Holomorphic function on a upper half plane that is scale invariant with respect to a positive real number.
$begingroup$
Let $theta in mathbb{R}^{+}$ be a fixed number and $theta neq 1$. Let $f:mathbb{H} to mathbb{C}$ be a holomorphic function on the upper half plane such that $forall z in mathbb{H}$, $f(z)=f(theta z)$.
Then, must $f$ be a constant? (I highly doubt it since $f$ is not defined at $0$, but I don't know how to rigorously construct a counterexample.)
complex-analysis
$endgroup$
add a comment |
$begingroup$
Let $theta in mathbb{R}^{+}$ be a fixed number and $theta neq 1$. Let $f:mathbb{H} to mathbb{C}$ be a holomorphic function on the upper half plane such that $forall z in mathbb{H}$, $f(z)=f(theta z)$.
Then, must $f$ be a constant? (I highly doubt it since $f$ is not defined at $0$, but I don't know how to rigorously construct a counterexample.)
complex-analysis
$endgroup$
$begingroup$
Do you mean to say that $f(z) = f(theta z), ; forall theta in Bbb R^+$, so that $f(z)$ constant on each ray $theta z_0$, $z_0 in Bbb H$?
$endgroup$
– Robert Lewis
Jan 7 at 22:33
1
$begingroup$
$f(z) = f(cz)$ iff $f(e^z)$ is $log c$ periodic
$endgroup$
– reuns
Jan 8 at 9:02
$begingroup$
@RobertLewis $theta$ is fixed. I am sorry for the confusion.
$endgroup$
– chbe
Jan 10 at 10:09
add a comment |
$begingroup$
Let $theta in mathbb{R}^{+}$ be a fixed number and $theta neq 1$. Let $f:mathbb{H} to mathbb{C}$ be a holomorphic function on the upper half plane such that $forall z in mathbb{H}$, $f(z)=f(theta z)$.
Then, must $f$ be a constant? (I highly doubt it since $f$ is not defined at $0$, but I don't know how to rigorously construct a counterexample.)
complex-analysis
$endgroup$
Let $theta in mathbb{R}^{+}$ be a fixed number and $theta neq 1$. Let $f:mathbb{H} to mathbb{C}$ be a holomorphic function on the upper half plane such that $forall z in mathbb{H}$, $f(z)=f(theta z)$.
Then, must $f$ be a constant? (I highly doubt it since $f$ is not defined at $0$, but I don't know how to rigorously construct a counterexample.)
complex-analysis
complex-analysis
edited Jan 10 at 10:02
chbe
asked Jan 7 at 22:08
chbechbe
83
83
$begingroup$
Do you mean to say that $f(z) = f(theta z), ; forall theta in Bbb R^+$, so that $f(z)$ constant on each ray $theta z_0$, $z_0 in Bbb H$?
$endgroup$
– Robert Lewis
Jan 7 at 22:33
1
$begingroup$
$f(z) = f(cz)$ iff $f(e^z)$ is $log c$ periodic
$endgroup$
– reuns
Jan 8 at 9:02
$begingroup$
@RobertLewis $theta$ is fixed. I am sorry for the confusion.
$endgroup$
– chbe
Jan 10 at 10:09
add a comment |
$begingroup$
Do you mean to say that $f(z) = f(theta z), ; forall theta in Bbb R^+$, so that $f(z)$ constant on each ray $theta z_0$, $z_0 in Bbb H$?
$endgroup$
– Robert Lewis
Jan 7 at 22:33
1
$begingroup$
$f(z) = f(cz)$ iff $f(e^z)$ is $log c$ periodic
$endgroup$
– reuns
Jan 8 at 9:02
$begingroup$
@RobertLewis $theta$ is fixed. I am sorry for the confusion.
$endgroup$
– chbe
Jan 10 at 10:09
$begingroup$
Do you mean to say that $f(z) = f(theta z), ; forall theta in Bbb R^+$, so that $f(z)$ constant on each ray $theta z_0$, $z_0 in Bbb H$?
$endgroup$
– Robert Lewis
Jan 7 at 22:33
$begingroup$
Do you mean to say that $f(z) = f(theta z), ; forall theta in Bbb R^+$, so that $f(z)$ constant on each ray $theta z_0$, $z_0 in Bbb H$?
$endgroup$
– Robert Lewis
Jan 7 at 22:33
1
1
$begingroup$
$f(z) = f(cz)$ iff $f(e^z)$ is $log c$ periodic
$endgroup$
– reuns
Jan 8 at 9:02
$begingroup$
$f(z) = f(cz)$ iff $f(e^z)$ is $log c$ periodic
$endgroup$
– reuns
Jan 8 at 9:02
$begingroup$
@RobertLewis $theta$ is fixed. I am sorry for the confusion.
$endgroup$
– chbe
Jan 10 at 10:09
$begingroup$
@RobertLewis $theta$ is fixed. I am sorry for the confusion.
$endgroup$
– chbe
Jan 10 at 10:09
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
No, $f$ does not need to be constant.
Define $g$ in the strip $S = { w mid 0 < operatorname{Im} w < i pi } $ as
$g(w) = f(e^w)$. Then for a given positive real number $theta ne 1$
$$
f(z) = f(theta z) quad text{for all } z in Bbb H
$$
if and only if
$$
g(w) = g(w + log theta) quad text{for all } w in S
$$
so that the problem reduces to find all $log theta$ - periodic functions in the strip $S$. The most simple example would be
$$
g(w) = exp left(frac{2 pi i w}{log theta} right)
$$
corresponding to
$$
f(z) = exp left(frac{2 pi i log z}{log theta} right)
$$
where $log z = log |z| + i operatorname{Arg} z$ is the main branch of the logarithm with $0 < operatorname{Arg} z < pi$ for $z in Bbb H$.
All $log theta$ - periodic functions in $S$ are obtained by composing $g$ with a holomorphic function, this leads to the general solution
$$
f(z) = h left(exp left(frac{2 pi i log z}{log theta} right) right)
$$
where $h$ is holomorphic in the annulus ${ w mid exp left(frac{-2 pi^2}{log theta}right) < |w| < 1 }$.
$endgroup$
$begingroup$
Thank you so much! It's a great help!
$endgroup$
– chbe
Jan 10 at 10:05
add a comment |
$begingroup$
If we write
$f(z) = f(x + iy) = u(x, y) + iv(x, y) = u(r, phi) + iv(r, phi), tag 1$
where $(r, phi)$ are standard polar coordinates on $Bbb H$, then we have
$dfrac{partial u}{partial r} + i dfrac{partial v}{partial r} = dfrac{partial f(z)}{partial r} = displaystyle lim_{Delta theta to 0} dfrac{f((1 + Delta theta)z) - f(z)}{Delta theta} = lim_{Delta theta to 0} dfrac{f(z) - f(z)}{Delta theta} = 0, tag 2$
since varying $theta$ moves $theta z$ in the radial direction; thus,
$dfrac{partial u}{partial r} = dfrac{partial v}{partial r} = 0; tag 3$
now according to the Cauchy-Riemann equations in the polar coordinates $(r, phi)$,
$ dfrac{partial v}{partial phi} = rdfrac{partial u}{partial r} = 0, tag 4$
$ dfrac{partial u}{partial phi} = -r dfrac{partial v}{partial r} = 0; tag 5$
these two equations together with (3) show that
$nabla u(z) = nabla v(z) = 0, forall z in Bbb H; tag 6$
it follows that $u(z)$ and $v(z)$ are constant in $Bbb H$, since it is a connected open set; thus so also $f(z)$ is constant on $Bbb H$.
The above derivation may be written in a slightly more compact fashion if we recall the polar form of the Cauchy-Riemann equations may be expressed in terms of the single equation in $f$,
$dfrac{partial f}{partial r} = dfrac{1}{ir} dfrac{partial f}{partial phi}; tag 7$
then (2) shows that
$dfrac{partial f}{partial r} = 0, tag 8$
from which we find that
$dfrac{partial f}{partial phi} = 0, tag 9$
and the constancy of $f(z)$ immediately follows.
$endgroup$
2
$begingroup$
I think that OP means that $f(theta z) = f(z)$ for all $z$ and some fixed $theta in Bbb R^+ - {1}$, not for all $z$ and all $theta$.
$endgroup$
– Travis
Jan 8 at 9:06
add a comment |
Your Answer
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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votes
$begingroup$
No, $f$ does not need to be constant.
Define $g$ in the strip $S = { w mid 0 < operatorname{Im} w < i pi } $ as
$g(w) = f(e^w)$. Then for a given positive real number $theta ne 1$
$$
f(z) = f(theta z) quad text{for all } z in Bbb H
$$
if and only if
$$
g(w) = g(w + log theta) quad text{for all } w in S
$$
so that the problem reduces to find all $log theta$ - periodic functions in the strip $S$. The most simple example would be
$$
g(w) = exp left(frac{2 pi i w}{log theta} right)
$$
corresponding to
$$
f(z) = exp left(frac{2 pi i log z}{log theta} right)
$$
where $log z = log |z| + i operatorname{Arg} z$ is the main branch of the logarithm with $0 < operatorname{Arg} z < pi$ for $z in Bbb H$.
All $log theta$ - periodic functions in $S$ are obtained by composing $g$ with a holomorphic function, this leads to the general solution
$$
f(z) = h left(exp left(frac{2 pi i log z}{log theta} right) right)
$$
where $h$ is holomorphic in the annulus ${ w mid exp left(frac{-2 pi^2}{log theta}right) < |w| < 1 }$.
$endgroup$
$begingroup$
Thank you so much! It's a great help!
$endgroup$
– chbe
Jan 10 at 10:05
add a comment |
$begingroup$
No, $f$ does not need to be constant.
Define $g$ in the strip $S = { w mid 0 < operatorname{Im} w < i pi } $ as
$g(w) = f(e^w)$. Then for a given positive real number $theta ne 1$
$$
f(z) = f(theta z) quad text{for all } z in Bbb H
$$
if and only if
$$
g(w) = g(w + log theta) quad text{for all } w in S
$$
so that the problem reduces to find all $log theta$ - periodic functions in the strip $S$. The most simple example would be
$$
g(w) = exp left(frac{2 pi i w}{log theta} right)
$$
corresponding to
$$
f(z) = exp left(frac{2 pi i log z}{log theta} right)
$$
where $log z = log |z| + i operatorname{Arg} z$ is the main branch of the logarithm with $0 < operatorname{Arg} z < pi$ for $z in Bbb H$.
All $log theta$ - periodic functions in $S$ are obtained by composing $g$ with a holomorphic function, this leads to the general solution
$$
f(z) = h left(exp left(frac{2 pi i log z}{log theta} right) right)
$$
where $h$ is holomorphic in the annulus ${ w mid exp left(frac{-2 pi^2}{log theta}right) < |w| < 1 }$.
$endgroup$
$begingroup$
Thank you so much! It's a great help!
$endgroup$
– chbe
Jan 10 at 10:05
add a comment |
$begingroup$
No, $f$ does not need to be constant.
Define $g$ in the strip $S = { w mid 0 < operatorname{Im} w < i pi } $ as
$g(w) = f(e^w)$. Then for a given positive real number $theta ne 1$
$$
f(z) = f(theta z) quad text{for all } z in Bbb H
$$
if and only if
$$
g(w) = g(w + log theta) quad text{for all } w in S
$$
so that the problem reduces to find all $log theta$ - periodic functions in the strip $S$. The most simple example would be
$$
g(w) = exp left(frac{2 pi i w}{log theta} right)
$$
corresponding to
$$
f(z) = exp left(frac{2 pi i log z}{log theta} right)
$$
where $log z = log |z| + i operatorname{Arg} z$ is the main branch of the logarithm with $0 < operatorname{Arg} z < pi$ for $z in Bbb H$.
All $log theta$ - periodic functions in $S$ are obtained by composing $g$ with a holomorphic function, this leads to the general solution
$$
f(z) = h left(exp left(frac{2 pi i log z}{log theta} right) right)
$$
where $h$ is holomorphic in the annulus ${ w mid exp left(frac{-2 pi^2}{log theta}right) < |w| < 1 }$.
$endgroup$
No, $f$ does not need to be constant.
Define $g$ in the strip $S = { w mid 0 < operatorname{Im} w < i pi } $ as
$g(w) = f(e^w)$. Then for a given positive real number $theta ne 1$
$$
f(z) = f(theta z) quad text{for all } z in Bbb H
$$
if and only if
$$
g(w) = g(w + log theta) quad text{for all } w in S
$$
so that the problem reduces to find all $log theta$ - periodic functions in the strip $S$. The most simple example would be
$$
g(w) = exp left(frac{2 pi i w}{log theta} right)
$$
corresponding to
$$
f(z) = exp left(frac{2 pi i log z}{log theta} right)
$$
where $log z = log |z| + i operatorname{Arg} z$ is the main branch of the logarithm with $0 < operatorname{Arg} z < pi$ for $z in Bbb H$.
All $log theta$ - periodic functions in $S$ are obtained by composing $g$ with a holomorphic function, this leads to the general solution
$$
f(z) = h left(exp left(frac{2 pi i log z}{log theta} right) right)
$$
where $h$ is holomorphic in the annulus ${ w mid exp left(frac{-2 pi^2}{log theta}right) < |w| < 1 }$.
edited Jan 8 at 9:17
answered Jan 8 at 8:59
Martin RMartin R
31k33561
31k33561
$begingroup$
Thank you so much! It's a great help!
$endgroup$
– chbe
Jan 10 at 10:05
add a comment |
$begingroup$
Thank you so much! It's a great help!
$endgroup$
– chbe
Jan 10 at 10:05
$begingroup$
Thank you so much! It's a great help!
$endgroup$
– chbe
Jan 10 at 10:05
$begingroup$
Thank you so much! It's a great help!
$endgroup$
– chbe
Jan 10 at 10:05
add a comment |
$begingroup$
If we write
$f(z) = f(x + iy) = u(x, y) + iv(x, y) = u(r, phi) + iv(r, phi), tag 1$
where $(r, phi)$ are standard polar coordinates on $Bbb H$, then we have
$dfrac{partial u}{partial r} + i dfrac{partial v}{partial r} = dfrac{partial f(z)}{partial r} = displaystyle lim_{Delta theta to 0} dfrac{f((1 + Delta theta)z) - f(z)}{Delta theta} = lim_{Delta theta to 0} dfrac{f(z) - f(z)}{Delta theta} = 0, tag 2$
since varying $theta$ moves $theta z$ in the radial direction; thus,
$dfrac{partial u}{partial r} = dfrac{partial v}{partial r} = 0; tag 3$
now according to the Cauchy-Riemann equations in the polar coordinates $(r, phi)$,
$ dfrac{partial v}{partial phi} = rdfrac{partial u}{partial r} = 0, tag 4$
$ dfrac{partial u}{partial phi} = -r dfrac{partial v}{partial r} = 0; tag 5$
these two equations together with (3) show that
$nabla u(z) = nabla v(z) = 0, forall z in Bbb H; tag 6$
it follows that $u(z)$ and $v(z)$ are constant in $Bbb H$, since it is a connected open set; thus so also $f(z)$ is constant on $Bbb H$.
The above derivation may be written in a slightly more compact fashion if we recall the polar form of the Cauchy-Riemann equations may be expressed in terms of the single equation in $f$,
$dfrac{partial f}{partial r} = dfrac{1}{ir} dfrac{partial f}{partial phi}; tag 7$
then (2) shows that
$dfrac{partial f}{partial r} = 0, tag 8$
from which we find that
$dfrac{partial f}{partial phi} = 0, tag 9$
and the constancy of $f(z)$ immediately follows.
$endgroup$
2
$begingroup$
I think that OP means that $f(theta z) = f(z)$ for all $z$ and some fixed $theta in Bbb R^+ - {1}$, not for all $z$ and all $theta$.
$endgroup$
– Travis
Jan 8 at 9:06
add a comment |
$begingroup$
If we write
$f(z) = f(x + iy) = u(x, y) + iv(x, y) = u(r, phi) + iv(r, phi), tag 1$
where $(r, phi)$ are standard polar coordinates on $Bbb H$, then we have
$dfrac{partial u}{partial r} + i dfrac{partial v}{partial r} = dfrac{partial f(z)}{partial r} = displaystyle lim_{Delta theta to 0} dfrac{f((1 + Delta theta)z) - f(z)}{Delta theta} = lim_{Delta theta to 0} dfrac{f(z) - f(z)}{Delta theta} = 0, tag 2$
since varying $theta$ moves $theta z$ in the radial direction; thus,
$dfrac{partial u}{partial r} = dfrac{partial v}{partial r} = 0; tag 3$
now according to the Cauchy-Riemann equations in the polar coordinates $(r, phi)$,
$ dfrac{partial v}{partial phi} = rdfrac{partial u}{partial r} = 0, tag 4$
$ dfrac{partial u}{partial phi} = -r dfrac{partial v}{partial r} = 0; tag 5$
these two equations together with (3) show that
$nabla u(z) = nabla v(z) = 0, forall z in Bbb H; tag 6$
it follows that $u(z)$ and $v(z)$ are constant in $Bbb H$, since it is a connected open set; thus so also $f(z)$ is constant on $Bbb H$.
The above derivation may be written in a slightly more compact fashion if we recall the polar form of the Cauchy-Riemann equations may be expressed in terms of the single equation in $f$,
$dfrac{partial f}{partial r} = dfrac{1}{ir} dfrac{partial f}{partial phi}; tag 7$
then (2) shows that
$dfrac{partial f}{partial r} = 0, tag 8$
from which we find that
$dfrac{partial f}{partial phi} = 0, tag 9$
and the constancy of $f(z)$ immediately follows.
$endgroup$
2
$begingroup$
I think that OP means that $f(theta z) = f(z)$ for all $z$ and some fixed $theta in Bbb R^+ - {1}$, not for all $z$ and all $theta$.
$endgroup$
– Travis
Jan 8 at 9:06
add a comment |
$begingroup$
If we write
$f(z) = f(x + iy) = u(x, y) + iv(x, y) = u(r, phi) + iv(r, phi), tag 1$
where $(r, phi)$ are standard polar coordinates on $Bbb H$, then we have
$dfrac{partial u}{partial r} + i dfrac{partial v}{partial r} = dfrac{partial f(z)}{partial r} = displaystyle lim_{Delta theta to 0} dfrac{f((1 + Delta theta)z) - f(z)}{Delta theta} = lim_{Delta theta to 0} dfrac{f(z) - f(z)}{Delta theta} = 0, tag 2$
since varying $theta$ moves $theta z$ in the radial direction; thus,
$dfrac{partial u}{partial r} = dfrac{partial v}{partial r} = 0; tag 3$
now according to the Cauchy-Riemann equations in the polar coordinates $(r, phi)$,
$ dfrac{partial v}{partial phi} = rdfrac{partial u}{partial r} = 0, tag 4$
$ dfrac{partial u}{partial phi} = -r dfrac{partial v}{partial r} = 0; tag 5$
these two equations together with (3) show that
$nabla u(z) = nabla v(z) = 0, forall z in Bbb H; tag 6$
it follows that $u(z)$ and $v(z)$ are constant in $Bbb H$, since it is a connected open set; thus so also $f(z)$ is constant on $Bbb H$.
The above derivation may be written in a slightly more compact fashion if we recall the polar form of the Cauchy-Riemann equations may be expressed in terms of the single equation in $f$,
$dfrac{partial f}{partial r} = dfrac{1}{ir} dfrac{partial f}{partial phi}; tag 7$
then (2) shows that
$dfrac{partial f}{partial r} = 0, tag 8$
from which we find that
$dfrac{partial f}{partial phi} = 0, tag 9$
and the constancy of $f(z)$ immediately follows.
$endgroup$
If we write
$f(z) = f(x + iy) = u(x, y) + iv(x, y) = u(r, phi) + iv(r, phi), tag 1$
where $(r, phi)$ are standard polar coordinates on $Bbb H$, then we have
$dfrac{partial u}{partial r} + i dfrac{partial v}{partial r} = dfrac{partial f(z)}{partial r} = displaystyle lim_{Delta theta to 0} dfrac{f((1 + Delta theta)z) - f(z)}{Delta theta} = lim_{Delta theta to 0} dfrac{f(z) - f(z)}{Delta theta} = 0, tag 2$
since varying $theta$ moves $theta z$ in the radial direction; thus,
$dfrac{partial u}{partial r} = dfrac{partial v}{partial r} = 0; tag 3$
now according to the Cauchy-Riemann equations in the polar coordinates $(r, phi)$,
$ dfrac{partial v}{partial phi} = rdfrac{partial u}{partial r} = 0, tag 4$
$ dfrac{partial u}{partial phi} = -r dfrac{partial v}{partial r} = 0; tag 5$
these two equations together with (3) show that
$nabla u(z) = nabla v(z) = 0, forall z in Bbb H; tag 6$
it follows that $u(z)$ and $v(z)$ are constant in $Bbb H$, since it is a connected open set; thus so also $f(z)$ is constant on $Bbb H$.
The above derivation may be written in a slightly more compact fashion if we recall the polar form of the Cauchy-Riemann equations may be expressed in terms of the single equation in $f$,
$dfrac{partial f}{partial r} = dfrac{1}{ir} dfrac{partial f}{partial phi}; tag 7$
then (2) shows that
$dfrac{partial f}{partial r} = 0, tag 8$
from which we find that
$dfrac{partial f}{partial phi} = 0, tag 9$
and the constancy of $f(z)$ immediately follows.
answered Jan 8 at 0:37
Robert LewisRobert Lewis
48.9k23168
48.9k23168
2
$begingroup$
I think that OP means that $f(theta z) = f(z)$ for all $z$ and some fixed $theta in Bbb R^+ - {1}$, not for all $z$ and all $theta$.
$endgroup$
– Travis
Jan 8 at 9:06
add a comment |
2
$begingroup$
I think that OP means that $f(theta z) = f(z)$ for all $z$ and some fixed $theta in Bbb R^+ - {1}$, not for all $z$ and all $theta$.
$endgroup$
– Travis
Jan 8 at 9:06
2
2
$begingroup$
I think that OP means that $f(theta z) = f(z)$ for all $z$ and some fixed $theta in Bbb R^+ - {1}$, not for all $z$ and all $theta$.
$endgroup$
– Travis
Jan 8 at 9:06
$begingroup$
I think that OP means that $f(theta z) = f(z)$ for all $z$ and some fixed $theta in Bbb R^+ - {1}$, not for all $z$ and all $theta$.
$endgroup$
– Travis
Jan 8 at 9:06
add a comment |
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Do you mean to say that $f(z) = f(theta z), ; forall theta in Bbb R^+$, so that $f(z)$ constant on each ray $theta z_0$, $z_0 in Bbb H$?
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– Robert Lewis
Jan 7 at 22:33
1
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$f(z) = f(cz)$ iff $f(e^z)$ is $log c$ periodic
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– reuns
Jan 8 at 9:02
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@RobertLewis $theta$ is fixed. I am sorry for the confusion.
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– chbe
Jan 10 at 10:09