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If $limlimits_{z to infty} |F(z)|=0$, can $limlimits_{z to infty} F(z) ne 0$?

1

$begingroup$

To prove the existence of limit of $F(z)$ at $z=infty$ in here (Eq. 1.8 page 3) it says that since $limlimits_{z to infty} |F(z)|=0$ thus $limlimits_{z to infty} F(z)=w_n$ exists; Of course exits, but also must $w_n=0$. But Nor this article neither the book Churchill Sec 117 Exercise 2 doesn't say $w_n=0$. Can $w_n ne 0$ be possible?

calculus complex-analysis limits proof-explanation

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edited Mar 26 at 11:08

rash

568216

asked Mar 26 at 11:02

72D72D

298117

$endgroup$

add a comment | 

1

$begingroup$

To prove the existence of limit of $F(z)$ at $z=infty$ in here (Eq. 1.8 page 3) it says that since $limlimits_{z to infty} |F(z)|=0$ thus $limlimits_{z to infty} F(z)=w_n$ exists; Of course exits, but also must $w_n=0$. But Nor this article neither the book Churchill Sec 117 Exercise 2 doesn't say $w_n=0$. Can $w_n ne 0$ be possible?

calculus complex-analysis limits proof-explanation

share|cite|improve this question

edited Mar 26 at 11:08

rash

568216

asked Mar 26 at 11:02

72D72D

298117

$endgroup$

add a comment | 

$begingroup$

To prove the existence of limit of $F(z)$ at $z=infty$ in here (Eq. 1.8 page 3) it says that since $limlimits_{z to infty} |F(z)|=0$ thus $limlimits_{z to infty} F(z)=w_n$ exists; Of course exits, but also must $w_n=0$. But Nor this article neither the book Churchill Sec 117 Exercise 2 doesn't say $w_n=0$. Can $w_n ne 0$ be possible?

calculus complex-analysis limits proof-explanation

share|cite|improve this question

edited Mar 26 at 11:08

rash

568216

asked Mar 26 at 11:02

72D72D

298117

$endgroup$

share|cite|improve this question

edited Mar 26 at 11:08

rash

568216

asked Mar 26 at 11:02

72D72D

298117

share|cite|improve this question

edited Mar 26 at 11:08

rash

568216

asked Mar 26 at 11:02

72D72D

298117

edited Mar 26 at 11:08

rash

568216

edited Mar 26 at 11:08

rash

568216

asked Mar 26 at 11:02

72D72D

298117

asked Mar 26 at 11:02

72D72D

298117

1 Answer
1

active

oldest

votes

1

$begingroup$

The limit must be zero. Suppose $lim F(z)=w_n$ exists. Then $lim | F(z)-w_n |=0$. However $|F(z)-w_n| geq big||F(z)|-|w_n| big|$, for any $z$ by the triangle inequality, so it follows that $lim big||F(z)|-|w_n| big|=0$. That is, $lim |F(z)|=|w_n|$.

Note that the convergence of $|F(z)|$ does not imply $F(z)$ converges unless $lim |F(z)|=0$.

share|cite|improve this answer

answered Mar 26 at 11:17

EricEric

37811

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $w_n=F(infty)$ for Schwarz-Christoffel transformations which is not zero always. That's why I am confused!
  $endgroup$
  – 72D
  Mar 26 at 11:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Honestly I think the document you linked is not terribly well written.The appearance of individual equations is messy, the spacing is all over the place, the alignment is bad. e.g. in equation 1.6 half of the equation is centered and half is not. The exponents are very far from the terms they exponetiate. Work like this is hard to follow, both for the reader and the author, and it leaves room for the author to have made a mistake. In equatioin 1.8 the author bounds the limit of $F(z)$ above by zero and uses that to conclude that $F(z)$ converges to some value.
  $endgroup$
  – Eric
  Mar 26 at 12:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There are two problems there. First, the author hasn't shown that the limit has a lower bound, so from what is shown they cannot conclude a limit exists. Now the equations can easily be applied to $-F(z)$ to get that the limit is bounded below by zero, but they should have at least commented on this. Now if the limit is both bounded above and below by zero then the limit must be zero. So why do they say it converges to some value and not emphasize zero? I didn't carefully check the details in the document, but I wouldn't trust what is written in without checking extremely carefully.
  $endgroup$
  – Eric
  Mar 26 at 12:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I also found that the journal which published the article you linked is on the predatory journal list found at predatoryjournals.com/journals/#I. In short, you should not trust that article or anything from that journal.
  $endgroup$
  – Eric
  Mar 26 at 12:21
  
  
  

  
  
  
  

add a comment | 

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1

$begingroup$

The limit must be zero. Suppose $lim F(z)=w_n$ exists. Then $lim | F(z)-w_n |=0$. However $|F(z)-w_n| geq big||F(z)|-|w_n| big|$, for any $z$ by the triangle inequality, so it follows that $lim big||F(z)|-|w_n| big|=0$. That is, $lim |F(z)|=|w_n|$.

Note that the convergence of $|F(z)|$ does not imply $F(z)$ converges unless $lim |F(z)|=0$.

share|cite|improve this answer

answered Mar 26 at 11:17

EricEric

37811

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $w_n=F(infty)$ for Schwarz-Christoffel transformations which is not zero always. That's why I am confused!
  $endgroup$
  – 72D
  Mar 26 at 11:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Honestly I think the document you linked is not terribly well written.The appearance of individual equations is messy, the spacing is all over the place, the alignment is bad. e.g. in equation 1.6 half of the equation is centered and half is not. The exponents are very far from the terms they exponetiate. Work like this is hard to follow, both for the reader and the author, and it leaves room for the author to have made a mistake. In equatioin 1.8 the author bounds the limit of $F(z)$ above by zero and uses that to conclude that $F(z)$ converges to some value.
  $endgroup$
  – Eric
  Mar 26 at 12:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There are two problems there. First, the author hasn't shown that the limit has a lower bound, so from what is shown they cannot conclude a limit exists. Now the equations can easily be applied to $-F(z)$ to get that the limit is bounded below by zero, but they should have at least commented on this. Now if the limit is both bounded above and below by zero then the limit must be zero. So why do they say it converges to some value and not emphasize zero? I didn't carefully check the details in the document, but I wouldn't trust what is written in without checking extremely carefully.
  $endgroup$
  – Eric
  Mar 26 at 12:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I also found that the journal which published the article you linked is on the predatory journal list found at predatoryjournals.com/journals/#I. In short, you should not trust that article or anything from that journal.
  $endgroup$
  – Eric
  Mar 26 at 12:21
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

The limit must be zero. Suppose $lim F(z)=w_n$ exists. Then $lim | F(z)-w_n |=0$. However $|F(z)-w_n| geq big||F(z)|-|w_n| big|$, for any $z$ by the triangle inequality, so it follows that $lim big||F(z)|-|w_n| big|=0$. That is, $lim |F(z)|=|w_n|$.

Note that the convergence of $|F(z)|$ does not imply $F(z)$ converges unless $lim |F(z)|=0$.

share|cite|improve this answer

answered Mar 26 at 11:17

EricEric

37811

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $w_n=F(infty)$ for Schwarz-Christoffel transformations which is not zero always. That's why I am confused!
  $endgroup$
  – 72D
  Mar 26 at 11:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Honestly I think the document you linked is not terribly well written.The appearance of individual equations is messy, the spacing is all over the place, the alignment is bad. e.g. in equation 1.6 half of the equation is centered and half is not. The exponents are very far from the terms they exponetiate. Work like this is hard to follow, both for the reader and the author, and it leaves room for the author to have made a mistake. In equatioin 1.8 the author bounds the limit of $F(z)$ above by zero and uses that to conclude that $F(z)$ converges to some value.
  $endgroup$
  – Eric
  Mar 26 at 12:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There are two problems there. First, the author hasn't shown that the limit has a lower bound, so from what is shown they cannot conclude a limit exists. Now the equations can easily be applied to $-F(z)$ to get that the limit is bounded below by zero, but they should have at least commented on this. Now if the limit is both bounded above and below by zero then the limit must be zero. So why do they say it converges to some value and not emphasize zero? I didn't carefully check the details in the document, but I wouldn't trust what is written in without checking extremely carefully.
  $endgroup$
  – Eric
  Mar 26 at 12:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I also found that the journal which published the article you linked is on the predatory journal list found at predatoryjournals.com/journals/#I. In short, you should not trust that article or anything from that journal.
  $endgroup$
  – Eric
  Mar 26 at 12:21
  
  
  

  
  
  
  

add a comment | 

$begingroup$

The limit must be zero. Suppose $lim F(z)=w_n$ exists. Then $lim | F(z)-w_n |=0$. However $|F(z)-w_n| geq big||F(z)|-|w_n| big|$, for any $z$ by the triangle inequality, so it follows that $lim big||F(z)|-|w_n| big|=0$. That is, $lim |F(z)|=|w_n|$.

Note that the convergence of $|F(z)|$ does not imply $F(z)$ converges unless $lim |F(z)|=0$.

share|cite|improve this answer

answered Mar 26 at 11:17

EricEric

37811

$endgroup$

share|cite|improve this answer

answered Mar 26 at 11:17

EricEric

37811

answered Mar 26 at 11:17

EricEric

37811

answered Mar 26 at 11:17

EricEric

37811