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If $f$ is an entire function that goes to infinity, $f$ is a polynomial

Entire function bounded below by a polynomialProve $f$ entire and with a pole at infinity to be polynomialEntire function (Complex Analysis)Show that an entire function is a proper if and only if it is a nonconstant polynomialTo show an entire function is zero$f$ entire but not polynomial then $lim_{ntoinfty}sup {|z|:p_n(z)=0}toinfty$, where $p_n$ is $n-th$ Taylor series of $f$Does an entire function $f$ with a pole at infinity implies $|Re(f)| to infty$ or $|Im(f)| to infty$?Entire function $f(z)$ that satisfies a given limit.Entire Functions with no zeros are equalOn proving the following function is entire

2

1

$begingroup$

I am asked to show that if $f$ is entire with the property that $lim_{ztoinfty} f(z)=infty$, then $f$ must be a polynomial.

However, I feel as if I am missing something.

$e^z$ is entire (holomorphic at all finite $z$), and clearly $lim_{ztoinfty} f(z)=infty$ holds. However, $e^z$ is not a polynomial.

This seems to be a disproof via counterexample. Is there something wrong with my reasoning, or with the problem?

complex-analysis holomorphic-functions entire-functions

share|cite|improve this question

edited Mar 11 at 20:03

José Carlos Santos

168k22132236

asked Mar 11 at 18:51

Tejas RaoTejas Rao

33711

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  $e^{-x}to0$ as $xtoinfty$.
  $endgroup$
  – Lord Shark the Unknown
  Mar 11 at 18:52
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  $e^{2pi in}rightarrow 1$ as $nrightarrowinfty$.
  $endgroup$
  – Robert Wolfe
  Mar 11 at 18:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $e^{in}$ is divergent, but not to infinity as $ntoinfty$
  $endgroup$
  – Hagen von Eitzen
  Mar 11 at 18:57
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @LordSharktheUnknown This problem makes me think about Liouville's theorem, which concerns bounded functions and constants. How can we get to the polynomial thing, sir?
  $endgroup$
  – Leaning
  Mar 11 at 19:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  By $Inf$ and $inf$ did you mean $infty$, i.e., $infty$?
  $endgroup$
  – J. W. Tanner
  Mar 11 at 20:01
  

  
  
  
  

add a comment | 

2

1

$begingroup$

I am asked to show that if $f$ is entire with the property that $lim_{ztoinfty} f(z)=infty$, then $f$ must be a polynomial.

However, I feel as if I am missing something.

$e^z$ is entire (holomorphic at all finite $z$), and clearly $lim_{ztoinfty} f(z)=infty$ holds. However, $e^z$ is not a polynomial.

This seems to be a disproof via counterexample. Is there something wrong with my reasoning, or with the problem?

complex-analysis holomorphic-functions entire-functions

share|cite|improve this question

edited Mar 11 at 20:03

José Carlos Santos

168k22132236

asked Mar 11 at 18:51

Tejas RaoTejas Rao

33711

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  $e^{-x}to0$ as $xtoinfty$.
  $endgroup$
  – Lord Shark the Unknown
  Mar 11 at 18:52
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  $e^{2pi in}rightarrow 1$ as $nrightarrowinfty$.
  $endgroup$
  – Robert Wolfe
  Mar 11 at 18:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $e^{in}$ is divergent, but not to infinity as $ntoinfty$
  $endgroup$
  – Hagen von Eitzen
  Mar 11 at 18:57
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @LordSharktheUnknown This problem makes me think about Liouville's theorem, which concerns bounded functions and constants. How can we get to the polynomial thing, sir?
  $endgroup$
  – Leaning
  Mar 11 at 19:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  By $Inf$ and $inf$ did you mean $infty$, i.e., $infty$?
  $endgroup$
  – J. W. Tanner
  Mar 11 at 20:01
  

  
  
  
  

add a comment | 

$begingroup$

I am asked to show that if $f$ is entire with the property that $lim_{ztoinfty} f(z)=infty$, then $f$ must be a polynomial.

However, I feel as if I am missing something.

$e^z$ is entire (holomorphic at all finite $z$), and clearly $lim_{ztoinfty} f(z)=infty$ holds. However, $e^z$ is not a polynomial.

This seems to be a disproof via counterexample. Is there something wrong with my reasoning, or with the problem?

complex-analysis holomorphic-functions entire-functions

share|cite|improve this question

edited Mar 11 at 20:03

José Carlos Santos

168k22132236

asked Mar 11 at 18:51

Tejas RaoTejas Rao

33711

$endgroup$

share|cite|improve this question

edited Mar 11 at 20:03

José Carlos Santos

168k22132236

asked Mar 11 at 18:51

Tejas RaoTejas Rao

33711

share|cite|improve this question

edited Mar 11 at 20:03

José Carlos Santos

168k22132236

asked Mar 11 at 18:51

Tejas RaoTejas Rao

33711

edited Mar 11 at 20:03

José Carlos Santos

168k22132236

edited Mar 11 at 20:03

José Carlos Santos

168k22132236

asked Mar 11 at 18:51

Tejas RaoTejas Rao

33711

asked Mar 11 at 18:51

Tejas RaoTejas Rao

33711

3 Answers
3

active

oldest

votes

5

$begingroup$

Let $g(z)=fleft(frac1zright)$. Then $lim_{zto0}g(z)=infty$ and therefore $g$ has a pole at $0$, because otherwise $g$ would have an essential singularity at $0$, which is impossible, by the Casorati-Weierstrass theorem. But if $f(z)=a_0+a_1z+a_2z^2+cdots$, then $g(z)=a_0+frac{a_1}z+frac{a_2}{z^2}+cdots$ and therefore asserting that $g$ has a pole at $0$ means that $a_n=0$ if $n$ is large enough. So, $f$ is a polynomial function.

share|cite|improve this answer

edited Mar 11 at 20:00

answered Mar 11 at 19:14

José Carlos SantosJosé Carlos Santos

168k22132236

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hi! I tried to give an answer too, would you mind giving it a quick read to check?
  $endgroup$
  – Ovi
  Mar 11 at 20:18
  

  
  
  
  

add a comment | 

2

$begingroup$

I believe you have mis-read or mis-interpreted the key part of the problem. For complex variables,
$$
lim_{zto infty} f(z) = X
$$
means that the limit is $X$ (whatever $X$ is) as $z$ approaches infinity in any direction.

I'm not certain, but I believe you could replace that with
$$
forall u in Bbb C : (|u|=1 implies lim_{rtoinfty_{rinBbb R}} f(r u) = X)
$$
that is, the limits, taken as the limit of the real length along a line from the origin, of the function of points on that line, are all the same value $X$ (which in this case is $infty$.

So you need to show that for any non-polynomial entire function, there is some direction along with the function does not go to infinity.

share|cite|improve this answer

answered Mar 11 at 19:11

Mark FischlerMark Fischler

33.5k12452

$endgroup$

add a comment | 

0

$begingroup$

One important feature about holomorphic functions is the number and position of their zeroes. The function going to $infty$ implies that outside of some compact disk $D$ centered at the origin, the function is never zero.

Now to prove that our function is a polynomial, it suffices by analytic continuation to prove that our $f$ has only a finite number of zeroes in this disk $D$. What would go wrong if our function had infinitely many zeroes in $D$? Let $S$ be the set of zeroes of $f$. Since $D$ is compact, $S$ would have a limit point in $D$; but again, by analytic continuation, this would imply that $f$ is the zero function, which is a contradiction.

share|cite|improve this answer

edited Mar 11 at 22:57

J. W. Tanner

3,3351320

answered Mar 11 at 20:18

OviOvi

12.4k1040113

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Asserting that a $lim_{ztoinfty}f(z)=infty$ does not mean that, outside a compact disk centered ar the origin, $f$ is never zero. For instance, it is not true that $lim_{ztoinfty}e^z=infty$ but nevertheless the exponential function has no zero whatsoever.
  $endgroup$
  – José Carlos Santos
  Mar 11 at 20:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JoséCarlosSantos Than you very much for the response! I've never seen $lim_{ztoinfty}f(z) = infty$, so I assumed it means $forall (M in mathbb{R}^+) exists (delta > 0) forall (z in mathbb{C}): |z|> delta implies |f(z)| > M$. I'm pretty sure that under this definition, outside of some compact disk $D$ centered at the origin, $f$ would not have any zeroes, no? Is there another definition of $lim_{ztoinfty}f(z) = infty$?
  $endgroup$
  – Ovi
  Mar 11 at 20:42
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are right about the definition. But it doesn't mean that it has no zeros outside a compact disk. It just implies it.
  $endgroup$
  – José Carlos Santos
  Mar 11 at 21:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JoséCarlosSantos Ahhh okay thank you, now I understand your example too. Yes, I used "means" to mean "imply". I fixed the answer.
  $endgroup$
  – Ovi
  Mar 11 at 21:31
  
  
  

  
  
  
  

add a comment | 

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5

$begingroup$

Let $g(z)=fleft(frac1zright)$. Then $lim_{zto0}g(z)=infty$ and therefore $g$ has a pole at $0$, because otherwise $g$ would have an essential singularity at $0$, which is impossible, by the Casorati-Weierstrass theorem. But if $f(z)=a_0+a_1z+a_2z^2+cdots$, then $g(z)=a_0+frac{a_1}z+frac{a_2}{z^2}+cdots$ and therefore asserting that $g$ has a pole at $0$ means that $a_n=0$ if $n$ is large enough. So, $f$ is a polynomial function.

share|cite|improve this answer

edited Mar 11 at 20:00

answered Mar 11 at 19:14

José Carlos SantosJosé Carlos Santos

168k22132236

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hi! I tried to give an answer too, would you mind giving it a quick read to check?
  $endgroup$
  – Ovi
  Mar 11 at 20:18
  

  
  
  
  

add a comment | 

5

$begingroup$

Let $g(z)=fleft(frac1zright)$. Then $lim_{zto0}g(z)=infty$ and therefore $g$ has a pole at $0$, because otherwise $g$ would have an essential singularity at $0$, which is impossible, by the Casorati-Weierstrass theorem. But if $f(z)=a_0+a_1z+a_2z^2+cdots$, then $g(z)=a_0+frac{a_1}z+frac{a_2}{z^2}+cdots$ and therefore asserting that $g$ has a pole at $0$ means that $a_n=0$ if $n$ is large enough. So, $f$ is a polynomial function.

share|cite|improve this answer

edited Mar 11 at 20:00

answered Mar 11 at 19:14

José Carlos SantosJosé Carlos Santos

168k22132236

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hi! I tried to give an answer too, would you mind giving it a quick read to check?
  $endgroup$
  – Ovi
  Mar 11 at 20:18
  

  
  
  
  

add a comment | 

$begingroup$

Let $g(z)=fleft(frac1zright)$. Then $lim_{zto0}g(z)=infty$ and therefore $g$ has a pole at $0$, because otherwise $g$ would have an essential singularity at $0$, which is impossible, by the Casorati-Weierstrass theorem. But if $f(z)=a_0+a_1z+a_2z^2+cdots$, then $g(z)=a_0+frac{a_1}z+frac{a_2}{z^2}+cdots$ and therefore asserting that $g$ has a pole at $0$ means that $a_n=0$ if $n$ is large enough. So, $f$ is a polynomial function.

share|cite|improve this answer

edited Mar 11 at 20:00

answered Mar 11 at 19:14

José Carlos SantosJosé Carlos Santos

168k22132236

$endgroup$

share|cite|improve this answer

edited Mar 11 at 20:00

answered Mar 11 at 19:14

José Carlos SantosJosé Carlos Santos

168k22132236

answered Mar 11 at 19:14

José Carlos SantosJosé Carlos Santos

168k22132236

answered Mar 11 at 19:14

José Carlos SantosJosé Carlos Santos

168k22132236

2

$begingroup$

I believe you have mis-read or mis-interpreted the key part of the problem. For complex variables,
$$
lim_{zto infty} f(z) = X
$$
means that the limit is $X$ (whatever $X$ is) as $z$ approaches infinity in any direction.

I'm not certain, but I believe you could replace that with
$$
forall u in Bbb C : (|u|=1 implies lim_{rtoinfty_{rinBbb R}} f(r u) = X)
$$
that is, the limits, taken as the limit of the real length along a line from the origin, of the function of points on that line, are all the same value $X$ (which in this case is $infty$.

So you need to show that for any non-polynomial entire function, there is some direction along with the function does not go to infinity.

share|cite|improve this answer

answered Mar 11 at 19:11

Mark FischlerMark Fischler

33.5k12452

$endgroup$

add a comment | 

2

$begingroup$

I believe you have mis-read or mis-interpreted the key part of the problem. For complex variables,
$$
lim_{zto infty} f(z) = X
$$
means that the limit is $X$ (whatever $X$ is) as $z$ approaches infinity in any direction.

I'm not certain, but I believe you could replace that with
$$
forall u in Bbb C : (|u|=1 implies lim_{rtoinfty_{rinBbb R}} f(r u) = X)
$$
that is, the limits, taken as the limit of the real length along a line from the origin, of the function of points on that line, are all the same value $X$ (which in this case is $infty$.

So you need to show that for any non-polynomial entire function, there is some direction along with the function does not go to infinity.

share|cite|improve this answer

answered Mar 11 at 19:11

Mark FischlerMark Fischler

33.5k12452

$endgroup$

add a comment | 

$begingroup$

I believe you have mis-read or mis-interpreted the key part of the problem. For complex variables,
$$
lim_{zto infty} f(z) = X
$$
means that the limit is $X$ (whatever $X$ is) as $z$ approaches infinity in any direction.

I'm not certain, but I believe you could replace that with
$$
forall u in Bbb C : (|u|=1 implies lim_{rtoinfty_{rinBbb R}} f(r u) = X)
$$
that is, the limits, taken as the limit of the real length along a line from the origin, of the function of points on that line, are all the same value $X$ (which in this case is $infty$.

So you need to show that for any non-polynomial entire function, there is some direction along with the function does not go to infinity.

share|cite|improve this answer

answered Mar 11 at 19:11

Mark FischlerMark Fischler

33.5k12452

$endgroup$

share|cite|improve this answer

answered Mar 11 at 19:11

Mark FischlerMark Fischler

33.5k12452

answered Mar 11 at 19:11

Mark FischlerMark Fischler

33.5k12452

answered Mar 11 at 19:11

Mark FischlerMark Fischler

33.5k12452

0

$begingroup$

One important feature about holomorphic functions is the number and position of their zeroes. The function going to $infty$ implies that outside of some compact disk $D$ centered at the origin, the function is never zero.

Now to prove that our function is a polynomial, it suffices by analytic continuation to prove that our $f$ has only a finite number of zeroes in this disk $D$. What would go wrong if our function had infinitely many zeroes in $D$? Let $S$ be the set of zeroes of $f$. Since $D$ is compact, $S$ would have a limit point in $D$; but again, by analytic continuation, this would imply that $f$ is the zero function, which is a contradiction.

share|cite|improve this answer

edited Mar 11 at 22:57

J. W. Tanner

3,3351320

answered Mar 11 at 20:18

OviOvi

12.4k1040113

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Asserting that a $lim_{ztoinfty}f(z)=infty$ does not mean that, outside a compact disk centered ar the origin, $f$ is never zero. For instance, it is not true that $lim_{ztoinfty}e^z=infty$ but nevertheless the exponential function has no zero whatsoever.
  $endgroup$
  – José Carlos Santos
  Mar 11 at 20:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JoséCarlosSantos Than you very much for the response! I've never seen $lim_{ztoinfty}f(z) = infty$, so I assumed it means $forall (M in mathbb{R}^+) exists (delta > 0) forall (z in mathbb{C}): |z|> delta implies |f(z)| > M$. I'm pretty sure that under this definition, outside of some compact disk $D$ centered at the origin, $f$ would not have any zeroes, no? Is there another definition of $lim_{ztoinfty}f(z) = infty$?
  $endgroup$
  – Ovi
  Mar 11 at 20:42
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are right about the definition. But it doesn't mean that it has no zeros outside a compact disk. It just implies it.
  $endgroup$
  – José Carlos Santos
  Mar 11 at 21:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JoséCarlosSantos Ahhh okay thank you, now I understand your example too. Yes, I used "means" to mean "imply". I fixed the answer.
  $endgroup$
  – Ovi
  Mar 11 at 21:31
  
  
  

  
  
  
  

add a comment | 

0

$begingroup$

One important feature about holomorphic functions is the number and position of their zeroes. The function going to $infty$ implies that outside of some compact disk $D$ centered at the origin, the function is never zero.

Now to prove that our function is a polynomial, it suffices by analytic continuation to prove that our $f$ has only a finite number of zeroes in this disk $D$. What would go wrong if our function had infinitely many zeroes in $D$? Let $S$ be the set of zeroes of $f$. Since $D$ is compact, $S$ would have a limit point in $D$; but again, by analytic continuation, this would imply that $f$ is the zero function, which is a contradiction.

share|cite|improve this answer

edited Mar 11 at 22:57

J. W. Tanner

3,3351320

answered Mar 11 at 20:18

OviOvi

12.4k1040113

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Asserting that a $lim_{ztoinfty}f(z)=infty$ does not mean that, outside a compact disk centered ar the origin, $f$ is never zero. For instance, it is not true that $lim_{ztoinfty}e^z=infty$ but nevertheless the exponential function has no zero whatsoever.
  $endgroup$
  – José Carlos Santos
  Mar 11 at 20:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JoséCarlosSantos Than you very much for the response! I've never seen $lim_{ztoinfty}f(z) = infty$, so I assumed it means $forall (M in mathbb{R}^+) exists (delta > 0) forall (z in mathbb{C}): |z|> delta implies |f(z)| > M$. I'm pretty sure that under this definition, outside of some compact disk $D$ centered at the origin, $f$ would not have any zeroes, no? Is there another definition of $lim_{ztoinfty}f(z) = infty$?
  $endgroup$
  – Ovi
  Mar 11 at 20:42
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are right about the definition. But it doesn't mean that it has no zeros outside a compact disk. It just implies it.
  $endgroup$
  – José Carlos Santos
  Mar 11 at 21:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JoséCarlosSantos Ahhh okay thank you, now I understand your example too. Yes, I used "means" to mean "imply". I fixed the answer.
  $endgroup$
  – Ovi
  Mar 11 at 21:31
  
  
  

  
  
  
  

add a comment | 

$begingroup$

One important feature about holomorphic functions is the number and position of their zeroes. The function going to $infty$ implies that outside of some compact disk $D$ centered at the origin, the function is never zero.

Now to prove that our function is a polynomial, it suffices by analytic continuation to prove that our $f$ has only a finite number of zeroes in this disk $D$. What would go wrong if our function had infinitely many zeroes in $D$? Let $S$ be the set of zeroes of $f$. Since $D$ is compact, $S$ would have a limit point in $D$; but again, by analytic continuation, this would imply that $f$ is the zero function, which is a contradiction.

share|cite|improve this answer

edited Mar 11 at 22:57

J. W. Tanner

3,3351320

answered Mar 11 at 20:18

OviOvi

12.4k1040113

$endgroup$

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edited Mar 11 at 22:57

J. W. Tanner

3,3351320

answered Mar 11 at 20:18

OviOvi

12.4k1040113

edited Mar 11 at 22:57

J. W. Tanner

3,3351320

edited Mar 11 at 22:57

J. W. Tanner

3,3351320

answered Mar 11 at 20:18

OviOvi

12.4k1040113

answered Mar 11 at 20:18

OviOvi

12.4k1040113