Mixing
Injective function on unit disc
$begingroup$
I'm looking for verification of my proof of the following statement:
Let $f colon mathbb{D} to mathbb{C}$ be an injective and holomorphic function on the unit disc which satisfies $f(0) = 0$ and $f'(0) = 1$. Show that
$$inf{vert w vert : w notin f(mathbb{D})} leq 1$$
with equality if and only if $f(z) = z$ for all $z in mathbb{D}$.
The given values of $f$ and $f'$ are hypotheses of the Schwarz lemma, so I tried to use that. If the claimed inequality is false or we have equality, then $f(mathbb{D})$ contains $mathbb{D}$, so we may set $U = f^{-1}(mathbb{D})$. Since $f$ is injective, $f$ has an inverse $g = f^{-1}$ on $U$. This gives us a holomorphic function $g colon mathbb{D} to U$ whose image $U$ lies in $mathbb{D}$. Moreover, since $f(0) = 0$ and $f'(0) = 1$, we have
$$g(0) = 0, g'(0) = frac{1}{f'(g(0))} = 1.$$
Therefore the Schwarz lemma says that $g(z) = alpha z$ with $vert alpha vert = 1$. By analytic continuation, the inverse of $f$ on $f(mathbb{D})$ must also be $alpha z$, thus $f(z) = alpha^{-1}z$, and $f'(z) = alpha^{-1}$, which forces $alpha = 1$. Equality is clearly satisfied for $f(z) = z$.
complex-analysis proof-verification
asked Jan 16 at 5:13
Ethan AlwaiseEthan Alwaise
6,193617
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add a comment |
$begingroup$
I'm looking for verification of my proof of the following statement:
Let $f colon mathbb{D} to mathbb{C}$ be an injective and holomorphic function on the unit disc which satisfies $f(0) = 0$ and $f'(0) = 1$. Show that
$$inf{vert w vert : w notin f(mathbb{D})} leq 1$$
with equality if and only if $f(z) = z$ for all $z in mathbb{D}$.
The given values of $f$ and $f'$ are hypotheses of the Schwarz lemma, so I tried to use that. If the claimed inequality is false or we have equality, then $f(mathbb{D})$ contains $mathbb{D}$, so we may set $U = f^{-1}(mathbb{D})$. Since $f$ is injective, $f$ has an inverse $g = f^{-1}$ on $U$. This gives us a holomorphic function $g colon mathbb{D} to U$ whose image $U$ lies in $mathbb{D}$. Moreover, since $f(0) = 0$ and $f'(0) = 1$, we have
$$g(0) = 0, g'(0) = frac{1}{f'(g(0))} = 1.$$
Therefore the Schwarz lemma says that $g(z) = alpha z$ with $vert alpha vert = 1$. By analytic continuation, the inverse of $f$ on $f(mathbb{D})$ must also be $alpha z$, thus $f(z) = alpha^{-1}z$, and $f'(z) = alpha^{-1}$, which forces $alpha = 1$. Equality is clearly satisfied for $f(z) = z$.
complex-analysis proof-verification
asked Jan 16 at 5:13
Ethan AlwaiseEthan Alwaise
6,193617
$endgroup$
1
$begingroup$
What is $w$? Does $w = f(z)$? Cheers!
$endgroup$
– Robert Lewis
Jan 16 at 5:24
$begingroup$
$w$ is a complex number not in the image of $f$. I edited the question.
$endgroup$
– Ethan Alwaise
Jan 16 at 5:31
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@EthanAlwaise Your argument looks fine.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 5:38
$begingroup$
That infimum is also $ge frac 14$, that is the contents of the en.wikipedia.org/wiki/Koebe_quarter_theorem.
$endgroup$
– Martin R
Jan 16 at 10:09
add a comment |
$begingroup$
I'm looking for verification of my proof of the following statement:
Let $f colon mathbb{D} to mathbb{C}$ be an injective and holomorphic function on the unit disc which satisfies $f(0) = 0$ and $f'(0) = 1$. Show that
$$inf{vert w vert : w notin f(mathbb{D})} leq 1$$
with equality if and only if $f(z) = z$ for all $z in mathbb{D}$.
The given values of $f$ and $f'$ are hypotheses of the Schwarz lemma, so I tried to use that. If the claimed inequality is false or we have equality, then $f(mathbb{D})$ contains $mathbb{D}$, so we may set $U = f^{-1}(mathbb{D})$. Since $f$ is injective, $f$ has an inverse $g = f^{-1}$ on $U$. This gives us a holomorphic function $g colon mathbb{D} to U$ whose image $U$ lies in $mathbb{D}$. Moreover, since $f(0) = 0$ and $f'(0) = 1$, we have
$$g(0) = 0, g'(0) = frac{1}{f'(g(0))} = 1.$$
Therefore the Schwarz lemma says that $g(z) = alpha z$ with $vert alpha vert = 1$. By analytic continuation, the inverse of $f$ on $f(mathbb{D})$ must also be $alpha z$, thus $f(z) = alpha^{-1}z$, and $f'(z) = alpha^{-1}$, which forces $alpha = 1$. Equality is clearly satisfied for $f(z) = z$.
complex-analysis proof-verification
asked Jan 16 at 5:13
Ethan AlwaiseEthan Alwaise
6,193617
$endgroup$
I'm looking for verification of my proof of the following statement:
Let $f colon mathbb{D} to mathbb{C}$ be an injective and holomorphic function on the unit disc which satisfies $f(0) = 0$ and $f'(0) = 1$. Show that
$$inf{vert w vert : w notin f(mathbb{D})} leq 1$$
with equality if and only if $f(z) = z$ for all $z in mathbb{D}$.
The given values of $f$ and $f'$ are hypotheses of the Schwarz lemma, so I tried to use that. If the claimed inequality is false or we have equality, then $f(mathbb{D})$ contains $mathbb{D}$, so we may set $U = f^{-1}(mathbb{D})$. Since $f$ is injective, $f$ has an inverse $g = f^{-1}$ on $U$. This gives us a holomorphic function $g colon mathbb{D} to U$ whose image $U$ lies in $mathbb{D}$. Moreover, since $f(0) = 0$ and $f'(0) = 1$, we have
$$g(0) = 0, g'(0) = frac{1}{f'(g(0))} = 1.$$
Therefore the Schwarz lemma says that $g(z) = alpha z$ with $vert alpha vert = 1$. By analytic continuation, the inverse of $f$ on $f(mathbb{D})$ must also be $alpha z$, thus $f(z) = alpha^{-1}z$, and $f'(z) = alpha^{-1}$, which forces $alpha = 1$. Equality is clearly satisfied for $f(z) = z$.
complex-analysis proof-verification
complex-analysis proof-verification
asked Jan 16 at 5:13
Ethan AlwaiseEthan Alwaise
6,193617
asked Jan 16 at 5:13
Ethan AlwaiseEthan Alwaise
6,193617
edited Jan 16 at 5:31
Ethan Alwaise
asked Jan 16 at 5:13
Ethan AlwaiseEthan Alwaise
6,193617
asked Jan 16 at 5:13
Ethan AlwaiseEthan Alwaise
6,193617
asked Jan 16 at 5:13
Ethan AlwaiseEthan Alwaise
6,193617
6,193617
1
$begingroup$
What is $w$? Does $w = f(z)$? Cheers!
$endgroup$
– Robert Lewis
Jan 16 at 5:24
$begingroup$
$w$ is a complex number not in the image of $f$. I edited the question.
$endgroup$
– Ethan Alwaise
Jan 16 at 5:31
$begingroup$
@EthanAlwaise Your argument looks fine.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 5:38
$begingroup$
That infimum is also $ge frac 14$, that is the contents of the en.wikipedia.org/wiki/Koebe_quarter_theorem.
$endgroup$
– Martin R
Jan 16 at 10:09
add a comment |
1
$begingroup$
What is $w$? Does $w = f(z)$? Cheers!
$endgroup$
– Robert Lewis
Jan 16 at 5:24
$begingroup$
$w$ is a complex number not in the image of $f$. I edited the question.
$endgroup$
– Ethan Alwaise
Jan 16 at 5:31
$begingroup$
@EthanAlwaise Your argument looks fine.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 5:38
$begingroup$
That infimum is also $ge frac 14$, that is the contents of the en.wikipedia.org/wiki/Koebe_quarter_theorem.
$endgroup$
– Martin R
Jan 16 at 10:09
1
1
$begingroup$
What is $w$? Does $w = f(z)$? Cheers!
$endgroup$
– Robert Lewis
Jan 16 at 5:24
$begingroup$
What is $w$? Does $w = f(z)$? Cheers!
$endgroup$
– Robert Lewis
Jan 16 at 5:24
$begingroup$
$w$ is a complex number not in the image of $f$. I edited the question.
$endgroup$
– Ethan Alwaise
Jan 16 at 5:31
$begingroup$
$w$ is a complex number not in the image of $f$. I edited the question.
$endgroup$
– Ethan Alwaise
Jan 16 at 5:31
$begingroup$
@EthanAlwaise Your argument looks fine.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 5:38
$begingroup$
@EthanAlwaise Your argument looks fine.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 5:38
$begingroup$
That infimum is also $ge frac 14$, that is the contents of the en.wikipedia.org/wiki/Koebe_quarter_theorem.
$endgroup$
– Martin R
Jan 16 at 10:09
$begingroup$
That infimum is also $ge frac 14$, that is the contents of the en.wikipedia.org/wiki/Koebe_quarter_theorem.
$endgroup$
– Martin R
Jan 16 at 10:09
add a comment |
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1
$begingroup$
What is $w$? Does $w = f(z)$? Cheers!
$endgroup$
– Robert Lewis
Jan 16 at 5:24
$begingroup$
$w$ is a complex number not in the image of $f$. I edited the question.
$endgroup$
– Ethan Alwaise
Jan 16 at 5:31
$begingroup$
@EthanAlwaise Your argument looks fine.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 5:38
$begingroup$
That infimum is also $ge frac 14$, that is the contents of the en.wikipedia.org/wiki/Koebe_quarter_theorem.
$endgroup$
– Martin R
Jan 16 at 10:09