Sequence of entire function that converges uniformly over on sets with empty interior
$begingroup$
I have to prove that the sequence of entire functions:
$$f_n(z)=frac 1n sin(nz)$$
converges uniformly over $mathbb{R}$ (and this I managed to verify) but doesn't on every set with non-empty interior of the complex plane $mathbb{C}$. I guess it has to do with Picard theorem but I'm not sure on how to proceed.
complex-analysis holomorphic-functions entire-functions sequence-of-function
asked Dec 9 '18 at 18:05
Renato FaraoneRenato Faraone
2,33111627
$endgroup$
|
show 1 more comment
$begingroup$
I have to prove that the sequence of entire functions:
$$f_n(z)=frac 1n sin(nz)$$
converges uniformly over $mathbb{R}$ (and this I managed to verify) but doesn't on every set with non-empty interior of the complex plane $mathbb{C}$. I guess it has to do with Picard theorem but I'm not sure on how to proceed.
complex-analysis holomorphic-functions entire-functions sequence-of-function
asked Dec 9 '18 at 18:05
Renato FaraoneRenato Faraone
2,33111627
$endgroup$
$begingroup$
Hint: Consider any open $S$ such that $Ssupset iBbb R= {ir:rin Bbb R}.$ If $rin Bbb R$ then $sin (nir)=sinh (nr).$ If $nin Bbb N$ is large and $r>0$ then $sinh (nr)approx e^{nr}/2$.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 18:48
$begingroup$
@DanielWainfleet what about open sets that don't contains the imaginary axis?
$endgroup$
– Renato Faraone
Dec 10 '18 at 8:20
$begingroup$
@RenatoFaraone the question seems to be to show that it doesn't converge uniformly on at least one open set in $mathbb{C},$ so we are done
$endgroup$
– Brevan Ellefsen
Dec 10 '18 at 12:11
$begingroup$
@BrevanEllefsen maybe this is a case of linguistic confusion, I want to prove: suppose $A$ is a set with non-empty interior, moreover such interior is not contained in the real axis, then such sequence doesn't converge uniformly over $A$.
$endgroup$
– Renato Faraone
Dec 10 '18 at 12:14
$begingroup$
@BrevanEllefsen That's makes a lot more sense. Probably the exercise were not well written or is my fault that I didn't quite understand it.
$endgroup$
– Renato Faraone
Dec 10 '18 at 12:26
|
show 1 more comment
$begingroup$
I have to prove that the sequence of entire functions:
$$f_n(z)=frac 1n sin(nz)$$
converges uniformly over $mathbb{R}$ (and this I managed to verify) but doesn't on every set with non-empty interior of the complex plane $mathbb{C}$. I guess it has to do with Picard theorem but I'm not sure on how to proceed.
complex-analysis holomorphic-functions entire-functions sequence-of-function
asked Dec 9 '18 at 18:05
Renato FaraoneRenato Faraone
2,33111627
$endgroup$
I have to prove that the sequence of entire functions:
$$f_n(z)=frac 1n sin(nz)$$
converges uniformly over $mathbb{R}$ (and this I managed to verify) but doesn't on every set with non-empty interior of the complex plane $mathbb{C}$. I guess it has to do with Picard theorem but I'm not sure on how to proceed.
complex-analysis holomorphic-functions entire-functions sequence-of-function
complex-analysis holomorphic-functions entire-functions sequence-of-function
asked Dec 9 '18 at 18:05
Renato FaraoneRenato Faraone
2,33111627
asked Dec 9 '18 at 18:05
Renato FaraoneRenato Faraone
2,33111627
asked Dec 9 '18 at 18:05
Renato FaraoneRenato Faraone
2,33111627
asked Dec 9 '18 at 18:05
Renato FaraoneRenato Faraone
2,33111627
asked Dec 9 '18 at 18:05
Renato FaraoneRenato Faraone
2,33111627
2,33111627
$begingroup$
Hint: Consider any open $S$ such that $Ssupset iBbb R= {ir:rin Bbb R}.$ If $rin Bbb R$ then $sin (nir)=sinh (nr).$ If $nin Bbb N$ is large and $r>0$ then $sinh (nr)approx e^{nr}/2$.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 18:48
$begingroup$
@DanielWainfleet what about open sets that don't contains the imaginary axis?
$endgroup$
– Renato Faraone
Dec 10 '18 at 8:20
$begingroup$
@RenatoFaraone the question seems to be to show that it doesn't converge uniformly on at least one open set in $mathbb{C},$ so we are done
$endgroup$
– Brevan Ellefsen
Dec 10 '18 at 12:11
$begingroup$
@BrevanEllefsen maybe this is a case of linguistic confusion, I want to prove: suppose $A$ is a set with non-empty interior, moreover such interior is not contained in the real axis, then such sequence doesn't converge uniformly over $A$.
$endgroup$
– Renato Faraone
Dec 10 '18 at 12:14
$begingroup$
@BrevanEllefsen That's makes a lot more sense. Probably the exercise were not well written or is my fault that I didn't quite understand it.
$endgroup$
– Renato Faraone
Dec 10 '18 at 12:26
|
show 1 more comment
$begingroup$
Hint: Consider any open $S$ such that $Ssupset iBbb R= {ir:rin Bbb R}.$ If $rin Bbb R$ then $sin (nir)=sinh (nr).$ If $nin Bbb N$ is large and $r>0$ then $sinh (nr)approx e^{nr}/2$.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 18:48
$begingroup$
@DanielWainfleet what about open sets that don't contains the imaginary axis?
$endgroup$
– Renato Faraone
Dec 10 '18 at 8:20
$begingroup$
@RenatoFaraone the question seems to be to show that it doesn't converge uniformly on at least one open set in $mathbb{C},$ so we are done
$endgroup$
– Brevan Ellefsen
Dec 10 '18 at 12:11
$begingroup$
@BrevanEllefsen maybe this is a case of linguistic confusion, I want to prove: suppose $A$ is a set with non-empty interior, moreover such interior is not contained in the real axis, then such sequence doesn't converge uniformly over $A$.
$endgroup$
– Renato Faraone
Dec 10 '18 at 12:14
$begingroup$
@BrevanEllefsen That's makes a lot more sense. Probably the exercise were not well written or is my fault that I didn't quite understand it.
$endgroup$
– Renato Faraone
Dec 10 '18 at 12:26
$begingroup$
Hint: Consider any open $S$ such that $Ssupset iBbb R= {ir:rin Bbb R}.$ If $rin Bbb R$ then $sin (nir)=sinh (nr).$ If $nin Bbb N$ is large and $r>0$ then $sinh (nr)approx e^{nr}/2$.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 18:48
$begingroup$
Hint: Consider any open $S$ such that $Ssupset iBbb R= {ir:rin Bbb R}.$ If $rin Bbb R$ then $sin (nir)=sinh (nr).$ If $nin Bbb N$ is large and $r>0$ then $sinh (nr)approx e^{nr}/2$.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 18:48
$begingroup$
@DanielWainfleet what about open sets that don't contains the imaginary axis?
$endgroup$
– Renato Faraone
Dec 10 '18 at 8:20
$begingroup$
@DanielWainfleet what about open sets that don't contains the imaginary axis?
$endgroup$
– Renato Faraone
Dec 10 '18 at 8:20
$begingroup$
@RenatoFaraone the question seems to be to show that it doesn't converge uniformly on at least one open set in $mathbb{C},$ so we are done
$endgroup$
– Brevan Ellefsen
Dec 10 '18 at 12:11
$begingroup$
@RenatoFaraone the question seems to be to show that it doesn't converge uniformly on at least one open set in $mathbb{C},$ so we are done
$endgroup$
– Brevan Ellefsen
Dec 10 '18 at 12:11
$begingroup$
@BrevanEllefsen maybe this is a case of linguistic confusion, I want to prove: suppose $A$ is a set with non-empty interior, moreover such interior is not contained in the real axis, then such sequence doesn't converge uniformly over $A$.
$endgroup$
– Renato Faraone
Dec 10 '18 at 12:14
$begingroup$
@BrevanEllefsen maybe this is a case of linguistic confusion, I want to prove: suppose $A$ is a set with non-empty interior, moreover such interior is not contained in the real axis, then such sequence doesn't converge uniformly over $A$.
$endgroup$
– Renato Faraone
Dec 10 '18 at 12:14
$begingroup$
@BrevanEllefsen That's makes a lot more sense. Probably the exercise were not well written or is my fault that I didn't quite understand it.
$endgroup$
– Renato Faraone
Dec 10 '18 at 12:26
$begingroup$
@BrevanEllefsen That's makes a lot more sense. Probably the exercise were not well written or is my fault that I didn't quite understand it.
$endgroup$
– Renato Faraone
Dec 10 '18 at 12:26
|
show 1 more comment
2 Answers
2
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oldest
votes
$begingroup$
We have $sin n(x+iy)=sin nz cosh ny +icos nz sinh ny.$
For convenience let $f(ny)=frac {e^{|ny|}-1}{2}.$
For $0ne yin Bbb R$ and $frac {1}{|y|}<nin Bbb N$ we have $$min(|sinh ny|,|cosh ny|)=|sinh ny|=frac {e^{|ny|}-e^{-|ny|}}{2}>frac {e^{|ny|}-1}{2}=f(ny).$$ We have $max (|sin nx|,|cos nx|)geq frac {1}{sqrt 2}$ because $|sin nx|^2+|cos nx|^2geq |sin^2 nx+cos^2 nx|=1.$
So if $z=x+iy$ with $x,yin Bbb R$ and $yne 0$, and if $frac {1}{|y|}<nin Bbb N$ then $$n^{-1}|sin nz|geq n^{-1} max (|Re (sin nz)|,|Im(sin nz|)=$$ $$=n^{-1}max (|sin nxcosh ny|, |cos nx sinh ny|)geq$$ $$> n^{-1}max (|sin nx|cdot f(ny),|cos nx|cdot f(ny))geq$$ $$geq n^{-1} frac {1}{sqrt 2}f(ny)=frac {e^{|ny|}-1}{2n sqrt 2}$$ which $to infty$ as $nto infty.$
answered Dec 11 '18 at 6:07
DanielWainfleetDanielWainfleet
34.7k31648
$endgroup$
add a comment |
$begingroup$
If $A$ is a subset of $Bbb C$ with nonempty interior, then there exists $zin A$ with $z=x+i,y$, $yne0$. Then
$$
sin(n,z)=sin(n,x)cosh(n,y)+icos(n,x)sinh(n,y).
$$
It is now easy to see that this sequence is unbounded.
answered Dec 10 '18 at 17:19
Julián AguirreJulián Aguirre
68.1k24094
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
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oldest
votes
$begingroup$
We have $sin n(x+iy)=sin nz cosh ny +icos nz sinh ny.$
For convenience let $f(ny)=frac {e^{|ny|}-1}{2}.$
For $0ne yin Bbb R$ and $frac {1}{|y|}<nin Bbb N$ we have $$min(|sinh ny|,|cosh ny|)=|sinh ny|=frac {e^{|ny|}-e^{-|ny|}}{2}>frac {e^{|ny|}-1}{2}=f(ny).$$ We have $max (|sin nx|,|cos nx|)geq frac {1}{sqrt 2}$ because $|sin nx|^2+|cos nx|^2geq |sin^2 nx+cos^2 nx|=1.$
So if $z=x+iy$ with $x,yin Bbb R$ and $yne 0$, and if $frac {1}{|y|}<nin Bbb N$ then $$n^{-1}|sin nz|geq n^{-1} max (|Re (sin nz)|,|Im(sin nz|)=$$ $$=n^{-1}max (|sin nxcosh ny|, |cos nx sinh ny|)geq$$ $$> n^{-1}max (|sin nx|cdot f(ny),|cos nx|cdot f(ny))geq$$ $$geq n^{-1} frac {1}{sqrt 2}f(ny)=frac {e^{|ny|}-1}{2n sqrt 2}$$ which $to infty$ as $nto infty.$
answered Dec 11 '18 at 6:07
DanielWainfleetDanielWainfleet
34.7k31648
$endgroup$
add a comment |
$begingroup$
We have $sin n(x+iy)=sin nz cosh ny +icos nz sinh ny.$
For convenience let $f(ny)=frac {e^{|ny|}-1}{2}.$
For $0ne yin Bbb R$ and $frac {1}{|y|}<nin Bbb N$ we have $$min(|sinh ny|,|cosh ny|)=|sinh ny|=frac {e^{|ny|}-e^{-|ny|}}{2}>frac {e^{|ny|}-1}{2}=f(ny).$$ We have $max (|sin nx|,|cos nx|)geq frac {1}{sqrt 2}$ because $|sin nx|^2+|cos nx|^2geq |sin^2 nx+cos^2 nx|=1.$
So if $z=x+iy$ with $x,yin Bbb R$ and $yne 0$, and if $frac {1}{|y|}<nin Bbb N$ then $$n^{-1}|sin nz|geq n^{-1} max (|Re (sin nz)|,|Im(sin nz|)=$$ $$=n^{-1}max (|sin nxcosh ny|, |cos nx sinh ny|)geq$$ $$> n^{-1}max (|sin nx|cdot f(ny),|cos nx|cdot f(ny))geq$$ $$geq n^{-1} frac {1}{sqrt 2}f(ny)=frac {e^{|ny|}-1}{2n sqrt 2}$$ which $to infty$ as $nto infty.$
answered Dec 11 '18 at 6:07
DanielWainfleetDanielWainfleet
34.7k31648
$endgroup$
add a comment |
$begingroup$
We have $sin n(x+iy)=sin nz cosh ny +icos nz sinh ny.$
For convenience let $f(ny)=frac {e^{|ny|}-1}{2}.$
For $0ne yin Bbb R$ and $frac {1}{|y|}<nin Bbb N$ we have $$min(|sinh ny|,|cosh ny|)=|sinh ny|=frac {e^{|ny|}-e^{-|ny|}}{2}>frac {e^{|ny|}-1}{2}=f(ny).$$ We have $max (|sin nx|,|cos nx|)geq frac {1}{sqrt 2}$ because $|sin nx|^2+|cos nx|^2geq |sin^2 nx+cos^2 nx|=1.$
So if $z=x+iy$ with $x,yin Bbb R$ and $yne 0$, and if $frac {1}{|y|}<nin Bbb N$ then $$n^{-1}|sin nz|geq n^{-1} max (|Re (sin nz)|,|Im(sin nz|)=$$ $$=n^{-1}max (|sin nxcosh ny|, |cos nx sinh ny|)geq$$ $$> n^{-1}max (|sin nx|cdot f(ny),|cos nx|cdot f(ny))geq$$ $$geq n^{-1} frac {1}{sqrt 2}f(ny)=frac {e^{|ny|}-1}{2n sqrt 2}$$ which $to infty$ as $nto infty.$
answered Dec 11 '18 at 6:07
DanielWainfleetDanielWainfleet
34.7k31648
$endgroup$
We have $sin n(x+iy)=sin nz cosh ny +icos nz sinh ny.$
For convenience let $f(ny)=frac {e^{|ny|}-1}{2}.$
For $0ne yin Bbb R$ and $frac {1}{|y|}<nin Bbb N$ we have $$min(|sinh ny|,|cosh ny|)=|sinh ny|=frac {e^{|ny|}-e^{-|ny|}}{2}>frac {e^{|ny|}-1}{2}=f(ny).$$ We have $max (|sin nx|,|cos nx|)geq frac {1}{sqrt 2}$ because $|sin nx|^2+|cos nx|^2geq |sin^2 nx+cos^2 nx|=1.$
So if $z=x+iy$ with $x,yin Bbb R$ and $yne 0$, and if $frac {1}{|y|}<nin Bbb N$ then $$n^{-1}|sin nz|geq n^{-1} max (|Re (sin nz)|,|Im(sin nz|)=$$ $$=n^{-1}max (|sin nxcosh ny|, |cos nx sinh ny|)geq$$ $$> n^{-1}max (|sin nx|cdot f(ny),|cos nx|cdot f(ny))geq$$ $$geq n^{-1} frac {1}{sqrt 2}f(ny)=frac {e^{|ny|}-1}{2n sqrt 2}$$ which $to infty$ as $nto infty.$
answered Dec 11 '18 at 6:07
DanielWainfleetDanielWainfleet
34.7k31648
answered Dec 11 '18 at 6:07
DanielWainfleetDanielWainfleet
34.7k31648
answered Dec 11 '18 at 6:07
DanielWainfleetDanielWainfleet
34.7k31648
answered Dec 11 '18 at 6:07
DanielWainfleetDanielWainfleet
34.7k31648
34.7k31648
add a comment |
add a comment |
$begingroup$
If $A$ is a subset of $Bbb C$ with nonempty interior, then there exists $zin A$ with $z=x+i,y$, $yne0$. Then
$$
sin(n,z)=sin(n,x)cosh(n,y)+icos(n,x)sinh(n,y).
$$
It is now easy to see that this sequence is unbounded.
answered Dec 10 '18 at 17:19
Julián AguirreJulián Aguirre
68.1k24094
$endgroup$
add a comment |
$begingroup$
If $A$ is a subset of $Bbb C$ with nonempty interior, then there exists $zin A$ with $z=x+i,y$, $yne0$. Then
$$
sin(n,z)=sin(n,x)cosh(n,y)+icos(n,x)sinh(n,y).
$$
It is now easy to see that this sequence is unbounded.
answered Dec 10 '18 at 17:19
Julián AguirreJulián Aguirre
68.1k24094
$endgroup$
add a comment |
$begingroup$
If $A$ is a subset of $Bbb C$ with nonempty interior, then there exists $zin A$ with $z=x+i,y$, $yne0$. Then
$$
sin(n,z)=sin(n,x)cosh(n,y)+icos(n,x)sinh(n,y).
$$
It is now easy to see that this sequence is unbounded.
answered Dec 10 '18 at 17:19
Julián AguirreJulián Aguirre
68.1k24094
$endgroup$
If $A$ is a subset of $Bbb C$ with nonempty interior, then there exists $zin A$ with $z=x+i,y$, $yne0$. Then
$$
sin(n,z)=sin(n,x)cosh(n,y)+icos(n,x)sinh(n,y).
$$
It is now easy to see that this sequence is unbounded.
answered Dec 10 '18 at 17:19
Julián AguirreJulián Aguirre
68.1k24094
answered Dec 10 '18 at 17:19
Julián AguirreJulián Aguirre
68.1k24094
answered Dec 10 '18 at 17:19
Julián AguirreJulián Aguirre
68.1k24094
answered Dec 10 '18 at 17:19
Julián AguirreJulián Aguirre
68.1k24094
68.1k24094
add a comment |
add a comment |
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$begingroup$
Hint: Consider any open $S$ such that $Ssupset iBbb R= {ir:rin Bbb R}.$ If $rin Bbb R$ then $sin (nir)=sinh (nr).$ If $nin Bbb N$ is large and $r>0$ then $sinh (nr)approx e^{nr}/2$.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 18:48
$begingroup$
@DanielWainfleet what about open sets that don't contains the imaginary axis?
$endgroup$
– Renato Faraone
Dec 10 '18 at 8:20
$begingroup$
@RenatoFaraone the question seems to be to show that it doesn't converge uniformly on at least one open set in $mathbb{C},$ so we are done
$endgroup$
– Brevan Ellefsen
Dec 10 '18 at 12:11
$begingroup$
@BrevanEllefsen maybe this is a case of linguistic confusion, I want to prove: suppose $A$ is a set with non-empty interior, moreover such interior is not contained in the real axis, then such sequence doesn't converge uniformly over $A$.
$endgroup$
– Renato Faraone
Dec 10 '18 at 12:14
$begingroup$
@BrevanEllefsen That's makes a lot more sense. Probably the exercise were not well written or is my fault that I didn't quite understand it.
$endgroup$
– Renato Faraone
Dec 10 '18 at 12:26