Open mapping Theorem and Rouches Theorem
$begingroup$
My question is related to the proof here: https://en.m.wikipedia.org/wiki/Open_mapping_theorem_(complex_analysis)
Consider the closed Ball $B$ with radius $d$ around $z_0$ and a holomorphic function $g(z)$ which vanishes at $z_0$. Then the closed ball is mapped to $g(B)$ which should be also closed. Let $e={rm min}_{z in partial B} |g(z)|$ be the minimal value of $|g(z)|$ on the boundary of $B$, so that the open disk $D$ at $0$ with radius $e$ is fully contained in $g(B)$, which is my assumption, since otherwise $|g(z)|$ would have a maximum inside $B$.
Now in the above link it seems to require Rouches theorem to claim that for any $win D$ there is a solution to $g(z)=w$ in $B$, but I feel like this statement is somehow trivial which it apparently is not. If $D$ is contained in $g(B)$ shouldn‘t this be automatically the case? What am I overlooking here?
So is the following argument valid?
Using the notation above: The boundary of $D$ - which has modulus $e$ - maps onto the boundary of $g^{-1}(D)$. This is because $g$ is holomorphic and by the maximum modulus principle attains its maximum modulus at the boundary of $g^{-1}(D)$, which however by construction is $e$ and therefore the boundary of $D$. Thus any value inside $g^{-1}(D)$ maps onto a value $g(z)$ with modulus $<e$ since $g$ is not constant.
The same argument is true for $B$ i.e. the maximum modulus is attained at the boundary of $B$. Since $|g(z_0)|=0$ and $|g(z)|$ does not have a local maximum inside $B$, but $|g(partial B)|geq e$ we must have $g^{-1}(D) subseteq B$ or $Dsubseteq g(B)$. (For any $phi in [0,2pi]$ and $r_1<r_2<d$ we have $left|gleft(z_0 + r_1, {rm e}^{iphi}right)right|<left|gleft(z_0 + r_2, {rm e}^{iphi}right)right|<left|gleft(z_0 + d, {rm e}^{iphi}right)right|$)
I'd be really thankful if somebody could look over this argument!
complex-analysis open-map
asked Nov 2 '18 at 0:50
DigerDiger
1,6021413
$endgroup$
|
show 1 more comment
$begingroup$
My question is related to the proof here: https://en.m.wikipedia.org/wiki/Open_mapping_theorem_(complex_analysis)
Consider the closed Ball $B$ with radius $d$ around $z_0$ and a holomorphic function $g(z)$ which vanishes at $z_0$. Then the closed ball is mapped to $g(B)$ which should be also closed. Let $e={rm min}_{z in partial B} |g(z)|$ be the minimal value of $|g(z)|$ on the boundary of $B$, so that the open disk $D$ at $0$ with radius $e$ is fully contained in $g(B)$, which is my assumption, since otherwise $|g(z)|$ would have a maximum inside $B$.
Now in the above link it seems to require Rouches theorem to claim that for any $win D$ there is a solution to $g(z)=w$ in $B$, but I feel like this statement is somehow trivial which it apparently is not. If $D$ is contained in $g(B)$ shouldn‘t this be automatically the case? What am I overlooking here?
So is the following argument valid?
Using the notation above: The boundary of $D$ - which has modulus $e$ - maps onto the boundary of $g^{-1}(D)$. This is because $g$ is holomorphic and by the maximum modulus principle attains its maximum modulus at the boundary of $g^{-1}(D)$, which however by construction is $e$ and therefore the boundary of $D$. Thus any value inside $g^{-1}(D)$ maps onto a value $g(z)$ with modulus $<e$ since $g$ is not constant.
The same argument is true for $B$ i.e. the maximum modulus is attained at the boundary of $B$. Since $|g(z_0)|=0$ and $|g(z)|$ does not have a local maximum inside $B$, but $|g(partial B)|geq e$ we must have $g^{-1}(D) subseteq B$ or $Dsubseteq g(B)$. (For any $phi in [0,2pi]$ and $r_1<r_2<d$ we have $left|gleft(z_0 + r_1, {rm e}^{iphi}right)right|<left|gleft(z_0 + r_2, {rm e}^{iphi}right)right|<left|gleft(z_0 + d, {rm e}^{iphi}right)right|$)
I'd be really thankful if somebody could look over this argument!
complex-analysis open-map
asked Nov 2 '18 at 0:50
DigerDiger
1,6021413
$endgroup$
$begingroup$
You don't automatically have the inclusion $Dsubseteq g(B)$, not without proving it.
$endgroup$
– quasi
Nov 2 '18 at 1:08
$begingroup$
So would it be also possible to proof this part with maximum modulus principle for $|g(z)|$ which is then attained at the boundary of $g^{-1}(D)$.
$endgroup$
– Diger
Nov 2 '18 at 11:41
$begingroup$
I don't see how that would immediately yield $Dsubseteq g(B)$. But feel free to post what you think is a proof, either by editing your question, or by posting an answer. That way someone (maybe not me) will be able to review your proposed argument.
$endgroup$
– quasi
Nov 2 '18 at 12:19
$begingroup$
I actually have issues to formulate it, but if the maximum is obtained at the boundary (which by definition is $e$), how can any $z$ inside $g^{-1}(D)$ attain a value $g(z)$ larger than $e$ in modulus.
$endgroup$
– Diger
Nov 2 '18 at 14:39
$begingroup$
I added a picture above. The point is that I think since B is simply connected so is g(B) and the minimal radius e defines D which then is fully contained in g(B). Also I think a boundary of B becomes a boundary of g(B).
$endgroup$
– Diger
Nov 2 '18 at 18:15
|
show 1 more comment
$begingroup$
My question is related to the proof here: https://en.m.wikipedia.org/wiki/Open_mapping_theorem_(complex_analysis)
Consider the closed Ball $B$ with radius $d$ around $z_0$ and a holomorphic function $g(z)$ which vanishes at $z_0$. Then the closed ball is mapped to $g(B)$ which should be also closed. Let $e={rm min}_{z in partial B} |g(z)|$ be the minimal value of $|g(z)|$ on the boundary of $B$, so that the open disk $D$ at $0$ with radius $e$ is fully contained in $g(B)$, which is my assumption, since otherwise $|g(z)|$ would have a maximum inside $B$.
Now in the above link it seems to require Rouches theorem to claim that for any $win D$ there is a solution to $g(z)=w$ in $B$, but I feel like this statement is somehow trivial which it apparently is not. If $D$ is contained in $g(B)$ shouldn‘t this be automatically the case? What am I overlooking here?
So is the following argument valid?
Using the notation above: The boundary of $D$ - which has modulus $e$ - maps onto the boundary of $g^{-1}(D)$. This is because $g$ is holomorphic and by the maximum modulus principle attains its maximum modulus at the boundary of $g^{-1}(D)$, which however by construction is $e$ and therefore the boundary of $D$. Thus any value inside $g^{-1}(D)$ maps onto a value $g(z)$ with modulus $<e$ since $g$ is not constant.
The same argument is true for $B$ i.e. the maximum modulus is attained at the boundary of $B$. Since $|g(z_0)|=0$ and $|g(z)|$ does not have a local maximum inside $B$, but $|g(partial B)|geq e$ we must have $g^{-1}(D) subseteq B$ or $Dsubseteq g(B)$. (For any $phi in [0,2pi]$ and $r_1<r_2<d$ we have $left|gleft(z_0 + r_1, {rm e}^{iphi}right)right|<left|gleft(z_0 + r_2, {rm e}^{iphi}right)right|<left|gleft(z_0 + d, {rm e}^{iphi}right)right|$)
I'd be really thankful if somebody could look over this argument!
complex-analysis open-map
asked Nov 2 '18 at 0:50
DigerDiger
1,6021413
$endgroup$
My question is related to the proof here: https://en.m.wikipedia.org/wiki/Open_mapping_theorem_(complex_analysis)
Consider the closed Ball $B$ with radius $d$ around $z_0$ and a holomorphic function $g(z)$ which vanishes at $z_0$. Then the closed ball is mapped to $g(B)$ which should be also closed. Let $e={rm min}_{z in partial B} |g(z)|$ be the minimal value of $|g(z)|$ on the boundary of $B$, so that the open disk $D$ at $0$ with radius $e$ is fully contained in $g(B)$, which is my assumption, since otherwise $|g(z)|$ would have a maximum inside $B$.
Now in the above link it seems to require Rouches theorem to claim that for any $win D$ there is a solution to $g(z)=w$ in $B$, but I feel like this statement is somehow trivial which it apparently is not. If $D$ is contained in $g(B)$ shouldn‘t this be automatically the case? What am I overlooking here?
So is the following argument valid?
Using the notation above: The boundary of $D$ - which has modulus $e$ - maps onto the boundary of $g^{-1}(D)$. This is because $g$ is holomorphic and by the maximum modulus principle attains its maximum modulus at the boundary of $g^{-1}(D)$, which however by construction is $e$ and therefore the boundary of $D$. Thus any value inside $g^{-1}(D)$ maps onto a value $g(z)$ with modulus $<e$ since $g$ is not constant.
The same argument is true for $B$ i.e. the maximum modulus is attained at the boundary of $B$. Since $|g(z_0)|=0$ and $|g(z)|$ does not have a local maximum inside $B$, but $|g(partial B)|geq e$ we must have $g^{-1}(D) subseteq B$ or $Dsubseteq g(B)$. (For any $phi in [0,2pi]$ and $r_1<r_2<d$ we have $left|gleft(z_0 + r_1, {rm e}^{iphi}right)right|<left|gleft(z_0 + r_2, {rm e}^{iphi}right)right|<left|gleft(z_0 + d, {rm e}^{iphi}right)right|$)
I'd be really thankful if somebody could look over this argument!
complex-analysis open-map
complex-analysis open-map
asked Nov 2 '18 at 0:50
DigerDiger
1,6021413
asked Nov 2 '18 at 0:50
DigerDiger
1,6021413
edited Dec 10 '18 at 14:48
Diger
asked Nov 2 '18 at 0:50
DigerDiger
1,6021413
asked Nov 2 '18 at 0:50
DigerDiger
1,6021413
asked Nov 2 '18 at 0:50
DigerDiger
1,6021413
1,6021413
$begingroup$
You don't automatically have the inclusion $Dsubseteq g(B)$, not without proving it.
$endgroup$
– quasi
Nov 2 '18 at 1:08
$begingroup$
So would it be also possible to proof this part with maximum modulus principle for $|g(z)|$ which is then attained at the boundary of $g^{-1}(D)$.
$endgroup$
– Diger
Nov 2 '18 at 11:41
$begingroup$
I don't see how that would immediately yield $Dsubseteq g(B)$. But feel free to post what you think is a proof, either by editing your question, or by posting an answer. That way someone (maybe not me) will be able to review your proposed argument.
$endgroup$
– quasi
Nov 2 '18 at 12:19
$begingroup$
I actually have issues to formulate it, but if the maximum is obtained at the boundary (which by definition is $e$), how can any $z$ inside $g^{-1}(D)$ attain a value $g(z)$ larger than $e$ in modulus.
$endgroup$
– Diger
Nov 2 '18 at 14:39
$begingroup$
I added a picture above. The point is that I think since B is simply connected so is g(B) and the minimal radius e defines D which then is fully contained in g(B). Also I think a boundary of B becomes a boundary of g(B).
$endgroup$
– Diger
Nov 2 '18 at 18:15
|
show 1 more comment
$begingroup$
You don't automatically have the inclusion $Dsubseteq g(B)$, not without proving it.
$endgroup$
– quasi
Nov 2 '18 at 1:08
$begingroup$
So would it be also possible to proof this part with maximum modulus principle for $|g(z)|$ which is then attained at the boundary of $g^{-1}(D)$.
$endgroup$
– Diger
Nov 2 '18 at 11:41
$begingroup$
I don't see how that would immediately yield $Dsubseteq g(B)$. But feel free to post what you think is a proof, either by editing your question, or by posting an answer. That way someone (maybe not me) will be able to review your proposed argument.
$endgroup$
– quasi
Nov 2 '18 at 12:19
$begingroup$
I actually have issues to formulate it, but if the maximum is obtained at the boundary (which by definition is $e$), how can any $z$ inside $g^{-1}(D)$ attain a value $g(z)$ larger than $e$ in modulus.
$endgroup$
– Diger
Nov 2 '18 at 14:39
$begingroup$
I added a picture above. The point is that I think since B is simply connected so is g(B) and the minimal radius e defines D which then is fully contained in g(B). Also I think a boundary of B becomes a boundary of g(B).
$endgroup$
– Diger
Nov 2 '18 at 18:15
$begingroup$
You don't automatically have the inclusion $Dsubseteq g(B)$, not without proving it.
$endgroup$
– quasi
Nov 2 '18 at 1:08
$begingroup$
You don't automatically have the inclusion $Dsubseteq g(B)$, not without proving it.
$endgroup$
– quasi
Nov 2 '18 at 1:08
$begingroup$
So would it be also possible to proof this part with maximum modulus principle for $|g(z)|$ which is then attained at the boundary of $g^{-1}(D)$.
$endgroup$
– Diger
Nov 2 '18 at 11:41
$begingroup$
So would it be also possible to proof this part with maximum modulus principle for $|g(z)|$ which is then attained at the boundary of $g^{-1}(D)$.
$endgroup$
– Diger
Nov 2 '18 at 11:41
$begingroup$
I don't see how that would immediately yield $Dsubseteq g(B)$. But feel free to post what you think is a proof, either by editing your question, or by posting an answer. That way someone (maybe not me) will be able to review your proposed argument.
$endgroup$
– quasi
Nov 2 '18 at 12:19
$begingroup$
I don't see how that would immediately yield $Dsubseteq g(B)$. But feel free to post what you think is a proof, either by editing your question, or by posting an answer. That way someone (maybe not me) will be able to review your proposed argument.
$endgroup$
– quasi
Nov 2 '18 at 12:19
$begingroup$
I actually have issues to formulate it, but if the maximum is obtained at the boundary (which by definition is $e$), how can any $z$ inside $g^{-1}(D)$ attain a value $g(z)$ larger than $e$ in modulus.
$endgroup$
– Diger
Nov 2 '18 at 14:39
$begingroup$
I actually have issues to formulate it, but if the maximum is obtained at the boundary (which by definition is $e$), how can any $z$ inside $g^{-1}(D)$ attain a value $g(z)$ larger than $e$ in modulus.
$endgroup$
– Diger
Nov 2 '18 at 14:39
$begingroup$
I added a picture above. The point is that I think since B is simply connected so is g(B) and the minimal radius e defines D which then is fully contained in g(B). Also I think a boundary of B becomes a boundary of g(B).
$endgroup$
– Diger
Nov 2 '18 at 18:15
$begingroup$
I added a picture above. The point is that I think since B is simply connected so is g(B) and the minimal radius e defines D which then is fully contained in g(B). Also I think a boundary of B becomes a boundary of g(B).
$endgroup$
– Diger
Nov 2 '18 at 18:15
|
show 1 more comment
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$begingroup$
You don't automatically have the inclusion $Dsubseteq g(B)$, not without proving it.
$endgroup$
– quasi
Nov 2 '18 at 1:08
$begingroup$
So would it be also possible to proof this part with maximum modulus principle for $|g(z)|$ which is then attained at the boundary of $g^{-1}(D)$.
$endgroup$
– Diger
Nov 2 '18 at 11:41
$begingroup$
I don't see how that would immediately yield $Dsubseteq g(B)$. But feel free to post what you think is a proof, either by editing your question, or by posting an answer. That way someone (maybe not me) will be able to review your proposed argument.
$endgroup$
– quasi
Nov 2 '18 at 12:19
$begingroup$
I actually have issues to formulate it, but if the maximum is obtained at the boundary (which by definition is $e$), how can any $z$ inside $g^{-1}(D)$ attain a value $g(z)$ larger than $e$ in modulus.
$endgroup$
– Diger
Nov 2 '18 at 14:39
$begingroup$
I added a picture above. The point is that I think since B is simply connected so is g(B) and the minimal radius e defines D which then is fully contained in g(B). Also I think a boundary of B becomes a boundary of g(B).
$endgroup$
– Diger
Nov 2 '18 at 18:15