non-constant entire functions (Liouville) [closed]
I have to show by using Liouville's theorem, whether there are non-constant entire functions such that:
$ f( mathbb{C}) $ is in an half-plane.
I found out, that this is not possible for non-constant functions, because this property must hold:
$ forall epsilon>0 forall w in mathbb{C} exists mathbb{C} : |f(z)-w|< epsilon$
Am I right?
complex-analysis entire-functions
edited Dec 12 '18 at 12:38
Arnaud D.
15.7k52443
asked Dec 12 '18 at 7:49
SvenMathSvenMath
104
closed as unclear what you're asking by Nosrati, Michael Hoppe, Arnaud D., Chris Custer, Did Dec 12 '18 at 21:01
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
I have to show by using Liouville's theorem, whether there are non-constant entire functions such that:
$ f( mathbb{C}) $ is in an half-plane.
I found out, that this is not possible for non-constant functions, because this property must hold:
$ forall epsilon>0 forall w in mathbb{C} exists mathbb{C} : |f(z)-w|< epsilon$
Am I right?
complex-analysis entire-functions
edited Dec 12 '18 at 12:38
Arnaud D.
15.7k52443
asked Dec 12 '18 at 7:49
SvenMathSvenMath
104
closed as unclear what you're asking by Nosrati, Michael Hoppe, Arnaud D., Chris Custer, Did Dec 12 '18 at 21:01
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
I have to show by using Liouville's theorem, whether there are non-constant entire functions such that:
$ f( mathbb{C}) $ is in an half-plane.
I found out, that this is not possible for non-constant functions, because this property must hold:
$ forall epsilon>0 forall w in mathbb{C} exists mathbb{C} : |f(z)-w|< epsilon$
Am I right?
complex-analysis entire-functions
edited Dec 12 '18 at 12:38
Arnaud D.
15.7k52443
asked Dec 12 '18 at 7:49
SvenMathSvenMath
104
I have to show by using Liouville's theorem, whether there are non-constant entire functions such that:
$ f( mathbb{C}) $ is in an half-plane.
I found out, that this is not possible for non-constant functions, because this property must hold:
$ forall epsilon>0 forall w in mathbb{C} exists mathbb{C} : |f(z)-w|< epsilon$
Am I right?
complex-analysis entire-functions
complex-analysis entire-functions
edited Dec 12 '18 at 12:38
Arnaud D.
15.7k52443
asked Dec 12 '18 at 7:49
SvenMathSvenMath
104
edited Dec 12 '18 at 12:38
Arnaud D.
15.7k52443
asked Dec 12 '18 at 7:49
SvenMathSvenMath
104
edited Dec 12 '18 at 12:38
Arnaud D.
15.7k52443
edited Dec 12 '18 at 12:38
Arnaud D.
15.7k52443
edited Dec 12 '18 at 12:38
Arnaud D.
15.7k52443
15.7k52443
asked Dec 12 '18 at 7:49
SvenMathSvenMath
104
asked Dec 12 '18 at 7:49
SvenMathSvenMath
104
asked Dec 12 '18 at 7:49
SvenMathSvenMath
104
104
closed as unclear what you're asking by Nosrati, Michael Hoppe, Arnaud D., Chris Custer, Did Dec 12 '18 at 21:01
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Nosrati, Michael Hoppe, Arnaud D., Chris Custer, Did Dec 12 '18 at 21:01
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
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active
oldest
votes
What theorem are you using to say that the range must be dense? The present question is an elementary application of Louiville's Theorem. Suppose your half plane is the upper half plane. Just consider $g(z)=e^{if(z)}$ and note that $|g(z)|=e^{- Im (f(z))} <1$ for all $z$. Hence, by Louiville's Theorem $g$ is a constant from which it follows that $f$ is also a constant. [ For this note that $frac d {dz} e^{if(z)}=0$ implies $f'(z)=0$ for all $z$ and hence $f$ is a constant]. This contradicts the hypothesis. For the lower half plane consider $e^{-if(z)}$, for the right half plane consider $e^{-f(z)}$ and for the left half plane consider $e^{f(z)}$. Conclusion: you cannot have a non-constant entire function whose range is a half plane.
answered Dec 12 '18 at 7:53
Kavi Rama MurthyKavi Rama Murthy
51.6k31955
So there arent such functions? In a diffferent task I showed that the range must be dense. It was elementary, but which deep theorem are you talking about?
– SvenMath
Dec 12 '18 at 7:57
@SvenMath Well, this result can be proved by a simple application of Louiville's Theorem and your argument looks somewhat circuitous. Unless you include the proof of the fact that the range is dense it is hard to say that your argument is OK. In any case, do you agree that you can answer this question without proving that the range is dense?
– Kavi Rama Murthy
Dec 12 '18 at 8:03
I dont get your argument, You define $g(z)= e^{i f(z)}$, but this is a special function. How can you conclude that this works for all functions?
– SvenMath
Dec 12 '18 at 8:06
@SvenMath Assume that there is a non-constant entire function whose range is the upper half plane and get a contradiction. For this purpose I apply Louville's Theorem to $e^{if(z)}$ not to $f$ itself.
– Kavi Rama Murthy
Dec 12 '18 at 8:09
So you get a constant function whose range is C?
– SvenMath
Dec 12 '18 at 8:18
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
What theorem are you using to say that the range must be dense? The present question is an elementary application of Louiville's Theorem. Suppose your half plane is the upper half plane. Just consider $g(z)=e^{if(z)}$ and note that $|g(z)|=e^{- Im (f(z))} <1$ for all $z$. Hence, by Louiville's Theorem $g$ is a constant from which it follows that $f$ is also a constant. [ For this note that $frac d {dz} e^{if(z)}=0$ implies $f'(z)=0$ for all $z$ and hence $f$ is a constant]. This contradicts the hypothesis. For the lower half plane consider $e^{-if(z)}$, for the right half plane consider $e^{-f(z)}$ and for the left half plane consider $e^{f(z)}$. Conclusion: you cannot have a non-constant entire function whose range is a half plane.
answered Dec 12 '18 at 7:53
Kavi Rama MurthyKavi Rama Murthy
51.6k31955
So there arent such functions? In a diffferent task I showed that the range must be dense. It was elementary, but which deep theorem are you talking about?
– SvenMath
Dec 12 '18 at 7:57
@SvenMath Well, this result can be proved by a simple application of Louiville's Theorem and your argument looks somewhat circuitous. Unless you include the proof of the fact that the range is dense it is hard to say that your argument is OK. In any case, do you agree that you can answer this question without proving that the range is dense?
– Kavi Rama Murthy
Dec 12 '18 at 8:03
I dont get your argument, You define $g(z)= e^{i f(z)}$, but this is a special function. How can you conclude that this works for all functions?
– SvenMath
Dec 12 '18 at 8:06
@SvenMath Assume that there is a non-constant entire function whose range is the upper half plane and get a contradiction. For this purpose I apply Louville's Theorem to $e^{if(z)}$ not to $f$ itself.
– Kavi Rama Murthy
Dec 12 '18 at 8:09
So you get a constant function whose range is C?
– SvenMath
Dec 12 '18 at 8:18
add a comment |
What theorem are you using to say that the range must be dense? The present question is an elementary application of Louiville's Theorem. Suppose your half plane is the upper half plane. Just consider $g(z)=e^{if(z)}$ and note that $|g(z)|=e^{- Im (f(z))} <1$ for all $z$. Hence, by Louiville's Theorem $g$ is a constant from which it follows that $f$ is also a constant. [ For this note that $frac d {dz} e^{if(z)}=0$ implies $f'(z)=0$ for all $z$ and hence $f$ is a constant]. This contradicts the hypothesis. For the lower half plane consider $e^{-if(z)}$, for the right half plane consider $e^{-f(z)}$ and for the left half plane consider $e^{f(z)}$. Conclusion: you cannot have a non-constant entire function whose range is a half plane.
answered Dec 12 '18 at 7:53
Kavi Rama MurthyKavi Rama Murthy
51.6k31955
So there arent such functions? In a diffferent task I showed that the range must be dense. It was elementary, but which deep theorem are you talking about?
– SvenMath
Dec 12 '18 at 7:57
@SvenMath Well, this result can be proved by a simple application of Louiville's Theorem and your argument looks somewhat circuitous. Unless you include the proof of the fact that the range is dense it is hard to say that your argument is OK. In any case, do you agree that you can answer this question without proving that the range is dense?
– Kavi Rama Murthy
Dec 12 '18 at 8:03
I dont get your argument, You define $g(z)= e^{i f(z)}$, but this is a special function. How can you conclude that this works for all functions?
– SvenMath
Dec 12 '18 at 8:06
@SvenMath Assume that there is a non-constant entire function whose range is the upper half plane and get a contradiction. For this purpose I apply Louville's Theorem to $e^{if(z)}$ not to $f$ itself.
– Kavi Rama Murthy
Dec 12 '18 at 8:09
So you get a constant function whose range is C?
– SvenMath
Dec 12 '18 at 8:18
add a comment |
What theorem are you using to say that the range must be dense? The present question is an elementary application of Louiville's Theorem. Suppose your half plane is the upper half plane. Just consider $g(z)=e^{if(z)}$ and note that $|g(z)|=e^{- Im (f(z))} <1$ for all $z$. Hence, by Louiville's Theorem $g$ is a constant from which it follows that $f$ is also a constant. [ For this note that $frac d {dz} e^{if(z)}=0$ implies $f'(z)=0$ for all $z$ and hence $f$ is a constant]. This contradicts the hypothesis. For the lower half plane consider $e^{-if(z)}$, for the right half plane consider $e^{-f(z)}$ and for the left half plane consider $e^{f(z)}$. Conclusion: you cannot have a non-constant entire function whose range is a half plane.
answered Dec 12 '18 at 7:53
Kavi Rama MurthyKavi Rama Murthy
51.6k31955
What theorem are you using to say that the range must be dense? The present question is an elementary application of Louiville's Theorem. Suppose your half plane is the upper half plane. Just consider $g(z)=e^{if(z)}$ and note that $|g(z)|=e^{- Im (f(z))} <1$ for all $z$. Hence, by Louiville's Theorem $g$ is a constant from which it follows that $f$ is also a constant. [ For this note that $frac d {dz} e^{if(z)}=0$ implies $f'(z)=0$ for all $z$ and hence $f$ is a constant]. This contradicts the hypothesis. For the lower half plane consider $e^{-if(z)}$, for the right half plane consider $e^{-f(z)}$ and for the left half plane consider $e^{f(z)}$. Conclusion: you cannot have a non-constant entire function whose range is a half plane.
answered Dec 12 '18 at 7:53
Kavi Rama MurthyKavi Rama Murthy
51.6k31955
edited Dec 12 '18 at 8:25
answered Dec 12 '18 at 7:53
Kavi Rama MurthyKavi Rama Murthy
51.6k31955
answered Dec 12 '18 at 7:53
Kavi Rama MurthyKavi Rama Murthy
51.6k31955
answered Dec 12 '18 at 7:53
Kavi Rama MurthyKavi Rama Murthy
51.6k31955
51.6k31955
So there arent such functions? In a diffferent task I showed that the range must be dense. It was elementary, but which deep theorem are you talking about?
– SvenMath
Dec 12 '18 at 7:57
@SvenMath Well, this result can be proved by a simple application of Louiville's Theorem and your argument looks somewhat circuitous. Unless you include the proof of the fact that the range is dense it is hard to say that your argument is OK. In any case, do you agree that you can answer this question without proving that the range is dense?
– Kavi Rama Murthy
Dec 12 '18 at 8:03
I dont get your argument, You define $g(z)= e^{i f(z)}$, but this is a special function. How can you conclude that this works for all functions?
– SvenMath
Dec 12 '18 at 8:06
@SvenMath Assume that there is a non-constant entire function whose range is the upper half plane and get a contradiction. For this purpose I apply Louville's Theorem to $e^{if(z)}$ not to $f$ itself.
– Kavi Rama Murthy
Dec 12 '18 at 8:09
So you get a constant function whose range is C?
– SvenMath
Dec 12 '18 at 8:18
add a comment |
So there arent such functions? In a diffferent task I showed that the range must be dense. It was elementary, but which deep theorem are you talking about?
– SvenMath
Dec 12 '18 at 7:57
@SvenMath Well, this result can be proved by a simple application of Louiville's Theorem and your argument looks somewhat circuitous. Unless you include the proof of the fact that the range is dense it is hard to say that your argument is OK. In any case, do you agree that you can answer this question without proving that the range is dense?
– Kavi Rama Murthy
Dec 12 '18 at 8:03
I dont get your argument, You define $g(z)= e^{i f(z)}$, but this is a special function. How can you conclude that this works for all functions?
– SvenMath
Dec 12 '18 at 8:06
@SvenMath Assume that there is a non-constant entire function whose range is the upper half plane and get a contradiction. For this purpose I apply Louville's Theorem to $e^{if(z)}$ not to $f$ itself.
– Kavi Rama Murthy
Dec 12 '18 at 8:09
So you get a constant function whose range is C?
– SvenMath
Dec 12 '18 at 8:18
So there arent such functions? In a diffferent task I showed that the range must be dense. It was elementary, but which deep theorem are you talking about?
– SvenMath
Dec 12 '18 at 7:57
So there arent such functions? In a diffferent task I showed that the range must be dense. It was elementary, but which deep theorem are you talking about?
– SvenMath
Dec 12 '18 at 7:57
@SvenMath Well, this result can be proved by a simple application of Louiville's Theorem and your argument looks somewhat circuitous. Unless you include the proof of the fact that the range is dense it is hard to say that your argument is OK. In any case, do you agree that you can answer this question without proving that the range is dense?
– Kavi Rama Murthy
Dec 12 '18 at 8:03
@SvenMath Well, this result can be proved by a simple application of Louiville's Theorem and your argument looks somewhat circuitous. Unless you include the proof of the fact that the range is dense it is hard to say that your argument is OK. In any case, do you agree that you can answer this question without proving that the range is dense?
– Kavi Rama Murthy
Dec 12 '18 at 8:03
I dont get your argument, You define $g(z)= e^{i f(z)}$, but this is a special function. How can you conclude that this works for all functions?
– SvenMath
Dec 12 '18 at 8:06
I dont get your argument, You define $g(z)= e^{i f(z)}$, but this is a special function. How can you conclude that this works for all functions?
– SvenMath
Dec 12 '18 at 8:06
@SvenMath Assume that there is a non-constant entire function whose range is the upper half plane and get a contradiction. For this purpose I apply Louville's Theorem to $e^{if(z)}$ not to $f$ itself.
– Kavi Rama Murthy
Dec 12 '18 at 8:09
@SvenMath Assume that there is a non-constant entire function whose range is the upper half plane and get a contradiction. For this purpose I apply Louville's Theorem to $e^{if(z)}$ not to $f$ itself.
– Kavi Rama Murthy
Dec 12 '18 at 8:09
So you get a constant function whose range is C?
– SvenMath
Dec 12 '18 at 8:18
So you get a constant function whose range is C?
– SvenMath
Dec 12 '18 at 8:18
add a comment |
