if $f_n(z)rightarrow f(z)$ uniformly, then $ frac{f'_n(z)}{f_n(z)}rightarrowfrac{f'(z)}{f(z)}$ uniformly?
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I am reading "The Taylor Series: An Introduction to the Theory of Functions of a Complex Variable", Dienes P.N., Dover (1957).
In the proof of a theorem (Hurwitz) on page 351 it says that, if $f_n(z)rightarrow f(z)$ uniformly, and given a circle C in wich $f(z)neq 0$ then
$$ frac{f'_n(z)}{f_n(z)}rightarrowfrac{f'(z)}{f(z)}$$ uniformly on C. And moreover,
$$frac{1}{2pi i}oint_C frac{f'_n(z)}{f_n(z)}dzrightarrowfrac{1}{2pi i}oint_C frac{f'(z)}{f(z)}dz$$
This statements are not exactly clear to me, is there any small proof to convince me?
complex-analysis convergence uniform-convergence
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Ajafca
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I am reading "The Taylor Series: An Introduction to the Theory of Functions of a Complex Variable", Dienes P.N., Dover (1957).
In the proof of a theorem (Hurwitz) on page 351 it says that, if $f_n(z)rightarrow f(z)$ uniformly, and given a circle C in wich $f(z)neq 0$ then
$$ frac{f'_n(z)}{f_n(z)}rightarrowfrac{f'(z)}{f(z)}$$ uniformly on C. And moreover,
$$frac{1}{2pi i}oint_C frac{f'_n(z)}{f_n(z)}dzrightarrowfrac{1}{2pi i}oint_C frac{f'(z)}{f(z)}dz$$
This statements are not exactly clear to me, is there any small proof to convince me?
complex-analysis convergence uniform-convergence
asked yesterday
Ajafca
61
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am reading "The Taylor Series: An Introduction to the Theory of Functions of a Complex Variable", Dienes P.N., Dover (1957).
In the proof of a theorem (Hurwitz) on page 351 it says that, if $f_n(z)rightarrow f(z)$ uniformly, and given a circle C in wich $f(z)neq 0$ then
$$ frac{f'_n(z)}{f_n(z)}rightarrowfrac{f'(z)}{f(z)}$$ uniformly on C. And moreover,
$$frac{1}{2pi i}oint_C frac{f'_n(z)}{f_n(z)}dzrightarrowfrac{1}{2pi i}oint_C frac{f'(z)}{f(z)}dz$$
This statements are not exactly clear to me, is there any small proof to convince me?
complex-analysis convergence uniform-convergence
asked yesterday
Ajafca
61
I am reading "The Taylor Series: An Introduction to the Theory of Functions of a Complex Variable", Dienes P.N., Dover (1957).
In the proof of a theorem (Hurwitz) on page 351 it says that, if $f_n(z)rightarrow f(z)$ uniformly, and given a circle C in wich $f(z)neq 0$ then
$$ frac{f'_n(z)}{f_n(z)}rightarrowfrac{f'(z)}{f(z)}$$ uniformly on C. And moreover,
$$frac{1}{2pi i}oint_C frac{f'_n(z)}{f_n(z)}dzrightarrowfrac{1}{2pi i}oint_C frac{f'(z)}{f(z)}dz$$
This statements are not exactly clear to me, is there any small proof to convince me?
complex-analysis convergence uniform-convergence
complex-analysis convergence uniform-convergence
asked yesterday
Ajafca
61
asked yesterday
Ajafca
61
asked yesterday
Ajafca
61
asked yesterday
Ajafca
61
asked yesterday
Ajafca
61
61
add a comment |
add a comment |
1 Answer
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This uses the fact that when a sequence $(f_n)_{ninmathbb N}$ of analytic functions converges uniformly to $f$ on each compact set, then $(f_n{,'})_{ninmathbb N}$ also converges uniformly to $f'$ on each compact. This, together with the fact that $f$ has no zeros on $C$ (which implies that $f_n$ has no zeros there if $ngg1$) is enough to prove that statement.
answered yesterday
José Carlos Santos
137k17110200
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
This uses the fact that when a sequence $(f_n)_{ninmathbb N}$ of analytic functions converges uniformly to $f$ on each compact set, then $(f_n{,'})_{ninmathbb N}$ also converges uniformly to $f'$ on each compact. This, together with the fact that $f$ has no zeros on $C$ (which implies that $f_n$ has no zeros there if $ngg1$) is enough to prove that statement.
answered yesterday
José Carlos Santos
137k17110200
add a comment |
up vote
2
down vote
This uses the fact that when a sequence $(f_n)_{ninmathbb N}$ of analytic functions converges uniformly to $f$ on each compact set, then $(f_n{,'})_{ninmathbb N}$ also converges uniformly to $f'$ on each compact. This, together with the fact that $f$ has no zeros on $C$ (which implies that $f_n$ has no zeros there if $ngg1$) is enough to prove that statement.
answered yesterday
José Carlos Santos
137k17110200
add a comment |
up vote
2
down vote
up vote
2
down vote
This uses the fact that when a sequence $(f_n)_{ninmathbb N}$ of analytic functions converges uniformly to $f$ on each compact set, then $(f_n{,'})_{ninmathbb N}$ also converges uniformly to $f'$ on each compact. This, together with the fact that $f$ has no zeros on $C$ (which implies that $f_n$ has no zeros there if $ngg1$) is enough to prove that statement.
answered yesterday
José Carlos Santos
137k17110200
This uses the fact that when a sequence $(f_n)_{ninmathbb N}$ of analytic functions converges uniformly to $f$ on each compact set, then $(f_n{,'})_{ninmathbb N}$ also converges uniformly to $f'$ on each compact. This, together with the fact that $f$ has no zeros on $C$ (which implies that $f_n$ has no zeros there if $ngg1$) is enough to prove that statement.
answered yesterday
José Carlos Santos
137k17110200
answered yesterday
José Carlos Santos
137k17110200
answered yesterday
José Carlos Santos
137k17110200
answered yesterday
José Carlos Santos
137k17110200
137k17110200
add a comment |
add a comment |
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