Analytic continuation of several complex variables












2












$begingroup$


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 $a_1,ldots,a_n,z$?










share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    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 $a_1,ldots,a_n,z$?










    share|cite|improve this question











    $endgroup$















      2












      2








      2


      2



      $begingroup$


      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 $a_1,ldots,a_n,z$?










      share|cite|improve this question











      $endgroup$




      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 $a_1,ldots,a_n,z$?







      complex-analysis ordinary-differential-equations several-complex-variables analytic-continuation






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 18 '18 at 8:26









      Daniele Tampieri

      2,3272922




      2,3272922










      asked Aug 12 '18 at 5:31









      user356126user356126

      1306




      1306






















          1 Answer
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          $begingroup$

          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}} $.
            Analytic continuation of $u$.
            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:
            Covering the entire domain where the function is analytically continuable.



          • $v$ 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 $D_v$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for your answer. I didn't hit upon the idea of using Taylor series.
            $endgroup$
            – user356126
            Aug 17 '18 at 10:03










          • $begingroup$
            You're welcome. Using this kind of argument is a standard way of proving results on analytic continuation in the elementary theory of holomorphic functions of several variables: and as such you probably already noticed its main problem. We have almost no information on the size of $D_v$, i.e. the analytic continuation results obtained in this way do not say much on the structure of the domain where the function can be continued.
            $endgroup$
            – Daniele Tampieri
            Aug 17 '18 at 14:56











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          1 Answer
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          1 Answer
          1






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          active

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          2












          $begingroup$

          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}} $.
            Analytic continuation of $u$.
            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:
            Covering the entire domain where the function is analytically continuable.



          • $v$ 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 $D_v$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for your answer. I didn't hit upon the idea of using Taylor series.
            $endgroup$
            – user356126
            Aug 17 '18 at 10:03










          • $begingroup$
            You're welcome. Using this kind of argument is a standard way of proving results on analytic continuation in the elementary theory of holomorphic functions of several variables: and as such you probably already noticed its main problem. We have almost no information on the size of $D_v$, i.e. the analytic continuation results obtained in this way do not say much on the structure of the domain where the function can be continued.
            $endgroup$
            – Daniele Tampieri
            Aug 17 '18 at 14:56
















          2












          $begingroup$

          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}} $.
            Analytic continuation of $u$.
            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:
            Covering the entire domain where the function is analytically continuable.



          • $v$ 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 $D_v$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for your answer. I didn't hit upon the idea of using Taylor series.
            $endgroup$
            – user356126
            Aug 17 '18 at 10:03










          • $begingroup$
            You're welcome. Using this kind of argument is a standard way of proving results on analytic continuation in the elementary theory of holomorphic functions of several variables: and as such you probably already noticed its main problem. We have almost no information on the size of $D_v$, i.e. the analytic continuation results obtained in this way do not say much on the structure of the domain where the function can be continued.
            $endgroup$
            – Daniele Tampieri
            Aug 17 '18 at 14:56














          2












          2








          2





          $begingroup$

          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}} $.
            Analytic continuation of $u$.
            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:
            Covering the entire domain where the function is analytically continuable.



          • $v$ 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 $D_v$.






          share|cite|improve this answer











          $endgroup$



          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}} $.
            Analytic continuation of $u$.
            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:
            Covering the entire domain where the function is analytically continuable.



          • $v$ 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 $D_v$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Aug 14 '18 at 9:42

























          answered Aug 12 '18 at 20:06









          Daniele TampieriDaniele Tampieri

          2,3272922




          2,3272922












          • $begingroup$
            Thank you for your answer. I didn't hit upon the idea of using Taylor series.
            $endgroup$
            – user356126
            Aug 17 '18 at 10:03










          • $begingroup$
            You're welcome. Using this kind of argument is a standard way of proving results on analytic continuation in the elementary theory of holomorphic functions of several variables: and as such you probably already noticed its main problem. We have almost no information on the size of $D_v$, i.e. the analytic continuation results obtained in this way do not say much on the structure of the domain where the function can be continued.
            $endgroup$
            – Daniele Tampieri
            Aug 17 '18 at 14:56


















          • $begingroup$
            Thank you for your answer. I didn't hit upon the idea of using Taylor series.
            $endgroup$
            – user356126
            Aug 17 '18 at 10:03










          • $begingroup$
            You're welcome. Using this kind of argument is a standard way of proving results on analytic continuation in the elementary theory of holomorphic functions of several variables: and as such you probably already noticed its main problem. We have almost no information on the size of $D_v$, i.e. the analytic continuation results obtained in this way do not say much on the structure of the domain where the function can be continued.
            $endgroup$
            – Daniele Tampieri
            Aug 17 '18 at 14:56
















          $begingroup$
          Thank you for your answer. I didn't hit upon the idea of using Taylor series.
          $endgroup$
          – user356126
          Aug 17 '18 at 10:03




          $begingroup$
          Thank you for your answer. I didn't hit upon the idea of using Taylor series.
          $endgroup$
          – user356126
          Aug 17 '18 at 10:03












          $begingroup$
          You're welcome. Using this kind of argument is a standard way of proving results on analytic continuation in the elementary theory of holomorphic functions of several variables: and as such you probably already noticed its main problem. We have almost no information on the size of $D_v$, i.e. the analytic continuation results obtained in this way do not say much on the structure of the domain where the function can be continued.
          $endgroup$
          – Daniele Tampieri
          Aug 17 '18 at 14:56




          $begingroup$
          You're welcome. Using this kind of argument is a standard way of proving results on analytic continuation in the elementary theory of holomorphic functions of several variables: and as such you probably already noticed its main problem. We have almost no information on the size of $D_v$, i.e. the analytic continuation results obtained in this way do not say much on the structure of the domain where the function can be continued.
          $endgroup$
          – Daniele Tampieri
          Aug 17 '18 at 14:56


















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