Prove that there exists $hin V$, such that $|f'(frac{1}{2})|leq|h'(frac{1}{2})|$ for all $fin V$.
$begingroup$
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
edited Dec 11 '18 at 20:18
the_candyman
9,14032145
asked Dec 11 '18 at 20:08
Ya GYa G
536211
$endgroup$
add a comment |
$begingroup$
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
edited Dec 11 '18 at 20:18
the_candyman
9,14032145
asked Dec 11 '18 at 20:08
Ya GYa G
536211
$endgroup$
add a comment |
$begingroup$
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
edited Dec 11 '18 at 20:18
the_candyman
9,14032145
asked Dec 11 '18 at 20:08
Ya GYa G
536211
$endgroup$
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
edited Dec 11 '18 at 20:18
the_candyman
9,14032145
asked Dec 11 '18 at 20:08
Ya GYa G
536211
edited Dec 11 '18 at 20:18
the_candyman
9,14032145
asked Dec 11 '18 at 20:08
Ya GYa G
536211
edited Dec 11 '18 at 20:18
the_candyman
9,14032145
edited Dec 11 '18 at 20:18
the_candyman
9,14032145
edited Dec 11 '18 at 20:18
the_candyman
9,14032145
9,14032145
asked Dec 11 '18 at 20:08
Ya GYa G
536211
asked Dec 11 '18 at 20:08
Ya GYa G
536211
asked Dec 11 '18 at 20:08
Ya GYa G
536211
536211
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Dec 11 '18 at 20:18
SongSong
18.5k21651
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Dec 11 '18 at 20:18
SongSong
18.5k21651
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 11 '18 at 20:18
SongSong
18.5k21651
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 11 '18 at 20:18
SongSong
18.5k21651
$endgroup$
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.
answered Dec 11 '18 at 20:18
SongSong
18.5k21651
edited Dec 11 '18 at 20:47
answered Dec 11 '18 at 20:18
SongSong
18.5k21651
answered Dec 11 '18 at 20:18
SongSong
18.5k21651
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
add a comment |
$begingroup$
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
add a comment |
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