Prove that there exists $hin V$, such that $|f'(frac{1}{2})|leq|h'(frac{1}{2})|$ for all $fin V$.












3












$begingroup$


Let $V={finmathcal{O}(mathbb{D}):f(z)=sum_{n=0}^{infty}a_{n}z^{n}text{ with }|a_{n}|leq n^{2}text{ for all }n}$. Prove that there exists $hin V$, such that $|f'(frac{1}{2})|leq|h'(frac{1}{2})|$ for all $fin V$.



I want to use Montel's Theorem:




Let $mathcal{F}$ be a family of holomorphic functions on Ω. If $mathcal{F}$ is uniformly bounded on every compact subset of Ω, then $mathcal{F}$ is equicontinuous on every compact subset of Ω, and hence $mathcal{F}$ is a normal family.




My initial thought is to first prove that the set $V$ is uniformly bounded on every compact subset, but I'm not quite sure how to show that. Also, How do I use the Montel's theorem to prove above? I guess my question is how is showing the existence of $h$ relate to $V$ is a normal family?






complex-analysis







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    Let $V={finmathcal{O}(mathbb{D}):f(z)=sum_{n=0}^{infty}a_{n}z^{n}text{ with }|a_{n}|leq n^{2}text{ for all }n}$. Prove that there exists $hin V$, such that $|f'(frac{1}{2})|leq|h'(frac{1}{2})|$ for all $fin V$.



    I want to use Montel's Theorem:




    Let $mathcal{F}$ be a family of holomorphic functions on Ω. If $mathcal{F}$ is uniformly bounded on every compact subset of Ω, then $mathcal{F}$ is equicontinuous on every compact subset of Ω, and hence $mathcal{F}$ is a normal family.




    My initial thought is to first prove that the set $V$ is uniformly bounded on every compact subset, but I'm not quite sure how to show that. Also, How do I use the Montel's theorem to prove above? I guess my question is how is showing the existence of $h$ relate to $V$ is a normal family?






    complex-analysis







    share|cite|improve this question











    $endgroup$















      3












      3








      3


      2



      $begingroup$


      Let $V={finmathcal{O}(mathbb{D}):f(z)=sum_{n=0}^{infty}a_{n}z^{n}text{ with }|a_{n}|leq n^{2}text{ for all }n}$. Prove that there exists $hin V$, such that $|f'(frac{1}{2})|leq|h'(frac{1}{2})|$ for all $fin V$.



      I want to use Montel's Theorem:




      Let $mathcal{F}$ be a family of holomorphic functions on Ω. If $mathcal{F}$ is uniformly bounded on every compact subset of Ω, then $mathcal{F}$ is equicontinuous on every compact subset of Ω, and hence $mathcal{F}$ is a normal family.




      My initial thought is to first prove that the set $V$ is uniformly bounded on every compact subset, but I'm not quite sure how to show that. Also, How do I use the Montel's theorem to prove above? I guess my question is how is showing the existence of $h$ relate to $V$ is a normal family?






      complex-analysis







      share|cite|improve this question











      $endgroup$




      Let $V={finmathcal{O}(mathbb{D}):f(z)=sum_{n=0}^{infty}a_{n}z^{n}text{ with }|a_{n}|leq n^{2}text{ for all }n}$. Prove that there exists $hin V$, such that $|f'(frac{1}{2})|leq|h'(frac{1}{2})|$ for all $fin V$.



      I want to use Montel's Theorem:




      Let $mathcal{F}$ be a family of holomorphic functions on Ω. If $mathcal{F}$ is uniformly bounded on every compact subset of Ω, then $mathcal{F}$ is equicontinuous on every compact subset of Ω, and hence $mathcal{F}$ is a normal family.




      My initial thought is to first prove that the set $V$ is uniformly bounded on every compact subset, but I'm not quite sure how to show that. Also, How do I use the Montel's theorem to prove above? I guess my question is how is showing the existence of $h$ relate to $V$ is a normal family?






      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 11 '18 at 20:18









      the_candyman

      9,14032145




      9,14032145










      asked Dec 11 '18 at 20:08









      Ya GYa G

      536211




      536211






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          In fact, the problem does not require Montel's theorem necessarily: simply let
          $$
          h(z) = sum_{n=0}^infty n^2 z^n,
          $$           and observe that $|f'(frac{1}{2})|leq |h'(frac{1}{2})|$ for all $fin V$.



          However, I'll present how Montel's theorem implies that such supremum is attained in $V$ as a maximum. For any compact set $Ksubsetmathbb{D}$, observe that there exists $rin (0,1)$ such that
          $$
          Ksubset D(0,r)={zinmathbb{D}:|z|<r}.
          $$           Therefore, this leads to a simple estimate that
          $$
          |f(z)|=|sum_{j=0}^infty a_j z^j|leq sum_{j=0}^infty j^2 r^j <infty,quadforall zin K.
          $$           This shows that $V$ is a normal family. Now, let $f_nin V$ be a sequence such that $$|f'_n(frac{1}{2})| to sup_{fin V}|f'(frac{1}{2})|.
          $$           Since $V$ is normal, by passing to a subsequence, we may assume that $f_n to h$ locally uniformly on $mathbb{D}$. By Cauchy's integral formula, this of course implies that each $k$-th derivative $f_n^{(k)}$ converges locally uniformly to $h^{(k)}$. Since $V$ is closed, we have $hin V$ and that $$
          sup_{fin V}|f'(frac{1}{2})| = lim_{ntoinfty}sup_{fin V}|f'(frac{1}{2})| = |h'(frac{1}{2})|,
          $$          as desired.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Would you please explain what it means by normal family?
            $endgroup$
            – Ya G
            Dec 11 '18 at 20:21






          • 1




            $begingroup$
            The precise definition is given in en.wikipedia.org/wiki/Normal_family. If $V$ is a normal family, then It happens that every sequence in $V$ has a locally uniformly convergent subsequence. Montel's theorem provides a sufficient condition for this. So $sup|f'(c)|$ is actually attained in $V$, if $V$ is closed, by a limit $hin V$ of approaching sequence $f_nin V$ such that $|f'_n(c)| to sup|f'(c)|$.
            $endgroup$
            – Song
            Dec 11 '18 at 20:25








          • 1




            $begingroup$
            This explanation is very rough, so if you are interested, see e.g. Rudin, Real and Complex Analysis.
            $endgroup$
            – Song
            Dec 11 '18 at 20:33












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          active

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          active

          oldest

          votes









          1












          $begingroup$

          In fact, the problem does not require Montel's theorem necessarily: simply let
          $$
          h(z) = sum_{n=0}^infty n^2 z^n,
          $$           and observe that $|f'(frac{1}{2})|leq |h'(frac{1}{2})|$ for all $fin V$.



          However, I'll present how Montel's theorem implies that such supremum is attained in $V$ as a maximum. For any compact set $Ksubsetmathbb{D}$, observe that there exists $rin (0,1)$ such that
          $$
          Ksubset D(0,r)={zinmathbb{D}:|z|<r}.
          $$           Therefore, this leads to a simple estimate that
          $$
          |f(z)|=|sum_{j=0}^infty a_j z^j|leq sum_{j=0}^infty j^2 r^j <infty,quadforall zin K.
          $$           This shows that $V$ is a normal family. Now, let $f_nin V$ be a sequence such that $$|f'_n(frac{1}{2})| to sup_{fin V}|f'(frac{1}{2})|.
          $$           Since $V$ is normal, by passing to a subsequence, we may assume that $f_n to h$ locally uniformly on $mathbb{D}$. By Cauchy's integral formula, this of course implies that each $k$-th derivative $f_n^{(k)}$ converges locally uniformly to $h^{(k)}$. Since $V$ is closed, we have $hin V$ and that $$
          sup_{fin V}|f'(frac{1}{2})| = lim_{ntoinfty}sup_{fin V}|f'(frac{1}{2})| = |h'(frac{1}{2})|,
          $$          as desired.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Would you please explain what it means by normal family?
            $endgroup$
            – Ya G
            Dec 11 '18 at 20:21






          • 1




            $begingroup$
            The precise definition is given in en.wikipedia.org/wiki/Normal_family. If $V$ is a normal family, then It happens that every sequence in $V$ has a locally uniformly convergent subsequence. Montel's theorem provides a sufficient condition for this. So $sup|f'(c)|$ is actually attained in $V$, if $V$ is closed, by a limit $hin V$ of approaching sequence $f_nin V$ such that $|f'_n(c)| to sup|f'(c)|$.
            $endgroup$
            – Song
            Dec 11 '18 at 20:25








          • 1




            $begingroup$
            This explanation is very rough, so if you are interested, see e.g. Rudin, Real and Complex Analysis.
            $endgroup$
            – Song
            Dec 11 '18 at 20:33
















          1












          $begingroup$

          In fact, the problem does not require Montel's theorem necessarily: simply let
          $$
          h(z) = sum_{n=0}^infty n^2 z^n,
          $$           and observe that $|f'(frac{1}{2})|leq |h'(frac{1}{2})|$ for all $fin V$.



          However, I'll present how Montel's theorem implies that such supremum is attained in $V$ as a maximum. For any compact set $Ksubsetmathbb{D}$, observe that there exists $rin (0,1)$ such that
          $$
          Ksubset D(0,r)={zinmathbb{D}:|z|<r}.
          $$           Therefore, this leads to a simple estimate that
          $$
          |f(z)|=|sum_{j=0}^infty a_j z^j|leq sum_{j=0}^infty j^2 r^j <infty,quadforall zin K.
          $$           This shows that $V$ is a normal family. Now, let $f_nin V$ be a sequence such that $$|f'_n(frac{1}{2})| to sup_{fin V}|f'(frac{1}{2})|.
          $$           Since $V$ is normal, by passing to a subsequence, we may assume that $f_n to h$ locally uniformly on $mathbb{D}$. By Cauchy's integral formula, this of course implies that each $k$-th derivative $f_n^{(k)}$ converges locally uniformly to $h^{(k)}$. Since $V$ is closed, we have $hin V$ and that $$
          sup_{fin V}|f'(frac{1}{2})| = lim_{ntoinfty}sup_{fin V}|f'(frac{1}{2})| = |h'(frac{1}{2})|,
          $$          as desired.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Would you please explain what it means by normal family?
            $endgroup$
            – Ya G
            Dec 11 '18 at 20:21






          • 1




            $begingroup$
            The precise definition is given in en.wikipedia.org/wiki/Normal_family. If $V$ is a normal family, then It happens that every sequence in $V$ has a locally uniformly convergent subsequence. Montel's theorem provides a sufficient condition for this. So $sup|f'(c)|$ is actually attained in $V$, if $V$ is closed, by a limit $hin V$ of approaching sequence $f_nin V$ such that $|f'_n(c)| to sup|f'(c)|$.
            $endgroup$
            – Song
            Dec 11 '18 at 20:25








          • 1




            $begingroup$
            This explanation is very rough, so if you are interested, see e.g. Rudin, Real and Complex Analysis.
            $endgroup$
            – Song
            Dec 11 '18 at 20:33














          1












          1








          1





          $begingroup$

          In fact, the problem does not require Montel's theorem necessarily: simply let
          $$
          h(z) = sum_{n=0}^infty n^2 z^n,
          $$           and observe that $|f'(frac{1}{2})|leq |h'(frac{1}{2})|$ for all $fin V$.



          However, I'll present how Montel's theorem implies that such supremum is attained in $V$ as a maximum. For any compact set $Ksubsetmathbb{D}$, observe that there exists $rin (0,1)$ such that
          $$
          Ksubset D(0,r)={zinmathbb{D}:|z|<r}.
          $$           Therefore, this leads to a simple estimate that
          $$
          |f(z)|=|sum_{j=0}^infty a_j z^j|leq sum_{j=0}^infty j^2 r^j <infty,quadforall zin K.
          $$           This shows that $V$ is a normal family. Now, let $f_nin V$ be a sequence such that $$|f'_n(frac{1}{2})| to sup_{fin V}|f'(frac{1}{2})|.
          $$           Since $V$ is normal, by passing to a subsequence, we may assume that $f_n to h$ locally uniformly on $mathbb{D}$. By Cauchy's integral formula, this of course implies that each $k$-th derivative $f_n^{(k)}$ converges locally uniformly to $h^{(k)}$. Since $V$ is closed, we have $hin V$ and that $$
          sup_{fin V}|f'(frac{1}{2})| = lim_{ntoinfty}sup_{fin V}|f'(frac{1}{2})| = |h'(frac{1}{2})|,
          $$          as desired.






          share|cite|improve this answer











          $endgroup$



          In fact, the problem does not require Montel's theorem necessarily: simply let
          $$
          h(z) = sum_{n=0}^infty n^2 z^n,
          $$           and observe that $|f'(frac{1}{2})|leq |h'(frac{1}{2})|$ for all $fin V$.



          However, I'll present how Montel's theorem implies that such supremum is attained in $V$ as a maximum. For any compact set $Ksubsetmathbb{D}$, observe that there exists $rin (0,1)$ such that
          $$
          Ksubset D(0,r)={zinmathbb{D}:|z|<r}.
          $$           Therefore, this leads to a simple estimate that
          $$
          |f(z)|=|sum_{j=0}^infty a_j z^j|leq sum_{j=0}^infty j^2 r^j <infty,quadforall zin K.
          $$           This shows that $V$ is a normal family. Now, let $f_nin V$ be a sequence such that $$|f'_n(frac{1}{2})| to sup_{fin V}|f'(frac{1}{2})|.
          $$           Since $V$ is normal, by passing to a subsequence, we may assume that $f_n to h$ locally uniformly on $mathbb{D}$. By Cauchy's integral formula, this of course implies that each $k$-th derivative $f_n^{(k)}$ converges locally uniformly to $h^{(k)}$. Since $V$ is closed, we have $hin V$ and that $$
          sup_{fin V}|f'(frac{1}{2})| = lim_{ntoinfty}sup_{fin V}|f'(frac{1}{2})| = |h'(frac{1}{2})|,
          $$          as desired.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 11 '18 at 20:47

























          answered Dec 11 '18 at 20:18









          SongSong

          18.5k21651




          18.5k21651












          • $begingroup$
            Would you please explain what it means by normal family?
            $endgroup$
            – Ya G
            Dec 11 '18 at 20:21






          • 1




            $begingroup$
            The precise definition is given in en.wikipedia.org/wiki/Normal_family. If $V$ is a normal family, then It happens that every sequence in $V$ has a locally uniformly convergent subsequence. Montel's theorem provides a sufficient condition for this. So $sup|f'(c)|$ is actually attained in $V$, if $V$ is closed, by a limit $hin V$ of approaching sequence $f_nin V$ such that $|f'_n(c)| to sup|f'(c)|$.
            $endgroup$
            – Song
            Dec 11 '18 at 20:25








          • 1




            $begingroup$
            This explanation is very rough, so if you are interested, see e.g. Rudin, Real and Complex Analysis.
            $endgroup$
            – Song
            Dec 11 '18 at 20:33


















          • $begingroup$
            Would you please explain what it means by normal family?
            $endgroup$
            – Ya G
            Dec 11 '18 at 20:21






          • 1




            $begingroup$
            The precise definition is given in en.wikipedia.org/wiki/Normal_family. If $V$ is a normal family, then It happens that every sequence in $V$ has a locally uniformly convergent subsequence. Montel's theorem provides a sufficient condition for this. So $sup|f'(c)|$ is actually attained in $V$, if $V$ is closed, by a limit $hin V$ of approaching sequence $f_nin V$ such that $|f'_n(c)| to sup|f'(c)|$.
            $endgroup$
            – Song
            Dec 11 '18 at 20:25








          • 1




            $begingroup$
            This explanation is very rough, so if you are interested, see e.g. Rudin, Real and Complex Analysis.
            $endgroup$
            – Song
            Dec 11 '18 at 20:33
















          $begingroup$
          Would you please explain what it means by normal family?
          $endgroup$
          – Ya G
          Dec 11 '18 at 20:21




          $begingroup$
          Would you please explain what it means by normal family?
          $endgroup$
          – Ya G
          Dec 11 '18 at 20:21




          1




          1




          $begingroup$
          The precise definition is given in en.wikipedia.org/wiki/Normal_family. If $V$ is a normal family, then It happens that every sequence in $V$ has a locally uniformly convergent subsequence. Montel's theorem provides a sufficient condition for this. So $sup|f'(c)|$ is actually attained in $V$, if $V$ is closed, by a limit $hin V$ of approaching sequence $f_nin V$ such that $|f'_n(c)| to sup|f'(c)|$.
          $endgroup$
          – Song
          Dec 11 '18 at 20:25






          $begingroup$
          The precise definition is given in en.wikipedia.org/wiki/Normal_family. If $V$ is a normal family, then It happens that every sequence in $V$ has a locally uniformly convergent subsequence. Montel's theorem provides a sufficient condition for this. So $sup|f'(c)|$ is actually attained in $V$, if $V$ is closed, by a limit $hin V$ of approaching sequence $f_nin V$ such that $|f'_n(c)| to sup|f'(c)|$.
          $endgroup$
          – Song
          Dec 11 '18 at 20:25






          1




          1




          $begingroup$
          This explanation is very rough, so if you are interested, see e.g. Rudin, Real and Complex Analysis.
          $endgroup$
          – Song
          Dec 11 '18 at 20:33




          $begingroup$
          This explanation is very rough, so if you are interested, see e.g. Rudin, Real and Complex Analysis.
          $endgroup$
          – Song
          Dec 11 '18 at 20:33


















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