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$ lim_ntoinftyu_n(x,y)=u(x,y) $ is true for a sequence of harmonic functions satisfying $ lim_ntoinftyint_B |u_n(x, y)-u(x,y)|dxdy=0 .$

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0

$begingroup$

(i)Let $ B= (x, y)inmathbb R^2:x^2+y^2<1 $, and let $ u(x, y) $ be a harmonic function defined on some open set $ U $ containing the closure of $ B $. Prove that
$$ u(0,0)=frac 1piint_Bu(x,y)dxdy .$$
(ii)Suppose, in addition to the above assumption, that $ u_n(x,y)_n=1^infty $ is a sequence of harmonic functions on $ U $ such that
$$ lim_ntoinftyint_B |u_n(x, y)-u(x,y)|dxdy=0 .$$
Show that $ lim_ntoinftyu_n(x,y)=u(x,y) $ for all $ (x,y)in B $.

---

My attempt:

(i) is rather straightforward:
beginalign
frac 1piint_Bu(x,y)dxdy&=frac 1 piint_B u(r,theta)rdrdtheta\
&=frac 1 piint_0^1 rleft( int_0^2piu(r, theta)dtheta right)dr\
&=frac 1 piint_0^1 r2pi u(0,0)dr\
&=2u(0,0)int_0^1 rdr\
&=u(0,0)
endalign
where we have used the mean value property of harmonic function over a unit ball.

For (ii)
beginalign
&&lim_ntoinftyint_B left|u_n(x,y)-u(x,y)right|dxdy&=0\
&implies&lim_ntoinftyleft|int_B left[u_n(x,y)-u(x,y)right]dxdyright|&=0\
&implies&lim_ntoinftyint_Bleft[ u_n(x,y)-u(x,y)right]dxdy&=0\
&implies&lim_ntoinftyint_Bu_n(x,y)dxdy&=int_Bu(x,y)dxdy\
&implies&pilim_ntoinftyu_n(0,0)&=pi u(0,0)\
&implies&lim_ntoinftyu_n(0,0)&= u(0,0)
endalign
So we have proved $ lim_ntoinftyu_n(x,y)=u(x,y) $ for $ (0,0) $. Can we apply the above technique to an arbitray ball centered at a point in $ B $ to conclude that $ lim_ntoinftyu_n(x,y)=u(x,y) $ is true for all $ (x,y)in B $ ?

complex-analysis proof-verification harmonic-functions

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asked Mar 29 at 2:57

user549397user549397

1,6591418

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. If $(a,b)$ is point in $B$, then the function $v(x,y) =u(a+x,b+y)$ is also harmonic. Therefore the results above hold for it as well.
  $endgroup$
  – Fnacool
  Mar 29 at 3:03
  

  
  
  
  

add a comment | 

0

$begingroup$

(i)Let $ B= (x, y)inmathbb R^2:x^2+y^2<1 $, and let $ u(x, y) $ be a harmonic function defined on some open set $ U $ containing the closure of $ B $. Prove that
$$ u(0,0)=frac 1piint_Bu(x,y)dxdy .$$
(ii)Suppose, in addition to the above assumption, that $ u_n(x,y)_n=1^infty $ is a sequence of harmonic functions on $ U $ such that
$$ lim_ntoinftyint_B |u_n(x, y)-u(x,y)|dxdy=0 .$$
Show that $ lim_ntoinftyu_n(x,y)=u(x,y) $ for all $ (x,y)in B $.

---

My attempt:

(i) is rather straightforward:
beginalign
frac 1piint_Bu(x,y)dxdy&=frac 1 piint_B u(r,theta)rdrdtheta\
&=frac 1 piint_0^1 rleft( int_0^2piu(r, theta)dtheta right)dr\
&=frac 1 piint_0^1 r2pi u(0,0)dr\
&=2u(0,0)int_0^1 rdr\
&=u(0,0)
endalign
where we have used the mean value property of harmonic function over a unit ball.

For (ii)
beginalign
&&lim_ntoinftyint_B left|u_n(x,y)-u(x,y)right|dxdy&=0\
&implies&lim_ntoinftyleft|int_B left[u_n(x,y)-u(x,y)right]dxdyright|&=0\
&implies&lim_ntoinftyint_Bleft[ u_n(x,y)-u(x,y)right]dxdy&=0\
&implies&lim_ntoinftyint_Bu_n(x,y)dxdy&=int_Bu(x,y)dxdy\
&implies&pilim_ntoinftyu_n(0,0)&=pi u(0,0)\
&implies&lim_ntoinftyu_n(0,0)&= u(0,0)
endalign
So we have proved $ lim_ntoinftyu_n(x,y)=u(x,y) $ for $ (0,0) $. Can we apply the above technique to an arbitray ball centered at a point in $ B $ to conclude that $ lim_ntoinftyu_n(x,y)=u(x,y) $ is true for all $ (x,y)in B $ ?

complex-analysis proof-verification harmonic-functions

share|cite|improve this question

asked Mar 29 at 2:57

user549397user549397

1,6591418

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. If $(a,b)$ is point in $B$, then the function $v(x,y) =u(a+x,b+y)$ is also harmonic. Therefore the results above hold for it as well.
  $endgroup$
  – Fnacool
  Mar 29 at 3:03
  

  
  
  
  

add a comment | 

$begingroup$

(i)Let $ B= (x, y)inmathbb R^2:x^2+y^2<1 $, and let $ u(x, y) $ be a harmonic function defined on some open set $ U $ containing the closure of $ B $. Prove that
$$ u(0,0)=frac 1piint_Bu(x,y)dxdy .$$
(ii)Suppose, in addition to the above assumption, that $ u_n(x,y)_n=1^infty $ is a sequence of harmonic functions on $ U $ such that
$$ lim_ntoinftyint_B |u_n(x, y)-u(x,y)|dxdy=0 .$$
Show that $ lim_ntoinftyu_n(x,y)=u(x,y) $ for all $ (x,y)in B $.

---

My attempt:

(i) is rather straightforward:
beginalign
frac 1piint_Bu(x,y)dxdy&=frac 1 piint_B u(r,theta)rdrdtheta\
&=frac 1 piint_0^1 rleft( int_0^2piu(r, theta)dtheta right)dr\
&=frac 1 piint_0^1 r2pi u(0,0)dr\
&=2u(0,0)int_0^1 rdr\
&=u(0,0)
endalign
where we have used the mean value property of harmonic function over a unit ball.

For (ii)
beginalign
&&lim_ntoinftyint_B left|u_n(x,y)-u(x,y)right|dxdy&=0\
&implies&lim_ntoinftyleft|int_B left[u_n(x,y)-u(x,y)right]dxdyright|&=0\
&implies&lim_ntoinftyint_Bleft[ u_n(x,y)-u(x,y)right]dxdy&=0\
&implies&lim_ntoinftyint_Bu_n(x,y)dxdy&=int_Bu(x,y)dxdy\
&implies&pilim_ntoinftyu_n(0,0)&=pi u(0,0)\
&implies&lim_ntoinftyu_n(0,0)&= u(0,0)
endalign
So we have proved $ lim_ntoinftyu_n(x,y)=u(x,y) $ for $ (0,0) $. Can we apply the above technique to an arbitray ball centered at a point in $ B $ to conclude that $ lim_ntoinftyu_n(x,y)=u(x,y) $ is true for all $ (x,y)in B $ ?

complex-analysis proof-verification harmonic-functions

share|cite|improve this question

asked Mar 29 at 2:57

user549397user549397

1,6591418

$endgroup$

share|cite|improve this question

asked Mar 29 at 2:57

user549397user549397

1,6591418

share|cite|improve this question

asked Mar 29 at 2:57

user549397user549397

1,6591418

asked Mar 29 at 2:57

user549397user549397

1,6591418

asked Mar 29 at 2:57

user549397user549397

1,6591418