If $A$ is a normal family, why does $bar A$ have to be?
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If $A$ is a normal family, why does $bar A$ have to be?
I am not sure how to prove this. Formally, given a sequence ${ f_n }$ of functions in $bar A$ we have to show that this has a subsequence which converges uniformly on evwry compact set in the domain. If ${f_n}$ has only finite number of functions from $bar A - A$ the ln this is true, but how can prove the general case?
normal-families
asked Nov 25 at 22:53
Cute Brownie
980316
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up vote
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down vote
favorite
If $A$ is a normal family, why does $bar A$ have to be?
I am not sure how to prove this. Formally, given a sequence ${ f_n }$ of functions in $bar A$ we have to show that this has a subsequence which converges uniformly on evwry compact set in the domain. If ${f_n}$ has only finite number of functions from $bar A - A$ the ln this is true, but how can prove the general case?
normal-families
asked Nov 25 at 22:53
Cute Brownie
980316
Great question! Keep it up.
– The Great Duck
Nov 26 at 0:02
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If $A$ is a normal family, why does $bar A$ have to be?
I am not sure how to prove this. Formally, given a sequence ${ f_n }$ of functions in $bar A$ we have to show that this has a subsequence which converges uniformly on evwry compact set in the domain. If ${f_n}$ has only finite number of functions from $bar A - A$ the ln this is true, but how can prove the general case?
normal-families
asked Nov 25 at 22:53
Cute Brownie
980316
If $A$ is a normal family, why does $bar A$ have to be?
I am not sure how to prove this. Formally, given a sequence ${ f_n }$ of functions in $bar A$ we have to show that this has a subsequence which converges uniformly on evwry compact set in the domain. If ${f_n}$ has only finite number of functions from $bar A - A$ the ln this is true, but how can prove the general case?
normal-families
normal-families
asked Nov 25 at 22:53
Cute Brownie
980316
asked Nov 25 at 22:53
Cute Brownie
980316
asked Nov 25 at 22:53
Cute Brownie
980316
asked Nov 25 at 22:53
Cute Brownie
980316
asked Nov 25 at 22:53
Cute Brownie
980316
980316
Great question! Keep it up.
– The Great Duck
Nov 26 at 0:02
add a comment |
Great question! Keep it up.
– The Great Duck
Nov 26 at 0:02
Great question! Keep it up.
– The Great Duck
Nov 26 at 0:02
Great question! Keep it up.
– The Great Duck
Nov 26 at 0:02
add a comment |
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A basic theorem about normal families says that a family of analytic functions is normal iff the the family is uniformly bounded on each compact subset of the domain. If $K$ is compact and $|f(z)| leq M$ for all $f in A$ for all $z in K$ then $|g(z)| leq M$ for all $z in K$ for any function $g$ in the closure of $A$, hence this closure is normal. [Note that closure here is w.r.t. uniform convergence on compact sets. So $g in overline {A}$ implies that $f_n to g$ uniformly on $K$ for some seqeunce ${f_n}$ in $A$].
answered Nov 25 at 23:46
Kavi Rama Murthy
46.3k31854
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A basic theorem about normal families says that a family of analytic functions is normal iff the the family is uniformly bounded on each compact subset of the domain. If $K$ is compact and $|f(z)| leq M$ for all $f in A$ for all $z in K$ then $|g(z)| leq M$ for all $z in K$ for any function $g$ in the closure of $A$, hence this closure is normal. [Note that closure here is w.r.t. uniform convergence on compact sets. So $g in overline {A}$ implies that $f_n to g$ uniformly on $K$ for some seqeunce ${f_n}$ in $A$].
answered Nov 25 at 23:46
Kavi Rama Murthy
46.3k31854
add a comment |
up vote
0
down vote
A basic theorem about normal families says that a family of analytic functions is normal iff the the family is uniformly bounded on each compact subset of the domain. If $K$ is compact and $|f(z)| leq M$ for all $f in A$ for all $z in K$ then $|g(z)| leq M$ for all $z in K$ for any function $g$ in the closure of $A$, hence this closure is normal. [Note that closure here is w.r.t. uniform convergence on compact sets. So $g in overline {A}$ implies that $f_n to g$ uniformly on $K$ for some seqeunce ${f_n}$ in $A$].
answered Nov 25 at 23:46
Kavi Rama Murthy
46.3k31854
add a comment |
up vote
0
down vote
up vote
0
down vote
A basic theorem about normal families says that a family of analytic functions is normal iff the the family is uniformly bounded on each compact subset of the domain. If $K$ is compact and $|f(z)| leq M$ for all $f in A$ for all $z in K$ then $|g(z)| leq M$ for all $z in K$ for any function $g$ in the closure of $A$, hence this closure is normal. [Note that closure here is w.r.t. uniform convergence on compact sets. So $g in overline {A}$ implies that $f_n to g$ uniformly on $K$ for some seqeunce ${f_n}$ in $A$].
answered Nov 25 at 23:46
Kavi Rama Murthy
46.3k31854
A basic theorem about normal families says that a family of analytic functions is normal iff the the family is uniformly bounded on each compact subset of the domain. If $K$ is compact and $|f(z)| leq M$ for all $f in A$ for all $z in K$ then $|g(z)| leq M$ for all $z in K$ for any function $g$ in the closure of $A$, hence this closure is normal. [Note that closure here is w.r.t. uniform convergence on compact sets. So $g in overline {A}$ implies that $f_n to g$ uniformly on $K$ for some seqeunce ${f_n}$ in $A$].
answered Nov 25 at 23:46
Kavi Rama Murthy
46.3k31854
answered Nov 25 at 23:46
Kavi Rama Murthy
46.3k31854
answered Nov 25 at 23:46
Kavi Rama Murthy
46.3k31854
answered Nov 25 at 23:46
Kavi Rama Murthy
46.3k31854
46.3k31854
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Great question! Keep it up.
– The Great Duck
Nov 26 at 0:02