= Solution
Fix $a\notin[\gamma]$ and let
$$
d=\operatorname{dist}(a,[\gamma])>0.
$$
For $|z-a|<d$ and $\lambda\in[\gamma]$, the <geometric series> gives
$$
\frac1{\lambda-z}
=\frac1{\lambda-a}\frac1{1-(z-a)/(\lambda-a)}
=\sum_{n=0}^{\infty}\frac{(z-a)^n}{(\lambda-a)^{n+1}}.
$$
The series converges uniformly on every smaller closed disc, so it may be integrated term by term along the curve. Hence
$$
\boxed{
f(z)=\sum_{n=0}^{\infty}c_n(z-a)^n,
\qquad
c_n=\int_\gamma\frac{\phi(\lambda)}
{(\lambda-a)^{n+1}}\,d\lambda
},
$$
which is a <power series> about $a$.
If $f$ is holomorphic on a neighbourhood of the closed disc $\overline D(a,r)$, the <Cauchy integral formula> says
$$
f(z)=\frac1{2\pi i}\int_{|\zeta-a|=r}
\frac{f(\zeta)}{\zeta-z}\,d\zeta,
\qquad |z-a|<r.
$$
Differentiation under the integral is valid uniformly on smaller discs and gives the <Cauchy derivative formula>
$$
\boxed{
f^{(n)}(z)=\frac{n!}{2\pi i}
\int_{|\zeta-a|=r}
\frac{f(\zeta)}{(\zeta-z)^{n+1}}\,d\zeta
}.
$$
This proves inductively that every holomorphic function has complex derivatives of every order. In particular,
$$
\boxed{
f'(z)=\frac1{2\pi i}
\int_{|\zeta-a|=r}\frac{f(\zeta)}{(\zeta-z)^2}\,d\zeta
}.
$$
Now suppose $f_n\to f$ <locally uniform convergence>[locally uniformly] on $U$. Given a compact set $K\subset U$, choose finitely many closed discs whose slightly larger concentric discs remain in $U$ and whose interiors cover $K$. Applying the derivative formula to $f_n-f$ on the larger boundary circles gives a uniform <Cauchy estimate>
$$
\sup_K|f_n'-f'|
\leq C\sup_L|f_n-f|,
$$
where $L\subset U$ is the compact union of those circles. The right-hand side tends to zero, proving that
$$
\boxed{f_n'\to f'\text{ locally uniformly}}.
$$
Finally, choose open neighbourhoods $U_j$ of the closed discs $D_j$ so small that
$$
U_1\cap U_2
$$
lies in the given neighbourhood on which $f$ is holomorphic. Inside that neighbourhood choose a positively oriented piecewise smooth contour $\Gamma$ surrounding $D_1\cap D_2$. It may be chosen as the boundary of a slightly enlarged lens and split into arcs
$$
\Gamma=\Gamma_1+\Gamma_2
$$
so that $\Gamma_1$ stays away from $D_1$ and $\Gamma_2$ stays away from $D_2$. Define
$$
f_1(z)=\frac1{2\pi i}\int_{\Gamma_1}
\frac{f(\zeta)}{\zeta-z}\,d\zeta
\quad\text{near }D_1,
$$
and
$$
f_2(z)=\frac1{2\pi i}\int_{\Gamma_2}
\frac{f(\zeta)}{\zeta-z}\,d\zeta
\quad\text{near }D_2.
$$
Because each arc avoids the corresponding disc, these formulas define holomorphic functions on possibly smaller neighbourhoods $U_1,U_2$. On their overlap, the <Cauchy integral formula> for the full contour gives
$$
\boxed{f=f_1+f_2}.
$$
Back to article page
