A generalization of maximum modulus principle












5












$begingroup$


Let $ emptyset neq U subset mathbb{C} $ be a bounded open connected set and let $ f_1, dots, f_n $ be analytic in $ overline{U} $. Prove that
$$ max_{z in overline{U}} sum_{j=1}^n |f_j(z) | = max_{z in partial U} sum_{j=1}^n |f_j(z) |. $$
Clearly we have "$geq$" but I don't know how to reduce to the $ n = 1 $ case to use the usual maximum modulus principle. Help is appreciated.






complex-analysis







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    5












    $begingroup$


    Let $ emptyset neq U subset mathbb{C} $ be a bounded open connected set and let $ f_1, dots, f_n $ be analytic in $ overline{U} $. Prove that
    $$ max_{z in overline{U}} sum_{j=1}^n |f_j(z) | = max_{z in partial U} sum_{j=1}^n |f_j(z) |. $$
    Clearly we have "$geq$" but I don't know how to reduce to the $ n = 1 $ case to use the usual maximum modulus principle. Help is appreciated.






    complex-analysis







    share|cite|improve this question









    $endgroup$















      5












      5








      5





      $begingroup$


      Let $ emptyset neq U subset mathbb{C} $ be a bounded open connected set and let $ f_1, dots, f_n $ be analytic in $ overline{U} $. Prove that
      $$ max_{z in overline{U}} sum_{j=1}^n |f_j(z) | = max_{z in partial U} sum_{j=1}^n |f_j(z) |. $$
      Clearly we have "$geq$" but I don't know how to reduce to the $ n = 1 $ case to use the usual maximum modulus principle. Help is appreciated.






      complex-analysis







      share|cite|improve this question









      $endgroup$




      Let $ emptyset neq U subset mathbb{C} $ be a bounded open connected set and let $ f_1, dots, f_n $ be analytic in $ overline{U} $. Prove that
      $$ max_{z in overline{U}} sum_{j=1}^n |f_j(z) | = max_{z in partial U} sum_{j=1}^n |f_j(z) |. $$
      Clearly we have "$geq$" but I don't know how to reduce to the $ n = 1 $ case to use the usual maximum modulus principle. Help is appreciated.






      complex-analysis




      complex-analysis






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      asked Jan 7 at 5:30









      lifeishard911lifeishard911

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          $begingroup$

          I will give a proof for $n=2$ because I have it typed out already. The same argument works for any $n$: given $f$ and $g$ analytic such that the maximum of $|f|+|g|$ on $overline {U}$ is attained at an interior point $a$ we have $leftvert f(a)rightvert +leftvert g(a)rightvert geq leftvert
          f(z)rightvert +leftvert g(z)rightvert $           $forall zin Omega .$
          Replace $f$ by $e^{is}f$ and $g$ by $e^{it}g$ where $s$ and $t$ are chosen
          such that $e^{is}f(a)$ and $e^{it}g(a)$ both belong to $[0,infty ).$ This
          reduces the proof to the case when $f(a)$ and $g(a)$ both belong to $%
          [0,infty ).$           We now have $$f(a)+g(a)geq leftvert f(z)rightvert
          +leftvert g(z)rightvert geq |f(z)+g(z)|$$           Maximum Modulus principle applied to $f+g$ shows that $f+g$ is
          a constant. Now $$f(a)+g(a)geq leftvert f(z)rightvert +leftvert
          g(z)rightvert geq Re f(z)+Re g(z)=Re(f(z)+g(z))$$           $$=Re (f(a)+g(a))$$ which implies that equality holds throughout. In particular $%
          leftvert f(z)rightvert =Re(f(z))$           and $leftvert g(z)rightvert
          =Re(g(z))$           for all z]. Hence $f$ and $g$ are both constants (because their imaginary parts vanish) in which case there is noting to prove.






          share|cite|improve this answer











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            $begingroup$

            I will give a proof for $n=2$ because I have it typed out already. The same argument works for any $n$: given $f$ and $g$ analytic such that the maximum of $|f|+|g|$ on $overline {U}$ is attained at an interior point $a$ we have $leftvert f(a)rightvert +leftvert g(a)rightvert geq leftvert
            f(z)rightvert +leftvert g(z)rightvert $             $forall zin Omega .$
            Replace $f$ by $e^{is}f$ and $g$ by $e^{it}g$ where $s$ and $t$ are chosen
            such that $e^{is}f(a)$ and $e^{it}g(a)$ both belong to $[0,infty ).$ This
            reduces the proof to the case when $f(a)$ and $g(a)$ both belong to $%
            [0,infty ).$             We now have $$f(a)+g(a)geq leftvert f(z)rightvert
            +leftvert g(z)rightvert geq |f(z)+g(z)|$$             Maximum Modulus principle applied to $f+g$ shows that $f+g$ is
            a constant. Now $$f(a)+g(a)geq leftvert f(z)rightvert +leftvert
            g(z)rightvert geq Re f(z)+Re g(z)=Re(f(z)+g(z))$$             $$=Re (f(a)+g(a))$$ which implies that equality holds throughout. In particular $%
            leftvert f(z)rightvert =Re(f(z))$             and $leftvert g(z)rightvert
            =Re(g(z))$             for all z]. Hence $f$ and $g$ are both constants (because their imaginary parts vanish) in which case there is noting to prove.






            share|cite|improve this answer











            $endgroup$


















              4












              $begingroup$

              I will give a proof for $n=2$ because I have it typed out already. The same argument works for any $n$: given $f$ and $g$ analytic such that the maximum of $|f|+|g|$ on $overline {U}$ is attained at an interior point $a$ we have $leftvert f(a)rightvert +leftvert g(a)rightvert geq leftvert
              f(z)rightvert +leftvert g(z)rightvert $               $forall zin Omega .$
              Replace $f$ by $e^{is}f$ and $g$ by $e^{it}g$ where $s$ and $t$ are chosen
              such that $e^{is}f(a)$ and $e^{it}g(a)$ both belong to $[0,infty ).$ This
              reduces the proof to the case when $f(a)$ and $g(a)$ both belong to $%
              [0,infty ).$               We now have $$f(a)+g(a)geq leftvert f(z)rightvert
              +leftvert g(z)rightvert geq |f(z)+g(z)|$$               Maximum Modulus principle applied to $f+g$ shows that $f+g$ is
              a constant. Now $$f(a)+g(a)geq leftvert f(z)rightvert +leftvert
              g(z)rightvert geq Re f(z)+Re g(z)=Re(f(z)+g(z))$$               $$=Re (f(a)+g(a))$$ which implies that equality holds throughout. In particular $%
              leftvert f(z)rightvert =Re(f(z))$               and $leftvert g(z)rightvert
              =Re(g(z))$               for all z]. Hence $f$ and $g$ are both constants (because their imaginary parts vanish) in which case there is noting to prove.






              share|cite|improve this answer











              $endgroup$
















                4












                4








                4





                $begingroup$

                I will give a proof for $n=2$ because I have it typed out already. The same argument works for any $n$: given $f$ and $g$ analytic such that the maximum of $|f|+|g|$ on $overline {U}$ is attained at an interior point $a$ we have $leftvert f(a)rightvert +leftvert g(a)rightvert geq leftvert
                f(z)rightvert +leftvert g(z)rightvert $                 $forall zin Omega .$
                Replace $f$ by $e^{is}f$ and $g$ by $e^{it}g$ where $s$ and $t$ are chosen
                such that $e^{is}f(a)$ and $e^{it}g(a)$ both belong to $[0,infty ).$ This
                reduces the proof to the case when $f(a)$ and $g(a)$ both belong to $%
                [0,infty ).$                 We now have $$f(a)+g(a)geq leftvert f(z)rightvert
                +leftvert g(z)rightvert geq |f(z)+g(z)|$$                 Maximum Modulus principle applied to $f+g$ shows that $f+g$ is
                a constant. Now $$f(a)+g(a)geq leftvert f(z)rightvert +leftvert
                g(z)rightvert geq Re f(z)+Re g(z)=Re(f(z)+g(z))$$                 $$=Re (f(a)+g(a))$$ which implies that equality holds throughout. In particular $%
                leftvert f(z)rightvert =Re(f(z))$                 and $leftvert g(z)rightvert
                =Re(g(z))$                 for all z]. Hence $f$ and $g$ are both constants (because their imaginary parts vanish) in which case there is noting to prove.






                share|cite|improve this answer











                $endgroup$



                I will give a proof for $n=2$ because I have it typed out already. The same argument works for any $n$: given $f$ and $g$ analytic such that the maximum of $|f|+|g|$ on $overline {U}$ is attained at an interior point $a$ we have $leftvert f(a)rightvert +leftvert g(a)rightvert geq leftvert
                f(z)rightvert +leftvert g(z)rightvert $                 $forall zin Omega .$
                Replace $f$ by $e^{is}f$ and $g$ by $e^{it}g$ where $s$ and $t$ are chosen
                such that $e^{is}f(a)$ and $e^{it}g(a)$ both belong to $[0,infty ).$ This
                reduces the proof to the case when $f(a)$ and $g(a)$ both belong to $%
                [0,infty ).$                 We now have $$f(a)+g(a)geq leftvert f(z)rightvert
                +leftvert g(z)rightvert geq |f(z)+g(z)|$$                 Maximum Modulus principle applied to $f+g$ shows that $f+g$ is
                a constant. Now $$f(a)+g(a)geq leftvert f(z)rightvert +leftvert
                g(z)rightvert geq Re f(z)+Re g(z)=Re(f(z)+g(z))$$                 $$=Re (f(a)+g(a))$$ which implies that equality holds throughout. In particular $%
                leftvert f(z)rightvert =Re(f(z))$                 and $leftvert g(z)rightvert
                =Re(g(z))$                 for all z]. Hence $f$ and $g$ are both constants (because their imaginary parts vanish) in which case there is noting to prove.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 7 at 6:41

























                answered Jan 7 at 6:28









                Kavi Rama MurthyKavi Rama Murthy

                56k42158




                56k42158






























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