if $f_n(z)rightarrow f(z)$ uniformly, then $ frac{f'_n(z)}{f_n(z)}rightarrowfrac{f'(z)}{f(z)}$ uniformly?











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I am reading "The Taylor Series: An Introduction to the Theory of Functions of a Complex Variable", Dienes P.N., Dover (1957).



In the proof of a theorem (Hurwitz) on page 351 it says that, if $f_n(z)rightarrow f(z)$ uniformly, and given a circle C in wich $f(z)neq 0$ then
$$ frac{f'_n(z)}{f_n(z)}rightarrowfrac{f'(z)}{f(z)}$$ uniformly on C. And moreover,
$$frac{1}{2pi i}oint_C frac{f'_n(z)}{f_n(z)}dzrightarrowfrac{1}{2pi i}oint_C frac{f'(z)}{f(z)}dz$$



This statements are not exactly clear to me, is there any small proof to convince me?






complex-analysis convergence uniform-convergence







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    up vote
    0
    down vote

    favorite












    I am reading "The Taylor Series: An Introduction to the Theory of Functions of a Complex Variable", Dienes P.N., Dover (1957).



    In the proof of a theorem (Hurwitz) on page 351 it says that, if $f_n(z)rightarrow f(z)$ uniformly, and given a circle C in wich $f(z)neq 0$ then
    $$ frac{f'_n(z)}{f_n(z)}rightarrowfrac{f'(z)}{f(z)}$$ uniformly on C. And moreover,
    $$frac{1}{2pi i}oint_C frac{f'_n(z)}{f_n(z)}dzrightarrowfrac{1}{2pi i}oint_C frac{f'(z)}{f(z)}dz$$



    This statements are not exactly clear to me, is there any small proof to convince me?






    complex-analysis convergence uniform-convergence







    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I am reading "The Taylor Series: An Introduction to the Theory of Functions of a Complex Variable", Dienes P.N., Dover (1957).



      In the proof of a theorem (Hurwitz) on page 351 it says that, if $f_n(z)rightarrow f(z)$ uniformly, and given a circle C in wich $f(z)neq 0$ then
      $$ frac{f'_n(z)}{f_n(z)}rightarrowfrac{f'(z)}{f(z)}$$ uniformly on C. And moreover,
      $$frac{1}{2pi i}oint_C frac{f'_n(z)}{f_n(z)}dzrightarrowfrac{1}{2pi i}oint_C frac{f'(z)}{f(z)}dz$$



      This statements are not exactly clear to me, is there any small proof to convince me?






      complex-analysis convergence uniform-convergence







      share|cite|improve this question













      I am reading "The Taylor Series: An Introduction to the Theory of Functions of a Complex Variable", Dienes P.N., Dover (1957).



      In the proof of a theorem (Hurwitz) on page 351 it says that, if $f_n(z)rightarrow f(z)$ uniformly, and given a circle C in wich $f(z)neq 0$ then
      $$ frac{f'_n(z)}{f_n(z)}rightarrowfrac{f'(z)}{f(z)}$$ uniformly on C. And moreover,
      $$frac{1}{2pi i}oint_C frac{f'_n(z)}{f_n(z)}dzrightarrowfrac{1}{2pi i}oint_C frac{f'(z)}{f(z)}dz$$



      This statements are not exactly clear to me, is there any small proof to convince me?






      complex-analysis convergence uniform-convergence




      complex-analysis convergence uniform-convergence






      share|cite|improve this question













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      Ajafca

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          This uses the fact that when a sequence $(f_n)_{ninmathbb N}$ of analytic functions converges uniformly to $f$ on each compact set, then $(f_n{,'})_{ninmathbb N}$ also converges uniformly to $f'$ on each compact. This, together with the fact that $f$ has no zeros on $C$ (which implies that $f_n$ has no zeros there if $ngg1$) is enough to prove that statement.






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            This uses the fact that when a sequence $(f_n)_{ninmathbb N}$ of analytic functions converges uniformly to $f$ on each compact set, then $(f_n{,'})_{ninmathbb N}$ also converges uniformly to $f'$ on each compact. This, together with the fact that $f$ has no zeros on $C$ (which implies that $f_n$ has no zeros there if $ngg1$) is enough to prove that statement.






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              This uses the fact that when a sequence $(f_n)_{ninmathbb N}$ of analytic functions converges uniformly to $f$ on each compact set, then $(f_n{,'})_{ninmathbb N}$ also converges uniformly to $f'$ on each compact. This, together with the fact that $f$ has no zeros on $C$ (which implies that $f_n$ has no zeros there if $ngg1$) is enough to prove that statement.






              share|cite|improve this answer























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                up vote
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                This uses the fact that when a sequence $(f_n)_{ninmathbb N}$ of analytic functions converges uniformly to $f$ on each compact set, then $(f_n{,'})_{ninmathbb N}$ also converges uniformly to $f'$ on each compact. This, together with the fact that $f$ has no zeros on $C$ (which implies that $f_n$ has no zeros there if $ngg1$) is enough to prove that statement.






                share|cite|improve this answer












                This uses the fact that when a sequence $(f_n)_{ninmathbb N}$ of analytic functions converges uniformly to $f$ on each compact set, then $(f_n{,'})_{ninmathbb N}$ also converges uniformly to $f'$ on each compact. This, together with the fact that $f$ has no zeros on $C$ (which implies that $f_n$ has no zeros there if $ngg1$) is enough to prove that statement.







                share|cite|improve this answer












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                answered yesterday









                José Carlos Santos

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