Krdytkyu

Let $f(w_1,ldots,w_n;z)$ be a holomorphic function of $n+1$ variables.
For every fixed $w_1ldots w_n$, let $g(w_1,ldots,w_n;z)$ be an analytic continuation of $f$ as a holomorphic function of $z$.
Of course, $g$ is holomorphic in $z$.
Now, is $g$ holomorphic in $w_1,ldots, w_n,z$ as well?

In the concrete, I am interested in the following situation.
Let $E(a_1,ldots,a_n;z)$ be an ODE, where $z$ is a complex variable and $a_1,ldots,a_n$ are complex parameters.
Assume that $E(a_1,ldots,a_n;z)$ has a local solution $u(a_1,ldots,a_n;z)$ which is holomorphic in $a_1,ldots,a_n,z$, and $v(a_1,ldots,a_n;z)$ is an analytic continuation of $u$.
Is $v$ holomorphic in $a_1,ldots,a_n,z$?

The answer is yes and it is a consequence of the classical Hartogs‘s theorem on separately holomorphic functions. Precisely, let $u(a_1,dots,a_n,z)$ be a holomorphic function on a domain $D_uinmathbb{C}^{n+1}$ which is analytically continuable to a larger domain respect to the variable $z$ as a function $v(a_1,dots,a_n,z)$: then

• 
  

  $v$ is separately holomorphic respect to $a_1,dots,a_n$ for every $znotin D_u$. To see this, choose $z_0in D_ucap{zinmathbb{C}}$ such that the Taylor series expansion of $v$ in $z_0$ has a convergence disk not entirely contained in $D_ucap{zinmathbb{C}}$ and whose of radius $R_{z_0}geq c>0$ does not depend on $a_1,dots,a_n$. Such $z_0$ exists, since assuming the contrary would deny the possibility of analytically continue $u$ respect to $z$ outside $D_ucap{zinmathbb{C}} $.
  
  Then, evaluating this Taylor series at fixed
  point $z_1notin D_ucap{zinmathbb{C}} $ inside its radius of convergence, we get
  $$
  begin{split}
  v(a_1,dots,a_n, z_1)&=sum_{k=0}^inftyfrac{1}{k!} frac{partial^k v}{partial z^k}(a_1,dots,a_n, z_0)(z_1-z_0)^k\
  &=sum_{k=0}^inftyfrac{1}{k!} frac{partial^k u}{partial z^k}(a_1,dots,a_n, z_0)(z_1-z_0)^k
  end{split}tag{1}label{1}
  $$
  since $v=u$ on $D_u$. Now eqref{1} implies that the $N$-th order Taylor polynomial
  $$
  v_N(dots,a_j,dots,z_1)=sum_{k=0}^Nfrac{1}{k!} frac{partial^k u}{partial z^k}(dots,a_j,dots, z_0)(z_1-z_0)^ktag{2}label{2}
  $$

  
  
  

  can be considered as a sequence of holomorphic functions in each single variable $a_j$, converging uniformly (thanks to the holomophicity of $u$ and to the uniform convergence of eqref{1}) to $v$. This suffices to prove the separate analyticity of $v$, analytic continuation of $u$, respect to all its variables in the domain
  $$
  D_ucup{(a_1,dots,a_n,z)|0<|z-z_0|leq |z_1-z_0|}.
  $$
  The same argument can be repeated to cover all the domain $D_v$, as quckly sketched in the following picture:
  

  
  


• $v$ is separately holomorphic respect to each of the variables $a_1,dots,a_n, zin D_vvarsupsetneq D_u$ thus, by Hartogs’s theorem, it is jointly holomorphic respect to all its variables on $D_v$.