Mixing
u harmonic on the open set $Dbackslash{z_{0}}$ then u harmonic on $D$
$begingroup$
My problem is:Let D be an open connected set,u continuous on D and u harmonic on $Dbackslash{z_{0}}$,$z_{0}in D$,then u harmonic on $D$.My plan was to prove that u is equal with an harmonic function v on a disk $D(z_{0},r)$.At this direction i took the poisson integral $v(z)=P[φ](z)$,where $φ$ defined on the boundary of the disk and $φ=u$ in there.I read somewhere a good technique,from a theorem,that may help.Theorem says that if u is an function and the second partial derivatives exists(you have not the assumption that the second partial derivatives are continuous) and satisfy $Delta(z)=0$ then u is harmonic.In the proof of this theorem you define $V(z)=u(z)-v(z)+epsilon(Re(z)-Re(z_{0}))^2$ and show that it has the maximal value on the boundary,and then $u(z)-v(z)+epsilon(Re(z)-Re(z_{0}))^2leqepsilon r^2,forall epsilon $,so $uleq v$.(and the other inequallity is similar).But in the proof of this inequallity you use the fact that $Delta(z_{0})=0$,a fact that you don't have in my problem.So i am wondering if we can reform this technique to solve the problem.Also,if my direction on solving the problem is wrong i would appreciate any other hint to this problem.
complex-analysis harmonic-analysis
asked Jan 13 at 15:32
T.KarawolfT.Karawolf
236
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add a comment |
$begingroup$
My problem is:Let D be an open connected set,u continuous on D and u harmonic on $Dbackslash{z_{0}}$,$z_{0}in D$,then u harmonic on $D$.My plan was to prove that u is equal with an harmonic function v on a disk $D(z_{0},r)$.At this direction i took the poisson integral $v(z)=P[φ](z)$,where $φ$ defined on the boundary of the disk and $φ=u$ in there.I read somewhere a good technique,from a theorem,that may help.Theorem says that if u is an function and the second partial derivatives exists(you have not the assumption that the second partial derivatives are continuous) and satisfy $Delta(z)=0$ then u is harmonic.In the proof of this theorem you define $V(z)=u(z)-v(z)+epsilon(Re(z)-Re(z_{0}))^2$ and show that it has the maximal value on the boundary,and then $u(z)-v(z)+epsilon(Re(z)-Re(z_{0}))^2leqepsilon r^2,forall epsilon $,so $uleq v$.(and the other inequallity is similar).But in the proof of this inequallity you use the fact that $Delta(z_{0})=0$,a fact that you don't have in my problem.So i am wondering if we can reform this technique to solve the problem.Also,if my direction on solving the problem is wrong i would appreciate any other hint to this problem.
complex-analysis harmonic-analysis
asked Jan 13 at 15:32
T.KarawolfT.Karawolf
236
$endgroup$
$begingroup$
math.stackexchange.com/a/184568/489194 maybe this answer the question.
$endgroup$
– T.Karawolf
Jan 14 at 0:32
add a comment |
$begingroup$
My problem is:Let D be an open connected set,u continuous on D and u harmonic on $Dbackslash{z_{0}}$,$z_{0}in D$,then u harmonic on $D$.My plan was to prove that u is equal with an harmonic function v on a disk $D(z_{0},r)$.At this direction i took the poisson integral $v(z)=P[φ](z)$,where $φ$ defined on the boundary of the disk and $φ=u$ in there.I read somewhere a good technique,from a theorem,that may help.Theorem says that if u is an function and the second partial derivatives exists(you have not the assumption that the second partial derivatives are continuous) and satisfy $Delta(z)=0$ then u is harmonic.In the proof of this theorem you define $V(z)=u(z)-v(z)+epsilon(Re(z)-Re(z_{0}))^2$ and show that it has the maximal value on the boundary,and then $u(z)-v(z)+epsilon(Re(z)-Re(z_{0}))^2leqepsilon r^2,forall epsilon $,so $uleq v$.(and the other inequallity is similar).But in the proof of this inequallity you use the fact that $Delta(z_{0})=0$,a fact that you don't have in my problem.So i am wondering if we can reform this technique to solve the problem.Also,if my direction on solving the problem is wrong i would appreciate any other hint to this problem.
complex-analysis harmonic-analysis
asked Jan 13 at 15:32
T.KarawolfT.Karawolf
236
$endgroup$
My problem is:Let D be an open connected set,u continuous on D and u harmonic on $Dbackslash{z_{0}}$,$z_{0}in D$,then u harmonic on $D$.My plan was to prove that u is equal with an harmonic function v on a disk $D(z_{0},r)$.At this direction i took the poisson integral $v(z)=P[φ](z)$,where $φ$ defined on the boundary of the disk and $φ=u$ in there.I read somewhere a good technique,from a theorem,that may help.Theorem says that if u is an function and the second partial derivatives exists(you have not the assumption that the second partial derivatives are continuous) and satisfy $Delta(z)=0$ then u is harmonic.In the proof of this theorem you define $V(z)=u(z)-v(z)+epsilon(Re(z)-Re(z_{0}))^2$ and show that it has the maximal value on the boundary,and then $u(z)-v(z)+epsilon(Re(z)-Re(z_{0}))^2leqepsilon r^2,forall epsilon $,so $uleq v$.(and the other inequallity is similar).But in the proof of this inequallity you use the fact that $Delta(z_{0})=0$,a fact that you don't have in my problem.So i am wondering if we can reform this technique to solve the problem.Also,if my direction on solving the problem is wrong i would appreciate any other hint to this problem.
complex-analysis harmonic-analysis
complex-analysis harmonic-analysis
asked Jan 13 at 15:32
T.KarawolfT.Karawolf
236
asked Jan 13 at 15:32
T.KarawolfT.Karawolf
236
edited Jan 13 at 18:17
T.Karawolf
asked Jan 13 at 15:32
T.KarawolfT.Karawolf
236
asked Jan 13 at 15:32
T.KarawolfT.Karawolf
236
asked Jan 13 at 15:32
T.KarawolfT.Karawolf
236
236
$begingroup$
math.stackexchange.com/a/184568/489194 maybe this answer the question.
$endgroup$
– T.Karawolf
Jan 14 at 0:32
add a comment |
$begingroup$
math.stackexchange.com/a/184568/489194 maybe this answer the question.
$endgroup$
– T.Karawolf
Jan 14 at 0:32
$begingroup$
math.stackexchange.com/a/184568/489194 maybe this answer the question.
$endgroup$
– T.Karawolf
Jan 14 at 0:32
$begingroup$
math.stackexchange.com/a/184568/489194 maybe this answer the question.
$endgroup$
– T.Karawolf
Jan 14 at 0:32
add a comment |
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$begingroup$
math.stackexchange.com/a/184568/489194 maybe this answer the question.
$endgroup$
– T.Karawolf
Jan 14 at 0:32