Proof of Schwarz lemma 5












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I dont understand the following proof of Schwarz lemma:
Schwarz Lemma
In the last section it says, "Moreover, suppose that $|f(z)| = |z|$ for some non-zero $z$ in $D$, or $|f'(0)| = 1$. Then, $|g(z)| = 1$ at some point of $D$".
Why does $g(z)$ have a local maximum then?






maxima-minima







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  • $begingroup$
    There is a proof following it, what is the question?
    $endgroup$
    – Hawk
    Dec 12 '18 at 11:52
















0












$begingroup$


I dont understand the following proof of Schwarz lemma:
Schwarz Lemma
In the last section it says, "Moreover, suppose that $|f(z)| = |z|$ for some non-zero $z$ in $D$, or $|f'(0)| = 1$. Then, $|g(z)| = 1$ at some point of $D$".
Why does $g(z)$ have a local maximum then?






maxima-minima







share|cite|improve this question











$endgroup$












  • $begingroup$
    There is a proof following it, what is the question?
    $endgroup$
    – Hawk
    Dec 12 '18 at 11:52














0












0








0





$begingroup$


I dont understand the following proof of Schwarz lemma:
Schwarz Lemma
In the last section it says, "Moreover, suppose that $|f(z)| = |z|$ for some non-zero $z$ in $D$, or $|f'(0)| = 1$. Then, $|g(z)| = 1$ at some point of $D$".
Why does $g(z)$ have a local maximum then?






maxima-minima







share|cite|improve this question











$endgroup$




I dont understand the following proof of Schwarz lemma:
Schwarz Lemma
In the last section it says, "Moreover, suppose that $|f(z)| = |z|$ for some non-zero $z$ in $D$, or $|f'(0)| = 1$. Then, $|g(z)| = 1$ at some point of $D$".
Why does $g(z)$ have a local maximum then?






maxima-minima




maxima-minima






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 11:28









amWhy

1




1










asked Dec 12 '18 at 10:51









Steven33Steven33

275




275












  • $begingroup$
    There is a proof following it, what is the question?
    $endgroup$
    – Hawk
    Dec 12 '18 at 11:52


















  • $begingroup$
    There is a proof following it, what is the question?
    $endgroup$
    – Hawk
    Dec 12 '18 at 11:52
















$begingroup$
There is a proof following it, what is the question?
$endgroup$
– Hawk
Dec 12 '18 at 11:52




$begingroup$
There is a proof following it, what is the question?
$endgroup$
– Hawk
Dec 12 '18 at 11:52










2 Answers
2






active

oldest

votes


















1












$begingroup$

Why does $g$ have a local maximum? We cannot even ask that question because $g$ is complex valued. Maximum Modulus Theorem says if $|g|$ attains its maximum at an interior point of the domain then $g$ is a constant.



It is already proved that $|g(z)|leq 1$. So $|g(z)|=1$ for some $z$ implies that $|g|$ attains its maximum at that point. This implies that $g$ is a constant $c$ so $f(z)=cz$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I wanted to make sure that that you fullfil all requirements to apply the Maximum Modulus Theorem. |g| =1 is a local maximum, because we proved $|g(z)| leq 1$. Thank you very much :)
    $endgroup$
    – Steven33
    Dec 12 '18 at 12:10












  • $begingroup$
    Sorry. I have another question. Why does |c| have to be 1?
    $endgroup$
    – Steven33
    Dec 12 '18 at 15:31










  • $begingroup$
    @Steven33 Just apply the hypothesis: either $|f(z)|=|z|$ (with $ zneq 0$) or $|f'(0)|=1$. In either case you get $|c|=1$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 12 '18 at 23:12










  • $begingroup$
    It is $ g(z)=c$ Then we get $ f'(0) =c $ and$ f(z)=cz. $. When you compare this expressions c must be 1?
    $endgroup$
    – Steven33
    Dec 13 '18 at 8:59



















0












$begingroup$

On that page, $D$ is defined to be the open complex unit disc: $D={zin {mathbb C}: |z|<1}$ and $g$ is defined as
$$left{ g = begin{eqnarray} frac{f(z)}{z} quad mbox{ if } f(z) not= 0 \ f'(0) quad mbox{ otherwise } end{eqnarray} right. $$



So by definition, either $|g(z)| = |f(z)/z| = |f(z)|/|z| = 1$ (since $|f(z)| = |z|$) or $|g(z)| = |f'(0)| = 1$.



Note that this isn't a local maximum really -- by the maximum modulus principle $g$ is now constant on $D$.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Why does $g$ have a local maximum? We cannot even ask that question because $g$ is complex valued. Maximum Modulus Theorem says if $|g|$ attains its maximum at an interior point of the domain then $g$ is a constant.



    It is already proved that $|g(z)|leq 1$. So $|g(z)|=1$ for some $z$ implies that $|g|$ attains its maximum at that point. This implies that $g$ is a constant $c$ so $f(z)=cz$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I wanted to make sure that that you fullfil all requirements to apply the Maximum Modulus Theorem. |g| =1 is a local maximum, because we proved $|g(z)| leq 1$. Thank you very much :)
      $endgroup$
      – Steven33
      Dec 12 '18 at 12:10












    • $begingroup$
      Sorry. I have another question. Why does |c| have to be 1?
      $endgroup$
      – Steven33
      Dec 12 '18 at 15:31










    • $begingroup$
      @Steven33 Just apply the hypothesis: either $|f(z)|=|z|$ (with $ zneq 0$) or $|f'(0)|=1$. In either case you get $|c|=1$.
      $endgroup$
      – Kavi Rama Murthy
      Dec 12 '18 at 23:12










    • $begingroup$
      It is $ g(z)=c$ Then we get $ f'(0) =c $ and$ f(z)=cz. $. When you compare this expressions c must be 1?
      $endgroup$
      – Steven33
      Dec 13 '18 at 8:59
















    1












    $begingroup$

    Why does $g$ have a local maximum? We cannot even ask that question because $g$ is complex valued. Maximum Modulus Theorem says if $|g|$ attains its maximum at an interior point of the domain then $g$ is a constant.



    It is already proved that $|g(z)|leq 1$. So $|g(z)|=1$ for some $z$ implies that $|g|$ attains its maximum at that point. This implies that $g$ is a constant $c$ so $f(z)=cz$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I wanted to make sure that that you fullfil all requirements to apply the Maximum Modulus Theorem. |g| =1 is a local maximum, because we proved $|g(z)| leq 1$. Thank you very much :)
      $endgroup$
      – Steven33
      Dec 12 '18 at 12:10












    • $begingroup$
      Sorry. I have another question. Why does |c| have to be 1?
      $endgroup$
      – Steven33
      Dec 12 '18 at 15:31










    • $begingroup$
      @Steven33 Just apply the hypothesis: either $|f(z)|=|z|$ (with $ zneq 0$) or $|f'(0)|=1$. In either case you get $|c|=1$.
      $endgroup$
      – Kavi Rama Murthy
      Dec 12 '18 at 23:12










    • $begingroup$
      It is $ g(z)=c$ Then we get $ f'(0) =c $ and$ f(z)=cz. $. When you compare this expressions c must be 1?
      $endgroup$
      – Steven33
      Dec 13 '18 at 8:59














    1












    1








    1





    $begingroup$

    Why does $g$ have a local maximum? We cannot even ask that question because $g$ is complex valued. Maximum Modulus Theorem says if $|g|$ attains its maximum at an interior point of the domain then $g$ is a constant.



    It is already proved that $|g(z)|leq 1$. So $|g(z)|=1$ for some $z$ implies that $|g|$ attains its maximum at that point. This implies that $g$ is a constant $c$ so $f(z)=cz$.






    share|cite|improve this answer











    $endgroup$



    Why does $g$ have a local maximum? We cannot even ask that question because $g$ is complex valued. Maximum Modulus Theorem says if $|g|$ attains its maximum at an interior point of the domain then $g$ is a constant.



    It is already proved that $|g(z)|leq 1$. So $|g(z)|=1$ for some $z$ implies that $|g|$ attains its maximum at that point. This implies that $g$ is a constant $c$ so $f(z)=cz$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 12 '18 at 11:50

























    answered Dec 12 '18 at 11:44









    Kavi Rama MurthyKavi Rama Murthy

    59.5k42161




    59.5k42161












    • $begingroup$
      I wanted to make sure that that you fullfil all requirements to apply the Maximum Modulus Theorem. |g| =1 is a local maximum, because we proved $|g(z)| leq 1$. Thank you very much :)
      $endgroup$
      – Steven33
      Dec 12 '18 at 12:10












    • $begingroup$
      Sorry. I have another question. Why does |c| have to be 1?
      $endgroup$
      – Steven33
      Dec 12 '18 at 15:31










    • $begingroup$
      @Steven33 Just apply the hypothesis: either $|f(z)|=|z|$ (with $ zneq 0$) or $|f'(0)|=1$. In either case you get $|c|=1$.
      $endgroup$
      – Kavi Rama Murthy
      Dec 12 '18 at 23:12










    • $begingroup$
      It is $ g(z)=c$ Then we get $ f'(0) =c $ and$ f(z)=cz. $. When you compare this expressions c must be 1?
      $endgroup$
      – Steven33
      Dec 13 '18 at 8:59


















    • $begingroup$
      I wanted to make sure that that you fullfil all requirements to apply the Maximum Modulus Theorem. |g| =1 is a local maximum, because we proved $|g(z)| leq 1$. Thank you very much :)
      $endgroup$
      – Steven33
      Dec 12 '18 at 12:10












    • $begingroup$
      Sorry. I have another question. Why does |c| have to be 1?
      $endgroup$
      – Steven33
      Dec 12 '18 at 15:31










    • $begingroup$
      @Steven33 Just apply the hypothesis: either $|f(z)|=|z|$ (with $ zneq 0$) or $|f'(0)|=1$. In either case you get $|c|=1$.
      $endgroup$
      – Kavi Rama Murthy
      Dec 12 '18 at 23:12










    • $begingroup$
      It is $ g(z)=c$ Then we get $ f'(0) =c $ and$ f(z)=cz. $. When you compare this expressions c must be 1?
      $endgroup$
      – Steven33
      Dec 13 '18 at 8:59
















    $begingroup$
    I wanted to make sure that that you fullfil all requirements to apply the Maximum Modulus Theorem. |g| =1 is a local maximum, because we proved $|g(z)| leq 1$. Thank you very much :)
    $endgroup$
    – Steven33
    Dec 12 '18 at 12:10






    $begingroup$
    I wanted to make sure that that you fullfil all requirements to apply the Maximum Modulus Theorem. |g| =1 is a local maximum, because we proved $|g(z)| leq 1$. Thank you very much :)
    $endgroup$
    – Steven33
    Dec 12 '18 at 12:10














    $begingroup$
    Sorry. I have another question. Why does |c| have to be 1?
    $endgroup$
    – Steven33
    Dec 12 '18 at 15:31




    $begingroup$
    Sorry. I have another question. Why does |c| have to be 1?
    $endgroup$
    – Steven33
    Dec 12 '18 at 15:31












    $begingroup$
    @Steven33 Just apply the hypothesis: either $|f(z)|=|z|$ (with $ zneq 0$) or $|f'(0)|=1$. In either case you get $|c|=1$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 12 '18 at 23:12




    $begingroup$
    @Steven33 Just apply the hypothesis: either $|f(z)|=|z|$ (with $ zneq 0$) or $|f'(0)|=1$. In either case you get $|c|=1$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 12 '18 at 23:12












    $begingroup$
    It is $ g(z)=c$ Then we get $ f'(0) =c $ and$ f(z)=cz. $. When you compare this expressions c must be 1?
    $endgroup$
    – Steven33
    Dec 13 '18 at 8:59




    $begingroup$
    It is $ g(z)=c$ Then we get $ f'(0) =c $ and$ f(z)=cz. $. When you compare this expressions c must be 1?
    $endgroup$
    – Steven33
    Dec 13 '18 at 8:59











    0












    $begingroup$

    On that page, $D$ is defined to be the open complex unit disc: $D={zin {mathbb C}: |z|<1}$ and $g$ is defined as
    $$left{ g = begin{eqnarray} frac{f(z)}{z} quad mbox{ if } f(z) not= 0 \ f'(0) quad mbox{ otherwise } end{eqnarray} right. $$



    So by definition, either $|g(z)| = |f(z)/z| = |f(z)|/|z| = 1$ (since $|f(z)| = |z|$) or $|g(z)| = |f'(0)| = 1$.



    Note that this isn't a local maximum really -- by the maximum modulus principle $g$ is now constant on $D$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      On that page, $D$ is defined to be the open complex unit disc: $D={zin {mathbb C}: |z|<1}$ and $g$ is defined as
      $$left{ g = begin{eqnarray} frac{f(z)}{z} quad mbox{ if } f(z) not= 0 \ f'(0) quad mbox{ otherwise } end{eqnarray} right. $$



      So by definition, either $|g(z)| = |f(z)/z| = |f(z)|/|z| = 1$ (since $|f(z)| = |z|$) or $|g(z)| = |f'(0)| = 1$.



      Note that this isn't a local maximum really -- by the maximum modulus principle $g$ is now constant on $D$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        On that page, $D$ is defined to be the open complex unit disc: $D={zin {mathbb C}: |z|<1}$ and $g$ is defined as
        $$left{ g = begin{eqnarray} frac{f(z)}{z} quad mbox{ if } f(z) not= 0 \ f'(0) quad mbox{ otherwise } end{eqnarray} right. $$



        So by definition, either $|g(z)| = |f(z)/z| = |f(z)|/|z| = 1$ (since $|f(z)| = |z|$) or $|g(z)| = |f'(0)| = 1$.



        Note that this isn't a local maximum really -- by the maximum modulus principle $g$ is now constant on $D$.






        share|cite|improve this answer









        $endgroup$



        On that page, $D$ is defined to be the open complex unit disc: $D={zin {mathbb C}: |z|<1}$ and $g$ is defined as
        $$left{ g = begin{eqnarray} frac{f(z)}{z} quad mbox{ if } f(z) not= 0 \ f'(0) quad mbox{ otherwise } end{eqnarray} right. $$



        So by definition, either $|g(z)| = |f(z)/z| = |f(z)|/|z| = 1$ (since $|f(z)| = |z|$) or $|g(z)| = |f'(0)| = 1$.



        Note that this isn't a local maximum really -- by the maximum modulus principle $g$ is now constant on $D$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 12 '18 at 11:43









        postmortespostmortes

        1,93621117




        1,93621117






























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