Why does not the Montel Theorem apply in this case?
I have proved that the family is not equicontinuous, but it is not understanding why the theorem 3.3 can not be applied. Can someone explain me?
complex-analysis
asked Nov 29 at 12:53
Ricardo Freire
392110
add a comment |
I have proved that the family is not equicontinuous, but it is not understanding why the theorem 3.3 can not be applied. Can someone explain me?
complex-analysis
asked Nov 29 at 12:53
Ricardo Freire
392110
2
Probably because $sin(nx)$ is not holomorphic.
– Lukas Kofler
Nov 29 at 12:55
add a comment |
I have proved that the family is not equicontinuous, but it is not understanding why the theorem 3.3 can not be applied. Can someone explain me?
complex-analysis
asked Nov 29 at 12:53
Ricardo Freire
392110
I have proved that the family is not equicontinuous, but it is not understanding why the theorem 3.3 can not be applied. Can someone explain me?
complex-analysis
complex-analysis
asked Nov 29 at 12:53
Ricardo Freire
392110
asked Nov 29 at 12:53
Ricardo Freire
392110
asked Nov 29 at 12:53
Ricardo Freire
392110
asked Nov 29 at 12:53
Ricardo Freire
392110
asked Nov 29 at 12:53
Ricardo Freire
392110
392110
2
Probably because $sin(nx)$ is not holomorphic.
– Lukas Kofler
Nov 29 at 12:55
add a comment |
2
Probably because $sin(nx)$ is not holomorphic.
– Lukas Kofler
Nov 29 at 12:55
2
2
Probably because $sin(nx)$ is not holomorphic.
– Lukas Kofler
Nov 29 at 12:55
Probably because $sin(nx)$ is not holomorphic.
– Lukas Kofler
Nov 29 at 12:55
add a comment |
1 Answer
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This is because if we take any open set $A$ in the complex plane such that $(0,1)subset A$ then $A$ must have an element $z=a+bi$ with $bneq 0.$
Therefore $|sin nz|=|(2i)^{-1} (e^{nia -nb} -e^{-nia +nb})|to infty $ as $nto infty.$
So the family is not uniformly bounded in any region $A$ in complex plane, such that $(0,1)subset A.$
answered Nov 29 at 13:11
MotylaNogaTomkaMazura
6,564917
The book itself says that the family is uniformly limited, do not you think it can be something else? Being uniformly limited does not refer to the open containing the set, but to the compacts that are contained in the set.
– Ricardo Freire
Nov 29 at 13:37
1
In the theorem $Omega$ is an open subset of the plane. If $Omega$ is an open subset of the plane and $(0,1)subsetOmega$ then there exists a compact $Ksubset Omega$ such that $\sin(nz)$ is not uniformly bounded on $K$. That's clear from the above, since singletons are compact.
– David C. Ullrich
Nov 29 at 15:08
add a comment |
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1 Answer
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1 Answer
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This is because if we take any open set $A$ in the complex plane such that $(0,1)subset A$ then $A$ must have an element $z=a+bi$ with $bneq 0.$
Therefore $|sin nz|=|(2i)^{-1} (e^{nia -nb} -e^{-nia +nb})|to infty $ as $nto infty.$
So the family is not uniformly bounded in any region $A$ in complex plane, such that $(0,1)subset A.$
answered Nov 29 at 13:11
MotylaNogaTomkaMazura
6,564917
The book itself says that the family is uniformly limited, do not you think it can be something else? Being uniformly limited does not refer to the open containing the set, but to the compacts that are contained in the set.
– Ricardo Freire
Nov 29 at 13:37
1
In the theorem $Omega$ is an open subset of the plane. If $Omega$ is an open subset of the plane and $(0,1)subsetOmega$ then there exists a compact $Ksubset Omega$ such that $\sin(nz)$ is not uniformly bounded on $K$. That's clear from the above, since singletons are compact.
– David C. Ullrich
Nov 29 at 15:08
add a comment |
This is because if we take any open set $A$ in the complex plane such that $(0,1)subset A$ then $A$ must have an element $z=a+bi$ with $bneq 0.$
Therefore $|sin nz|=|(2i)^{-1} (e^{nia -nb} -e^{-nia +nb})|to infty $ as $nto infty.$
So the family is not uniformly bounded in any region $A$ in complex plane, such that $(0,1)subset A.$
answered Nov 29 at 13:11
MotylaNogaTomkaMazura
6,564917
The book itself says that the family is uniformly limited, do not you think it can be something else? Being uniformly limited does not refer to the open containing the set, but to the compacts that are contained in the set.
– Ricardo Freire
Nov 29 at 13:37
1
In the theorem $Omega$ is an open subset of the plane. If $Omega$ is an open subset of the plane and $(0,1)subsetOmega$ then there exists a compact $Ksubset Omega$ such that $\sin(nz)$ is not uniformly bounded on $K$. That's clear from the above, since singletons are compact.
– David C. Ullrich
Nov 29 at 15:08
add a comment |
This is because if we take any open set $A$ in the complex plane such that $(0,1)subset A$ then $A$ must have an element $z=a+bi$ with $bneq 0.$
Therefore $|sin nz|=|(2i)^{-1} (e^{nia -nb} -e^{-nia +nb})|to infty $ as $nto infty.$
So the family is not uniformly bounded in any region $A$ in complex plane, such that $(0,1)subset A.$
answered Nov 29 at 13:11
MotylaNogaTomkaMazura
6,564917
This is because if we take any open set $A$ in the complex plane such that $(0,1)subset A$ then $A$ must have an element $z=a+bi$ with $bneq 0.$
Therefore $|sin nz|=|(2i)^{-1} (e^{nia -nb} -e^{-nia +nb})|to infty $ as $nto infty.$
So the family is not uniformly bounded in any region $A$ in complex plane, such that $(0,1)subset A.$
answered Nov 29 at 13:11
MotylaNogaTomkaMazura
6,564917
answered Nov 29 at 13:11
MotylaNogaTomkaMazura
6,564917
answered Nov 29 at 13:11
MotylaNogaTomkaMazura
6,564917
answered Nov 29 at 13:11
MotylaNogaTomkaMazura
6,564917
6,564917
The book itself says that the family is uniformly limited, do not you think it can be something else? Being uniformly limited does not refer to the open containing the set, but to the compacts that are contained in the set.
– Ricardo Freire
Nov 29 at 13:37
1
In the theorem $Omega$ is an open subset of the plane. If $Omega$ is an open subset of the plane and $(0,1)subsetOmega$ then there exists a compact $Ksubset Omega$ such that $\sin(nz)$ is not uniformly bounded on $K$. That's clear from the above, since singletons are compact.
– David C. Ullrich
Nov 29 at 15:08
add a comment |
The book itself says that the family is uniformly limited, do not you think it can be something else? Being uniformly limited does not refer to the open containing the set, but to the compacts that are contained in the set.
– Ricardo Freire
Nov 29 at 13:37
1
In the theorem $Omega$ is an open subset of the plane. If $Omega$ is an open subset of the plane and $(0,1)subsetOmega$ then there exists a compact $Ksubset Omega$ such that $\sin(nz)$ is not uniformly bounded on $K$. That's clear from the above, since singletons are compact.
– David C. Ullrich
Nov 29 at 15:08
The book itself says that the family is uniformly limited, do not you think it can be something else? Being uniformly limited does not refer to the open containing the set, but to the compacts that are contained in the set.
– Ricardo Freire
Nov 29 at 13:37
The book itself says that the family is uniformly limited, do not you think it can be something else? Being uniformly limited does not refer to the open containing the set, but to the compacts that are contained in the set.
– Ricardo Freire
Nov 29 at 13:37
1
1
In the theorem $Omega$ is an open subset of the plane. If $Omega$ is an open subset of the plane and $(0,1)subsetOmega$ then there exists a compact $Ksubset Omega$ such that $\sin(nz)$ is not uniformly bounded on $K$. That's clear from the above, since singletons are compact.
– David C. Ullrich
Nov 29 at 15:08
In the theorem $Omega$ is an open subset of the plane. If $Omega$ is an open subset of the plane and $(0,1)subsetOmega$ then there exists a compact $Ksubset Omega$ such that $\sin(nz)$ is not uniformly bounded on $K$. That's clear from the above, since singletons are compact.
– David C. Ullrich
Nov 29 at 15:08
add a comment |
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2
Probably because $sin(nx)$ is not holomorphic.
– Lukas Kofler
Nov 29 at 12:55