local minimum of $|f|$

1

$begingroup$

Suppose $f in H(Omega)$, where $Omegasubsetmathbb C$ is an open set. Under what condition can $|f|$ have a local minimum?

Here $|f| = u^2 +v^2 = g$ say. We assumed $f(x,y)= u(x,y) +i v(x,y)$.
Then $g$ has local minimum if $g_{xx} > 0$ and $g_{xx}= 2[u_x^2 +uu_{xx} +v_x ^2 +vv_{xx}]$. So as square terms are positive always, the required condition is $uu_{xx}+vv_{xx} >0$.

I am asking if this a correct answer; if not then please guide me in the right way.

Thanks in advance.

complex-analysis

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edited Jan 7 at 16:27

amWhy

1

asked Feb 17 '14 at 7:51

GermainGermain

784820

$endgroup$

add a comment | 

1

$begingroup$

Suppose $f in H(Omega)$, where $Omegasubsetmathbb C$ is an open set. Under what condition can $|f|$ have a local minimum?

Here $|f| = u^2 +v^2 = g$ say. We assumed $f(x,y)= u(x,y) +i v(x,y)$.
Then $g$ has local minimum if $g_{xx} > 0$ and $g_{xx}= 2[u_x^2 +uu_{xx} +v_x ^2 +vv_{xx}]$. So as square terms are positive always, the required condition is $uu_{xx}+vv_{xx} >0$.

I am asking if this a correct answer; if not then please guide me in the right way.

Thanks in advance.

complex-analysis

share|cite|improve this question

edited Jan 7 at 16:27

amWhy

1

asked Feb 17 '14 at 7:51

GermainGermain

784820

$endgroup$

add a comment | 

1

1

1

$begingroup$

Suppose $f in H(Omega)$, where $Omegasubsetmathbb C$ is an open set. Under what condition can $|f|$ have a local minimum?

Here $|f| = u^2 +v^2 = g$ say. We assumed $f(x,y)= u(x,y) +i v(x,y)$.
Then $g$ has local minimum if $g_{xx} > 0$ and $g_{xx}= 2[u_x^2 +uu_{xx} +v_x ^2 +vv_{xx}]$. So as square terms are positive always, the required condition is $uu_{xx}+vv_{xx} >0$.

I am asking if this a correct answer; if not then please guide me in the right way.

Thanks in advance.

complex-analysis

share|cite|improve this question

edited Jan 7 at 16:27

amWhy

1

asked Feb 17 '14 at 7:51

GermainGermain

784820

$endgroup$

Suppose $f in H(Omega)$, where $Omegasubsetmathbb C$ is an open set. Under what condition can $|f|$ have a local minimum?

Here $|f| = u^2 +v^2 = g$ say. We assumed $f(x,y)= u(x,y) +i v(x,y)$.
Then $g$ has local minimum if $g_{xx} > 0$ and $g_{xx}= 2[u_x^2 +uu_{xx} +v_x ^2 +vv_{xx}]$. So as square terms are positive always, the required condition is $uu_{xx}+vv_{xx} >0$.

I am asking if this a correct answer; if not then please guide me in the right way.

Thanks in advance.

complex-analysis

complex-analysis

share|cite|improve this question

edited Jan 7 at 16:27

amWhy

1

asked Feb 17 '14 at 7:51

GermainGermain

784820

share|cite|improve this question

edited Jan 7 at 16:27

amWhy

1

asked Feb 17 '14 at 7:51

GermainGermain

784820

share|cite|improve this question

share|cite|improve this question

edited Jan 7 at 16:27

amWhy

1

edited Jan 7 at 16:27

amWhy

1

edited Jan 7 at 16:27

amWhy

1

1

asked Feb 17 '14 at 7:51

GermainGermain

784820

asked Feb 17 '14 at 7:51

GermainGermain

784820

asked Feb 17 '14 at 7:51

GermainGermain

784820

784820

add a comment | 

add a comment | 

3 Answers
3

active

oldest

votes

2

$begingroup$

If $f$ has a zero on $Omega$, then clearly $|f|$ has a local minimum at those points. Otherwise, the open mapping principle prevents $|f|$ from having a local minimum. (If $a in Omega$ then $f$ maps open discs centered at $a$ to open sets. In particular if $f(a) neq 0$ then there are nearby points $z$ where $|f(z)| < |f(a)|$.)

share|cite|improve this answer

answered Feb 17 '14 at 10:16

mrfmrf

37.4k54685

$endgroup$

add a comment | 

0

$begingroup$

All the isolated zeros of $f$ in $Omega$ will be the local minimum of $f$. If the set of zeros of $f$ has a limit point in $Omega$ then it is identically zero, being holomorphic. If the value of $f$ at a point $z_0$ is non zero then we can use the open set property to prove that it is not a point of minimum value.

share|cite|improve this answer

edited Feb 17 '14 at 10:31

answered Feb 17 '14 at 10:07

viplov_jainviplov_jain

723311

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is not correct. Look at one of the simplest possible examples, $f(z) = z$ on the unit disc.
  $endgroup$
  – mrf
  Feb 17 '14 at 10:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was using the wrong principle, Thanks for pointing out. I've corrected it.
  $endgroup$
  – viplov_jain
  Feb 17 '14 at 10:32
  

  
  
  
  

add a comment | 

0

$begingroup$

If $f$ is a constant, $|f|$ has a local minimum at any point of $Omega.$

If $f$ is not a constant and there exists $zinOmega$ s.t. $f(z)=0,$ then the minimum occurs at those points $zinOmega.$

If $f$ is not a constant and there isn't any $zinOmega$ s.t. $f(z)=0,$ then $frac{1}{f}$ doesn't have singularities on $Omega.$ So, $fin h(Omega)$ and $frac{1}{f}in h(Omega).$ Let suppose that $|f|$ occurs minimum at $ainOmega.$ Then, by the Principle of Maximum Modulus for $frac{1}{f}$ we have that $vertfrac{1}{f(a)}vert<{vertfrac{1}{f(z)}vertvert zinpartial D}$ for any neighborhood $bar{D}subsetOmega$ of $a,$ i.e. $|f(a)|>|f(z)|$ for a $zinpartial DsubsetOmega$ witch contradicts the supposition that $a$ was a local minimum. This means that under there conditions, |f| does not occur a local minimum on $Omega.$

share|cite|improve this answer

answered Jan 7 at 16:24

EminEmin

1,36021330

$endgroup$

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

2

$begingroup$

If $f$ has a zero on $Omega$, then clearly $|f|$ has a local minimum at those points. Otherwise, the open mapping principle prevents $|f|$ from having a local minimum. (If $a in Omega$ then $f$ maps open discs centered at $a$ to open sets. In particular if $f(a) neq 0$ then there are nearby points $z$ where $|f(z)| < |f(a)|$.)

share|cite|improve this answer

answered Feb 17 '14 at 10:16

mrfmrf

37.4k54685

$endgroup$

add a comment | 

2

$begingroup$

If $f$ has a zero on $Omega$, then clearly $|f|$ has a local minimum at those points. Otherwise, the open mapping principle prevents $|f|$ from having a local minimum. (If $a in Omega$ then $f$ maps open discs centered at $a$ to open sets. In particular if $f(a) neq 0$ then there are nearby points $z$ where $|f(z)| < |f(a)|$.)

share|cite|improve this answer

answered Feb 17 '14 at 10:16

mrfmrf

37.4k54685

$endgroup$

add a comment | 

2

2

2

$begingroup$

If $f$ has a zero on $Omega$, then clearly $|f|$ has a local minimum at those points. Otherwise, the open mapping principle prevents $|f|$ from having a local minimum. (If $a in Omega$ then $f$ maps open discs centered at $a$ to open sets. In particular if $f(a) neq 0$ then there are nearby points $z$ where $|f(z)| < |f(a)|$.)

share|cite|improve this answer

answered Feb 17 '14 at 10:16

mrfmrf

37.4k54685

$endgroup$

If $f$ has a zero on $Omega$, then clearly $|f|$ has a local minimum at those points. Otherwise, the open mapping principle prevents $|f|$ from having a local minimum. (If $a in Omega$ then $f$ maps open discs centered at $a$ to open sets. In particular if $f(a) neq 0$ then there are nearby points $z$ where $|f(z)| < |f(a)|$.)

share|cite|improve this answer

answered Feb 17 '14 at 10:16

mrfmrf

37.4k54685

share|cite|improve this answer

share|cite|improve this answer

answered Feb 17 '14 at 10:16

mrfmrf

37.4k54685

answered Feb 17 '14 at 10:16

mrfmrf

37.4k54685

answered Feb 17 '14 at 10:16

mrfmrf

37.4k54685

37.4k54685

add a comment | 

add a comment | 

0

$begingroup$

All the isolated zeros of $f$ in $Omega$ will be the local minimum of $f$. If the set of zeros of $f$ has a limit point in $Omega$ then it is identically zero, being holomorphic. If the value of $f$ at a point $z_0$ is non zero then we can use the open set property to prove that it is not a point of minimum value.

share|cite|improve this answer

edited Feb 17 '14 at 10:31

answered Feb 17 '14 at 10:07

viplov_jainviplov_jain

723311

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is not correct. Look at one of the simplest possible examples, $f(z) = z$ on the unit disc.
  $endgroup$
  – mrf
  Feb 17 '14 at 10:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was using the wrong principle, Thanks for pointing out. I've corrected it.
  $endgroup$
  – viplov_jain
  Feb 17 '14 at 10:32
  

  
  
  
  

add a comment | 

0

$begingroup$

All the isolated zeros of $f$ in $Omega$ will be the local minimum of $f$. If the set of zeros of $f$ has a limit point in $Omega$ then it is identically zero, being holomorphic. If the value of $f$ at a point $z_0$ is non zero then we can use the open set property to prove that it is not a point of minimum value.

share|cite|improve this answer

edited Feb 17 '14 at 10:31

answered Feb 17 '14 at 10:07

viplov_jainviplov_jain

723311

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is not correct. Look at one of the simplest possible examples, $f(z) = z$ on the unit disc.
  $endgroup$
  – mrf
  Feb 17 '14 at 10:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was using the wrong principle, Thanks for pointing out. I've corrected it.
  $endgroup$
  – viplov_jain
  Feb 17 '14 at 10:32
  

  
  
  
  

add a comment | 

0

0

0

$begingroup$

All the isolated zeros of $f$ in $Omega$ will be the local minimum of $f$. If the set of zeros of $f$ has a limit point in $Omega$ then it is identically zero, being holomorphic. If the value of $f$ at a point $z_0$ is non zero then we can use the open set property to prove that it is not a point of minimum value.

share|cite|improve this answer

edited Feb 17 '14 at 10:31

answered Feb 17 '14 at 10:07

viplov_jainviplov_jain

723311

$endgroup$

All the isolated zeros of $f$ in $Omega$ will be the local minimum of $f$. If the set of zeros of $f$ has a limit point in $Omega$ then it is identically zero, being holomorphic. If the value of $f$ at a point $z_0$ is non zero then we can use the open set property to prove that it is not a point of minimum value.

share|cite|improve this answer

edited Feb 17 '14 at 10:31

answered Feb 17 '14 at 10:07

viplov_jainviplov_jain

723311

share|cite|improve this answer

share|cite|improve this answer

edited Feb 17 '14 at 10:31

edited Feb 17 '14 at 10:31

edited Feb 17 '14 at 10:31

answered Feb 17 '14 at 10:07

viplov_jainviplov_jain

723311

answered Feb 17 '14 at 10:07

viplov_jainviplov_jain

723311

answered Feb 17 '14 at 10:07

viplov_jainviplov_jain

723311

723311

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is not correct. Look at one of the simplest possible examples, $f(z) = z$ on the unit disc.
  $endgroup$
  – mrf
  Feb 17 '14 at 10:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was using the wrong principle, Thanks for pointing out. I've corrected it.
  $endgroup$
  – viplov_jain
  Feb 17 '14 at 10:32
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is not correct. Look at one of the simplest possible examples, $f(z) = z$ on the unit disc.
  $endgroup$
  – mrf
  Feb 17 '14 at 10:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was using the wrong principle, Thanks for pointing out. I've corrected it.
  $endgroup$
  – viplov_jain
  Feb 17 '14 at 10:32
  

  
  
  
  

$begingroup$
This is not correct. Look at one of the simplest possible examples, $f(z) = z$ on the unit disc.
$endgroup$
– mrf
Feb 17 '14 at 10:16

$begingroup$
This is not correct. Look at one of the simplest possible examples, $f(z) = z$ on the unit disc.
$endgroup$
– mrf
Feb 17 '14 at 10:16

$begingroup$
I was using the wrong principle, Thanks for pointing out. I've corrected it.
$endgroup$
– viplov_jain
Feb 17 '14 at 10:32

$begingroup$
I was using the wrong principle, Thanks for pointing out. I've corrected it.
$endgroup$
– viplov_jain
Feb 17 '14 at 10:32

add a comment | 

0

$begingroup$

If $f$ is a constant, $|f|$ has a local minimum at any point of $Omega.$

If $f$ is not a constant and there exists $zinOmega$ s.t. $f(z)=0,$ then the minimum occurs at those points $zinOmega.$

If $f$ is not a constant and there isn't any $zinOmega$ s.t. $f(z)=0,$ then $frac{1}{f}$ doesn't have singularities on $Omega.$ So, $fin h(Omega)$ and $frac{1}{f}in h(Omega).$ Let suppose that $|f|$ occurs minimum at $ainOmega.$ Then, by the Principle of Maximum Modulus for $frac{1}{f}$ we have that $vertfrac{1}{f(a)}vert<{vertfrac{1}{f(z)}vertvert zinpartial D}$ for any neighborhood $bar{D}subsetOmega$ of $a,$ i.e. $|f(a)|>|f(z)|$ for a $zinpartial DsubsetOmega$ witch contradicts the supposition that $a$ was a local minimum. This means that under there conditions, |f| does not occur a local minimum on $Omega.$

share|cite|improve this answer

answered Jan 7 at 16:24

EminEmin

1,36021330

$endgroup$

add a comment | 

0

$begingroup$

If $f$ is a constant, $|f|$ has a local minimum at any point of $Omega.$

If $f$ is not a constant and there exists $zinOmega$ s.t. $f(z)=0,$ then the minimum occurs at those points $zinOmega.$

If $f$ is not a constant and there isn't any $zinOmega$ s.t. $f(z)=0,$ then $frac{1}{f}$ doesn't have singularities on $Omega.$ So, $fin h(Omega)$ and $frac{1}{f}in h(Omega).$ Let suppose that $|f|$ occurs minimum at $ainOmega.$ Then, by the Principle of Maximum Modulus for $frac{1}{f}$ we have that $vertfrac{1}{f(a)}vert<{vertfrac{1}{f(z)}vertvert zinpartial D}$ for any neighborhood $bar{D}subsetOmega$ of $a,$ i.e. $|f(a)|>|f(z)|$ for a $zinpartial DsubsetOmega$ witch contradicts the supposition that $a$ was a local minimum. This means that under there conditions, |f| does not occur a local minimum on $Omega.$

share|cite|improve this answer

answered Jan 7 at 16:24

EminEmin

1,36021330

$endgroup$

add a comment | 

0

0

0

$begingroup$

If $f$ is a constant, $|f|$ has a local minimum at any point of $Omega.$

If $f$ is not a constant and there exists $zinOmega$ s.t. $f(z)=0,$ then the minimum occurs at those points $zinOmega.$

If $f$ is not a constant and there isn't any $zinOmega$ s.t. $f(z)=0,$ then $frac{1}{f}$ doesn't have singularities on $Omega.$ So, $fin h(Omega)$ and $frac{1}{f}in h(Omega).$ Let suppose that $|f|$ occurs minimum at $ainOmega.$ Then, by the Principle of Maximum Modulus for $frac{1}{f}$ we have that $vertfrac{1}{f(a)}vert<{vertfrac{1}{f(z)}vertvert zinpartial D}$ for any neighborhood $bar{D}subsetOmega$ of $a,$ i.e. $|f(a)|>|f(z)|$ for a $zinpartial DsubsetOmega$ witch contradicts the supposition that $a$ was a local minimum. This means that under there conditions, |f| does not occur a local minimum on $Omega.$

share|cite|improve this answer

answered Jan 7 at 16:24

EminEmin

1,36021330

$endgroup$

If $f$ is a constant, $|f|$ has a local minimum at any point of $Omega.$

If $f$ is not a constant and there exists $zinOmega$ s.t. $f(z)=0,$ then the minimum occurs at those points $zinOmega.$

If $f$ is not a constant and there isn't any $zinOmega$ s.t. $f(z)=0,$ then $frac{1}{f}$ doesn't have singularities on $Omega.$ So, $fin h(Omega)$ and $frac{1}{f}in h(Omega).$ Let suppose that $|f|$ occurs minimum at $ainOmega.$ Then, by the Principle of Maximum Modulus for $frac{1}{f}$ we have that $vertfrac{1}{f(a)}vert<{vertfrac{1}{f(z)}vertvert zinpartial D}$ for any neighborhood $bar{D}subsetOmega$ of $a,$ i.e. $|f(a)|>|f(z)|$ for a $zinpartial DsubsetOmega$ witch contradicts the supposition that $a$ was a local minimum. This means that under there conditions, |f| does not occur a local minimum on $Omega.$

share|cite|improve this answer

answered Jan 7 at 16:24

EminEmin

1,36021330

share|cite|improve this answer

share|cite|improve this answer

answered Jan 7 at 16:24

EminEmin

1,36021330

answered Jan 7 at 16:24

EminEmin

1,36021330

answered Jan 7 at 16:24

EminEmin

1,36021330

1,36021330

add a comment | 

add a comment | 

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