construct a continuous function which vanishes at infinity
up vote
-1
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I want to construct a continuous function $f$ on $mathbb{C}$ such that $f(0)=0 $ and $f$ vanishes at infinity.Can anyone give me some examples,thanks!
calculus complex-analysis functional-analysis functions
asked Nov 18 at 10:48
mathrookie
709512
add a comment |
up vote
-1
down vote
favorite
I want to construct a continuous function $f$ on $mathbb{C}$ such that $f(0)=0 $ and $f$ vanishes at infinity.Can anyone give me some examples,thanks!
calculus complex-analysis functional-analysis functions
asked Nov 18 at 10:48
mathrookie
709512
3
$f(z)=dfrac{z}{e^z}$
– Yadati Kiran
Nov 18 at 10:51
This example is wrong. $f(2npi i)=2npi i $ which does not tend to $0$.
– Kavi Rama Murthy
Nov 18 at 11:53
What about $f(z) = 0$?
– gerw
Nov 18 at 19:28
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I want to construct a continuous function $f$ on $mathbb{C}$ such that $f(0)=0 $ and $f$ vanishes at infinity.Can anyone give me some examples,thanks!
calculus complex-analysis functional-analysis functions
asked Nov 18 at 10:48
mathrookie
709512
I want to construct a continuous function $f$ on $mathbb{C}$ such that $f(0)=0 $ and $f$ vanishes at infinity.Can anyone give me some examples,thanks!
calculus complex-analysis functional-analysis functions
calculus complex-analysis functional-analysis functions
asked Nov 18 at 10:48
mathrookie
709512
asked Nov 18 at 10:48
mathrookie
709512
asked Nov 18 at 10:48
mathrookie
709512
asked Nov 18 at 10:48
mathrookie
709512
asked Nov 18 at 10:48
mathrookie
709512
709512
3
$f(z)=dfrac{z}{e^z}$
– Yadati Kiran
Nov 18 at 10:51
This example is wrong. $f(2npi i)=2npi i $ which does not tend to $0$.
– Kavi Rama Murthy
Nov 18 at 11:53
What about $f(z) = 0$?
– gerw
Nov 18 at 19:28
add a comment |
3
$f(z)=dfrac{z}{e^z}$
– Yadati Kiran
Nov 18 at 10:51
This example is wrong. $f(2npi i)=2npi i $ which does not tend to $0$.
– Kavi Rama Murthy
Nov 18 at 11:53
What about $f(z) = 0$?
– gerw
Nov 18 at 19:28
3
3
$f(z)=dfrac{z}{e^z}$
– Yadati Kiran
Nov 18 at 10:51
$f(z)=dfrac{z}{e^z}$
– Yadati Kiran
Nov 18 at 10:51
This example is wrong. $f(2npi i)=2npi i $ which does not tend to $0$.
– Kavi Rama Murthy
Nov 18 at 11:53
This example is wrong. $f(2npi i)=2npi i $ which does not tend to $0$.
– Kavi Rama Murthy
Nov 18 at 11:53
What about $f(z) = 0$?
– gerw
Nov 18 at 19:28
What about $f(z) = 0$?
– gerw
Nov 18 at 19:28
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
$f(z)=z$ for $|z| <1$, $f(z)=frac z {|z|^{2}}$ if $|z| geq 1$.
answered Nov 18 at 11:56
Kavi Rama Murthy
41.7k31751
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
$f(z)=z$ for $|z| <1$, $f(z)=frac z {|z|^{2}}$ if $|z| geq 1$.
answered Nov 18 at 11:56
Kavi Rama Murthy
41.7k31751
add a comment |
up vote
2
down vote
accepted
$f(z)=z$ for $|z| <1$, $f(z)=frac z {|z|^{2}}$ if $|z| geq 1$.
answered Nov 18 at 11:56
Kavi Rama Murthy
41.7k31751
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
$f(z)=z$ for $|z| <1$, $f(z)=frac z {|z|^{2}}$ if $|z| geq 1$.
answered Nov 18 at 11:56
Kavi Rama Murthy
41.7k31751
$f(z)=z$ for $|z| <1$, $f(z)=frac z {|z|^{2}}$ if $|z| geq 1$.
answered Nov 18 at 11:56
Kavi Rama Murthy
41.7k31751
answered Nov 18 at 11:56
Kavi Rama Murthy
41.7k31751
answered Nov 18 at 11:56
Kavi Rama Murthy
41.7k31751
answered Nov 18 at 11:56
Kavi Rama Murthy
41.7k31751
41.7k31751
add a comment |
add a comment |
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3
$f(z)=dfrac{z}{e^z}$
– Yadati Kiran
Nov 18 at 10:51
This example is wrong. $f(2npi i)=2npi i $ which does not tend to $0$.
– Kavi Rama Murthy
Nov 18 at 11:53
What about $f(z) = 0$?
– gerw
Nov 18 at 19:28