Cauchy's formula proof: why can we say $iint_0^{2pi} f(z_0+epsilon e^{itheta})dtheta rightarrow2pi i f(z_0)$...












1












$begingroup$


For $n=1$ the proof of Cauchy's formula



$$f^{(n)}(z_0)={n!over2pi i}intlimits_gammafrac{f(z)}{(z-z_0)^{n+1}}dz$$



defines $g(z)={f(z)over z-z_0}$ given the closed path $gamma$ around $z_0$ in the open subset $OmegasubsetBbb C$ and a holomorphic function $f:OmegamapstoBbb C$ with $epsilon>0$ small enough so that $overline{B_epsilon(z_0)}$ is inside $Omega$ using another theorem and since $mathcal O=Omegabackslash B_epsilon(z_0)$ is an open domain and $g$ is holomorphic on it, $intlimits_{mathcal O}g(z)dz=0$



From that we eventually get $intlimits_gamma frac{f(z)}{z-z_0}dz=iintlimits_0^{2pi} f(z_0+epsilon e^{itheta})dtheta xrightarrow[epsilonto0]{}2pi i f(z_0)$



Why can we move the limit $xrightarrow[epsilonto0]{}$ inside the integral?






complex-analysis limits







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    For $n=1$ the proof of Cauchy's formula



    $$f^{(n)}(z_0)={n!over2pi i}intlimits_gammafrac{f(z)}{(z-z_0)^{n+1}}dz$$



    defines $g(z)={f(z)over z-z_0}$ given the closed path $gamma$ around $z_0$ in the open subset $OmegasubsetBbb C$ and a holomorphic function $f:OmegamapstoBbb C$ with $epsilon>0$ small enough so that $overline{B_epsilon(z_0)}$ is inside $Omega$ using another theorem and since $mathcal O=Omegabackslash B_epsilon(z_0)$ is an open domain and $g$ is holomorphic on it, $intlimits_{mathcal O}g(z)dz=0$



    From that we eventually get $intlimits_gamma frac{f(z)}{z-z_0}dz=iintlimits_0^{2pi} f(z_0+epsilon e^{itheta})dtheta xrightarrow[epsilonto0]{}2pi i f(z_0)$



    Why can we move the limit $xrightarrow[epsilonto0]{}$ inside the integral?






    complex-analysis limits







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      For $n=1$ the proof of Cauchy's formula



      $$f^{(n)}(z_0)={n!over2pi i}intlimits_gammafrac{f(z)}{(z-z_0)^{n+1}}dz$$



      defines $g(z)={f(z)over z-z_0}$ given the closed path $gamma$ around $z_0$ in the open subset $OmegasubsetBbb C$ and a holomorphic function $f:OmegamapstoBbb C$ with $epsilon>0$ small enough so that $overline{B_epsilon(z_0)}$ is inside $Omega$ using another theorem and since $mathcal O=Omegabackslash B_epsilon(z_0)$ is an open domain and $g$ is holomorphic on it, $intlimits_{mathcal O}g(z)dz=0$



      From that we eventually get $intlimits_gamma frac{f(z)}{z-z_0}dz=iintlimits_0^{2pi} f(z_0+epsilon e^{itheta})dtheta xrightarrow[epsilonto0]{}2pi i f(z_0)$



      Why can we move the limit $xrightarrow[epsilonto0]{}$ inside the integral?






      complex-analysis limits







      share|cite|improve this question











      $endgroup$




      For $n=1$ the proof of Cauchy's formula



      $$f^{(n)}(z_0)={n!over2pi i}intlimits_gammafrac{f(z)}{(z-z_0)^{n+1}}dz$$



      defines $g(z)={f(z)over z-z_0}$ given the closed path $gamma$ around $z_0$ in the open subset $OmegasubsetBbb C$ and a holomorphic function $f:OmegamapstoBbb C$ with $epsilon>0$ small enough so that $overline{B_epsilon(z_0)}$ is inside $Omega$ using another theorem and since $mathcal O=Omegabackslash B_epsilon(z_0)$ is an open domain and $g$ is holomorphic on it, $intlimits_{mathcal O}g(z)dz=0$



      From that we eventually get $intlimits_gamma frac{f(z)}{z-z_0}dz=iintlimits_0^{2pi} f(z_0+epsilon e^{itheta})dtheta xrightarrow[epsilonto0]{}2pi i f(z_0)$



      Why can we move the limit $xrightarrow[epsilonto0]{}$ inside the integral?






      complex-analysis limits




      complex-analysis limits






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 30 '18 at 14:33









      Henning Makholm

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      241k17307546










      asked Dec 30 '18 at 12:50









      John CataldoJohn Cataldo

      1,1881316




      1,1881316






















          3 Answers
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          active

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          1












          $begingroup$

          You want to show that: $iintlimits_0^{2pi} f(z_0+epsilon e^{itheta})dtheta xrightarrow[epsilonto0]{}2pi i f(z_0)$. That is:



          For any $delta >0$. There is $eta>0 $ such that $left | int_{0}^{2pi}f(z_0+epsilon e^{itheta })dtheta -2pi f(z_0)right |< delta $ whenever $left | epsilon right |leq eta $.



          But:
          $left | int_{0}^{2pi}f(z_0+epsilon e^{itheta })dtheta -2pi f(z_0)right |=left | int_{0}^{2pi}(f(z_0+epsilon e^{itheta }) - f(z_0))dthetaright | leq int_{0}^{2pi}left |f(z_0+epsilon e^{itheta }) - f(z_0) right |dtheta $



          Note that: $left | z_0+epsilon e^{itheta }-z_0 right |=epsilon $



          Now by continuity of $f$ at $z_0$, We can find $eta$ such that:



          $left |f(z) - f(z_0) right |leq frac{delta }{2pi}$ when $left | z-z_0 right |leq eta $



          Can you take it from here?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you this is very helpful
            $endgroup$
            – John Cataldo
            Dec 30 '18 at 15:24



















          2












          $begingroup$

          We can use Lebesgue dominated convergence theorem.






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            By the continuity of $f$, as $epsilonto 0$ $$f(z_0+epsilon e^{itheta}) to f(z_0)$$ uniformly over $thetain [0,2pi]$.






            share|cite|improve this answer









            $endgroup$













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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              1












              $begingroup$

              You want to show that: $iintlimits_0^{2pi} f(z_0+epsilon e^{itheta})dtheta xrightarrow[epsilonto0]{}2pi i f(z_0)$. That is:



              For any $delta >0$. There is $eta>0 $ such that $left | int_{0}^{2pi}f(z_0+epsilon e^{itheta })dtheta -2pi f(z_0)right |< delta $ whenever $left | epsilon right |leq eta $.



              But:
              $left | int_{0}^{2pi}f(z_0+epsilon e^{itheta })dtheta -2pi f(z_0)right |=left | int_{0}^{2pi}(f(z_0+epsilon e^{itheta }) - f(z_0))dthetaright | leq int_{0}^{2pi}left |f(z_0+epsilon e^{itheta }) - f(z_0) right |dtheta $



              Note that: $left | z_0+epsilon e^{itheta }-z_0 right |=epsilon $



              Now by continuity of $f$ at $z_0$, We can find $eta$ such that:



              $left |f(z) - f(z_0) right |leq frac{delta }{2pi}$ when $left | z-z_0 right |leq eta $



              Can you take it from here?






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Thank you this is very helpful
                $endgroup$
                – John Cataldo
                Dec 30 '18 at 15:24
















              1












              $begingroup$

              You want to show that: $iintlimits_0^{2pi} f(z_0+epsilon e^{itheta})dtheta xrightarrow[epsilonto0]{}2pi i f(z_0)$. That is:



              For any $delta >0$. There is $eta>0 $ such that $left | int_{0}^{2pi}f(z_0+epsilon e^{itheta })dtheta -2pi f(z_0)right |< delta $ whenever $left | epsilon right |leq eta $.



              But:
              $left | int_{0}^{2pi}f(z_0+epsilon e^{itheta })dtheta -2pi f(z_0)right |=left | int_{0}^{2pi}(f(z_0+epsilon e^{itheta }) - f(z_0))dthetaright | leq int_{0}^{2pi}left |f(z_0+epsilon e^{itheta }) - f(z_0) right |dtheta $



              Note that: $left | z_0+epsilon e^{itheta }-z_0 right |=epsilon $



              Now by continuity of $f$ at $z_0$, We can find $eta$ such that:



              $left |f(z) - f(z_0) right |leq frac{delta }{2pi}$ when $left | z-z_0 right |leq eta $



              Can you take it from here?






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Thank you this is very helpful
                $endgroup$
                – John Cataldo
                Dec 30 '18 at 15:24














              1












              1








              1





              $begingroup$

              You want to show that: $iintlimits_0^{2pi} f(z_0+epsilon e^{itheta})dtheta xrightarrow[epsilonto0]{}2pi i f(z_0)$. That is:



              For any $delta >0$. There is $eta>0 $ such that $left | int_{0}^{2pi}f(z_0+epsilon e^{itheta })dtheta -2pi f(z_0)right |< delta $ whenever $left | epsilon right |leq eta $.



              But:
              $left | int_{0}^{2pi}f(z_0+epsilon e^{itheta })dtheta -2pi f(z_0)right |=left | int_{0}^{2pi}(f(z_0+epsilon e^{itheta }) - f(z_0))dthetaright | leq int_{0}^{2pi}left |f(z_0+epsilon e^{itheta }) - f(z_0) right |dtheta $



              Note that: $left | z_0+epsilon e^{itheta }-z_0 right |=epsilon $



              Now by continuity of $f$ at $z_0$, We can find $eta$ such that:



              $left |f(z) - f(z_0) right |leq frac{delta }{2pi}$ when $left | z-z_0 right |leq eta $



              Can you take it from here?






              share|cite|improve this answer









              $endgroup$



              You want to show that: $iintlimits_0^{2pi} f(z_0+epsilon e^{itheta})dtheta xrightarrow[epsilonto0]{}2pi i f(z_0)$. That is:



              For any $delta >0$. There is $eta>0 $ such that $left | int_{0}^{2pi}f(z_0+epsilon e^{itheta })dtheta -2pi f(z_0)right |< delta $ whenever $left | epsilon right |leq eta $.



              But:
              $left | int_{0}^{2pi}f(z_0+epsilon e^{itheta })dtheta -2pi f(z_0)right |=left | int_{0}^{2pi}(f(z_0+epsilon e^{itheta }) - f(z_0))dthetaright | leq int_{0}^{2pi}left |f(z_0+epsilon e^{itheta }) - f(z_0) right |dtheta $



              Note that: $left | z_0+epsilon e^{itheta }-z_0 right |=epsilon $



              Now by continuity of $f$ at $z_0$, We can find $eta$ such that:



              $left |f(z) - f(z_0) right |leq frac{delta }{2pi}$ when $left | z-z_0 right |leq eta $



              Can you take it from here?







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 30 '18 at 14:29









              John11John11

              1,0321821




              1,0321821












              • $begingroup$
                Thank you this is very helpful
                $endgroup$
                – John Cataldo
                Dec 30 '18 at 15:24


















              • $begingroup$
                Thank you this is very helpful
                $endgroup$
                – John Cataldo
                Dec 30 '18 at 15:24
















              $begingroup$
              Thank you this is very helpful
              $endgroup$
              – John Cataldo
              Dec 30 '18 at 15:24




              $begingroup$
              Thank you this is very helpful
              $endgroup$
              – John Cataldo
              Dec 30 '18 at 15:24











              2












              $begingroup$

              We can use Lebesgue dominated convergence theorem.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                We can use Lebesgue dominated convergence theorem.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  We can use Lebesgue dominated convergence theorem.






                  share|cite|improve this answer









                  $endgroup$



                  We can use Lebesgue dominated convergence theorem.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 30 '18 at 12:55









                  Trần Quang MinhTrần Quang Minh

                  3337




                  3337























                      2












                      $begingroup$

                      By the continuity of $f$, as $epsilonto 0$ $$f(z_0+epsilon e^{itheta}) to f(z_0)$$ uniformly over $thetain [0,2pi]$.






                      share|cite|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        By the continuity of $f$, as $epsilonto 0$ $$f(z_0+epsilon e^{itheta}) to f(z_0)$$ uniformly over $thetain [0,2pi]$.






                        share|cite|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          By the continuity of $f$, as $epsilonto 0$ $$f(z_0+epsilon e^{itheta}) to f(z_0)$$ uniformly over $thetain [0,2pi]$.






                          share|cite|improve this answer









                          $endgroup$



                          By the continuity of $f$, as $epsilonto 0$ $$f(z_0+epsilon e^{itheta}) to f(z_0)$$ uniformly over $thetain [0,2pi]$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 30 '18 at 13:01









                          SongSong

                          15.4k1736




                          15.4k1736






























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