$f$ is an entire function, show that $ sum_n=0^infty fraca_nn+1 leq pi int_0^2 pi|f(e^i theta)| ,dtheta. $

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Let $f(z) = displaystyle sum_n=0^infty a_n z^n$ be an entire
function. Show that $$ sum_n=0^infty fracn+1 leq pi
int_0^2 pi|f(e^i theta)| ,dtheta. $$




The Cauchy integral formula gives
$$ a_n = frac12 pi i int_0^2 pi fracf(e^i theta)e^i (n+1) theta i e^i theta , d theta = frac12 pi int_0^2 pi f(e^i theta) e^-i n theta , d theta. $$
Then
$$
beginaligned
sum_n=0^infty fracn+1
&= frac12 pi sum_n=0^infty frac1n+1 left|int_0^2 pi f(e^i theta) e^-i n theta , d theta right| \
&leq frac12 pi sum_n=0^infty frac1n+1 int_0^2 pi left| f(e^i theta) right| left| e^-i n theta right| , d theta \
&= frac12 pi sum_n=0^infty frac1n+1 int_0^2 pi left| f(e^i theta) right| , d theta.
endaligned
$$
It seems that this method doesn't work. Is there any idea to estimate it?




complex-analysis




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  • Are you sure that's a $pi$ out in front of the integral and not a $frac1pi$ or even $frac12pi$?
    – Carl Schildkraut
    Aug 3 at 3:29














up vote
2
down vote

favorite
2













Let $f(z) = displaystyle sum_n=0^infty a_n z^n$ be an entire
function. Show that $$ sum_n=0^infty fracn+1 leq pi
int_0^2 pi|f(e^i theta)| ,dtheta. $$




The Cauchy integral formula gives
$$ a_n = frac12 pi i int_0^2 pi fracf(e^i theta)e^i (n+1) theta i e^i theta , d theta = frac12 pi int_0^2 pi f(e^i theta) e^-i n theta , d theta. $$
Then
$$
beginaligned
sum_n=0^infty fracn+1
&= frac12 pi sum_n=0^infty frac1n+1 left|int_0^2 pi f(e^i theta) e^-i n theta , d theta right| \
&leq frac12 pi sum_n=0^infty frac1n+1 int_0^2 pi left| f(e^i theta) right| left| e^-i n theta right| , d theta \
&= frac12 pi sum_n=0^infty frac1n+1 int_0^2 pi left| f(e^i theta) right| , d theta.
endaligned
$$
It seems that this method doesn't work. Is there any idea to estimate it?




complex-analysis




share|cite|improve this question



















  • Are you sure that's a $pi$ out in front of the integral and not a $frac1pi$ or even $frac12pi$?
    – Carl Schildkraut
    Aug 3 at 3:29












up vote
2
down vote

favorite
2









up vote
2
down vote

favorite
2






2






Let $f(z) = displaystyle sum_n=0^infty a_n z^n$ be an entire
function. Show that $$ sum_n=0^infty fracn+1 leq pi
int_0^2 pi|f(e^i theta)| ,dtheta. $$




The Cauchy integral formula gives
$$ a_n = frac12 pi i int_0^2 pi fracf(e^i theta)e^i (n+1) theta i e^i theta , d theta = frac12 pi int_0^2 pi f(e^i theta) e^-i n theta , d theta. $$
Then
$$
beginaligned
sum_n=0^infty fracn+1
&= frac12 pi sum_n=0^infty frac1n+1 left|int_0^2 pi f(e^i theta) e^-i n theta , d theta right| \
&leq frac12 pi sum_n=0^infty frac1n+1 int_0^2 pi left| f(e^i theta) right| left| e^-i n theta right| , d theta \
&= frac12 pi sum_n=0^infty frac1n+1 int_0^2 pi left| f(e^i theta) right| , d theta.
endaligned
$$
It seems that this method doesn't work. Is there any idea to estimate it?




complex-analysis




share|cite|improve this question












Let $f(z) = displaystyle sum_n=0^infty a_n z^n$ be an entire
function. Show that $$ sum_n=0^infty fracn+1 leq pi
int_0^2 pi|f(e^i theta)| ,dtheta. $$




The Cauchy integral formula gives
$$ a_n = frac12 pi i int_0^2 pi fracf(e^i theta)e^i (n+1) theta i e^i theta , d theta = frac12 pi int_0^2 pi f(e^i theta) e^-i n theta , d theta. $$
Then
$$
beginaligned
sum_n=0^infty fracn+1
&= frac12 pi sum_n=0^infty frac1n+1 left|int_0^2 pi f(e^i theta) e^-i n theta , d theta right| \
&leq frac12 pi sum_n=0^infty frac1n+1 int_0^2 pi left| f(e^i theta) right| left| e^-i n theta right| , d theta \
&= frac12 pi sum_n=0^infty frac1n+1 int_0^2 pi left| f(e^i theta) right| , d theta.
endaligned
$$
It seems that this method doesn't work. Is there any idea to estimate it?





complex-analysis





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asked Aug 3 at 3:22









Star Chou

36410




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  • Are you sure that's a $pi$ out in front of the integral and not a $frac1pi$ or even $frac12pi$?
    – Carl Schildkraut
    Aug 3 at 3:29
















  • Are you sure that's a $pi$ out in front of the integral and not a $frac1pi$ or even $frac12pi$?
    – Carl Schildkraut
    Aug 3 at 3:29















Are you sure that's a $pi$ out in front of the integral and not a $frac1pi$ or even $frac12pi$?
– Carl Schildkraut
Aug 3 at 3:29




Are you sure that's a $pi$ out in front of the integral and not a $frac1pi$ or even $frac12pi$?
– Carl Schildkraut
Aug 3 at 3:29










1 Answer
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up vote
3
down vote



accepted










It is a theorem due to Hardy. See for example:



  • Kenneth Hoffman, Banach space of analytic functions, the theorem at page 70

  • Peter Duren, Theory of $H^p$ spaces, corollary to theorem 3.15 at page 48





share|cite|improve this answer





















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    1 Answer
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    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    3
    down vote



    accepted










    It is a theorem due to Hardy. See for example:



    • Kenneth Hoffman, Banach space of analytic functions, the theorem at page 70

    • Peter Duren, Theory of $H^p$ spaces, corollary to theorem 3.15 at page 48





    share|cite|improve this answer

























      up vote
      3
      down vote



      accepted










      It is a theorem due to Hardy. See for example:



      • Kenneth Hoffman, Banach space of analytic functions, the theorem at page 70

      • Peter Duren, Theory of $H^p$ spaces, corollary to theorem 3.15 at page 48





      share|cite|improve this answer























        up vote
        3
        down vote



        accepted







        up vote
        3
        down vote



        accepted






        It is a theorem due to Hardy. See for example:



        • Kenneth Hoffman, Banach space of analytic functions, the theorem at page 70

        • Peter Duren, Theory of $H^p$ spaces, corollary to theorem 3.15 at page 48





        share|cite|improve this answer













        It is a theorem due to Hardy. See for example:



        • Kenneth Hoffman, Banach space of analytic functions, the theorem at page 70

        • Peter Duren, Theory of $H^p$ spaces, corollary to theorem 3.15 at page 48






        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Aug 3 at 5:28









        Bob

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