Ahlfors Proof of Cauchy-Riemann

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In proving the sufficient condition, Ahlfors claims that if $u(x,y)$ has a continuous, first order partials, then we can write $$u(x+h,y+k) - u(x,y) = fracpartialupartial xh +fracpartialupartial yk + epsilon$$ where $epsilon$ is little-o of $h +ik$. Now I am familiar with this definition for real functions of one variable (in fact I attempted a proof of it on a previous question), however I'm having difficulties seeing it given the limit definition of partials. Can anyone show a quick proof of this for 2 or n-variables.

complex-analysis proof-explanation

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edited Jul 30 at 21:16

asked Jul 30 at 20:30

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450417

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  You lost a $k$ in there.
  – GEdgar
  Jul 30 at 20:56
  

  
  
  
  

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In proving the sufficient condition, Ahlfors claims that if $u(x,y)$ has a continuous, first order partials, then we can write $$u(x+h,y+k) - u(x,y) = fracpartialupartial xh +fracpartialupartial yk + epsilon$$ where $epsilon$ is little-o of $h +ik$. Now I am familiar with this definition for real functions of one variable (in fact I attempted a proof of it on a previous question), however I'm having difficulties seeing it given the limit definition of partials. Can anyone show a quick proof of this for 2 or n-variables.

complex-analysis proof-explanation

share|cite|improve this question

edited Jul 30 at 21:16

asked Jul 30 at 20:30

Good Morning Captain

450417

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You lost a $k$ in there.
  – GEdgar
  Jul 30 at 20:56
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

In proving the sufficient condition, Ahlfors claims that if $u(x,y)$ has a continuous, first order partials, then we can write $$u(x+h,y+k) - u(x,y) = fracpartialupartial xh +fracpartialupartial yk + epsilon$$ where $epsilon$ is little-o of $h +ik$. Now I am familiar with this definition for real functions of one variable (in fact I attempted a proof of it on a previous question), however I'm having difficulties seeing it given the limit definition of partials. Can anyone show a quick proof of this for 2 or n-variables.

complex-analysis proof-explanation

share|cite|improve this question

edited Jul 30 at 21:16

asked Jul 30 at 20:30

Good Morning Captain

450417

In proving the sufficient condition, Ahlfors claims that if $u(x,y)$ has a continuous, first order partials, then we can write $$u(x+h,y+k) - u(x,y) = fracpartialupartial xh +fracpartialupartial yk + epsilon$$ where $epsilon$ is little-o of $h +ik$. Now I am familiar with this definition for real functions of one variable (in fact I attempted a proof of it on a previous question), however I'm having difficulties seeing it given the limit definition of partials. Can anyone show a quick proof of this for 2 or n-variables.

complex-analysis proof-explanation

share|cite|improve this question

edited Jul 30 at 21:16

asked Jul 30 at 20:30

Good Morning Captain

450417

share|cite|improve this question

share|cite|improve this question

edited Jul 30 at 21:16

edited Jul 30 at 21:16

edited Jul 30 at 21:16

asked Jul 30 at 20:30

Good Morning Captain

450417

asked Jul 30 at 20:30

Good Morning Captain

450417

asked Jul 30 at 20:30

Good Morning Captain

450417

450417

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You lost a $k$ in there.
  – GEdgar
  Jul 30 at 20:56
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You lost a $k$ in there.
  – GEdgar
  Jul 30 at 20:56
  

  
  
  
  

1

1

You lost a $k$ in there.
– GEdgar
Jul 30 at 20:56

You lost a $k$ in there.
– GEdgar
Jul 30 at 20:56

add a comment | 

1 Answer
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Continuous partial derivatives implies differentiability, can be found in many books of multivariable calculus, and has been asked and proved here before.

Differentiability implies that it can be written in such a form, with some linear function on $(h,k)$ in place of $fracpartial upartial xh+fracpartial upartial yk$.

$$beginalignu(x+h,y+k)-u(x,y)&=L_(x,y)(h,k)+epsilon\L_(x,y)(h,k)&=ah+bkendalign$$

Restricting the limit in the definition of differentiability to the $X$-axis and then to the $Y$-axis you get that the linear function has coefficients $a=fracpartial upartial x$ and $b=fracpartial upartial y$.

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answered Jul 30 at 21:15

plasticConnection

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1 Answer
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1 Answer
1

active

oldest

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oldest

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active

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votes

up vote
1
down vote



accepted

Continuous partial derivatives implies differentiability, can be found in many books of multivariable calculus, and has been asked and proved here before.

Differentiability implies that it can be written in such a form, with some linear function on $(h,k)$ in place of $fracpartial upartial xh+fracpartial upartial yk$.

$$beginalignu(x+h,y+k)-u(x,y)&=L_(x,y)(h,k)+epsilon\L_(x,y)(h,k)&=ah+bkendalign$$

Restricting the limit in the definition of differentiability to the $X$-axis and then to the $Y$-axis you get that the linear function has coefficients $a=fracpartial upartial x$ and $b=fracpartial upartial y$.

share|cite|improve this answer

answered Jul 30 at 21:15

plasticConnection

362

add a comment | 

up vote
1
down vote



accepted

Continuous partial derivatives implies differentiability, can be found in many books of multivariable calculus, and has been asked and proved here before.

Differentiability implies that it can be written in such a form, with some linear function on $(h,k)$ in place of $fracpartial upartial xh+fracpartial upartial yk$.

$$beginalignu(x+h,y+k)-u(x,y)&=L_(x,y)(h,k)+epsilon\L_(x,y)(h,k)&=ah+bkendalign$$

Restricting the limit in the definition of differentiability to the $X$-axis and then to the $Y$-axis you get that the linear function has coefficients $a=fracpartial upartial x$ and $b=fracpartial upartial y$.

share|cite|improve this answer

answered Jul 30 at 21:15

plasticConnection

362

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

Continuous partial derivatives implies differentiability, can be found in many books of multivariable calculus, and has been asked and proved here before.

Differentiability implies that it can be written in such a form, with some linear function on $(h,k)$ in place of $fracpartial upartial xh+fracpartial upartial yk$.

$$beginalignu(x+h,y+k)-u(x,y)&=L_(x,y)(h,k)+epsilon\L_(x,y)(h,k)&=ah+bkendalign$$

Restricting the limit in the definition of differentiability to the $X$-axis and then to the $Y$-axis you get that the linear function has coefficients $a=fracpartial upartial x$ and $b=fracpartial upartial y$.

share|cite|improve this answer

answered Jul 30 at 21:15

plasticConnection

362

Continuous partial derivatives implies differentiability, can be found in many books of multivariable calculus, and has been asked and proved here before.

Differentiability implies that it can be written in such a form, with some linear function on $(h,k)$ in place of $fracpartial upartial xh+fracpartial upartial yk$.

$$beginalignu(x+h,y+k)-u(x,y)&=L_(x,y)(h,k)+epsilon\L_(x,y)(h,k)&=ah+bkendalign$$

Restricting the limit in the definition of differentiability to the $X$-axis and then to the $Y$-axis you get that the linear function has coefficients $a=fracpartial upartial x$ and $b=fracpartial upartial y$.

share|cite|improve this answer

answered Jul 30 at 21:15

plasticConnection

362

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 21:15

plasticConnection

362

answered Jul 30 at 21:15

plasticConnection

362

answered Jul 30 at 21:15

plasticConnection

362

362

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