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Find the supremum of holomorphic functions from semiplane to disk.
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Consider all the holomorphic functions $f:Pto B_1(0)$ with $f(1)=0$ where $zin P$ iff $textRe z>0$.
I want to find the supremum of $|f(2)|$ over all such $f$.
I thought about using a Möbius transformation sending $B_1(0)$ to $P$ and $0mapsto 1$ (or unrestricted) and then applying Schwarz Lemma (or Schwarz-Pick) to the composite to find some bound. Then finding a function which takes that value. But is this the right approach? I don't know if such a function exists.
complex-analysis
asked Jul 18 at 22:39
David Molano
1,363618
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up vote
0
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favorite
Consider all the holomorphic functions $f:Pto B_1(0)$ with $f(1)=0$ where $zin P$ iff $textRe z>0$.
I want to find the supremum of $|f(2)|$ over all such $f$.
I thought about using a Möbius transformation sending $B_1(0)$ to $P$ and $0mapsto 1$ (or unrestricted) and then applying Schwarz Lemma (or Schwarz-Pick) to the composite to find some bound. Then finding a function which takes that value. But is this the right approach? I don't know if such a function exists.
complex-analysis
asked Jul 18 at 22:39
David Molano
1,363618
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Consider all the holomorphic functions $f:Pto B_1(0)$ with $f(1)=0$ where $zin P$ iff $textRe z>0$.
I want to find the supremum of $|f(2)|$ over all such $f$.
I thought about using a Möbius transformation sending $B_1(0)$ to $P$ and $0mapsto 1$ (or unrestricted) and then applying Schwarz Lemma (or Schwarz-Pick) to the composite to find some bound. Then finding a function which takes that value. But is this the right approach? I don't know if such a function exists.
complex-analysis
asked Jul 18 at 22:39
David Molano
1,363618
Consider all the holomorphic functions $f:Pto B_1(0)$ with $f(1)=0$ where $zin P$ iff $textRe z>0$.
I want to find the supremum of $|f(2)|$ over all such $f$.
I thought about using a Möbius transformation sending $B_1(0)$ to $P$ and $0mapsto 1$ (or unrestricted) and then applying Schwarz Lemma (or Schwarz-Pick) to the composite to find some bound. Then finding a function which takes that value. But is this the right approach? I don't know if such a function exists.
complex-analysis
asked Jul 18 at 22:39
David Molano
1,363618
edited Jul 18 at 22:51
asked Jul 18 at 22:39
David Molano
1,363618
asked Jul 18 at 22:39
David Molano
1,363618
asked Jul 18 at 22:39
David Molano
1,363618
1,363618
add a comment |Â
add a comment |Â
1 Answer
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$tau(z) = tfrac1+z1-z$ is a biholomorphic transformation mapping $B_1(0)$ bijectively to $P$. As $tau(tfrac 1 3) = 2$, you get $|f(2)| = |f(tau(tfrac 1 3))| letfrac 1 3$ by the Schwarz lemma. Since $tau^-1 : Pto B_1(0)$ and $tau^-1(2) = tfrac 1 3$, the supremum is $tfrac 1 3$.
answered Jul 18 at 23:19
amsmath
1,613114
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$tau(z) = tfrac1+z1-z$ is a biholomorphic transformation mapping $B_1(0)$ bijectively to $P$. As $tau(tfrac 1 3) = 2$, you get $|f(2)| = |f(tau(tfrac 1 3))| letfrac 1 3$ by the Schwarz lemma. Since $tau^-1 : Pto B_1(0)$ and $tau^-1(2) = tfrac 1 3$, the supremum is $tfrac 1 3$.
answered Jul 18 at 23:19
amsmath
1,613114
add a comment |Â
up vote
1
down vote
accepted
$tau(z) = tfrac1+z1-z$ is a biholomorphic transformation mapping $B_1(0)$ bijectively to $P$. As $tau(tfrac 1 3) = 2$, you get $|f(2)| = |f(tau(tfrac 1 3))| letfrac 1 3$ by the Schwarz lemma. Since $tau^-1 : Pto B_1(0)$ and $tau^-1(2) = tfrac 1 3$, the supremum is $tfrac 1 3$.
answered Jul 18 at 23:19
amsmath
1,613114
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$tau(z) = tfrac1+z1-z$ is a biholomorphic transformation mapping $B_1(0)$ bijectively to $P$. As $tau(tfrac 1 3) = 2$, you get $|f(2)| = |f(tau(tfrac 1 3))| letfrac 1 3$ by the Schwarz lemma. Since $tau^-1 : Pto B_1(0)$ and $tau^-1(2) = tfrac 1 3$, the supremum is $tfrac 1 3$.
answered Jul 18 at 23:19
amsmath
1,613114
$tau(z) = tfrac1+z1-z$ is a biholomorphic transformation mapping $B_1(0)$ bijectively to $P$. As $tau(tfrac 1 3) = 2$, you get $|f(2)| = |f(tau(tfrac 1 3))| letfrac 1 3$ by the Schwarz lemma. Since $tau^-1 : Pto B_1(0)$ and $tau^-1(2) = tfrac 1 3$, the supremum is $tfrac 1 3$.
answered Jul 18 at 23:19
amsmath
1,613114
answered Jul 18 at 23:19
amsmath
1,613114
answered Jul 18 at 23:19
amsmath
1,613114
answered Jul 18 at 23:19
amsmath
1,613114
1,613114
add a comment |Â
add a comment |Â
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