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A harmonic function cannot attain a strict local max in the domain
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I feel like I understand how the max principle works, but I don't understand the hint that was given on this problem. I am working on studying for an exam in a couple weeks. Any help would be much appreciated on how to use the hint.
The proof that I currently have is that by the mean value theorem, $$|f(a)|leq |frac12 pi i|int_0^2pi |f(a+Re^itheta)|dtheta$$
Suppose for the sake on contradiction that a is a strict local max in the domain D. Then $$|f(a)|> |frac12 pi i|int_0^2pi |f(a+Re^itheta)|dtheta$$ which contradicts the above inequality. So we have that $|f(z)|$ cannot attain a strict local max in D.
complex-analysis
asked Aug 2 at 19:13
MathIsHard
1,122415
add a comment |Â
up vote
1
down vote
favorite
I feel like I understand how the max principle works, but I don't understand the hint that was given on this problem. I am working on studying for an exam in a couple weeks. Any help would be much appreciated on how to use the hint.
The proof that I currently have is that by the mean value theorem, $$|f(a)|leq |frac12 pi i|int_0^2pi |f(a+Re^itheta)|dtheta$$
Suppose for the sake on contradiction that a is a strict local max in the domain D. Then $$|f(a)|> |frac12 pi i|int_0^2pi |f(a+Re^itheta)|dtheta$$ which contradicts the above inequality. So we have that $|f(z)|$ cannot attain a strict local max in D.
complex-analysis
asked Aug 2 at 19:13
MathIsHard
1,122415
2
Do you know that a real harmonic function is, locally, the real part of a holomorphic function?
– zhw.
Aug 2 at 20:59
1
The proof you have using the mean value theorem is correct and easy to understand. I don't know why you need a more complicated proof.
– Somos
Aug 2 at 21:44
@Somos thank you for the praise on my proof. Just always wanting to understand more math in different ways :)
– MathIsHard
Aug 2 at 23:00
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I feel like I understand how the max principle works, but I don't understand the hint that was given on this problem. I am working on studying for an exam in a couple weeks. Any help would be much appreciated on how to use the hint.
The proof that I currently have is that by the mean value theorem, $$|f(a)|leq |frac12 pi i|int_0^2pi |f(a+Re^itheta)|dtheta$$
Suppose for the sake on contradiction that a is a strict local max in the domain D. Then $$|f(a)|> |frac12 pi i|int_0^2pi |f(a+Re^itheta)|dtheta$$ which contradicts the above inequality. So we have that $|f(z)|$ cannot attain a strict local max in D.
complex-analysis
asked Aug 2 at 19:13
MathIsHard
1,122415
I feel like I understand how the max principle works, but I don't understand the hint that was given on this problem. I am working on studying for an exam in a couple weeks. Any help would be much appreciated on how to use the hint.
The proof that I currently have is that by the mean value theorem, $$|f(a)|leq |frac12 pi i|int_0^2pi |f(a+Re^itheta)|dtheta$$
Suppose for the sake on contradiction that a is a strict local max in the domain D. Then $$|f(a)|> |frac12 pi i|int_0^2pi |f(a+Re^itheta)|dtheta$$ which contradicts the above inequality. So we have that $|f(z)|$ cannot attain a strict local max in D.
complex-analysis
asked Aug 2 at 19:13
MathIsHard
1,122415
asked Aug 2 at 19:13
MathIsHard
1,122415
asked Aug 2 at 19:13
MathIsHard
1,122415
asked Aug 2 at 19:13
MathIsHard
1,122415
1,122415
2
Do you know that a real harmonic function is, locally, the real part of a holomorphic function?
– zhw.
Aug 2 at 20:59
1
The proof you have using the mean value theorem is correct and easy to understand. I don't know why you need a more complicated proof.
– Somos
Aug 2 at 21:44
@Somos thank you for the praise on my proof. Just always wanting to understand more math in different ways :)
– MathIsHard
Aug 2 at 23:00
add a comment |Â
2
Do you know that a real harmonic function is, locally, the real part of a holomorphic function?
– zhw.
Aug 2 at 20:59
1
The proof you have using the mean value theorem is correct and easy to understand. I don't know why you need a more complicated proof.
– Somos
Aug 2 at 21:44
@Somos thank you for the praise on my proof. Just always wanting to understand more math in different ways :)
– MathIsHard
Aug 2 at 23:00
2
2
Do you know that a real harmonic function is, locally, the real part of a holomorphic function?
– zhw.
Aug 2 at 20:59
Do you know that a real harmonic function is, locally, the real part of a holomorphic function?
– zhw.
Aug 2 at 20:59
1
1
The proof you have using the mean value theorem is correct and easy to understand. I don't know why you need a more complicated proof.
– Somos
Aug 2 at 21:44
The proof you have using the mean value theorem is correct and easy to understand. I don't know why you need a more complicated proof.
– Somos
Aug 2 at 21:44
@Somos thank you for the praise on my proof. Just always wanting to understand more math in different ways :)
– MathIsHard
Aug 2 at 23:00
@Somos thank you for the praise on my proof. Just always wanting to understand more math in different ways :)
– MathIsHard
Aug 2 at 23:00
add a comment |Â
1 Answer
1
active
oldest
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up vote
1
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accepted
Let $v(x, y)$ be a harmonic conjugate of $u(x, y)$ in $D$; then
$f(z) = f(x + iy) = u(x, y) + i v(x, y) tag 1$
is holomorphic in $D$, as is well-known. Thus, so
$g(z) = e^f(z) = e^u(x, y) + i v(x, y) tag 2$
is also holomorphic in $D$; this implies, by the maximum modulus principle, that $vert g(z) vert$ cannot attain a strict local maximum in $D$; but
$vert g(z) vert = vert e^f(z) vert = vert e^u(x, y) + i v(x, y) vert = vert e^u(x, y) e^i v(x, y) vert = vert e^u(x, y) vert vert e^i v(x, y) vert = e^u(x, y), tag 3$
since
$vert e^i v(x, y) vert = vert cos v(x, y) + i sin v(x, y) vert = 1 tag 4$
and, since $e^u(x, y)$ is positive real,
$vert e^u(x, y) vert = e^u(x, y); tag 5$
since $vert g(z) vert$ cannot attain a strict local maximum in $D$, neither can the function $u(x, y)$, since it is, by (3), the log modulus of the holomorphic function $g(z)$:
$u(x, y) = ln vert g(z) vert, tag 6$
and $ln$ is strictly monotonically increasing on the positive reals.
answered Aug 2 at 21:22
Robert Lewis
36.7k22155
1
Great proof. Thank you
– MathIsHard
Aug 2 at 22:51
@MathIsHard: You are most welcome. Thanks for the good word, and for the "acceptance"! Cheers!
– Robert Lewis
Aug 2 at 22:53
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Let $v(x, y)$ be a harmonic conjugate of $u(x, y)$ in $D$; then
$f(z) = f(x + iy) = u(x, y) + i v(x, y) tag 1$
is holomorphic in $D$, as is well-known. Thus, so
$g(z) = e^f(z) = e^u(x, y) + i v(x, y) tag 2$
is also holomorphic in $D$; this implies, by the maximum modulus principle, that $vert g(z) vert$ cannot attain a strict local maximum in $D$; but
$vert g(z) vert = vert e^f(z) vert = vert e^u(x, y) + i v(x, y) vert = vert e^u(x, y) e^i v(x, y) vert = vert e^u(x, y) vert vert e^i v(x, y) vert = e^u(x, y), tag 3$
since
$vert e^i v(x, y) vert = vert cos v(x, y) + i sin v(x, y) vert = 1 tag 4$
and, since $e^u(x, y)$ is positive real,
$vert e^u(x, y) vert = e^u(x, y); tag 5$
since $vert g(z) vert$ cannot attain a strict local maximum in $D$, neither can the function $u(x, y)$, since it is, by (3), the log modulus of the holomorphic function $g(z)$:
$u(x, y) = ln vert g(z) vert, tag 6$
and $ln$ is strictly monotonically increasing on the positive reals.
answered Aug 2 at 21:22
Robert Lewis
36.7k22155
1
Great proof. Thank you
– MathIsHard
Aug 2 at 22:51
@MathIsHard: You are most welcome. Thanks for the good word, and for the "acceptance"! Cheers!
– Robert Lewis
Aug 2 at 22:53
add a comment |Â
up vote
1
down vote
accepted
Let $v(x, y)$ be a harmonic conjugate of $u(x, y)$ in $D$; then
$f(z) = f(x + iy) = u(x, y) + i v(x, y) tag 1$
is holomorphic in $D$, as is well-known. Thus, so
$g(z) = e^f(z) = e^u(x, y) + i v(x, y) tag 2$
is also holomorphic in $D$; this implies, by the maximum modulus principle, that $vert g(z) vert$ cannot attain a strict local maximum in $D$; but
$vert g(z) vert = vert e^f(z) vert = vert e^u(x, y) + i v(x, y) vert = vert e^u(x, y) e^i v(x, y) vert = vert e^u(x, y) vert vert e^i v(x, y) vert = e^u(x, y), tag 3$
since
$vert e^i v(x, y) vert = vert cos v(x, y) + i sin v(x, y) vert = 1 tag 4$
and, since $e^u(x, y)$ is positive real,
$vert e^u(x, y) vert = e^u(x, y); tag 5$
since $vert g(z) vert$ cannot attain a strict local maximum in $D$, neither can the function $u(x, y)$, since it is, by (3), the log modulus of the holomorphic function $g(z)$:
$u(x, y) = ln vert g(z) vert, tag 6$
and $ln$ is strictly monotonically increasing on the positive reals.
answered Aug 2 at 21:22
Robert Lewis
36.7k22155
1
Great proof. Thank you
– MathIsHard
Aug 2 at 22:51
@MathIsHard: You are most welcome. Thanks for the good word, and for the "acceptance"! Cheers!
– Robert Lewis
Aug 2 at 22:53
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let $v(x, y)$ be a harmonic conjugate of $u(x, y)$ in $D$; then
$f(z) = f(x + iy) = u(x, y) + i v(x, y) tag 1$
is holomorphic in $D$, as is well-known. Thus, so
$g(z) = e^f(z) = e^u(x, y) + i v(x, y) tag 2$
is also holomorphic in $D$; this implies, by the maximum modulus principle, that $vert g(z) vert$ cannot attain a strict local maximum in $D$; but
$vert g(z) vert = vert e^f(z) vert = vert e^u(x, y) + i v(x, y) vert = vert e^u(x, y) e^i v(x, y) vert = vert e^u(x, y) vert vert e^i v(x, y) vert = e^u(x, y), tag 3$
since
$vert e^i v(x, y) vert = vert cos v(x, y) + i sin v(x, y) vert = 1 tag 4$
and, since $e^u(x, y)$ is positive real,
$vert e^u(x, y) vert = e^u(x, y); tag 5$
since $vert g(z) vert$ cannot attain a strict local maximum in $D$, neither can the function $u(x, y)$, since it is, by (3), the log modulus of the holomorphic function $g(z)$:
$u(x, y) = ln vert g(z) vert, tag 6$
and $ln$ is strictly monotonically increasing on the positive reals.
answered Aug 2 at 21:22
Robert Lewis
36.7k22155
Let $v(x, y)$ be a harmonic conjugate of $u(x, y)$ in $D$; then
$f(z) = f(x + iy) = u(x, y) + i v(x, y) tag 1$
is holomorphic in $D$, as is well-known. Thus, so
$g(z) = e^f(z) = e^u(x, y) + i v(x, y) tag 2$
is also holomorphic in $D$; this implies, by the maximum modulus principle, that $vert g(z) vert$ cannot attain a strict local maximum in $D$; but
$vert g(z) vert = vert e^f(z) vert = vert e^u(x, y) + i v(x, y) vert = vert e^u(x, y) e^i v(x, y) vert = vert e^u(x, y) vert vert e^i v(x, y) vert = e^u(x, y), tag 3$
since
$vert e^i v(x, y) vert = vert cos v(x, y) + i sin v(x, y) vert = 1 tag 4$
and, since $e^u(x, y)$ is positive real,
$vert e^u(x, y) vert = e^u(x, y); tag 5$
since $vert g(z) vert$ cannot attain a strict local maximum in $D$, neither can the function $u(x, y)$, since it is, by (3), the log modulus of the holomorphic function $g(z)$:
$u(x, y) = ln vert g(z) vert, tag 6$
and $ln$ is strictly monotonically increasing on the positive reals.
answered Aug 2 at 21:22
Robert Lewis
36.7k22155
edited Aug 2 at 21:29
answered Aug 2 at 21:22
Robert Lewis
36.7k22155
answered Aug 2 at 21:22
Robert Lewis
36.7k22155
answered Aug 2 at 21:22
Robert Lewis
36.7k22155
36.7k22155
1
Great proof. Thank you
– MathIsHard
Aug 2 at 22:51
@MathIsHard: You are most welcome. Thanks for the good word, and for the "acceptance"! Cheers!
– Robert Lewis
Aug 2 at 22:53
add a comment |Â
1
Great proof. Thank you
– MathIsHard
Aug 2 at 22:51
@MathIsHard: You are most welcome. Thanks for the good word, and for the "acceptance"! Cheers!
– Robert Lewis
Aug 2 at 22:53
1
1
Great proof. Thank you
– MathIsHard
Aug 2 at 22:51
Great proof. Thank you
– MathIsHard
Aug 2 at 22:51
@MathIsHard: You are most welcome. Thanks for the good word, and for the "acceptance"! Cheers!
– Robert Lewis
Aug 2 at 22:53
@MathIsHard: You are most welcome. Thanks for the good word, and for the "acceptance"! Cheers!
– Robert Lewis
Aug 2 at 22:53
add a comment |Â
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2
Do you know that a real harmonic function is, locally, the real part of a holomorphic function?
– zhw.
Aug 2 at 20:59
1
The proof you have using the mean value theorem is correct and easy to understand. I don't know why you need a more complicated proof.
– Somos
Aug 2 at 21:44
@Somos thank you for the praise on my proof. Just always wanting to understand more math in different ways :)
– MathIsHard
Aug 2 at 23:00