Weierstrass double series theorem

If the complex functions  $f_0,f_1,f_2,\mathrm{\dots }$  are holomorphic in the disc    and thus

$f_n(z)={\displaystyle \sum _{\nu =0}^{\mathrm{\infty }}}{a}_{n\nu }{(z-z_0)}^{\nu },a_{n\nu }={\displaystyle \frac{f_n^{(\nu )}(z_0)}{\nu !}}\mathit{\hspace{1em}}\forall n,\nu$(1)

in this disc, and if the function series

${\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{f}_n=f_0+f_1+f_2+\mathrm{\dots }$(2)

converges uniformly to the function $F$ in each disc  $|z-z_0|\leqq \varrho$  where  ,  then also all the series

${\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{a}_{n\nu }=a_{0\nu }+a_{1\nu }+a_{2\nu }+\mathrm{\dots }\mathit{\hspace{1em}}(\nu =0,\mathrm{\hspace{0.17em}1},\mathrm{\hspace{0.17em}2},\mathrm{\dots })$(3)

converge, and in the disc    one has

$F(z)={\displaystyle \sum _{\nu =0}^{\mathrm{\infty }}}{A}_{\nu }{(z-z_0)}^{\nu }$(4)

where the $A_{\nu }$s are the sums of the series (3).

Proof.  Apparently, the series (2) converges uniformly also in every closed sub-disc of the open disc   .  Therefore the theorem 2 in the entry “theorems on complex function series (http://planetmath.org/TheoremsOnComplexFunctionSeries)” says that the sum $F(z)$ is holomorphic in    and

$$F^{(\nu )}(z)=f_0^{(\nu )}(z_0)+f_1^{(\nu )}(z_0)+f_2^{(\nu )}(z_0)+\mathrm{\dots }\mathit{\hspace{1em}}(\nu =0,\mathrm{\hspace{0.17em}1},\mathrm{\hspace{0.17em}2},\mathrm{\dots }).$$

Theorem 3 in the same entry thus guarantees that $F(z)$ has the Taylor expansion of the form (4) wherein

$$A_{\nu }=\frac{1}{\nu !}{F}^{(\nu )}(z_0)\mathit{\hspace{1em}}(\nu =0,\mathrm{\hspace{0.17em}1},\mathrm{\hspace{0.17em}2},\mathrm{\dots }).$$

According to theorem 2 in the same entry the series (2) may be differentiated termwise,

$$A_{\nu }=\frac{1}{\nu !}\sum _{n=0}^{\mathrm{\infty }}{f}_n^{(\nu )}(z_0)=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{\nu !}{f}_n^{(\nu )}(z_0)=\sum _{n=0}^{\mathrm{\infty }}{a}_{n\nu }$$

Q.E.D.

Note.  In Weierstrass double series theorem it’s a question of changing the summing :

$$\begin{array}{c}F(z)=f_0(z)+f_1(z)+\mathrm{\dots }+f_n(z)+\mathrm{\dots }=\hfill \\ =[a_{00}+a_{01}(z-z_0)+a_{02}{(z-z_0)}^2+\mathrm{\dots }+a_{0\nu }{(z-z_0)}^{\nu }+\mathrm{\dots }]\hfill \\ +[a_{10}+a_{11}(z-z_0)+a_{12}{(z-z_0)}^2+\mathrm{\dots }+a_{1\nu }{(z-z_0)}^{\nu }+\mathrm{\dots }]\hfill \\ +[a_{20}+a_{21}(z-z_0)+a_{22}{(z-z_0)}^2+\mathrm{\dots }+a_{2\nu }{(z-z_0)}^{\nu }+\mathrm{\dots }]\hfill \\ \mathrm{\dots }\mathrm{\dots }\hfill \\ +[a_{n0}+a_{n1}(z-z_0)+a_{n2}{(z-z_0)}^2+\mathrm{\dots }+a_{n\nu }{(z-z_0)}^{\nu }+\mathrm{\dots }]\hfill \\ \underset{¯}{\mathrm{\cdots }\mathrm{\cdots }\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}\hfill \\ =A_0+A_1(z-z_0)+A_2{(z-z_0)}^2+\mathrm{\dots }+A_{\nu }{(z-z_0)}^{\nu }+\mathrm{\dots }\hfill \end{array}$$