Mixing
A non-constant harmonic function must have a zero
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
I have the question:
If $u$ is a non-constant harmonic function, it must have a zero. Where is $u$ defined from the complex to the real set?
My approach has been as follows:
Since $u$ is non-constant harmonic, it must have a harmonic conjugate $v$ such that $f = u + iv$ is analytic. For a $z$ in a circle, we can use the Cauchy Integral formula to define $f(z)$ and then separate it into real and imaginary parts.
Doing this, we get the formula from the mean value property for harmonic functions. But I am very confused about the mean value property and how it ensures a zero. Since $u$ is non-constant, does it take both negative and positive values and hence by IVP, it must take zero as well? Any pointers on how to proceed?
complex-analysis harmonic-functions
asked Jul 16 at 0:20
deadcode
228
add a comment |Â
up vote
1
down vote
favorite
I have the question:
If $u$ is a non-constant harmonic function, it must have a zero. Where is $u$ defined from the complex to the real set?
My approach has been as follows:
Since $u$ is non-constant harmonic, it must have a harmonic conjugate $v$ such that $f = u + iv$ is analytic. For a $z$ in a circle, we can use the Cauchy Integral formula to define $f(z)$ and then separate it into real and imaginary parts.
Doing this, we get the formula from the mean value property for harmonic functions. But I am very confused about the mean value property and how it ensures a zero. Since $u$ is non-constant, does it take both negative and positive values and hence by IVP, it must take zero as well? Any pointers on how to proceed?
complex-analysis harmonic-functions
asked Jul 16 at 0:20
deadcode
228
You seem to be asking, when you write "Where is $u$ defined from the complex to the real set?", what is the preimage of the reals. That is, what is $u^-1(mathbbR)$, which is the same as asking: what is the set $z in mathbbC : u(z) in mathbbR$. Is this what you want to know? I ask because the work you've shown seems to be trying to find a zero, not finding the preimage of the reals.
– Eric Towers
Jul 16 at 1:53
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have the question:
If $u$ is a non-constant harmonic function, it must have a zero. Where is $u$ defined from the complex to the real set?
My approach has been as follows:
Since $u$ is non-constant harmonic, it must have a harmonic conjugate $v$ such that $f = u + iv$ is analytic. For a $z$ in a circle, we can use the Cauchy Integral formula to define $f(z)$ and then separate it into real and imaginary parts.
Doing this, we get the formula from the mean value property for harmonic functions. But I am very confused about the mean value property and how it ensures a zero. Since $u$ is non-constant, does it take both negative and positive values and hence by IVP, it must take zero as well? Any pointers on how to proceed?
complex-analysis harmonic-functions
asked Jul 16 at 0:20
deadcode
228
I have the question:
If $u$ is a non-constant harmonic function, it must have a zero. Where is $u$ defined from the complex to the real set?
My approach has been as follows:
Since $u$ is non-constant harmonic, it must have a harmonic conjugate $v$ such that $f = u + iv$ is analytic. For a $z$ in a circle, we can use the Cauchy Integral formula to define $f(z)$ and then separate it into real and imaginary parts.
Doing this, we get the formula from the mean value property for harmonic functions. But I am very confused about the mean value property and how it ensures a zero. Since $u$ is non-constant, does it take both negative and positive values and hence by IVP, it must take zero as well? Any pointers on how to proceed?
complex-analysis harmonic-functions
asked Jul 16 at 0:20
deadcode
228
edited Jul 16 at 1:47
asked Jul 16 at 0:20
deadcode
228
asked Jul 16 at 0:20
deadcode
228
asked Jul 16 at 0:20
deadcode
228
228
You seem to be asking, when you write "Where is $u$ defined from the complex to the real set?", what is the preimage of the reals. That is, what is $u^-1(mathbbR)$, which is the same as asking: what is the set $z in mathbbC : u(z) in mathbbR$. Is this what you want to know? I ask because the work you've shown seems to be trying to find a zero, not finding the preimage of the reals.
– Eric Towers
Jul 16 at 1:53
add a comment |Â
You seem to be asking, when you write "Where is $u$ defined from the complex to the real set?", what is the preimage of the reals. That is, what is $u^-1(mathbbR)$, which is the same as asking: what is the set $z in mathbbC : u(z) in mathbbR$. Is this what you want to know? I ask because the work you've shown seems to be trying to find a zero, not finding the preimage of the reals.
– Eric Towers
Jul 16 at 1:53
You seem to be asking, when you write "Where is $u$ defined from the complex to the real set?", what is the preimage of the reals. That is, what is $u^-1(mathbbR)$, which is the same as asking: what is the set $z in mathbbC : u(z) in mathbbR$. Is this what you want to know? I ask because the work you've shown seems to be trying to find a zero, not finding the preimage of the reals.
– Eric Towers
Jul 16 at 1:53
You seem to be asking, when you write "Where is $u$ defined from the complex to the real set?", what is the preimage of the reals. That is, what is $u^-1(mathbbR)$, which is the same as asking: what is the set $z in mathbbC : u(z) in mathbbR$. Is this what you want to know? I ask because the work you've shown seems to be trying to find a zero, not finding the preimage of the reals.
– Eric Towers
Jul 16 at 1:53
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Suppose $u:mathbb Cto mathbb R$ is a nonconstant harmonic function, and that $u$ is never $0.$ Because $u$ is continuous, $u(mathbb C)$ is a connected subset of $mathbb R.$ That implies $u(mathbb C)$ is an interval that doesn't contain $0.$ Thus either $u(mathbb C)subset (0,infty)$ or $u(mathbb C)subset (-infty,0).$ Suppose it's the second case. Let $v$ be a harmonic conjugate of $u$ on $mathbb C.$ Then $f(z) = e^u+iv$ is an entire function such that $|f(z)| = e^u(z) le e^0=1$ everywhere. This implies $f$ is a bounded entire function. By Liouville, $f$ is constant. That implies $u$ is constant, contradiction.
answered Jul 16 at 2:36
zhw.
65.8k42870
Nice! Very beautiful!
– Hugocito
Jul 16 at 2:42
thank you for the elegant solution!
– deadcode
Jul 16 at 3:14
add a comment |Â
up vote
0
down vote
The result is true because of the Little Picard Theorem.
Since $u: mathbb C to mathbb R$ is harmonic and non-constant, there is a non-constant holomorphic function $f:mathbb C to mathbb C$ such that $textRef = u$. The Little Picard Theorem says that $f(mathbb C) = mathbb C$ or $f(mathbb C) = mathbb Csetminus p$ for some $pin mathbb C$. So, you can take some point of the form $0+ieta$ on $f(mathbb C)$. Therefore, there is $z=x+iy$ such that $f(z) = 0+ieta$ and from that we conclude $u(z) = 0$.
answered Jul 16 at 2:21
Hugocito
1,6451019
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Suppose $u:mathbb Cto mathbb R$ is a nonconstant harmonic function, and that $u$ is never $0.$ Because $u$ is continuous, $u(mathbb C)$ is a connected subset of $mathbb R.$ That implies $u(mathbb C)$ is an interval that doesn't contain $0.$ Thus either $u(mathbb C)subset (0,infty)$ or $u(mathbb C)subset (-infty,0).$ Suppose it's the second case. Let $v$ be a harmonic conjugate of $u$ on $mathbb C.$ Then $f(z) = e^u+iv$ is an entire function such that $|f(z)| = e^u(z) le e^0=1$ everywhere. This implies $f$ is a bounded entire function. By Liouville, $f$ is constant. That implies $u$ is constant, contradiction.
answered Jul 16 at 2:36
zhw.
65.8k42870
Nice! Very beautiful!
– Hugocito
Jul 16 at 2:42
thank you for the elegant solution!
– deadcode
Jul 16 at 3:14
add a comment |Â
up vote
1
down vote
accepted
Suppose $u:mathbb Cto mathbb R$ is a nonconstant harmonic function, and that $u$ is never $0.$ Because $u$ is continuous, $u(mathbb C)$ is a connected subset of $mathbb R.$ That implies $u(mathbb C)$ is an interval that doesn't contain $0.$ Thus either $u(mathbb C)subset (0,infty)$ or $u(mathbb C)subset (-infty,0).$ Suppose it's the second case. Let $v$ be a harmonic conjugate of $u$ on $mathbb C.$ Then $f(z) = e^u+iv$ is an entire function such that $|f(z)| = e^u(z) le e^0=1$ everywhere. This implies $f$ is a bounded entire function. By Liouville, $f$ is constant. That implies $u$ is constant, contradiction.
answered Jul 16 at 2:36
zhw.
65.8k42870
Nice! Very beautiful!
– Hugocito
Jul 16 at 2:42
thank you for the elegant solution!
– deadcode
Jul 16 at 3:14
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Suppose $u:mathbb Cto mathbb R$ is a nonconstant harmonic function, and that $u$ is never $0.$ Because $u$ is continuous, $u(mathbb C)$ is a connected subset of $mathbb R.$ That implies $u(mathbb C)$ is an interval that doesn't contain $0.$ Thus either $u(mathbb C)subset (0,infty)$ or $u(mathbb C)subset (-infty,0).$ Suppose it's the second case. Let $v$ be a harmonic conjugate of $u$ on $mathbb C.$ Then $f(z) = e^u+iv$ is an entire function such that $|f(z)| = e^u(z) le e^0=1$ everywhere. This implies $f$ is a bounded entire function. By Liouville, $f$ is constant. That implies $u$ is constant, contradiction.
answered Jul 16 at 2:36
zhw.
65.8k42870
Suppose $u:mathbb Cto mathbb R$ is a nonconstant harmonic function, and that $u$ is never $0.$ Because $u$ is continuous, $u(mathbb C)$ is a connected subset of $mathbb R.$ That implies $u(mathbb C)$ is an interval that doesn't contain $0.$ Thus either $u(mathbb C)subset (0,infty)$ or $u(mathbb C)subset (-infty,0).$ Suppose it's the second case. Let $v$ be a harmonic conjugate of $u$ on $mathbb C.$ Then $f(z) = e^u+iv$ is an entire function such that $|f(z)| = e^u(z) le e^0=1$ everywhere. This implies $f$ is a bounded entire function. By Liouville, $f$ is constant. That implies $u$ is constant, contradiction.
answered Jul 16 at 2:36
zhw.
65.8k42870
answered Jul 16 at 2:36
zhw.
65.8k42870
answered Jul 16 at 2:36
zhw.
65.8k42870
answered Jul 16 at 2:36
zhw.
65.8k42870
65.8k42870
Nice! Very beautiful!
– Hugocito
Jul 16 at 2:42
thank you for the elegant solution!
– deadcode
Jul 16 at 3:14
add a comment |Â
Nice! Very beautiful!
– Hugocito
Jul 16 at 2:42
thank you for the elegant solution!
– deadcode
Jul 16 at 3:14
Nice! Very beautiful!
– Hugocito
Jul 16 at 2:42
Nice! Very beautiful!
– Hugocito
Jul 16 at 2:42
thank you for the elegant solution!
– deadcode
Jul 16 at 3:14
thank you for the elegant solution!
– deadcode
Jul 16 at 3:14
add a comment |Â
up vote
0
down vote
The result is true because of the Little Picard Theorem.
Since $u: mathbb C to mathbb R$ is harmonic and non-constant, there is a non-constant holomorphic function $f:mathbb C to mathbb C$ such that $textRef = u$. The Little Picard Theorem says that $f(mathbb C) = mathbb C$ or $f(mathbb C) = mathbb Csetminus p$ for some $pin mathbb C$. So, you can take some point of the form $0+ieta$ on $f(mathbb C)$. Therefore, there is $z=x+iy$ such that $f(z) = 0+ieta$ and from that we conclude $u(z) = 0$.
answered Jul 16 at 2:21
Hugocito
1,6451019
add a comment |Â
up vote
0
down vote
The result is true because of the Little Picard Theorem.
Since $u: mathbb C to mathbb R$ is harmonic and non-constant, there is a non-constant holomorphic function $f:mathbb C to mathbb C$ such that $textRef = u$. The Little Picard Theorem says that $f(mathbb C) = mathbb C$ or $f(mathbb C) = mathbb Csetminus p$ for some $pin mathbb C$. So, you can take some point of the form $0+ieta$ on $f(mathbb C)$. Therefore, there is $z=x+iy$ such that $f(z) = 0+ieta$ and from that we conclude $u(z) = 0$.
answered Jul 16 at 2:21
Hugocito
1,6451019
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The result is true because of the Little Picard Theorem.
Since $u: mathbb C to mathbb R$ is harmonic and non-constant, there is a non-constant holomorphic function $f:mathbb C to mathbb C$ such that $textRef = u$. The Little Picard Theorem says that $f(mathbb C) = mathbb C$ or $f(mathbb C) = mathbb Csetminus p$ for some $pin mathbb C$. So, you can take some point of the form $0+ieta$ on $f(mathbb C)$. Therefore, there is $z=x+iy$ such that $f(z) = 0+ieta$ and from that we conclude $u(z) = 0$.
answered Jul 16 at 2:21
Hugocito
1,6451019
The result is true because of the Little Picard Theorem.
Since $u: mathbb C to mathbb R$ is harmonic and non-constant, there is a non-constant holomorphic function $f:mathbb C to mathbb C$ such that $textRef = u$. The Little Picard Theorem says that $f(mathbb C) = mathbb C$ or $f(mathbb C) = mathbb Csetminus p$ for some $pin mathbb C$. So, you can take some point of the form $0+ieta$ on $f(mathbb C)$. Therefore, there is $z=x+iy$ such that $f(z) = 0+ieta$ and from that we conclude $u(z) = 0$.
answered Jul 16 at 2:21
Hugocito
1,6451019
edited Jul 16 at 2:40
answered Jul 16 at 2:21
Hugocito
1,6451019
answered Jul 16 at 2:21
Hugocito
1,6451019
answered Jul 16 at 2:21
Hugocito
1,6451019
1,6451019
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2852962%2fa-non-constant-harmonic-function-must-have-a-zero%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
You seem to be asking, when you write "Where is $u$ defined from the complex to the real set?", what is the preimage of the reals. That is, what is $u^-1(mathbbR)$, which is the same as asking: what is the set $z in mathbbC : u(z) in mathbbR$. Is this what you want to know? I ask because the work you've shown seems to be trying to find a zero, not finding the preimage of the reals.
– Eric Towers
Jul 16 at 1:53