Meromorphic on unit disc with absolute value 1 on the circle is a rational function.












6












$begingroup$


Let $f$ be a meromorphic function on the open unit disk such that $f$ has a continuous extension to the boundary circle.Suppose $f$ has only poles in the open unit disc and suppose $|f(z)|=1$ for all $z$ with $|z|=1$.Prove that $f$ is a rational function.






complex-analysis







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$endgroup$












  • $begingroup$
    Have you heard of Blaschke products? Alternatively: Which forms of the reflection principle do you know?
    $endgroup$
    – Daniel Fischer♦
    May 6 '14 at 15:33










  • $begingroup$
    I know about Schwarz reflection principle.And I can show zeros and poles of $f$ are reflected by the unit circle.I do know about Blaschke product,but I don't know how to show the quotient of this function and the Blaschke product is a polynomial.
    $endgroup$
    – Daniel S.
    May 6 '14 at 15:44










  • $begingroup$
    @DanielFischer I can reach to an entire function without poles or zeros,but this is not necessarily a polynomial,say,exponential functions.
    $endgroup$
    – Daniel S.
    May 6 '14 at 15:46
















6












$begingroup$


Let $f$ be a meromorphic function on the open unit disk such that $f$ has a continuous extension to the boundary circle.Suppose $f$ has only poles in the open unit disc and suppose $|f(z)|=1$ for all $z$ with $|z|=1$.Prove that $f$ is a rational function.






complex-analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    Have you heard of Blaschke products? Alternatively: Which forms of the reflection principle do you know?
    $endgroup$
    – Daniel Fischer♦
    May 6 '14 at 15:33










  • $begingroup$
    I know about Schwarz reflection principle.And I can show zeros and poles of $f$ are reflected by the unit circle.I do know about Blaschke product,but I don't know how to show the quotient of this function and the Blaschke product is a polynomial.
    $endgroup$
    – Daniel S.
    May 6 '14 at 15:44










  • $begingroup$
    @DanielFischer I can reach to an entire function without poles or zeros,but this is not necessarily a polynomial,say,exponential functions.
    $endgroup$
    – Daniel S.
    May 6 '14 at 15:46














6












6








6


5



$begingroup$


Let $f$ be a meromorphic function on the open unit disk such that $f$ has a continuous extension to the boundary circle.Suppose $f$ has only poles in the open unit disc and suppose $|f(z)|=1$ for all $z$ with $|z|=1$.Prove that $f$ is a rational function.






complex-analysis







share|cite|improve this question









$endgroup$




Let $f$ be a meromorphic function on the open unit disk such that $f$ has a continuous extension to the boundary circle.Suppose $f$ has only poles in the open unit disc and suppose $|f(z)|=1$ for all $z$ with $|z|=1$.Prove that $f$ is a rational function.






complex-analysis




complex-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked May 6 '14 at 15:28









Daniel S.Daniel S.

260111




260111












  • $begingroup$
    Have you heard of Blaschke products? Alternatively: Which forms of the reflection principle do you know?
    $endgroup$
    – Daniel Fischer♦
    May 6 '14 at 15:33










  • $begingroup$
    I know about Schwarz reflection principle.And I can show zeros and poles of $f$ are reflected by the unit circle.I do know about Blaschke product,but I don't know how to show the quotient of this function and the Blaschke product is a polynomial.
    $endgroup$
    – Daniel S.
    May 6 '14 at 15:44










  • $begingroup$
    @DanielFischer I can reach to an entire function without poles or zeros,but this is not necessarily a polynomial,say,exponential functions.
    $endgroup$
    – Daniel S.
    May 6 '14 at 15:46


















  • $begingroup$
    Have you heard of Blaschke products? Alternatively: Which forms of the reflection principle do you know?
    $endgroup$
    – Daniel Fischer♦
    May 6 '14 at 15:33










  • $begingroup$
    I know about Schwarz reflection principle.And I can show zeros and poles of $f$ are reflected by the unit circle.I do know about Blaschke product,but I don't know how to show the quotient of this function and the Blaschke product is a polynomial.
    $endgroup$
    – Daniel S.
    May 6 '14 at 15:44










  • $begingroup$
    @DanielFischer I can reach to an entire function without poles or zeros,but this is not necessarily a polynomial,say,exponential functions.
    $endgroup$
    – Daniel S.
    May 6 '14 at 15:46
















$begingroup$
Have you heard of Blaschke products? Alternatively: Which forms of the reflection principle do you know?
$endgroup$
– Daniel Fischer♦
May 6 '14 at 15:33




$begingroup$
Have you heard of Blaschke products? Alternatively: Which forms of the reflection principle do you know?
$endgroup$
– Daniel Fischer♦
May 6 '14 at 15:33












$begingroup$
I know about Schwarz reflection principle.And I can show zeros and poles of $f$ are reflected by the unit circle.I do know about Blaschke product,but I don't know how to show the quotient of this function and the Blaschke product is a polynomial.
$endgroup$
– Daniel S.
May 6 '14 at 15:44




$begingroup$
I know about Schwarz reflection principle.And I can show zeros and poles of $f$ are reflected by the unit circle.I do know about Blaschke product,but I don't know how to show the quotient of this function and the Blaschke product is a polynomial.
$endgroup$
– Daniel S.
May 6 '14 at 15:44












$begingroup$
@DanielFischer I can reach to an entire function without poles or zeros,but this is not necessarily a polynomial,say,exponential functions.
$endgroup$
– Daniel S.
May 6 '14 at 15:46




$begingroup$
@DanielFischer I can reach to an entire function without poles or zeros,but this is not necessarily a polynomial,say,exponential functions.
$endgroup$
– Daniel S.
May 6 '14 at 15:46










1 Answer
1






active

oldest

votes


















5












$begingroup$

Since the closed unit disk is compact, $f$ can have only finitely many zeros and poles in the unit disk. Let $k$ be the order of $f$ in $0$, that is, $f(z) = z^kcdot g(z)$ where $g$ is holomorphic in a neighbourhood of $0$ with $g(0) neq 0$. Let $zeta_1,dotsc, zeta_k$ be the zeros of $f$ in the punctured unit disk, with multiplicities $mu_1,dotsc,mu_k$. Let $pi_1,dotsc,pi_r$ be the poles of $f$ in the punctured unit disk with orders $nu_1,dotsc,nu_r$. Consider the finite Blaschke products



$$Z(z) = prod_{kappa=1}^k left(frac{z - zeta_kappa}{1 - overline{zeta}_kappa z}right)^{mu_kappa}$$



and



$$P(z) = prod_{rho = 1}^r left(frac{z-pi_rho}{1-overline{pi}_rho z}right)^{nu_rho}.$$



Evidently,



$$h(z) = z^kfrac{Z(z)}{P(z)}$$



is a rational function, and it has the same zeros and poles in the unit disk as $f$.



Since every factor in $h$ has modulus $1$ on the unit circle, we have $lvert h(z)rvert = 1$ for all $z$ with $lvert zrvert = 1$, and hence



$$frac{f(z)}{h(z)}$$



is a zero-free holomorphic function on the unit disk with $leftlvert frac{f(z)}{h(z)}rightrvert = 1$ for $lvert zrvert = 1$, thus constant.




An alternative way to obtain the result is by using the reflection principle:



$$F(z) = begin{cases} f(z) &, lvert zrvert leqslant 1 \ dfrac{1}{overline{f(1/overline{z})}} &, lvert zrvert > 1end{cases}$$



defines a function that is meromorphic on $widehat{mathbb{C}}$ by the reflection principle. A function that is meromorphic on the entire sphere is a rational function, so $F$, and hence $f$ is rational.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Final step use the maximum principle to show it's constant.Thanks!
    $endgroup$
    – Daniel S.
    May 6 '14 at 16:02











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









5












$begingroup$

Since the closed unit disk is compact, $f$ can have only finitely many zeros and poles in the unit disk. Let $k$ be the order of $f$ in $0$, that is, $f(z) = z^kcdot g(z)$ where $g$ is holomorphic in a neighbourhood of $0$ with $g(0) neq 0$. Let $zeta_1,dotsc, zeta_k$ be the zeros of $f$ in the punctured unit disk, with multiplicities $mu_1,dotsc,mu_k$. Let $pi_1,dotsc,pi_r$ be the poles of $f$ in the punctured unit disk with orders $nu_1,dotsc,nu_r$. Consider the finite Blaschke products



$$Z(z) = prod_{kappa=1}^k left(frac{z - zeta_kappa}{1 - overline{zeta}_kappa z}right)^{mu_kappa}$$



and



$$P(z) = prod_{rho = 1}^r left(frac{z-pi_rho}{1-overline{pi}_rho z}right)^{nu_rho}.$$



Evidently,



$$h(z) = z^kfrac{Z(z)}{P(z)}$$



is a rational function, and it has the same zeros and poles in the unit disk as $f$.



Since every factor in $h$ has modulus $1$ on the unit circle, we have $lvert h(z)rvert = 1$ for all $z$ with $lvert zrvert = 1$, and hence



$$frac{f(z)}{h(z)}$$



is a zero-free holomorphic function on the unit disk with $leftlvert frac{f(z)}{h(z)}rightrvert = 1$ for $lvert zrvert = 1$, thus constant.




An alternative way to obtain the result is by using the reflection principle:



$$F(z) = begin{cases} f(z) &, lvert zrvert leqslant 1 \ dfrac{1}{overline{f(1/overline{z})}} &, lvert zrvert > 1end{cases}$$



defines a function that is meromorphic on $widehat{mathbb{C}}$ by the reflection principle. A function that is meromorphic on the entire sphere is a rational function, so $F$, and hence $f$ is rational.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Final step use the maximum principle to show it's constant.Thanks!
    $endgroup$
    – Daniel S.
    May 6 '14 at 16:02
















5












$begingroup$

Since the closed unit disk is compact, $f$ can have only finitely many zeros and poles in the unit disk. Let $k$ be the order of $f$ in $0$, that is, $f(z) = z^kcdot g(z)$ where $g$ is holomorphic in a neighbourhood of $0$ with $g(0) neq 0$. Let $zeta_1,dotsc, zeta_k$ be the zeros of $f$ in the punctured unit disk, with multiplicities $mu_1,dotsc,mu_k$. Let $pi_1,dotsc,pi_r$ be the poles of $f$ in the punctured unit disk with orders $nu_1,dotsc,nu_r$. Consider the finite Blaschke products



$$Z(z) = prod_{kappa=1}^k left(frac{z - zeta_kappa}{1 - overline{zeta}_kappa z}right)^{mu_kappa}$$



and



$$P(z) = prod_{rho = 1}^r left(frac{z-pi_rho}{1-overline{pi}_rho z}right)^{nu_rho}.$$



Evidently,



$$h(z) = z^kfrac{Z(z)}{P(z)}$$



is a rational function, and it has the same zeros and poles in the unit disk as $f$.



Since every factor in $h$ has modulus $1$ on the unit circle, we have $lvert h(z)rvert = 1$ for all $z$ with $lvert zrvert = 1$, and hence



$$frac{f(z)}{h(z)}$$



is a zero-free holomorphic function on the unit disk with $leftlvert frac{f(z)}{h(z)}rightrvert = 1$ for $lvert zrvert = 1$, thus constant.




An alternative way to obtain the result is by using the reflection principle:



$$F(z) = begin{cases} f(z) &, lvert zrvert leqslant 1 \ dfrac{1}{overline{f(1/overline{z})}} &, lvert zrvert > 1end{cases}$$



defines a function that is meromorphic on $widehat{mathbb{C}}$ by the reflection principle. A function that is meromorphic on the entire sphere is a rational function, so $F$, and hence $f$ is rational.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Final step use the maximum principle to show it's constant.Thanks!
    $endgroup$
    – Daniel S.
    May 6 '14 at 16:02














5












5








5





$begingroup$

Since the closed unit disk is compact, $f$ can have only finitely many zeros and poles in the unit disk. Let $k$ be the order of $f$ in $0$, that is, $f(z) = z^kcdot g(z)$ where $g$ is holomorphic in a neighbourhood of $0$ with $g(0) neq 0$. Let $zeta_1,dotsc, zeta_k$ be the zeros of $f$ in the punctured unit disk, with multiplicities $mu_1,dotsc,mu_k$. Let $pi_1,dotsc,pi_r$ be the poles of $f$ in the punctured unit disk with orders $nu_1,dotsc,nu_r$. Consider the finite Blaschke products



$$Z(z) = prod_{kappa=1}^k left(frac{z - zeta_kappa}{1 - overline{zeta}_kappa z}right)^{mu_kappa}$$



and



$$P(z) = prod_{rho = 1}^r left(frac{z-pi_rho}{1-overline{pi}_rho z}right)^{nu_rho}.$$



Evidently,



$$h(z) = z^kfrac{Z(z)}{P(z)}$$



is a rational function, and it has the same zeros and poles in the unit disk as $f$.



Since every factor in $h$ has modulus $1$ on the unit circle, we have $lvert h(z)rvert = 1$ for all $z$ with $lvert zrvert = 1$, and hence



$$frac{f(z)}{h(z)}$$



is a zero-free holomorphic function on the unit disk with $leftlvert frac{f(z)}{h(z)}rightrvert = 1$ for $lvert zrvert = 1$, thus constant.




An alternative way to obtain the result is by using the reflection principle:



$$F(z) = begin{cases} f(z) &, lvert zrvert leqslant 1 \ dfrac{1}{overline{f(1/overline{z})}} &, lvert zrvert > 1end{cases}$$



defines a function that is meromorphic on $widehat{mathbb{C}}$ by the reflection principle. A function that is meromorphic on the entire sphere is a rational function, so $F$, and hence $f$ is rational.






share|cite|improve this answer











$endgroup$



Since the closed unit disk is compact, $f$ can have only finitely many zeros and poles in the unit disk. Let $k$ be the order of $f$ in $0$, that is, $f(z) = z^kcdot g(z)$ where $g$ is holomorphic in a neighbourhood of $0$ with $g(0) neq 0$. Let $zeta_1,dotsc, zeta_k$ be the zeros of $f$ in the punctured unit disk, with multiplicities $mu_1,dotsc,mu_k$. Let $pi_1,dotsc,pi_r$ be the poles of $f$ in the punctured unit disk with orders $nu_1,dotsc,nu_r$. Consider the finite Blaschke products



$$Z(z) = prod_{kappa=1}^k left(frac{z - zeta_kappa}{1 - overline{zeta}_kappa z}right)^{mu_kappa}$$



and



$$P(z) = prod_{rho = 1}^r left(frac{z-pi_rho}{1-overline{pi}_rho z}right)^{nu_rho}.$$



Evidently,



$$h(z) = z^kfrac{Z(z)}{P(z)}$$



is a rational function, and it has the same zeros and poles in the unit disk as $f$.



Since every factor in $h$ has modulus $1$ on the unit circle, we have $lvert h(z)rvert = 1$ for all $z$ with $lvert zrvert = 1$, and hence



$$frac{f(z)}{h(z)}$$



is a zero-free holomorphic function on the unit disk with $leftlvert frac{f(z)}{h(z)}rightrvert = 1$ for $lvert zrvert = 1$, thus constant.




An alternative way to obtain the result is by using the reflection principle:



$$F(z) = begin{cases} f(z) &, lvert zrvert leqslant 1 \ dfrac{1}{overline{f(1/overline{z})}} &, lvert zrvert > 1end{cases}$$



defines a function that is meromorphic on $widehat{mathbb{C}}$ by the reflection principle. A function that is meromorphic on the entire sphere is a rational function, so $F$, and hence $f$ is rational.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited May 6 '14 at 17:50

























answered May 6 '14 at 15:55









Daniel Fischer♦Daniel Fischer

174k16167287




174k16167287












  • $begingroup$
    Final step use the maximum principle to show it's constant.Thanks!
    $endgroup$
    – Daniel S.
    May 6 '14 at 16:02


















  • $begingroup$
    Final step use the maximum principle to show it's constant.Thanks!
    $endgroup$
    – Daniel S.
    May 6 '14 at 16:02
















$begingroup$
Final step use the maximum principle to show it's constant.Thanks!
$endgroup$
– Daniel S.
May 6 '14 at 16:02




$begingroup$
Final step use the maximum principle to show it's constant.Thanks!
$endgroup$
– Daniel S.
May 6 '14 at 16:02


















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