+iy) \biggl[ \frac{1}{y-(a-z)i}- \frac{1}{y-(\overline{z}-a)i} \biggr]\,\mathrm{d}y. \end{aligned}$$
For \(z\in\Omega\) and \(z\rightarrow t\in(a,\mu)\), we obtain \((a-z)i\in \mathbb{C}^{+}\) and \(\rightarrow(a-t)i\in(0,b)\). Thus, from the classical Schwarz kernel theory (23), the first limit in Lemma 3.1 equals \(\rho(t)\).
For \(\xi\in[\mu,\nu]\), we take \(\xi=y+ib\), then \(y\in[a,0]\) and the second equation is
$$\lim_{z\rightarrow t, t\in(\mu,\nu)} \frac{1}{2\pi i} \int_{0}^{a}\rho(y+ib) \biggl[ \frac{1}{y-(\overline{z}+i b)}- \frac{1}{y-(z-i b)} \biggr]\,\mathrm{d}y. $$
When \(z\in\Omega\) and \(z\rightarrow t\in(\mu,\nu)\), we obtain \(\overline{z}+i b\in\mathbb{C}^{+}\) and \(\rightarrow\overline{t}+i b\in(0,a)\). Hence, from (23), the second equation is true. In the same way, when \(\xi\in[\nu,0]\), suppose \(\xi=iy\), then \(y\in[b,0]\) and the third limit equals
$$\lim_{z\rightarrow t\in(\nu,0)} \frac{1}{2\pi i} \int _{0}^{b}\rho(i y) \biggl[ \frac{1}{y-i\overline{z}}- \frac{1}{ y+iz} \biggr]\,\mathrm{d}y=\rho(t). $$
The proof is completed. □
Define the following Pompeiu-type operator:
$$\begin{aligned} \begin{aligned}[b] A_{\alpha}[f](z)={}&- \frac{1}{\pi} \int_{\Omega}\biggl\{ f(\zeta) \sum_{m,n} \biggl[q_{m,n}(\zeta ,z)- \frac{q_{m,n}(\zeta,\alpha)+q_{m,n}(\zeta,\overline {\alpha})}{2} \biggr] \\ &{}-\overline{f(\zeta)} \sum_{m,n} \biggl[q_{m,n}(\overline{\zeta},z)- \frac {q_{m,n}(\overline{\zeta},\alpha) +q_{m,n}(\overline{\zeta},\overline{\alpha})}{2} \biggr] \biggr\} \,\mathrm{d}\xi\,\mathrm{d}\eta, \end{aligned} \end{aligned}$$
(24)
for \(f\in L_{p}(\Omega,\mathbb{C})\), \(p>2\) with \(q_{m,n}\) given by (4) and \(\alpha\in\Omega\).
Lemma 3.2
If
\(f\in L_{p}(\Omega,\mathbb{C})\), \(p>2\), then
\(A_{\alpha}[f](z)\in C(\overline{\Omega},\mathbb{C})\)
and
\(\frac{\partial A_{\alpha}[f](z)}{\partial\overline{z}}=f(z)\)
for
\(z\in\Omega\).
Proof
From the classical Pompeiu-type operator in [14],
$$T[f](z)=- \frac{1}{\pi} \int_{\Omega} \frac {f(\zeta)}{\zeta-z}\,\mathrm{d}\xi\,\mathrm{d}\eta\in C(\overline{\Omega },\mathbb{C}) $$
and \(\frac{\partial T[f](z)}{\partial\overline{z}}=f(z)\) for \(z\in\Omega\). When \(z\in\Omega\), then \(z+\omega_{mn}\notin\Omega\) for \((m,n)\neq(0,0)\) and \(\pm\overline{z}+\omega_{mn}, -z+\omega_{mn}\notin\Omega\). Thus by (4) and (24), we have \(A_{\alpha}[f](z)\in C(\overline {\Omega},\mathbb{C})\) and the integral in (24) is analytic for \(z\in\Omega\) except for one term \(T[f](z)\), therefore,
$$\frac{\partial A_{\alpha}[f](z)}{\partial\overline {z}}= \frac{\partial (- \frac{1}{\pi}\int _{\Omega} \frac{f(\zeta)}{\zeta-z}\,\mathrm{d}\xi\,\mathrm{d}\eta )}{\partial\overline{z}}=f(z). $$
□
Lemma 3.3
For
\(f\in L_{p}(\Omega,\mathbb{C})\), \(p>2\),
$$\lim_{z\in\Omega, z\rightarrow t}\operatorname{Re}A_{\alpha}[f](z)=0, \quad t \in\partial\Omega. $$
Proof
By (24), we obtain
$$\operatorname{Re}A_{\alpha}[f](z)=- \frac{1}{2\pi} \int_{\Omega}\bigl\{ f(\zeta) \bigl[P(\zeta,z)+\Phi(\zeta,\alpha) \bigr] -\overline{f(\zeta)} \bigl[P(\overline{\zeta},z)+\Phi(\overline{\zeta }, \alpha) \bigr] \bigr\} \,\mathrm{d}\xi\,\mathrm{d}\eta $$
with
$$P(\zeta,z)=\sum_{m,n} \bigl[q_{m,n}( \zeta,z)-q_{m,-n}(\zeta ,\overline{z}) \bigr] $$
and
$$\Phi(\zeta,z)= \frac{1}{2}\sum_{m,n} \bigl[q_{m,-n}(\zeta,\overline{z})-q_{m,n}(\zeta,\overline{z}) +q_{m,-n}(\zeta,z)-q_{m,n}(\zeta,z) \bigr]. $$
Obviously, we know \(\Phi(\zeta,\alpha)=\Phi(\overline{\zeta},\alpha)=0\) for \(\zeta\in\Omega\). Therefore,
$$\begin{aligned} \operatorname{Re}A_{\alpha}[f](z) =&- \frac{1}{2\pi} \int_{\Omega}\biggl\{ f(\zeta)\sum_{m,n} \bigl[q_{m,n}(\zeta,z)-q_{m,-n}(\zeta,\overline {z}) \bigr] \\ &{}-\overline{f(\zeta)} \sum_{m,n} \bigl[q_{m,n}(\overline{\zeta},z)-q_{m,-n}(\overline{\zeta }, \overline{z}) \bigr] \biggr\} \,\mathrm{d}\xi\,\mathrm{d}\eta. \end{aligned}$$
Furthermore, when \(z\in\partial\Omega\), it satisfies equation (17) by replacing ζ with z. For \(\zeta\in\Omega\), \(z\in[\mu ,\nu]\), we have \(\overline{z}=z-2bi\), then
$$\sum_{m,n} \bigl[q_{m,n}( \zeta,z)-q_{m,-n}(\zeta,\overline{z}) \bigr]=\sum _{m,n} \bigl[q_{m,n}(\zeta,z)-q_{m,-n}( \zeta,z-2bi) \bigr]=0. $$
Similarly,
$$\sum_{m,n} \bigl[q_{m,n}(\overline{ \zeta},z)-q_{m,-n}(\overline {\zeta},\overline{z}) \bigr]=0, \quad z\in[ \mu,\nu], \zeta\in\Omega. $$
Also, for \(\zeta\in\Omega\) and \(z\in\partial\Omega\setminus[\mu,\nu]\),
$$\sum_{m,n} \bigl[q_{m,n}( \zeta,z)-q_{m,-n}(\zeta,\overline{z}) \bigr] =\sum _{m,n} \bigl[q_{m,n}(\overline{\zeta},z)-q_{m,-n}( \overline {\zeta},\overline{z}) \bigr]=0, $$
thus the proof is completed. □
Consider the Schwarz-type operator
$$ S_{\alpha}[\gamma](z)= \frac{1}{\pi i} \int_{\partial\Omega }\gamma(\zeta) \sum_{m,n} \biggl[q_{m,n}(\zeta,z)- \frac {q_{m,n}(\zeta,\alpha)+q_{m,n}(\zeta,\overline{\alpha})}{2} \biggr] \,\mathrm{d}\zeta, $$
(25)
where \(q_{m,n}\) is given by (4) and \(\gamma\in C(\partial\Omega ,\mathbb{R})\). Then we have
$$\begin{aligned} \operatorname{Re}S_{\alpha}[\gamma](z) =& \frac{1}{2\pi i} \int _{\partial\Omega}\gamma(\zeta) \biggl\{ \sum _{m,n} \biggl[q_{m,n}(\zeta ,z)- \frac{q_{m,n}(\zeta,\alpha)+q_{m,n}(\zeta,\overline {\alpha})}{2} \biggr]\,\mathrm{d}\zeta \\ &{}- \sum_{m,n} \biggl[\overline {q_{m,n}( \zeta,z)}- \frac{\overline{q_{m,n}(\zeta,\alpha)} +\overline{q_{m,n}(\zeta,\overline{\alpha})}}{2} \biggr]\,\mathrm{d} {\overline{\zeta}} \biggr\} . \end{aligned}$$
By (18),
$$\begin{aligned} \operatorname{Re}S_{\alpha}[\gamma](z) =& \frac{1}{2\pi i} \int _{\partial\Omega}\gamma(\zeta) \biggl\{ \sum _{m,n} \bigl[q_{m,n}(\zeta,z)-g_{m,n}(\zeta , \overline{z}) \bigr] \\ &{}- \frac{1}{2} \sum_{m,n} \bigl[q_{m,n}(\zeta,\alpha)-g_{m,n}(\zeta,\overline{\alpha })+q_{m,n}(\zeta,\overline{\alpha})-g_{m,n}(\zeta,\alpha) \bigr] \biggr\} \,\mathrm{d} {\zeta}, \end{aligned}$$
where \(g_{m,n}\) is given by (16). From (16) and Lemma 2.1, for \(\zeta\in\partial\Omega\),
$$\sum_{m,n} \bigl[q_{m,n}(\zeta, \alpha)-g_{m,n}(\zeta ,\overline{\alpha})+q_{m,n}(\zeta, \overline{\alpha})-g_{m,n}(\zeta ,\alpha) \bigr]=0 $$
and
$$\sum_{m,n} \bigl[q_{m,n}( \zeta,z)-g_{m,n}(\zeta,\overline{z}) \bigr]=\sum _{m,n} \bigl[q_{m,n}(\zeta,z)-q_{m,n}(\zeta, \overline{z}) \bigr], $$
therefore,
$$ \operatorname{Re}S_{\alpha}[\gamma](z)= \frac{1}{2\pi i} \int _{\partial\Omega}\gamma(\zeta) \sum_{m,n} \bigl[q_{m,n}(\zeta,z)-q_{m,n}(\zeta,\overline{z}) \bigr]\, \mathrm{d} {\zeta}. $$
(26)
Lemma 3.4
For
\(\gamma\in C(\partial\Omega,\mathbb{R})\), \(S_{\alpha}[\gamma](z)\)
is analytic in Ω, i.e.,
$$\frac{\partial S_{\alpha}[\gamma](z)}{\partial\overline{z}}=0. $$
Proof
From (25), the sum in integrand can be rewritten as
$$\frac{1}{2}\sum_{m,n} \bigl\{ \bigl[q_{m,n}(\zeta ,z)-q_{m,n}(\zeta,\alpha) \bigr] + \bigl[q_{m,n}(\zeta,z)-q_{m,n}(\zeta,\overline{\alpha}) \bigr] \bigr\} , $$
which is convergent for \(\zeta\in\partial\Omega\), \(z\in\Omega\) by Lemma 2.1. Obviously, from the expression of \(q_{m,n}\) in (4), the integrand in \(S_{\alpha}[\gamma](z)\) is analytic for \(z\in\Omega\), hence the proof is completed. □
Lemma 3.5
For
\(z\in\Omega\), \(\gamma\in C(\partial\Omega,\mathbb{R})\),
$$ \lim_{z\rightarrow t} \frac{1}{2\pi i} \int_{L} \bigl[\gamma(\zeta) -\gamma(t_{0}) \bigr] \sum_{m,n} \bigl[q_{m,n}( \zeta,z)-q_{m,n}(\zeta ,\overline{z}) \bigr]\,\mathrm{d} {\zeta}= \gamma(t)-\gamma(t_{0}), $$
(27)
where
\(t\in L^{0}\cup\{t_{0}\}\), \(L^{0}\)
is
L
except for two endpoints and
$$L=\left \{ \textstyle\begin{array}{l@{\quad}l} {[\mu,\nu]},& \textit{for } t_{0}=\mu \textit{ or } \nu, \\ {[\nu,0]},& \textit{for } t_{0}=0 \textit{ or } \nu, \\ {[0,a]},& \textit{for } t_{0}=0 \textit{ or } a, \\ {[a,\mu]},& \textit{for } t_{0}=a \textit{ or } \mu. \end{array}\displaystyle \right . $$
Furthermore, for
\(t\in\partial\Omega\setminus L^{0}\),
$$ \lim_{z\rightarrow t} \frac{1}{2\pi i} \int_{L} \bigl[\gamma(\zeta)-\gamma(t) \bigr] \sum _{m,n} \bigl[q_{m,n}(\zeta,z)-q_{m,n}(\zeta, \overline{z}) \bigr]\,\mathrm{d} {\zeta}=0. $$
(28)
Proof
When \(L=[\mu,\nu]\), \(t_{0}=\nu\) and \(t\in L^{0}\cup\{t_{0}\}=(\mu,\nu]\),
$$\begin{aligned}& \lim_{z\rightarrow t} \frac{1}{2\pi i} \int_{[\mu,\nu ]} \bigl[\gamma(\zeta) -\gamma(\nu) \bigr]\sum _{m,n} \bigl[q_{m,n}(\zeta,z)-q_{m,n}(\zeta , \overline{z}) \bigr]\,\mathrm{d} {\zeta} \\& \quad = \lim_{z\rightarrow t} \frac{1}{2\pi i} \int_{[\mu,\nu ]} \bigl[\gamma(\zeta) -\gamma(\nu) \bigr] \biggl[ \frac{1}{\zeta-z}-\frac{1}{\zeta -\overline{z}-2bi}+\frac{1}{\zeta+z-2bi}-\frac{1}{\zeta+\overline {z}} \biggr]\,\mathrm{d} {\zeta} \\& \quad =\lim_{z\rightarrow t} \frac{1}{2\pi i} \int_{[\mu,\nu ]\cup[\nu,-a+i b]}\Gamma(\zeta) \biggl[ \frac{1}{\zeta-z}- \frac {1}{\zeta-\overline{z}-2bi} \biggr]\,\mathrm{d} {\zeta}, \end{aligned}$$
where
$$\Gamma(\zeta)=\left \{ \textstyle\begin{array}{l@{\quad}l} \gamma(\zeta)-\gamma(\nu),&\zeta\in[\mu,\nu], \\ \gamma(\nu)-\gamma(2bi-\zeta),&\zeta\in[\nu,-a+i b]. \end{array}\displaystyle \right . $$
By \(\Gamma\in C([\mu,\nu]\cup[\nu,-a+i b],\mathbb{R})\) and from the second equation in Lemma 3.1, the above limit is \(\Gamma(t)=\gamma (t)-\gamma(\nu)\). That is,
$$\lim_{z\rightarrow t} \frac{1}{2\pi i} \int_{[\mu,\nu ]} \bigl[\gamma(\zeta) -\gamma(\nu) \bigr]\sum _{m,n} \bigl[q_{m,n}(\zeta,z)-q_{m,n}(\zeta , \overline{z}) \bigr]\,\mathrm{d} {\zeta}=\gamma(t)-\gamma(\nu), \quad t\in (\mu, \nu]. $$
Similarly,
$$\lim_{z\rightarrow t} \frac{1}{2\pi i} \int_{[\mu,\nu ]} \bigl[\gamma(\zeta) -\gamma(\mu) \bigr]\sum _{m,n} \bigl[q_{m,n}(\zeta,z)-q_{m,n}(\zeta , \overline{z}) \bigr]\,\mathrm{d} {\zeta}=\gamma(t)-\gamma(\mu), \quad t\in [\mu, \nu). $$
In the same way, when L and \(t_{0}\) are in all other cases of (27), the result (27) is also true.
Furthermore, by (17), we obtain for \(z\in[0,a]\cup[a,\mu)\cup (\nu,0]\) and \(\zeta\in[\mu,\nu]\), \(\sum_{m,n} [q_{m,n}(\zeta, z)-q_{m,n}(\zeta,\overline{z}) ]=0\), therefore, for \(t\in[0,a]\cup[a,\mu)\cup(\nu,0]\),
$$ \lim_{z\rightarrow t} \frac{1}{2\pi i} \int_{[\mu,\nu ]} \bigl[\gamma(\zeta)-\gamma(t) \bigr] \sum _{m,n} \bigl[q_{m,n}(\zeta,z)-q_{m,n}(\zeta, \overline{z}) \bigr]\,\mathrm{d} {\zeta}=0. $$
(29)
Combining the result (27), we obtain (29) is also true for corner points \(t=\mu,\nu\), hence (28) holds for \(L=[\mu ,\nu]\) and \(t\in\partial\Omega\setminus(\mu,\nu)\). Similarly, (28) is also true for L, t in the other cases. The proof is completed. □
Lemma 3.6
For
\(\gamma\in C(\partial\Omega,\mathbb{R})\)
and
\(t\in\partial\Omega\),
$$\lim_{ z\in\Omega, z\rightarrow t}\operatorname{Re}S_{\alpha}[\gamma ](z)= \gamma(t). $$
Proof
From (26), we only need to prove
$$ \lim_{ z\in\Omega, z\rightarrow t} \frac{1}{2\pi i} \int_{\partial\Omega}\gamma(\zeta) \sum_{m,n} \bigl[q_{m,n}(\zeta,z)-q_{m,n}(\zeta,\overline{z}) \bigr]\, \mathrm{d} {\zeta}=\gamma(t), \quad t\in\partial\Omega. $$
(30)
First of all, taking \(w(z)\equiv1\) and by (12), (13), we know
$$\frac{1}{2\pi i} \int_{\partial\Omega} Q_{M,N}(\zeta ,z)\,\mathrm{d}\zeta=1, \quad z\in\Omega $$
and
$$\frac{1}{2\pi i} \int_{\partial\Omega} Q_{M,N}(\zeta,\alpha )\,\mathrm{d}\zeta=1. $$
In particular, for \(\alpha\in\Omega\), \(\pm\overline{\alpha}+\omega _{mn}\) do not belong to Ω, so we have
$$\frac{1}{2\pi i} \int_{\partial\Omega}Q_{M,N}(\zeta ,\overline{\alpha})\,\mathrm{d} \zeta=0. $$
Combining the above three equations and taking \((M,N)\rightarrow(\infty ,\infty)\), thus
$$S_{\alpha}[1](z)= \frac{1}{\pi i} \int_{\partial\Omega} \sum_{m,n} \biggl[q_{m,n}(\zeta,z)- \frac{q_{m,n}(\zeta ,\alpha)+q_{m,n}(\zeta,\overline{\alpha})}{2} \biggr] \,\mathrm{d}\zeta=1. $$
Then from (26),
$$ \operatorname{Re}S_{\alpha}[1](z)= \frac{1}{2\pi i} \int_{\partial\Omega} \sum_{m,n} \bigl[q_{m,n}(\zeta,z)-q_{m,n}(\zeta,\overline{z}) \bigr]\, \mathrm{d} {\zeta}=1, \quad z\in\Omega, $$
(31)
hence,
$$\begin{aligned}& \lim_{ z\in\Omega, z\rightarrow t}\operatorname{Re}S_{\alpha}[ \gamma ](z) \\& \quad =\lim_{ z\in\Omega, z\rightarrow t} \frac{1}{2\pi i} \int_{\partial\Omega} \bigl[\gamma(\zeta)-\gamma(t) \bigr] \sum _{m,n} \bigl[q_{m,n}(\zeta,z)-q_{m,n}(\zeta, \overline{z}) \bigr]\,\mathrm{d} {\zeta}+\gamma(t). \end{aligned}$$
(32)
When \(t\in(\mu,\nu)\), we rewrite \(\operatorname{Re}S_{\alpha}[\gamma](z)\) as
$$\frac{1}{2\pi i} \int_{\partial\Omega} \bigl[\gamma(\zeta )-\gamma(\mu) \bigr] \sum _{m,n} \bigl[q_{m,n}(\zeta,z)-q_{m,n}(\zeta, \overline{z}) \bigr]\,\mathrm{d} {\zeta}+\gamma(\mu), $$
thus from the results for \(L=[\mu,\nu]\), \(t_{0}=\mu\) in (27) and for \(L=[0,a]\), \([a,\mu]\), and \([\nu,0]\) in (28), we get \(\lim_{ z\in\Omega, z\rightarrow t}\operatorname{Re}S_{\alpha}[\gamma ](z)=\gamma(t)\), \(t\in(\mu,\nu)\). Furthermore, by (32) and Lemma 3.5, (30) obviously holds for \(t=\mu,\nu\). Similarly, when \(t\in \partial\Omega\setminus[\mu,\nu]\), the result is also true. Then the proof is completed. □
Theorem 3.1
The Schwarz problem
$$ \left \{ \textstyle\begin{array}{l}\partial_{\overline{z}}w=f\quad \textit{in } \Omega,\qquad \operatorname{Re}w=\gamma\quad \textit{on } \partial\Omega, \\ \operatorname{Im}w(\alpha)=c, \quad c\in \mathbb{R}, \end{array}\displaystyle \right . $$
(33)
for
\(f\in L_{p}(\Omega;\mathbb{C})\), \(p>2\), \(\gamma\in C(\partial\Omega;\mathbb{R}) \), and
α
is a fixed point in Ω, is uniquely solvable by
$$ w(z)=S_{\alpha}[\gamma](z)+ A_{\alpha}[f](z)+ic, $$
(34)
where
\(A_{\alpha}\), \(S_{\alpha}\)
are defined by (24) and (25), respectively.
Proof
By Lemmas 3.2-3.4 and Lemma 3.6, \(\varphi(z)=S_{\alpha}[\gamma](z)+ A_{\alpha}[f](z)\) satisfies the first two conditions in (33). Suppose \(\phi (z)=w(z)-\varphi(z)\), then \(\phi(z)\) satisfies
$$\left \{ \textstyle\begin{array}{l} \partial_{\overline{z}}\phi=0 \quad \text{in } \Omega, \\ \operatorname{Re}\phi=0 \quad \text{on } \partial\Omega. \end{array}\displaystyle \right . $$
Then from Theorem 2.1, we know \(\phi(z)=i\operatorname{Im}w(\alpha)=ic\), thus, the proof is completed. □