Show that $Lgeq2pi|f'(0)|$












1












$begingroup$


Suppose $f$ is analytic on and inside the unit circle $partialmathbb{D}$. Let $L$ be the length of the curve $f(partialmathbb{D})$. Show that $Lgeq2pi|f'(0)|$.



I have tried different approaches from Schwarz Lemma, and maximum modulus principle, but got nowhere. Any guidance is appreciated.



Edit:
I think I may have found it (please confirm):
By maximum modulus principle, $f$ attains it's max value on $partialmathbb{D}$. By Cauchy Estimates:
$$begin{align*}
|f'(0)|leq|sup f|
end{align*}$$
Note that since $L=2pi |sup f|$, we conclude that
$$begin{align*}|f'(0)|&leq frac{L}{2pi}\
2pi|f'(0)|&leq L
end{align*}$$






complex-analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Why does $L=2pilvertsup frvert$? What does it mean to take sup of (strictly) complex numbers?
    $endgroup$
    – user10354138
    Dec 27 '18 at 19:17












  • $begingroup$
    @user10354138 Circumference of a circle is defined by $2picdot r$. Here, radius is $f(z)$ where $|z|=1$. Since, by maximum modulus principle, that's where $f$ attains its maximum, i.e. $|sup f|$. I may be totally wrong.
    $endgroup$
    – Ya G
    Dec 27 '18 at 19:20












  • $begingroup$
    @YaG But $f(partial mathbb{D})$ need not be a circle
    $endgroup$
    – angryavian
    Dec 27 '18 at 19:24






  • 1




    $begingroup$
    @YaG Also, I think user10354138 is trying to get you notice that you probably meant $sup |f|$ rather than $|sup f|$.
    $endgroup$
    – angryavian
    Dec 27 '18 at 19:30






  • 1




    $begingroup$
    That is one aspect of it. The other part is indeed $f(partialmathbb{D})$ need not be a circle and $f$ need not be univalent, so it doesn't make sense to multiply by $2pi$ and say it is $L$.
    $endgroup$
    – user10354138
    Dec 27 '18 at 19:33


















1












$begingroup$


Suppose $f$ is analytic on and inside the unit circle $partialmathbb{D}$. Let $L$ be the length of the curve $f(partialmathbb{D})$. Show that $Lgeq2pi|f'(0)|$.



I have tried different approaches from Schwarz Lemma, and maximum modulus principle, but got nowhere. Any guidance is appreciated.



Edit:
I think I may have found it (please confirm):
By maximum modulus principle, $f$ attains it's max value on $partialmathbb{D}$. By Cauchy Estimates:
$$begin{align*}
|f'(0)|leq|sup f|
end{align*}$$
Note that since $L=2pi |sup f|$, we conclude that
$$begin{align*}|f'(0)|&leq frac{L}{2pi}\
2pi|f'(0)|&leq L
end{align*}$$






complex-analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Why does $L=2pilvertsup frvert$? What does it mean to take sup of (strictly) complex numbers?
    $endgroup$
    – user10354138
    Dec 27 '18 at 19:17












  • $begingroup$
    @user10354138 Circumference of a circle is defined by $2picdot r$. Here, radius is $f(z)$ where $|z|=1$. Since, by maximum modulus principle, that's where $f$ attains its maximum, i.e. $|sup f|$. I may be totally wrong.
    $endgroup$
    – Ya G
    Dec 27 '18 at 19:20












  • $begingroup$
    @YaG But $f(partial mathbb{D})$ need not be a circle
    $endgroup$
    – angryavian
    Dec 27 '18 at 19:24






  • 1




    $begingroup$
    @YaG Also, I think user10354138 is trying to get you notice that you probably meant $sup |f|$ rather than $|sup f|$.
    $endgroup$
    – angryavian
    Dec 27 '18 at 19:30






  • 1




    $begingroup$
    That is one aspect of it. The other part is indeed $f(partialmathbb{D})$ need not be a circle and $f$ need not be univalent, so it doesn't make sense to multiply by $2pi$ and say it is $L$.
    $endgroup$
    – user10354138
    Dec 27 '18 at 19:33
















1












1








1





$begingroup$


Suppose $f$ is analytic on and inside the unit circle $partialmathbb{D}$. Let $L$ be the length of the curve $f(partialmathbb{D})$. Show that $Lgeq2pi|f'(0)|$.



I have tried different approaches from Schwarz Lemma, and maximum modulus principle, but got nowhere. Any guidance is appreciated.



Edit:
I think I may have found it (please confirm):
By maximum modulus principle, $f$ attains it's max value on $partialmathbb{D}$. By Cauchy Estimates:
$$begin{align*}
|f'(0)|leq|sup f|
end{align*}$$
Note that since $L=2pi |sup f|$, we conclude that
$$begin{align*}|f'(0)|&leq frac{L}{2pi}\
2pi|f'(0)|&leq L
end{align*}$$






complex-analysis







share|cite|improve this question











$endgroup$




Suppose $f$ is analytic on and inside the unit circle $partialmathbb{D}$. Let $L$ be the length of the curve $f(partialmathbb{D})$. Show that $Lgeq2pi|f'(0)|$.



I have tried different approaches from Schwarz Lemma, and maximum modulus principle, but got nowhere. Any guidance is appreciated.



Edit:
I think I may have found it (please confirm):
By maximum modulus principle, $f$ attains it's max value on $partialmathbb{D}$. By Cauchy Estimates:
$$begin{align*}
|f'(0)|leq|sup f|
end{align*}$$
Note that since $L=2pi |sup f|$, we conclude that
$$begin{align*}|f'(0)|&leq frac{L}{2pi}\
2pi|f'(0)|&leq L
end{align*}$$






complex-analysis




complex-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 27 '18 at 18:57







Ya G

















asked Dec 27 '18 at 18:28









Ya GYa G

536211




536211








  • 1




    $begingroup$
    Why does $L=2pilvertsup frvert$? What does it mean to take sup of (strictly) complex numbers?
    $endgroup$
    – user10354138
    Dec 27 '18 at 19:17












  • $begingroup$
    @user10354138 Circumference of a circle is defined by $2picdot r$. Here, radius is $f(z)$ where $|z|=1$. Since, by maximum modulus principle, that's where $f$ attains its maximum, i.e. $|sup f|$. I may be totally wrong.
    $endgroup$
    – Ya G
    Dec 27 '18 at 19:20












  • $begingroup$
    @YaG But $f(partial mathbb{D})$ need not be a circle
    $endgroup$
    – angryavian
    Dec 27 '18 at 19:24






  • 1




    $begingroup$
    @YaG Also, I think user10354138 is trying to get you notice that you probably meant $sup |f|$ rather than $|sup f|$.
    $endgroup$
    – angryavian
    Dec 27 '18 at 19:30






  • 1




    $begingroup$
    That is one aspect of it. The other part is indeed $f(partialmathbb{D})$ need not be a circle and $f$ need not be univalent, so it doesn't make sense to multiply by $2pi$ and say it is $L$.
    $endgroup$
    – user10354138
    Dec 27 '18 at 19:33
















  • 1




    $begingroup$
    Why does $L=2pilvertsup frvert$? What does it mean to take sup of (strictly) complex numbers?
    $endgroup$
    – user10354138
    Dec 27 '18 at 19:17












  • $begingroup$
    @user10354138 Circumference of a circle is defined by $2picdot r$. Here, radius is $f(z)$ where $|z|=1$. Since, by maximum modulus principle, that's where $f$ attains its maximum, i.e. $|sup f|$. I may be totally wrong.
    $endgroup$
    – Ya G
    Dec 27 '18 at 19:20












  • $begingroup$
    @YaG But $f(partial mathbb{D})$ need not be a circle
    $endgroup$
    – angryavian
    Dec 27 '18 at 19:24






  • 1




    $begingroup$
    @YaG Also, I think user10354138 is trying to get you notice that you probably meant $sup |f|$ rather than $|sup f|$.
    $endgroup$
    – angryavian
    Dec 27 '18 at 19:30






  • 1




    $begingroup$
    That is one aspect of it. The other part is indeed $f(partialmathbb{D})$ need not be a circle and $f$ need not be univalent, so it doesn't make sense to multiply by $2pi$ and say it is $L$.
    $endgroup$
    – user10354138
    Dec 27 '18 at 19:33










1




1




$begingroup$
Why does $L=2pilvertsup frvert$? What does it mean to take sup of (strictly) complex numbers?
$endgroup$
– user10354138
Dec 27 '18 at 19:17






$begingroup$
Why does $L=2pilvertsup frvert$? What does it mean to take sup of (strictly) complex numbers?
$endgroup$
– user10354138
Dec 27 '18 at 19:17














$begingroup$
@user10354138 Circumference of a circle is defined by $2picdot r$. Here, radius is $f(z)$ where $|z|=1$. Since, by maximum modulus principle, that's where $f$ attains its maximum, i.e. $|sup f|$. I may be totally wrong.
$endgroup$
– Ya G
Dec 27 '18 at 19:20






$begingroup$
@user10354138 Circumference of a circle is defined by $2picdot r$. Here, radius is $f(z)$ where $|z|=1$. Since, by maximum modulus principle, that's where $f$ attains its maximum, i.e. $|sup f|$. I may be totally wrong.
$endgroup$
– Ya G
Dec 27 '18 at 19:20














$begingroup$
@YaG But $f(partial mathbb{D})$ need not be a circle
$endgroup$
– angryavian
Dec 27 '18 at 19:24




$begingroup$
@YaG But $f(partial mathbb{D})$ need not be a circle
$endgroup$
– angryavian
Dec 27 '18 at 19:24




1




1




$begingroup$
@YaG Also, I think user10354138 is trying to get you notice that you probably meant $sup |f|$ rather than $|sup f|$.
$endgroup$
– angryavian
Dec 27 '18 at 19:30




$begingroup$
@YaG Also, I think user10354138 is trying to get you notice that you probably meant $sup |f|$ rather than $|sup f|$.
$endgroup$
– angryavian
Dec 27 '18 at 19:30




1




1




$begingroup$
That is one aspect of it. The other part is indeed $f(partialmathbb{D})$ need not be a circle and $f$ need not be univalent, so it doesn't make sense to multiply by $2pi$ and say it is $L$.
$endgroup$
– user10354138
Dec 27 '18 at 19:33






$begingroup$
That is one aspect of it. The other part is indeed $f(partialmathbb{D})$ need not be a circle and $f$ need not be univalent, so it doesn't make sense to multiply by $2pi$ and say it is $L$.
$endgroup$
– user10354138
Dec 27 '18 at 19:33












1 Answer
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From definition of length of a curve and mean value property:
$$L = int_0^{2pi} |(f(e^{it}))'| dt = int_0^{2pi} |f'(e^{it})|dt geqleft|int_0^{2pi}f'(e^{it})dt right| = 2pi |f'(0)| $$






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    3












    $begingroup$

    From definition of length of a curve and mean value property:
    $$L = int_0^{2pi} |(f(e^{it}))'| dt = int_0^{2pi} |f'(e^{it})|dt geqleft|int_0^{2pi}f'(e^{it})dt right| = 2pi |f'(0)| $$






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      From definition of length of a curve and mean value property:
      $$L = int_0^{2pi} |(f(e^{it}))'| dt = int_0^{2pi} |f'(e^{it})|dt geqleft|int_0^{2pi}f'(e^{it})dt right| = 2pi |f'(0)| $$






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        From definition of length of a curve and mean value property:
        $$L = int_0^{2pi} |(f(e^{it}))'| dt = int_0^{2pi} |f'(e^{it})|dt geqleft|int_0^{2pi}f'(e^{it})dt right| = 2pi |f'(0)| $$






        share|cite|improve this answer









        $endgroup$



        From definition of length of a curve and mean value property:
        $$L = int_0^{2pi} |(f(e^{it}))'| dt = int_0^{2pi} |f'(e^{it})|dt geqleft|int_0^{2pi}f'(e^{it})dt right| = 2pi |f'(0)| $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 27 '18 at 19:34









        JakobianJakobian

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