A maximum principle for bounded functions in unbounded domain
$begingroup$
Let $U subsetneq mathbb{R}^2$ be a domain.
Suppose that $u in C^2(U) cap C(bar{U})$ is a bounded harmonic function such that $u leq 0$ on $partial U$.
If $U$ is bounded, then the maximum principle yields that $uleq 0$ in all of $U$.
Is it possible to conclude that $u leq 0$ in all of $U$ without the assumption that $U$ is bounded? Does anyone have an idea of how to proceed with this?
Thanks!
Update: If $U$ is such that $U^complement$ contains an open ball, then using the fundamental solution in $mathbb{R}^2$ and following strategy outlined by @user254433 in the comments, I was able to prove the statement.
Any ideas on how to proceed if $U^complement$ does not contain an open ball? In case it's helpful: I know that if $U = mathbb{R}^2setminus{p}$ for some point $p$ then any bounded harmonic function on $U$ is constant.
real-analysis pde harmonic-functions maximum-principle
asked Dec 5 '18 at 5:58
QuokaQuoka
1,240212
$endgroup$
add a comment |
$begingroup$
Let $U subsetneq mathbb{R}^2$ be a domain.
Suppose that $u in C^2(U) cap C(bar{U})$ is a bounded harmonic function such that $u leq 0$ on $partial U$.
If $U$ is bounded, then the maximum principle yields that $uleq 0$ in all of $U$.
Is it possible to conclude that $u leq 0$ in all of $U$ without the assumption that $U$ is bounded? Does anyone have an idea of how to proceed with this?
Thanks!
Update: If $U$ is such that $U^complement$ contains an open ball, then using the fundamental solution in $mathbb{R}^2$ and following strategy outlined by @user254433 in the comments, I was able to prove the statement.
Any ideas on how to proceed if $U^complement$ does not contain an open ball? In case it's helpful: I know that if $U = mathbb{R}^2setminus{p}$ for some point $p$ then any bounded harmonic function on $U$ is constant.
real-analysis pde harmonic-functions maximum-principle
asked Dec 5 '18 at 5:58
QuokaQuoka
1,240212
$endgroup$
add a comment |
$begingroup$
Let $U subsetneq mathbb{R}^2$ be a domain.
Suppose that $u in C^2(U) cap C(bar{U})$ is a bounded harmonic function such that $u leq 0$ on $partial U$.
If $U$ is bounded, then the maximum principle yields that $uleq 0$ in all of $U$.
Is it possible to conclude that $u leq 0$ in all of $U$ without the assumption that $U$ is bounded? Does anyone have an idea of how to proceed with this?
Thanks!
Update: If $U$ is such that $U^complement$ contains an open ball, then using the fundamental solution in $mathbb{R}^2$ and following strategy outlined by @user254433 in the comments, I was able to prove the statement.
Any ideas on how to proceed if $U^complement$ does not contain an open ball? In case it's helpful: I know that if $U = mathbb{R}^2setminus{p}$ for some point $p$ then any bounded harmonic function on $U$ is constant.
real-analysis pde harmonic-functions maximum-principle
asked Dec 5 '18 at 5:58
QuokaQuoka
1,240212
$endgroup$
Let $U subsetneq mathbb{R}^2$ be a domain.
Suppose that $u in C^2(U) cap C(bar{U})$ is a bounded harmonic function such that $u leq 0$ on $partial U$.
If $U$ is bounded, then the maximum principle yields that $uleq 0$ in all of $U$.
Is it possible to conclude that $u leq 0$ in all of $U$ without the assumption that $U$ is bounded? Does anyone have an idea of how to proceed with this?
Thanks!
Update: If $U$ is such that $U^complement$ contains an open ball, then using the fundamental solution in $mathbb{R}^2$ and following strategy outlined by @user254433 in the comments, I was able to prove the statement.
Any ideas on how to proceed if $U^complement$ does not contain an open ball? In case it's helpful: I know that if $U = mathbb{R}^2setminus{p}$ for some point $p$ then any bounded harmonic function on $U$ is constant.
real-analysis pde harmonic-functions maximum-principle
real-analysis pde harmonic-functions maximum-principle
asked Dec 5 '18 at 5:58
QuokaQuoka
1,240212
asked Dec 5 '18 at 5:58
QuokaQuoka
1,240212
edited Dec 5 '18 at 19:33
Quoka
asked Dec 5 '18 at 5:58
QuokaQuoka
1,240212
asked Dec 5 '18 at 5:58
QuokaQuoka
1,240212
asked Dec 5 '18 at 5:58
QuokaQuoka
1,240212
1,240212
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
One problem is that $partial U$ is "smaller" in the unbounded case, so $ule 0$ on $partial U$ becomes less restrictive. For example, if $U=mathbb R^n$, then $partial U=varnothing$, so the condition $u|_{partial U}le 0$ implies no restrictions on $u$. So although Liouville's theorem for bounded harmonic functions on $mathbb R^n$ implies $uequiv a$ is constant, this constant could be positive.
To make sense of $u|_{|x|=infty}le 0$, we could replace it with a limiting condition, like $limsup_{|x|toinfty} u(x)le 0$. Once we do this, it becomes natural to apply the maximum principle on bounded approximations of $U$:
For each $R>0$, let $U_R=Ucap B_R$ be a large subdomain of $U$. Then for each small $epsilon>0$, we can find a large radius $R=R(epsilon)$ so that $u|_{partial B_R}le epsilon$, which, since $epsilon>0$, implies $u|_{partial U_R}le epsilon$. By the maximum principle, $u|_{U_R}le epsilon$. Sending $epsilonto 0$ gives the desired conclusion.
answered Dec 5 '18 at 6:41
user254433user254433
2,531712
$endgroup$
$begingroup$
Sorry I should have been clearer in my question. So this is inspired from this question. I was wondering if it would be sufficient to only have the condition on the boundary for $Usubsetneq mathbb{R}^n$ (i.e. without assuming $umid_infty leq 0$). I remember seeing that this holds for $n=2$ but even in that case I'm not sure how to prove it
$endgroup$
– Quoka
Dec 5 '18 at 6:47
$begingroup$
I see. I think you need more conditions on $U$: if we take $U=mathbb R^nsetminus B_1(0)$, then an inverted fundamental solution $u(x)=1−|x|^{2−n}$ would be a counterexample. Maybe assuming $U$ is convex would work, or less generally, maybe assume that $U$ is a cone. In the half-space case, the proof is like this: for each $epsilon>0$, observe that $u(x)-epsilon x_nto-infty$ as $x_ntoinfty$, so by the maximum principle, $ule epsilon x_n$ for all $epsilon>0$.
$endgroup$
– user254433
Dec 5 '18 at 7:04
$begingroup$
I edited the question. Thanks for the counterexample. Do you see how to generalize the proof for domains in $mathbb{R}^2$?
$endgroup$
– Quoka
Dec 5 '18 at 18:49
$begingroup$
An other counterexample, the function $ln(x^2 + y^2)$ is harmonic and positive outside the unit disk, and it vanishes on the domain’s boundary.
$endgroup$
– Wang
Dec 5 '18 at 18:59
$begingroup$
@Wang True, but $ln(x^2+y^2)$ is not bounded
$endgroup$
– Quoka
Dec 5 '18 at 19:22
|
show 1 more comment
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$begingroup$
One problem is that $partial U$ is "smaller" in the unbounded case, so $ule 0$ on $partial U$ becomes less restrictive. For example, if $U=mathbb R^n$, then $partial U=varnothing$, so the condition $u|_{partial U}le 0$ implies no restrictions on $u$. So although Liouville's theorem for bounded harmonic functions on $mathbb R^n$ implies $uequiv a$ is constant, this constant could be positive.
To make sense of $u|_{|x|=infty}le 0$, we could replace it with a limiting condition, like $limsup_{|x|toinfty} u(x)le 0$. Once we do this, it becomes natural to apply the maximum principle on bounded approximations of $U$:
For each $R>0$, let $U_R=Ucap B_R$ be a large subdomain of $U$. Then for each small $epsilon>0$, we can find a large radius $R=R(epsilon)$ so that $u|_{partial B_R}le epsilon$, which, since $epsilon>0$, implies $u|_{partial U_R}le epsilon$. By the maximum principle, $u|_{U_R}le epsilon$. Sending $epsilonto 0$ gives the desired conclusion.
answered Dec 5 '18 at 6:41
user254433user254433
2,531712
$endgroup$
$begingroup$
Sorry I should have been clearer in my question. So this is inspired from this question. I was wondering if it would be sufficient to only have the condition on the boundary for $Usubsetneq mathbb{R}^n$ (i.e. without assuming $umid_infty leq 0$). I remember seeing that this holds for $n=2$ but even in that case I'm not sure how to prove it
$endgroup$
– Quoka
Dec 5 '18 at 6:47
$begingroup$
I see. I think you need more conditions on $U$: if we take $U=mathbb R^nsetminus B_1(0)$, then an inverted fundamental solution $u(x)=1−|x|^{2−n}$ would be a counterexample. Maybe assuming $U$ is convex would work, or less generally, maybe assume that $U$ is a cone. In the half-space case, the proof is like this: for each $epsilon>0$, observe that $u(x)-epsilon x_nto-infty$ as $x_ntoinfty$, so by the maximum principle, $ule epsilon x_n$ for all $epsilon>0$.
$endgroup$
– user254433
Dec 5 '18 at 7:04
$begingroup$
I edited the question. Thanks for the counterexample. Do you see how to generalize the proof for domains in $mathbb{R}^2$?
$endgroup$
– Quoka
Dec 5 '18 at 18:49
$begingroup$
An other counterexample, the function $ln(x^2 + y^2)$ is harmonic and positive outside the unit disk, and it vanishes on the domain’s boundary.
$endgroup$
– Wang
Dec 5 '18 at 18:59
$begingroup$
@Wang True, but $ln(x^2+y^2)$ is not bounded
$endgroup$
– Quoka
Dec 5 '18 at 19:22
|
show 1 more comment
$begingroup$
One problem is that $partial U$ is "smaller" in the unbounded case, so $ule 0$ on $partial U$ becomes less restrictive. For example, if $U=mathbb R^n$, then $partial U=varnothing$, so the condition $u|_{partial U}le 0$ implies no restrictions on $u$. So although Liouville's theorem for bounded harmonic functions on $mathbb R^n$ implies $uequiv a$ is constant, this constant could be positive.
To make sense of $u|_{|x|=infty}le 0$, we could replace it with a limiting condition, like $limsup_{|x|toinfty} u(x)le 0$. Once we do this, it becomes natural to apply the maximum principle on bounded approximations of $U$:
For each $R>0$, let $U_R=Ucap B_R$ be a large subdomain of $U$. Then for each small $epsilon>0$, we can find a large radius $R=R(epsilon)$ so that $u|_{partial B_R}le epsilon$, which, since $epsilon>0$, implies $u|_{partial U_R}le epsilon$. By the maximum principle, $u|_{U_R}le epsilon$. Sending $epsilonto 0$ gives the desired conclusion.
answered Dec 5 '18 at 6:41
user254433user254433
2,531712
$endgroup$
$begingroup$
Sorry I should have been clearer in my question. So this is inspired from this question. I was wondering if it would be sufficient to only have the condition on the boundary for $Usubsetneq mathbb{R}^n$ (i.e. without assuming $umid_infty leq 0$). I remember seeing that this holds for $n=2$ but even in that case I'm not sure how to prove it
$endgroup$
– Quoka
Dec 5 '18 at 6:47
$begingroup$
I see. I think you need more conditions on $U$: if we take $U=mathbb R^nsetminus B_1(0)$, then an inverted fundamental solution $u(x)=1−|x|^{2−n}$ would be a counterexample. Maybe assuming $U$ is convex would work, or less generally, maybe assume that $U$ is a cone. In the half-space case, the proof is like this: for each $epsilon>0$, observe that $u(x)-epsilon x_nto-infty$ as $x_ntoinfty$, so by the maximum principle, $ule epsilon x_n$ for all $epsilon>0$.
$endgroup$
– user254433
Dec 5 '18 at 7:04
$begingroup$
I edited the question. Thanks for the counterexample. Do you see how to generalize the proof for domains in $mathbb{R}^2$?
$endgroup$
– Quoka
Dec 5 '18 at 18:49
$begingroup$
An other counterexample, the function $ln(x^2 + y^2)$ is harmonic and positive outside the unit disk, and it vanishes on the domain’s boundary.
$endgroup$
– Wang
Dec 5 '18 at 18:59
$begingroup$
@Wang True, but $ln(x^2+y^2)$ is not bounded
$endgroup$
– Quoka
Dec 5 '18 at 19:22
|
show 1 more comment
$begingroup$
One problem is that $partial U$ is "smaller" in the unbounded case, so $ule 0$ on $partial U$ becomes less restrictive. For example, if $U=mathbb R^n$, then $partial U=varnothing$, so the condition $u|_{partial U}le 0$ implies no restrictions on $u$. So although Liouville's theorem for bounded harmonic functions on $mathbb R^n$ implies $uequiv a$ is constant, this constant could be positive.
To make sense of $u|_{|x|=infty}le 0$, we could replace it with a limiting condition, like $limsup_{|x|toinfty} u(x)le 0$. Once we do this, it becomes natural to apply the maximum principle on bounded approximations of $U$:
For each $R>0$, let $U_R=Ucap B_R$ be a large subdomain of $U$. Then for each small $epsilon>0$, we can find a large radius $R=R(epsilon)$ so that $u|_{partial B_R}le epsilon$, which, since $epsilon>0$, implies $u|_{partial U_R}le epsilon$. By the maximum principle, $u|_{U_R}le epsilon$. Sending $epsilonto 0$ gives the desired conclusion.
answered Dec 5 '18 at 6:41
user254433user254433
2,531712
$endgroup$
One problem is that $partial U$ is "smaller" in the unbounded case, so $ule 0$ on $partial U$ becomes less restrictive. For example, if $U=mathbb R^n$, then $partial U=varnothing$, so the condition $u|_{partial U}le 0$ implies no restrictions on $u$. So although Liouville's theorem for bounded harmonic functions on $mathbb R^n$ implies $uequiv a$ is constant, this constant could be positive.
To make sense of $u|_{|x|=infty}le 0$, we could replace it with a limiting condition, like $limsup_{|x|toinfty} u(x)le 0$. Once we do this, it becomes natural to apply the maximum principle on bounded approximations of $U$:
For each $R>0$, let $U_R=Ucap B_R$ be a large subdomain of $U$. Then for each small $epsilon>0$, we can find a large radius $R=R(epsilon)$ so that $u|_{partial B_R}le epsilon$, which, since $epsilon>0$, implies $u|_{partial U_R}le epsilon$. By the maximum principle, $u|_{U_R}le epsilon$. Sending $epsilonto 0$ gives the desired conclusion.
answered Dec 5 '18 at 6:41
user254433user254433
2,531712
answered Dec 5 '18 at 6:41
user254433user254433
2,531712
answered Dec 5 '18 at 6:41
user254433user254433
2,531712
answered Dec 5 '18 at 6:41
user254433user254433
2,531712
2,531712
$begingroup$
Sorry I should have been clearer in my question. So this is inspired from this question. I was wondering if it would be sufficient to only have the condition on the boundary for $Usubsetneq mathbb{R}^n$ (i.e. without assuming $umid_infty leq 0$). I remember seeing that this holds for $n=2$ but even in that case I'm not sure how to prove it
$endgroup$
– Quoka
Dec 5 '18 at 6:47
$begingroup$
I see. I think you need more conditions on $U$: if we take $U=mathbb R^nsetminus B_1(0)$, then an inverted fundamental solution $u(x)=1−|x|^{2−n}$ would be a counterexample. Maybe assuming $U$ is convex would work, or less generally, maybe assume that $U$ is a cone. In the half-space case, the proof is like this: for each $epsilon>0$, observe that $u(x)-epsilon x_nto-infty$ as $x_ntoinfty$, so by the maximum principle, $ule epsilon x_n$ for all $epsilon>0$.
$endgroup$
– user254433
Dec 5 '18 at 7:04
$begingroup$
I edited the question. Thanks for the counterexample. Do you see how to generalize the proof for domains in $mathbb{R}^2$?
$endgroup$
– Quoka
Dec 5 '18 at 18:49
$begingroup$
An other counterexample, the function $ln(x^2 + y^2)$ is harmonic and positive outside the unit disk, and it vanishes on the domain’s boundary.
$endgroup$
– Wang
Dec 5 '18 at 18:59
$begingroup$
@Wang True, but $ln(x^2+y^2)$ is not bounded
$endgroup$
– Quoka
Dec 5 '18 at 19:22
|
show 1 more comment
$begingroup$
Sorry I should have been clearer in my question. So this is inspired from this question. I was wondering if it would be sufficient to only have the condition on the boundary for $Usubsetneq mathbb{R}^n$ (i.e. without assuming $umid_infty leq 0$). I remember seeing that this holds for $n=2$ but even in that case I'm not sure how to prove it
$endgroup$
– Quoka
Dec 5 '18 at 6:47
$begingroup$
I see. I think you need more conditions on $U$: if we take $U=mathbb R^nsetminus B_1(0)$, then an inverted fundamental solution $u(x)=1−|x|^{2−n}$ would be a counterexample. Maybe assuming $U$ is convex would work, or less generally, maybe assume that $U$ is a cone. In the half-space case, the proof is like this: for each $epsilon>0$, observe that $u(x)-epsilon x_nto-infty$ as $x_ntoinfty$, so by the maximum principle, $ule epsilon x_n$ for all $epsilon>0$.
$endgroup$
– user254433
Dec 5 '18 at 7:04
$begingroup$
I edited the question. Thanks for the counterexample. Do you see how to generalize the proof for domains in $mathbb{R}^2$?
$endgroup$
– Quoka
Dec 5 '18 at 18:49
$begingroup$
An other counterexample, the function $ln(x^2 + y^2)$ is harmonic and positive outside the unit disk, and it vanishes on the domain’s boundary.
$endgroup$
– Wang
Dec 5 '18 at 18:59
$begingroup$
@Wang True, but $ln(x^2+y^2)$ is not bounded
$endgroup$
– Quoka
Dec 5 '18 at 19:22
$begingroup$
Sorry I should have been clearer in my question. So this is inspired from this question. I was wondering if it would be sufficient to only have the condition on the boundary for $Usubsetneq mathbb{R}^n$ (i.e. without assuming $umid_infty leq 0$). I remember seeing that this holds for $n=2$ but even in that case I'm not sure how to prove it
$endgroup$
– Quoka
Dec 5 '18 at 6:47
$begingroup$
Sorry I should have been clearer in my question. So this is inspired from this question. I was wondering if it would be sufficient to only have the condition on the boundary for $Usubsetneq mathbb{R}^n$ (i.e. without assuming $umid_infty leq 0$). I remember seeing that this holds for $n=2$ but even in that case I'm not sure how to prove it
$endgroup$
– Quoka
Dec 5 '18 at 6:47
$begingroup$
I see. I think you need more conditions on $U$: if we take $U=mathbb R^nsetminus B_1(0)$, then an inverted fundamental solution $u(x)=1−|x|^{2−n}$ would be a counterexample. Maybe assuming $U$ is convex would work, or less generally, maybe assume that $U$ is a cone. In the half-space case, the proof is like this: for each $epsilon>0$, observe that $u(x)-epsilon x_nto-infty$ as $x_ntoinfty$, so by the maximum principle, $ule epsilon x_n$ for all $epsilon>0$.
$endgroup$
– user254433
Dec 5 '18 at 7:04
$begingroup$
I see. I think you need more conditions on $U$: if we take $U=mathbb R^nsetminus B_1(0)$, then an inverted fundamental solution $u(x)=1−|x|^{2−n}$ would be a counterexample. Maybe assuming $U$ is convex would work, or less generally, maybe assume that $U$ is a cone. In the half-space case, the proof is like this: for each $epsilon>0$, observe that $u(x)-epsilon x_nto-infty$ as $x_ntoinfty$, so by the maximum principle, $ule epsilon x_n$ for all $epsilon>0$.
$endgroup$
– user254433
Dec 5 '18 at 7:04
$begingroup$
I edited the question. Thanks for the counterexample. Do you see how to generalize the proof for domains in $mathbb{R}^2$?
$endgroup$
– Quoka
Dec 5 '18 at 18:49
$begingroup$
I edited the question. Thanks for the counterexample. Do you see how to generalize the proof for domains in $mathbb{R}^2$?
$endgroup$
– Quoka
Dec 5 '18 at 18:49
$begingroup$
An other counterexample, the function $ln(x^2 + y^2)$ is harmonic and positive outside the unit disk, and it vanishes on the domain’s boundary.
$endgroup$
– Wang
Dec 5 '18 at 18:59
$begingroup$
An other counterexample, the function $ln(x^2 + y^2)$ is harmonic and positive outside the unit disk, and it vanishes on the domain’s boundary.
$endgroup$
– Wang
Dec 5 '18 at 18:59
$begingroup$
@Wang True, but $ln(x^2+y^2)$ is not bounded
$endgroup$
– Quoka
Dec 5 '18 at 19:22
$begingroup$
@Wang True, but $ln(x^2+y^2)$ is not bounded
$endgroup$
– Quoka
Dec 5 '18 at 19:22
|
show 1 more comment
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