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How to complexify a harmonic function with an isolated singularity?
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I would really appreciate if you could direct me to a reference for the following fact.
Given a harmonic function $h$ defined in $R^Nbackslash0$ we can find a holomorphic function $g$ of $N$ complex variables, such that $g$ coincides with $h$ at real points, and $g$ has a holomorphic extension to $C^Nbackslash E$ where $E=zinmathbbC^N:z_1^2+z_2^2+dots+z_N^2=0.$
Here holomorphic extention to $C^Nbackslash E$ includes the case where there are two functions $g_1$ holomorphic on a domain $D_1$ and $g_2$ holomorphic on a domain $D_2,$ both coinciding with $h$ for real values, and $D_1cup D_2 =C^Nbackslash E.$
Thank you!
harmonic-functions several-complex-variables
asked Jul 19 at 15:22
May
164
add a comment |Â
up vote
0
down vote
favorite
I would really appreciate if you could direct me to a reference for the following fact.
Given a harmonic function $h$ defined in $R^Nbackslash0$ we can find a holomorphic function $g$ of $N$ complex variables, such that $g$ coincides with $h$ at real points, and $g$ has a holomorphic extension to $C^Nbackslash E$ where $E=zinmathbbC^N:z_1^2+z_2^2+dots+z_N^2=0.$
Here holomorphic extention to $C^Nbackslash E$ includes the case where there are two functions $g_1$ holomorphic on a domain $D_1$ and $g_2$ holomorphic on a domain $D_2,$ both coinciding with $h$ for real values, and $D_1cup D_2 =C^Nbackslash E.$
Thank you!
harmonic-functions several-complex-variables
asked Jul 19 at 15:22
May
164
I doubt it is true: Suppose $N=2$, then take the harmonic function $(x,y) mapsto e^1/(x+iy) + e^1/(x-iy)$, that is real valued harmonic on the punctured plane. The complexification is to assume that $x$ and $y$ are complex, but that complexification blows up (and cannot extend to) the points where $x=iy$ or $x=-iy$. Am I not understanding the question?
– Jiri Lebl
Aug 7 at 3:44
Hi, thank you for looking into the question! If x=iy it satisfies x^2+y^2=0, so it is in the exceptional set and I don't need the extension there. Could you advise me if I should make the question clearer?
– May
Aug 8 at 10:55
@JiriLebl Hi again, I just answered your comment, but I forgot to tag you there;)
– May
Aug 8 at 12:20
Ahhh, I did misunderstand. You are correct. So my example is a nonexample. I assume this is also a nonexample: $log (x^2+y^2)$ this does not complexify as a single valued function to outside of $x^2+y^2=0$. If this is allowed, I think it might be true if $N=2$. But I wonder about $N > 2$. In higher dimensions harmonic functions are not real values of holomorphic functions, though that doesn't even make sense in $N=3$. In fact, a harmonic function can be essentially arbitrary (as long as it's real analytic) on a hyperplane, so it seems like it shouldn't be true if $N > 2$.
– Jiri Lebl
Aug 9 at 21:30
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I would really appreciate if you could direct me to a reference for the following fact.
Given a harmonic function $h$ defined in $R^Nbackslash0$ we can find a holomorphic function $g$ of $N$ complex variables, such that $g$ coincides with $h$ at real points, and $g$ has a holomorphic extension to $C^Nbackslash E$ where $E=zinmathbbC^N:z_1^2+z_2^2+dots+z_N^2=0.$
Here holomorphic extention to $C^Nbackslash E$ includes the case where there are two functions $g_1$ holomorphic on a domain $D_1$ and $g_2$ holomorphic on a domain $D_2,$ both coinciding with $h$ for real values, and $D_1cup D_2 =C^Nbackslash E.$
Thank you!
harmonic-functions several-complex-variables
asked Jul 19 at 15:22
May
164
I would really appreciate if you could direct me to a reference for the following fact.
Given a harmonic function $h$ defined in $R^Nbackslash0$ we can find a holomorphic function $g$ of $N$ complex variables, such that $g$ coincides with $h$ at real points, and $g$ has a holomorphic extension to $C^Nbackslash E$ where $E=zinmathbbC^N:z_1^2+z_2^2+dots+z_N^2=0.$
Here holomorphic extention to $C^Nbackslash E$ includes the case where there are two functions $g_1$ holomorphic on a domain $D_1$ and $g_2$ holomorphic on a domain $D_2,$ both coinciding with $h$ for real values, and $D_1cup D_2 =C^Nbackslash E.$
Thank you!
harmonic-functions several-complex-variables
asked Jul 19 at 15:22
May
164
edited Aug 8 at 12:24
asked Jul 19 at 15:22
May
164
asked Jul 19 at 15:22
May
164
asked Jul 19 at 15:22
May
164
164
I doubt it is true: Suppose $N=2$, then take the harmonic function $(x,y) mapsto e^1/(x+iy) + e^1/(x-iy)$, that is real valued harmonic on the punctured plane. The complexification is to assume that $x$ and $y$ are complex, but that complexification blows up (and cannot extend to) the points where $x=iy$ or $x=-iy$. Am I not understanding the question?
– Jiri Lebl
Aug 7 at 3:44
Hi, thank you for looking into the question! If x=iy it satisfies x^2+y^2=0, so it is in the exceptional set and I don't need the extension there. Could you advise me if I should make the question clearer?
– May
Aug 8 at 10:55
@JiriLebl Hi again, I just answered your comment, but I forgot to tag you there;)
– May
Aug 8 at 12:20
Ahhh, I did misunderstand. You are correct. So my example is a nonexample. I assume this is also a nonexample: $log (x^2+y^2)$ this does not complexify as a single valued function to outside of $x^2+y^2=0$. If this is allowed, I think it might be true if $N=2$. But I wonder about $N > 2$. In higher dimensions harmonic functions are not real values of holomorphic functions, though that doesn't even make sense in $N=3$. In fact, a harmonic function can be essentially arbitrary (as long as it's real analytic) on a hyperplane, so it seems like it shouldn't be true if $N > 2$.
– Jiri Lebl
Aug 9 at 21:30
add a comment |Â
I doubt it is true: Suppose $N=2$, then take the harmonic function $(x,y) mapsto e^1/(x+iy) + e^1/(x-iy)$, that is real valued harmonic on the punctured plane. The complexification is to assume that $x$ and $y$ are complex, but that complexification blows up (and cannot extend to) the points where $x=iy$ or $x=-iy$. Am I not understanding the question?
– Jiri Lebl
Aug 7 at 3:44
Hi, thank you for looking into the question! If x=iy it satisfies x^2+y^2=0, so it is in the exceptional set and I don't need the extension there. Could you advise me if I should make the question clearer?
– May
Aug 8 at 10:55
@JiriLebl Hi again, I just answered your comment, but I forgot to tag you there;)
– May
Aug 8 at 12:20
Ahhh, I did misunderstand. You are correct. So my example is a nonexample. I assume this is also a nonexample: $log (x^2+y^2)$ this does not complexify as a single valued function to outside of $x^2+y^2=0$. If this is allowed, I think it might be true if $N=2$. But I wonder about $N > 2$. In higher dimensions harmonic functions are not real values of holomorphic functions, though that doesn't even make sense in $N=3$. In fact, a harmonic function can be essentially arbitrary (as long as it's real analytic) on a hyperplane, so it seems like it shouldn't be true if $N > 2$.
– Jiri Lebl
Aug 9 at 21:30
I doubt it is true: Suppose $N=2$, then take the harmonic function $(x,y) mapsto e^1/(x+iy) + e^1/(x-iy)$, that is real valued harmonic on the punctured plane. The complexification is to assume that $x$ and $y$ are complex, but that complexification blows up (and cannot extend to) the points where $x=iy$ or $x=-iy$. Am I not understanding the question?
– Jiri Lebl
Aug 7 at 3:44
I doubt it is true: Suppose $N=2$, then take the harmonic function $(x,y) mapsto e^1/(x+iy) + e^1/(x-iy)$, that is real valued harmonic on the punctured plane. The complexification is to assume that $x$ and $y$ are complex, but that complexification blows up (and cannot extend to) the points where $x=iy$ or $x=-iy$. Am I not understanding the question?
– Jiri Lebl
Aug 7 at 3:44
Hi, thank you for looking into the question! If x=iy it satisfies x^2+y^2=0, so it is in the exceptional set and I don't need the extension there. Could you advise me if I should make the question clearer?
– May
Aug 8 at 10:55
Hi, thank you for looking into the question! If x=iy it satisfies x^2+y^2=0, so it is in the exceptional set and I don't need the extension there. Could you advise me if I should make the question clearer?
– May
Aug 8 at 10:55
@JiriLebl Hi again, I just answered your comment, but I forgot to tag you there;)
– May
Aug 8 at 12:20
@JiriLebl Hi again, I just answered your comment, but I forgot to tag you there;)
– May
Aug 8 at 12:20
Ahhh, I did misunderstand. You are correct. So my example is a nonexample. I assume this is also a nonexample: $log (x^2+y^2)$ this does not complexify as a single valued function to outside of $x^2+y^2=0$. If this is allowed, I think it might be true if $N=2$. But I wonder about $N > 2$. In higher dimensions harmonic functions are not real values of holomorphic functions, though that doesn't even make sense in $N=3$. In fact, a harmonic function can be essentially arbitrary (as long as it's real analytic) on a hyperplane, so it seems like it shouldn't be true if $N > 2$.
– Jiri Lebl
Aug 9 at 21:30
Ahhh, I did misunderstand. You are correct. So my example is a nonexample. I assume this is also a nonexample: $log (x^2+y^2)$ this does not complexify as a single valued function to outside of $x^2+y^2=0$. If this is allowed, I think it might be true if $N=2$. But I wonder about $N > 2$. In higher dimensions harmonic functions are not real values of holomorphic functions, though that doesn't even make sense in $N=3$. In fact, a harmonic function can be essentially arbitrary (as long as it's real analytic) on a hyperplane, so it seems like it shouldn't be true if $N > 2$.
– Jiri Lebl
Aug 9 at 21:30
add a comment |Â
1 Answer
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Let's try. First $N=2$ and suppose that $h$ is real-valued (it should work for complex valued too with a bit more work). First we complexify $h$ to some neighborhood of $mathbbR^2 setminus 0 $ in $mathbbC^2$. Locally near a point $h(z) = f(z) + barf(barz)$ ($h$ is a real value of a holomorphic function), where $z = x+iy$. Starting at a point $(x_0,y_0)$ the function $f$ can be path continued to any $(x_1,y_1)$ in $mathbbR^2$ outside the origin. Pick a $(xi_0,eta_0) in mathbbC^2$ outside of $xi^2+eta^2=0$. Near $(x_0,y_0)$ $h$ complexifies to $h(xi + i eta) = f(xi+ieta) + barf(xi-ieta)$, there exists a path $xi(t),eta(t)$ space from $(x_0,y_1)$ to $(xi_0,eta_0)$ which does not pass through $xi^2+eta^2 = 0$ (as the complement of this set is connected). Continue $f$ along the path $xi(t)+ieta(t)$ and $barf$ along the path $xi(t)-ieta(t)$ (both paths are in $mathbbC setminus 0 = mathbbR^2 setminus 0 $). So you can continue the complexified $h$ along any path to any point as long as you stay away from $xi^2+eta^2=0$. That you cannot do this in a single-valued manner, the counterexample is $log(x^2+y^2)$ as in the comment above.
For higher $N$, I assume you want to use Laurent series. As long as $N > 2$, then there exist harmonic homogeneous polynomials $p_m$ and $q_m$ such that
$$h(x) = sum_m=0^infty p_m(x) + sum_m=0^infty fracq_m(x)x$$
See for example Axler-Bourdon-Ramey (http://www.axler.net/HFT.html), page 209. Convergence is uniform on compact subsets of $mathbbR^N setminus 0 $ as usual. The proof would follow then similarly as above. The $p_m$ and the $q_m$ shouldn't present any problems. I haven't gone through the calculation, so I won't give details, but I'd assume it involves writing the denominator as $(x_1^2 + cdots + x_N^2)^(2m+N-2)/2$, then find a path from a point in the real punctured plane to your desired point (as in the $N=2$ case) where we stay well away from (the complexified) $x_1^2 + cdots + x_N^2 = 0$ and you prove that along this path the complexified series still converges. Comparison test using the fact that the original series converges uniformly and absolutely on any annulus should do the trick.
answered Aug 9 at 22:13
Jiri Lebl
1,405312
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Let's try. First $N=2$ and suppose that $h$ is real-valued (it should work for complex valued too with a bit more work). First we complexify $h$ to some neighborhood of $mathbbR^2 setminus 0 $ in $mathbbC^2$. Locally near a point $h(z) = f(z) + barf(barz)$ ($h$ is a real value of a holomorphic function), where $z = x+iy$. Starting at a point $(x_0,y_0)$ the function $f$ can be path continued to any $(x_1,y_1)$ in $mathbbR^2$ outside the origin. Pick a $(xi_0,eta_0) in mathbbC^2$ outside of $xi^2+eta^2=0$. Near $(x_0,y_0)$ $h$ complexifies to $h(xi + i eta) = f(xi+ieta) + barf(xi-ieta)$, there exists a path $xi(t),eta(t)$ space from $(x_0,y_1)$ to $(xi_0,eta_0)$ which does not pass through $xi^2+eta^2 = 0$ (as the complement of this set is connected). Continue $f$ along the path $xi(t)+ieta(t)$ and $barf$ along the path $xi(t)-ieta(t)$ (both paths are in $mathbbC setminus 0 = mathbbR^2 setminus 0 $). So you can continue the complexified $h$ along any path to any point as long as you stay away from $xi^2+eta^2=0$. That you cannot do this in a single-valued manner, the counterexample is $log(x^2+y^2)$ as in the comment above.
For higher $N$, I assume you want to use Laurent series. As long as $N > 2$, then there exist harmonic homogeneous polynomials $p_m$ and $q_m$ such that
$$h(x) = sum_m=0^infty p_m(x) + sum_m=0^infty fracq_m(x)x$$
See for example Axler-Bourdon-Ramey (http://www.axler.net/HFT.html), page 209. Convergence is uniform on compact subsets of $mathbbR^N setminus 0 $ as usual. The proof would follow then similarly as above. The $p_m$ and the $q_m$ shouldn't present any problems. I haven't gone through the calculation, so I won't give details, but I'd assume it involves writing the denominator as $(x_1^2 + cdots + x_N^2)^(2m+N-2)/2$, then find a path from a point in the real punctured plane to your desired point (as in the $N=2$ case) where we stay well away from (the complexified) $x_1^2 + cdots + x_N^2 = 0$ and you prove that along this path the complexified series still converges. Comparison test using the fact that the original series converges uniformly and absolutely on any annulus should do the trick.
answered Aug 9 at 22:13
Jiri Lebl
1,405312
add a comment |Â
up vote
0
down vote
Let's try. First $N=2$ and suppose that $h$ is real-valued (it should work for complex valued too with a bit more work). First we complexify $h$ to some neighborhood of $mathbbR^2 setminus 0 $ in $mathbbC^2$. Locally near a point $h(z) = f(z) + barf(barz)$ ($h$ is a real value of a holomorphic function), where $z = x+iy$. Starting at a point $(x_0,y_0)$ the function $f$ can be path continued to any $(x_1,y_1)$ in $mathbbR^2$ outside the origin. Pick a $(xi_0,eta_0) in mathbbC^2$ outside of $xi^2+eta^2=0$. Near $(x_0,y_0)$ $h$ complexifies to $h(xi + i eta) = f(xi+ieta) + barf(xi-ieta)$, there exists a path $xi(t),eta(t)$ space from $(x_0,y_1)$ to $(xi_0,eta_0)$ which does not pass through $xi^2+eta^2 = 0$ (as the complement of this set is connected). Continue $f$ along the path $xi(t)+ieta(t)$ and $barf$ along the path $xi(t)-ieta(t)$ (both paths are in $mathbbC setminus 0 = mathbbR^2 setminus 0 $). So you can continue the complexified $h$ along any path to any point as long as you stay away from $xi^2+eta^2=0$. That you cannot do this in a single-valued manner, the counterexample is $log(x^2+y^2)$ as in the comment above.
For higher $N$, I assume you want to use Laurent series. As long as $N > 2$, then there exist harmonic homogeneous polynomials $p_m$ and $q_m$ such that
$$h(x) = sum_m=0^infty p_m(x) + sum_m=0^infty fracq_m(x)x$$
See for example Axler-Bourdon-Ramey (http://www.axler.net/HFT.html), page 209. Convergence is uniform on compact subsets of $mathbbR^N setminus 0 $ as usual. The proof would follow then similarly as above. The $p_m$ and the $q_m$ shouldn't present any problems. I haven't gone through the calculation, so I won't give details, but I'd assume it involves writing the denominator as $(x_1^2 + cdots + x_N^2)^(2m+N-2)/2$, then find a path from a point in the real punctured plane to your desired point (as in the $N=2$ case) where we stay well away from (the complexified) $x_1^2 + cdots + x_N^2 = 0$ and you prove that along this path the complexified series still converges. Comparison test using the fact that the original series converges uniformly and absolutely on any annulus should do the trick.
answered Aug 9 at 22:13
Jiri Lebl
1,405312
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let's try. First $N=2$ and suppose that $h$ is real-valued (it should work for complex valued too with a bit more work). First we complexify $h$ to some neighborhood of $mathbbR^2 setminus 0 $ in $mathbbC^2$. Locally near a point $h(z) = f(z) + barf(barz)$ ($h$ is a real value of a holomorphic function), where $z = x+iy$. Starting at a point $(x_0,y_0)$ the function $f$ can be path continued to any $(x_1,y_1)$ in $mathbbR^2$ outside the origin. Pick a $(xi_0,eta_0) in mathbbC^2$ outside of $xi^2+eta^2=0$. Near $(x_0,y_0)$ $h$ complexifies to $h(xi + i eta) = f(xi+ieta) + barf(xi-ieta)$, there exists a path $xi(t),eta(t)$ space from $(x_0,y_1)$ to $(xi_0,eta_0)$ which does not pass through $xi^2+eta^2 = 0$ (as the complement of this set is connected). Continue $f$ along the path $xi(t)+ieta(t)$ and $barf$ along the path $xi(t)-ieta(t)$ (both paths are in $mathbbC setminus 0 = mathbbR^2 setminus 0 $). So you can continue the complexified $h$ along any path to any point as long as you stay away from $xi^2+eta^2=0$. That you cannot do this in a single-valued manner, the counterexample is $log(x^2+y^2)$ as in the comment above.
For higher $N$, I assume you want to use Laurent series. As long as $N > 2$, then there exist harmonic homogeneous polynomials $p_m$ and $q_m$ such that
$$h(x) = sum_m=0^infty p_m(x) + sum_m=0^infty fracq_m(x)x$$
See for example Axler-Bourdon-Ramey (http://www.axler.net/HFT.html), page 209. Convergence is uniform on compact subsets of $mathbbR^N setminus 0 $ as usual. The proof would follow then similarly as above. The $p_m$ and the $q_m$ shouldn't present any problems. I haven't gone through the calculation, so I won't give details, but I'd assume it involves writing the denominator as $(x_1^2 + cdots + x_N^2)^(2m+N-2)/2$, then find a path from a point in the real punctured plane to your desired point (as in the $N=2$ case) where we stay well away from (the complexified) $x_1^2 + cdots + x_N^2 = 0$ and you prove that along this path the complexified series still converges. Comparison test using the fact that the original series converges uniformly and absolutely on any annulus should do the trick.
answered Aug 9 at 22:13
Jiri Lebl
1,405312
Let's try. First $N=2$ and suppose that $h$ is real-valued (it should work for complex valued too with a bit more work). First we complexify $h$ to some neighborhood of $mathbbR^2 setminus 0 $ in $mathbbC^2$. Locally near a point $h(z) = f(z) + barf(barz)$ ($h$ is a real value of a holomorphic function), where $z = x+iy$. Starting at a point $(x_0,y_0)$ the function $f$ can be path continued to any $(x_1,y_1)$ in $mathbbR^2$ outside the origin. Pick a $(xi_0,eta_0) in mathbbC^2$ outside of $xi^2+eta^2=0$. Near $(x_0,y_0)$ $h$ complexifies to $h(xi + i eta) = f(xi+ieta) + barf(xi-ieta)$, there exists a path $xi(t),eta(t)$ space from $(x_0,y_1)$ to $(xi_0,eta_0)$ which does not pass through $xi^2+eta^2 = 0$ (as the complement of this set is connected). Continue $f$ along the path $xi(t)+ieta(t)$ and $barf$ along the path $xi(t)-ieta(t)$ (both paths are in $mathbbC setminus 0 = mathbbR^2 setminus 0 $). So you can continue the complexified $h$ along any path to any point as long as you stay away from $xi^2+eta^2=0$. That you cannot do this in a single-valued manner, the counterexample is $log(x^2+y^2)$ as in the comment above.
For higher $N$, I assume you want to use Laurent series. As long as $N > 2$, then there exist harmonic homogeneous polynomials $p_m$ and $q_m$ such that
$$h(x) = sum_m=0^infty p_m(x) + sum_m=0^infty fracq_m(x)x$$
See for example Axler-Bourdon-Ramey (http://www.axler.net/HFT.html), page 209. Convergence is uniform on compact subsets of $mathbbR^N setminus 0 $ as usual. The proof would follow then similarly as above. The $p_m$ and the $q_m$ shouldn't present any problems. I haven't gone through the calculation, so I won't give details, but I'd assume it involves writing the denominator as $(x_1^2 + cdots + x_N^2)^(2m+N-2)/2$, then find a path from a point in the real punctured plane to your desired point (as in the $N=2$ case) where we stay well away from (the complexified) $x_1^2 + cdots + x_N^2 = 0$ and you prove that along this path the complexified series still converges. Comparison test using the fact that the original series converges uniformly and absolutely on any annulus should do the trick.
answered Aug 9 at 22:13
Jiri Lebl
1,405312
answered Aug 9 at 22:13
Jiri Lebl
1,405312
answered Aug 9 at 22:13
Jiri Lebl
1,405312
answered Aug 9 at 22:13
Jiri Lebl
1,405312
1,405312
add a comment |Â
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I doubt it is true: Suppose $N=2$, then take the harmonic function $(x,y) mapsto e^1/(x+iy) + e^1/(x-iy)$, that is real valued harmonic on the punctured plane. The complexification is to assume that $x$ and $y$ are complex, but that complexification blows up (and cannot extend to) the points where $x=iy$ or $x=-iy$. Am I not understanding the question?
– Jiri Lebl
Aug 7 at 3:44
Hi, thank you for looking into the question! If x=iy it satisfies x^2+y^2=0, so it is in the exceptional set and I don't need the extension there. Could you advise me if I should make the question clearer?
– May
Aug 8 at 10:55
@JiriLebl Hi again, I just answered your comment, but I forgot to tag you there;)
– May
Aug 8 at 12:20
Ahhh, I did misunderstand. You are correct. So my example is a nonexample. I assume this is also a nonexample: $log (x^2+y^2)$ this does not complexify as a single valued function to outside of $x^2+y^2=0$. If this is allowed, I think it might be true if $N=2$. But I wonder about $N > 2$. In higher dimensions harmonic functions are not real values of holomorphic functions, though that doesn't even make sense in $N=3$. In fact, a harmonic function can be essentially arbitrary (as long as it's real analytic) on a hyperplane, so it seems like it shouldn't be true if $N > 2$.
– Jiri Lebl
Aug 9 at 21:30