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A sequence of holomorphic functions in D(0,1)
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Let $f_n$ be a sequence of holomorphic functions in the open unit disk $D(0,1)$ such that $Re f_n geq 0$.
How can I show that either $|f_n(z)| rightarrow infty$ as $ n rightarrow infty$ for all $z in D(0,1)$ or $f_n$ has a subsequence which converges uniformly on every compact subset of $D(0,1)$?
complex-analysis holomorphic-functions
asked Jul 16 at 16:11
Rachel.L
375
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Let $f_n$ be a sequence of holomorphic functions in the open unit disk $D(0,1)$ such that $Re f_n geq 0$.
How can I show that either $|f_n(z)| rightarrow infty$ as $ n rightarrow infty$ for all $z in D(0,1)$ or $f_n$ has a subsequence which converges uniformly on every compact subset of $D(0,1)$?
complex-analysis holomorphic-functions
asked Jul 16 at 16:11
Rachel.L
375
Hint: en.wikipedia.org/wiki/Harnack%27s_inequality
– David C. Ullrich
Jul 16 at 16:33
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up vote
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down vote
favorite
up vote
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down vote
favorite
Let $f_n$ be a sequence of holomorphic functions in the open unit disk $D(0,1)$ such that $Re f_n geq 0$.
How can I show that either $|f_n(z)| rightarrow infty$ as $ n rightarrow infty$ for all $z in D(0,1)$ or $f_n$ has a subsequence which converges uniformly on every compact subset of $D(0,1)$?
complex-analysis holomorphic-functions
asked Jul 16 at 16:11
Rachel.L
375
Let $f_n$ be a sequence of holomorphic functions in the open unit disk $D(0,1)$ such that $Re f_n geq 0$.
How can I show that either $|f_n(z)| rightarrow infty$ as $ n rightarrow infty$ for all $z in D(0,1)$ or $f_n$ has a subsequence which converges uniformly on every compact subset of $D(0,1)$?
complex-analysis holomorphic-functions
asked Jul 16 at 16:11
Rachel.L
375
edited Jul 26 at 21:19
asked Jul 16 at 16:11
Rachel.L
375
asked Jul 16 at 16:11
Rachel.L
375
asked Jul 16 at 16:11
Rachel.L
375
375
Hint: en.wikipedia.org/wiki/Harnack%27s_inequality
– David C. Ullrich
Jul 16 at 16:33
add a comment |Â
Hint: en.wikipedia.org/wiki/Harnack%27s_inequality
– David C. Ullrich
Jul 16 at 16:33
Hint: en.wikipedia.org/wiki/Harnack%27s_inequality
– David C. Ullrich
Jul 16 at 16:33
Hint: en.wikipedia.org/wiki/Harnack%27s_inequality
– David C. Ullrich
Jul 16 at 16:33
add a comment |Â
1 Answer
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It follows easily from the version of Harnak's inequality here that if $K$ is a compact subset of $Bbb D=D(0,1)$ there exist $c_K$ and $C_K$ such that
$$c_Ku(0)le u(z)le C_Ku(0)quad(zin K)$$for every positive harmonic function $u$ in $Bbb D$.
So if $Re f_n(0)toinfty$ then $|f_n(z)|toinfty$ uniformly on compact sets. Otoh if $Re f_n(0)nottoinfty$ then restricting to a subsequence we can assume that $Re f_n(0)$ is bounded; now the inequality above shows that $e^f_n$ is uniformly bounded on compact sets...
Heh, a slick way to get around unpleasantness with the logarithm for the punchline: So again passing to a subequence we can assume that $e^f_nto g$ uniformly on compact sets. Since $left|e^f_nright|ge1$ it follows that $g$ has no zero, so $g=e^f$. How to get from there to $f_nto f$: Since normal convergence implies normal convergence of the derivatives we have $f_n'e^f_nto f'e^f$, and now since $left|e^f_nright|ge1$ it follows that $f_n'to f'$ uniformly on compact sets. So if we choose our logarithms so that $f_n(0)to f(0)$ it follows that $f_nto f$ uniformly on compact sets.
answered Jul 16 at 23:34
David C. Ullrich
54.3k33583
Why not just go directly to $e^-f_n $ is uniformly bounded?
– zhw.
Jul 16 at 23:46
@zhw. This is what they call a dense argument...
– David C. Ullrich
Jul 16 at 23:48
@zhw. Actually I have a better reason than I did when I wrote the first version. Saw a cute way to handle the technicalities at the end; for that to work with the $e^-f_n$ version I need to know that $|e^-f_n|ge c>0$ on compact sets.
– David C. Ullrich
Jul 16 at 23:58
@Rachel.L Seriously? $g$ is holomorphic in a disk and has no zero, so it has a holomorphic logarithm...
– David C. Ullrich
Jul 24 at 20:57
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
It follows easily from the version of Harnak's inequality here that if $K$ is a compact subset of $Bbb D=D(0,1)$ there exist $c_K$ and $C_K$ such that
$$c_Ku(0)le u(z)le C_Ku(0)quad(zin K)$$for every positive harmonic function $u$ in $Bbb D$.
So if $Re f_n(0)toinfty$ then $|f_n(z)|toinfty$ uniformly on compact sets. Otoh if $Re f_n(0)nottoinfty$ then restricting to a subsequence we can assume that $Re f_n(0)$ is bounded; now the inequality above shows that $e^f_n$ is uniformly bounded on compact sets...
Heh, a slick way to get around unpleasantness with the logarithm for the punchline: So again passing to a subequence we can assume that $e^f_nto g$ uniformly on compact sets. Since $left|e^f_nright|ge1$ it follows that $g$ has no zero, so $g=e^f$. How to get from there to $f_nto f$: Since normal convergence implies normal convergence of the derivatives we have $f_n'e^f_nto f'e^f$, and now since $left|e^f_nright|ge1$ it follows that $f_n'to f'$ uniformly on compact sets. So if we choose our logarithms so that $f_n(0)to f(0)$ it follows that $f_nto f$ uniformly on compact sets.
answered Jul 16 at 23:34
David C. Ullrich
54.3k33583
Why not just go directly to $e^-f_n $ is uniformly bounded?
– zhw.
Jul 16 at 23:46
@zhw. This is what they call a dense argument...
– David C. Ullrich
Jul 16 at 23:48
@zhw. Actually I have a better reason than I did when I wrote the first version. Saw a cute way to handle the technicalities at the end; for that to work with the $e^-f_n$ version I need to know that $|e^-f_n|ge c>0$ on compact sets.
– David C. Ullrich
Jul 16 at 23:58
@Rachel.L Seriously? $g$ is holomorphic in a disk and has no zero, so it has a holomorphic logarithm...
– David C. Ullrich
Jul 24 at 20:57
add a comment |Â
up vote
0
down vote
accepted
It follows easily from the version of Harnak's inequality here that if $K$ is a compact subset of $Bbb D=D(0,1)$ there exist $c_K$ and $C_K$ such that
$$c_Ku(0)le u(z)le C_Ku(0)quad(zin K)$$for every positive harmonic function $u$ in $Bbb D$.
So if $Re f_n(0)toinfty$ then $|f_n(z)|toinfty$ uniformly on compact sets. Otoh if $Re f_n(0)nottoinfty$ then restricting to a subsequence we can assume that $Re f_n(0)$ is bounded; now the inequality above shows that $e^f_n$ is uniformly bounded on compact sets...
Heh, a slick way to get around unpleasantness with the logarithm for the punchline: So again passing to a subequence we can assume that $e^f_nto g$ uniformly on compact sets. Since $left|e^f_nright|ge1$ it follows that $g$ has no zero, so $g=e^f$. How to get from there to $f_nto f$: Since normal convergence implies normal convergence of the derivatives we have $f_n'e^f_nto f'e^f$, and now since $left|e^f_nright|ge1$ it follows that $f_n'to f'$ uniformly on compact sets. So if we choose our logarithms so that $f_n(0)to f(0)$ it follows that $f_nto f$ uniformly on compact sets.
answered Jul 16 at 23:34
David C. Ullrich
54.3k33583
Why not just go directly to $e^-f_n $ is uniformly bounded?
– zhw.
Jul 16 at 23:46
@zhw. This is what they call a dense argument...
– David C. Ullrich
Jul 16 at 23:48
@zhw. Actually I have a better reason than I did when I wrote the first version. Saw a cute way to handle the technicalities at the end; for that to work with the $e^-f_n$ version I need to know that $|e^-f_n|ge c>0$ on compact sets.
– David C. Ullrich
Jul 16 at 23:58
@Rachel.L Seriously? $g$ is holomorphic in a disk and has no zero, so it has a holomorphic logarithm...
– David C. Ullrich
Jul 24 at 20:57
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
It follows easily from the version of Harnak's inequality here that if $K$ is a compact subset of $Bbb D=D(0,1)$ there exist $c_K$ and $C_K$ such that
$$c_Ku(0)le u(z)le C_Ku(0)quad(zin K)$$for every positive harmonic function $u$ in $Bbb D$.
So if $Re f_n(0)toinfty$ then $|f_n(z)|toinfty$ uniformly on compact sets. Otoh if $Re f_n(0)nottoinfty$ then restricting to a subsequence we can assume that $Re f_n(0)$ is bounded; now the inequality above shows that $e^f_n$ is uniformly bounded on compact sets...
Heh, a slick way to get around unpleasantness with the logarithm for the punchline: So again passing to a subequence we can assume that $e^f_nto g$ uniformly on compact sets. Since $left|e^f_nright|ge1$ it follows that $g$ has no zero, so $g=e^f$. How to get from there to $f_nto f$: Since normal convergence implies normal convergence of the derivatives we have $f_n'e^f_nto f'e^f$, and now since $left|e^f_nright|ge1$ it follows that $f_n'to f'$ uniformly on compact sets. So if we choose our logarithms so that $f_n(0)to f(0)$ it follows that $f_nto f$ uniformly on compact sets.
answered Jul 16 at 23:34
David C. Ullrich
54.3k33583
It follows easily from the version of Harnak's inequality here that if $K$ is a compact subset of $Bbb D=D(0,1)$ there exist $c_K$ and $C_K$ such that
$$c_Ku(0)le u(z)le C_Ku(0)quad(zin K)$$for every positive harmonic function $u$ in $Bbb D$.
So if $Re f_n(0)toinfty$ then $|f_n(z)|toinfty$ uniformly on compact sets. Otoh if $Re f_n(0)nottoinfty$ then restricting to a subsequence we can assume that $Re f_n(0)$ is bounded; now the inequality above shows that $e^f_n$ is uniformly bounded on compact sets...
Heh, a slick way to get around unpleasantness with the logarithm for the punchline: So again passing to a subequence we can assume that $e^f_nto g$ uniformly on compact sets. Since $left|e^f_nright|ge1$ it follows that $g$ has no zero, so $g=e^f$. How to get from there to $f_nto f$: Since normal convergence implies normal convergence of the derivatives we have $f_n'e^f_nto f'e^f$, and now since $left|e^f_nright|ge1$ it follows that $f_n'to f'$ uniformly on compact sets. So if we choose our logarithms so that $f_n(0)to f(0)$ it follows that $f_nto f$ uniformly on compact sets.
answered Jul 16 at 23:34
David C. Ullrich
54.3k33583
edited Jul 16 at 23:55
answered Jul 16 at 23:34
David C. Ullrich
54.3k33583
answered Jul 16 at 23:34
David C. Ullrich
54.3k33583
answered Jul 16 at 23:34
David C. Ullrich
54.3k33583
54.3k33583
Why not just go directly to $e^-f_n $ is uniformly bounded?
– zhw.
Jul 16 at 23:46
@zhw. This is what they call a dense argument...
– David C. Ullrich
Jul 16 at 23:48
@zhw. Actually I have a better reason than I did when I wrote the first version. Saw a cute way to handle the technicalities at the end; for that to work with the $e^-f_n$ version I need to know that $|e^-f_n|ge c>0$ on compact sets.
– David C. Ullrich
Jul 16 at 23:58
@Rachel.L Seriously? $g$ is holomorphic in a disk and has no zero, so it has a holomorphic logarithm...
– David C. Ullrich
Jul 24 at 20:57
add a comment |Â
Why not just go directly to $e^-f_n $ is uniformly bounded?
– zhw.
Jul 16 at 23:46
@zhw. This is what they call a dense argument...
– David C. Ullrich
Jul 16 at 23:48
@zhw. Actually I have a better reason than I did when I wrote the first version. Saw a cute way to handle the technicalities at the end; for that to work with the $e^-f_n$ version I need to know that $|e^-f_n|ge c>0$ on compact sets.
– David C. Ullrich
Jul 16 at 23:58
@Rachel.L Seriously? $g$ is holomorphic in a disk and has no zero, so it has a holomorphic logarithm...
– David C. Ullrich
Jul 24 at 20:57
Why not just go directly to $e^-f_n $ is uniformly bounded?
– zhw.
Jul 16 at 23:46
Why not just go directly to $e^-f_n $ is uniformly bounded?
– zhw.
Jul 16 at 23:46
@zhw. This is what they call a dense argument...
– David C. Ullrich
Jul 16 at 23:48
@zhw. This is what they call a dense argument...
– David C. Ullrich
Jul 16 at 23:48
@zhw. Actually I have a better reason than I did when I wrote the first version. Saw a cute way to handle the technicalities at the end; for that to work with the $e^-f_n$ version I need to know that $|e^-f_n|ge c>0$ on compact sets.
– David C. Ullrich
Jul 16 at 23:58
@zhw. Actually I have a better reason than I did when I wrote the first version. Saw a cute way to handle the technicalities at the end; for that to work with the $e^-f_n$ version I need to know that $|e^-f_n|ge c>0$ on compact sets.
– David C. Ullrich
Jul 16 at 23:58
@Rachel.L Seriously? $g$ is holomorphic in a disk and has no zero, so it has a holomorphic logarithm...
– David C. Ullrich
Jul 24 at 20:57
@Rachel.L Seriously? $g$ is holomorphic in a disk and has no zero, so it has a holomorphic logarithm...
– David C. Ullrich
Jul 24 at 20:57
add a comment |Â
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Hint: en.wikipedia.org/wiki/Harnack%27s_inequality
– David C. Ullrich
Jul 16 at 16:33