Holomorphic function on a upper half plane that is scale invariant with respect to a positive real number.












1












$begingroup$


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.)










share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















1












$begingroup$


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.)










share|cite|improve this question











$endgroup$












  • $begingroup$
    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














1












1








1


1



$begingroup$


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.)










share|cite|improve this question











$endgroup$




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






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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


















  • $begingroup$
    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










2 Answers
2






active

oldest

votes


















1












$begingroup$

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 ${ w mid exp left(frac{-2 pi^2}{log theta}right) < |w| < 1 }$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you so much! It's a great help!
    $endgroup$
    – chbe
    Jan 10 at 10:05



















0












$begingroup$

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.






share|cite|improve this answer









$endgroup$









  • 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












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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 ${ w mid exp left(frac{-2 pi^2}{log theta}right) < |w| < 1 }$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you so much! It's a great help!
    $endgroup$
    – chbe
    Jan 10 at 10:05
















1












$begingroup$

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 ${ w mid exp left(frac{-2 pi^2}{log theta}right) < |w| < 1 }$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you so much! It's a great help!
    $endgroup$
    – chbe
    Jan 10 at 10:05














1












1








1





$begingroup$

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 ${ w mid exp left(frac{-2 pi^2}{log theta}right) < |w| < 1 }$.






share|cite|improve this answer











$endgroup$



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 ${ w mid exp left(frac{-2 pi^2}{log theta}right) < |w| < 1 }$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








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


















  • $begingroup$
    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











0












$begingroup$

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.






share|cite|improve this answer









$endgroup$









  • 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
















0












$begingroup$

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.






share|cite|improve this answer









$endgroup$









  • 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














0












0








0





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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














  • 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


















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