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Rado–Kneser–Choquet Theorem proof

Conformal map of doubly connected domain into annulus.understanding topological argument in rado-kneser theorem“Geometric” proof of Rouche's theorem on the number of zeros?Doubt in a step of the proof of Rado-Kneser-Choquet theoremRiemann's proof of his mapping theorem(Rudin's) Definition of a harmonic functionPositive harmonic function tending to infinity on the integers tends to infinity everywhereDifficulty in understanding the proof of open mapping theorem and maximum modulus principleCaccioppoli type inequality for harmonic functionA proof of the Open Mapping Theorem

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I was reading the proof of the Rado–Kneser–Choquet Theorem. The statement is there in the image (taken from the book "Harmonic Mapping in the Plane, Duren page-$30$":

In the proof, he shows that the Poisson integral function $$f(z)=dfrac{1}{2pi}int_0^{2pi}dfrac{1-|z|^2}{|e^{it}-z|^2}phi(e^{it}~dt $$ is locally univalent and then he finishes his proof by using the Argument Principle. Could you please tell me how it can be shown by using the argument principle that the function $f$ is univalent in $mathbb{D}.$

complex-analysis harmonic-functions poisson-integrals

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asked Mar 18 at 13:17

I am piI am pi

1909

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0

$begingroup$

I was reading the proof of the Rado–Kneser–Choquet Theorem. The statement is there in the image (taken from the book "Harmonic Mapping in the Plane, Duren page-$30$":

In the proof, he shows that the Poisson integral function $$f(z)=dfrac{1}{2pi}int_0^{2pi}dfrac{1-|z|^2}{|e^{it}-z|^2}phi(e^{it}~dt $$ is locally univalent and then he finishes his proof by using the Argument Principle. Could you please tell me how it can be shown by using the argument principle that the function $f$ is univalent in $mathbb{D}.$

complex-analysis harmonic-functions poisson-integrals

share|cite|improve this question

asked Mar 18 at 13:17

I am piI am pi

1909

$endgroup$

add a comment | 

$begingroup$

I was reading the proof of the Rado–Kneser–Choquet Theorem. The statement is there in the image (taken from the book "Harmonic Mapping in the Plane, Duren page-$30$":

In the proof, he shows that the Poisson integral function $$f(z)=dfrac{1}{2pi}int_0^{2pi}dfrac{1-|z|^2}{|e^{it}-z|^2}phi(e^{it}~dt $$ is locally univalent and then he finishes his proof by using the Argument Principle. Could you please tell me how it can be shown by using the argument principle that the function $f$ is univalent in $mathbb{D}.$

complex-analysis harmonic-functions poisson-integrals

share|cite|improve this question

asked Mar 18 at 13:17

I am piI am pi

1909

$endgroup$

share|cite|improve this question

asked Mar 18 at 13:17

I am piI am pi

1909

share|cite|improve this question

asked Mar 18 at 13:17

I am piI am pi

1909

asked Mar 18 at 13:17

I am piI am pi

1909

asked Mar 18 at 13:17

I am piI am pi

1909

1 Answer
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oldest

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1

$begingroup$

The theorem you want is on page 9 - the extension of the analytic argument principle to sense preserving harmonic ones; then you just apply it to $f-a$ for any $a$ not in the image of the Jordan curve $f(mathbb{T})$ since then $f-a$ satisfies the same properties as $f$ (non-vanishing homeomorphism on the boundary etc) and then by the argument principle and discretness of $2pi N$, $f-a$ has the same number of zeros inside $mathbb{D}$ when $a$ belongs to the same connected component of $C- f(mathbb{T})$, so no zeros if $a$ is outside and precisely one zero if $a$ is inside since otherwise $f$ cannot be a homeomorphism at the boundary.

There are two subtle points here, both implied by the sense-preserving property of $f$, first being that zeros are discrete (there are harmonic functions, even harmonic polynomials like $Re(z)$, with non-discrete zero set) and second is that they all have positive index (otherwise you could have two zeros with index 1 and one with index -1, say since only zeros of analytic functions are apriori guaranteed to have positive index), so if the argument change is $2pi$ it means there is a unique zero.

share|cite|improve this answer

edited Mar 18 at 15:17

answered Mar 18 at 15:11

ConradConrad

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1

$begingroup$

The theorem you want is on page 9 - the extension of the analytic argument principle to sense preserving harmonic ones; then you just apply it to $f-a$ for any $a$ not in the image of the Jordan curve $f(mathbb{T})$ since then $f-a$ satisfies the same properties as $f$ (non-vanishing homeomorphism on the boundary etc) and then by the argument principle and discretness of $2pi N$, $f-a$ has the same number of zeros inside $mathbb{D}$ when $a$ belongs to the same connected component of $C- f(mathbb{T})$, so no zeros if $a$ is outside and precisely one zero if $a$ is inside since otherwise $f$ cannot be a homeomorphism at the boundary.

There are two subtle points here, both implied by the sense-preserving property of $f$, first being that zeros are discrete (there are harmonic functions, even harmonic polynomials like $Re(z)$, with non-discrete zero set) and second is that they all have positive index (otherwise you could have two zeros with index 1 and one with index -1, say since only zeros of analytic functions are apriori guaranteed to have positive index), so if the argument change is $2pi$ it means there is a unique zero.

share|cite|improve this answer

edited Mar 18 at 15:17

answered Mar 18 at 15:11

ConradConrad

1,32745

$endgroup$

add a comment | 

1

$begingroup$

The theorem you want is on page 9 - the extension of the analytic argument principle to sense preserving harmonic ones; then you just apply it to $f-a$ for any $a$ not in the image of the Jordan curve $f(mathbb{T})$ since then $f-a$ satisfies the same properties as $f$ (non-vanishing homeomorphism on the boundary etc) and then by the argument principle and discretness of $2pi N$, $f-a$ has the same number of zeros inside $mathbb{D}$ when $a$ belongs to the same connected component of $C- f(mathbb{T})$, so no zeros if $a$ is outside and precisely one zero if $a$ is inside since otherwise $f$ cannot be a homeomorphism at the boundary.

There are two subtle points here, both implied by the sense-preserving property of $f$, first being that zeros are discrete (there are harmonic functions, even harmonic polynomials like $Re(z)$, with non-discrete zero set) and second is that they all have positive index (otherwise you could have two zeros with index 1 and one with index -1, say since only zeros of analytic functions are apriori guaranteed to have positive index), so if the argument change is $2pi$ it means there is a unique zero.

share|cite|improve this answer

edited Mar 18 at 15:17

answered Mar 18 at 15:11

ConradConrad

1,32745

$endgroup$

add a comment | 

$begingroup$

The theorem you want is on page 9 - the extension of the analytic argument principle to sense preserving harmonic ones; then you just apply it to $f-a$ for any $a$ not in the image of the Jordan curve $f(mathbb{T})$ since then $f-a$ satisfies the same properties as $f$ (non-vanishing homeomorphism on the boundary etc) and then by the argument principle and discretness of $2pi N$, $f-a$ has the same number of zeros inside $mathbb{D}$ when $a$ belongs to the same connected component of $C- f(mathbb{T})$, so no zeros if $a$ is outside and precisely one zero if $a$ is inside since otherwise $f$ cannot be a homeomorphism at the boundary.

There are two subtle points here, both implied by the sense-preserving property of $f$, first being that zeros are discrete (there are harmonic functions, even harmonic polynomials like $Re(z)$, with non-discrete zero set) and second is that they all have positive index (otherwise you could have two zeros with index 1 and one with index -1, say since only zeros of analytic functions are apriori guaranteed to have positive index), so if the argument change is $2pi$ it means there is a unique zero.

share|cite|improve this answer

edited Mar 18 at 15:17

answered Mar 18 at 15:11

ConradConrad

1,32745

$endgroup$

share|cite|improve this answer

edited Mar 18 at 15:17

answered Mar 18 at 15:11

ConradConrad

1,32745

answered Mar 18 at 15:11

ConradConrad

1,32745

answered Mar 18 at 15:11

ConradConrad

1,32745