Let $f$ be analytic on $mathbb{D}$, $f(0)=-1$ and $|1+f(z)| < 1+|f(z)|$. Prove $|f'(0)| leq 4$.
$begingroup$
I am trying to solve the following problem:
Let $f$ be analytic on $mathbb{D} = {z in mathbb{C}: |z|<1}$, $f(0)=-1$ and $|1+f(z)| < 1+|f(z)|$ for $|z|<1$. Prove $|f'(0)| leq 4$.
I have an idea of how to do it, but I am stuck at one of the steps.
I'd like to use Schwarz' lemma. So I wish to find a function $h = g circ f, $ such that $h(0) = 0$, and where $g: f(mathbb{D}) rightarrow mathbb{D}$. Then $h$ would be a function from $mathbb{D}$ to $mathbb{D}$, and Schwarz' lemma would apply, so that $|h'(0)| = |g'(f(0))f'(0)| leq 1.$ If I have then managed to find a $g$ such that $g'(-1) = 1/4$, I believe I would be done with the proof.
Right now, however, I am stuck on what $f(mathbb{D})$ is. I suppose I should use the assumption that $|1+f(z)| < 1 + |f(z)|$. Simply by drawing a picture I've been able to conclude that if $Im(z) = 0$, then $Re(z)$ cannot be in $left[0, inftyright].$ But that's all I have at this point...
complex-analysis inequality
asked Dec 15 '18 at 17:04
j.eeej.eee
787
$endgroup$
add a comment |
$begingroup$
I am trying to solve the following problem:
Let $f$ be analytic on $mathbb{D} = {z in mathbb{C}: |z|<1}$, $f(0)=-1$ and $|1+f(z)| < 1+|f(z)|$ for $|z|<1$. Prove $|f'(0)| leq 4$.
I have an idea of how to do it, but I am stuck at one of the steps.
I'd like to use Schwarz' lemma. So I wish to find a function $h = g circ f, $ such that $h(0) = 0$, and where $g: f(mathbb{D}) rightarrow mathbb{D}$. Then $h$ would be a function from $mathbb{D}$ to $mathbb{D}$, and Schwarz' lemma would apply, so that $|h'(0)| = |g'(f(0))f'(0)| leq 1.$ If I have then managed to find a $g$ such that $g'(-1) = 1/4$, I believe I would be done with the proof.
Right now, however, I am stuck on what $f(mathbb{D})$ is. I suppose I should use the assumption that $|1+f(z)| < 1 + |f(z)|$. Simply by drawing a picture I've been able to conclude that if $Im(z) = 0$, then $Re(z)$ cannot be in $left[0, inftyright].$ But that's all I have at this point...
complex-analysis inequality
asked Dec 15 '18 at 17:04
j.eeej.eee
787
$endgroup$
add a comment |
$begingroup$
I am trying to solve the following problem:
Let $f$ be analytic on $mathbb{D} = {z in mathbb{C}: |z|<1}$, $f(0)=-1$ and $|1+f(z)| < 1+|f(z)|$ for $|z|<1$. Prove $|f'(0)| leq 4$.
I have an idea of how to do it, but I am stuck at one of the steps.
I'd like to use Schwarz' lemma. So I wish to find a function $h = g circ f, $ such that $h(0) = 0$, and where $g: f(mathbb{D}) rightarrow mathbb{D}$. Then $h$ would be a function from $mathbb{D}$ to $mathbb{D}$, and Schwarz' lemma would apply, so that $|h'(0)| = |g'(f(0))f'(0)| leq 1.$ If I have then managed to find a $g$ such that $g'(-1) = 1/4$, I believe I would be done with the proof.
Right now, however, I am stuck on what $f(mathbb{D})$ is. I suppose I should use the assumption that $|1+f(z)| < 1 + |f(z)|$. Simply by drawing a picture I've been able to conclude that if $Im(z) = 0$, then $Re(z)$ cannot be in $left[0, inftyright].$ But that's all I have at this point...
complex-analysis inequality
asked Dec 15 '18 at 17:04
j.eeej.eee
787
$endgroup$
I am trying to solve the following problem:
Let $f$ be analytic on $mathbb{D} = {z in mathbb{C}: |z|<1}$, $f(0)=-1$ and $|1+f(z)| < 1+|f(z)|$ for $|z|<1$. Prove $|f'(0)| leq 4$.
I have an idea of how to do it, but I am stuck at one of the steps.
I'd like to use Schwarz' lemma. So I wish to find a function $h = g circ f, $ such that $h(0) = 0$, and where $g: f(mathbb{D}) rightarrow mathbb{D}$. Then $h$ would be a function from $mathbb{D}$ to $mathbb{D}$, and Schwarz' lemma would apply, so that $|h'(0)| = |g'(f(0))f'(0)| leq 1.$ If I have then managed to find a $g$ such that $g'(-1) = 1/4$, I believe I would be done with the proof.
Right now, however, I am stuck on what $f(mathbb{D})$ is. I suppose I should use the assumption that $|1+f(z)| < 1 + |f(z)|$. Simply by drawing a picture I've been able to conclude that if $Im(z) = 0$, then $Re(z)$ cannot be in $left[0, inftyright].$ But that's all I have at this point...
complex-analysis inequality
complex-analysis inequality
asked Dec 15 '18 at 17:04
j.eeej.eee
787
asked Dec 15 '18 at 17:04
j.eeej.eee
787
edited Dec 15 '18 at 17:14
j.eee
asked Dec 15 '18 at 17:04
j.eeej.eee
787
asked Dec 15 '18 at 17:04
j.eeej.eee
787
asked Dec 15 '18 at 17:04
j.eeej.eee
787
787
add a comment |
add a comment |
1 Answer
1
active
oldest
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$begingroup$
$|1+f(z)| < 1+|f(z)|$ means that equality does not hold in the triangle inequality $$|1+f(z)| le 1+|f(z)|, ,$$ and that means that $f(z)$ is not a non-negative real multiple of $1$. In other words:
$$
f(z) in Bbb C setminus [0, infty) , .
$$
It is a bit more convenient to consider
$$
-f(z) in Bbb C setminus (-infty, 0] = U , .
$$
The “main branch” of the square root maps the slit domain $U$ conformally to the right-half plain, and that is mapped conformally to the unit disk with the Möbius transformation $T(w) = frac{w-1}{w+1}$.
Then
$$
h(z) = frac{sqrt{-f(z)}-1}{sqrt{-f(z)}+1}
$$
maps $Bbb D$ into $Bbb D$ with $h(0) = 0$, and the Schwarz Lemma can be applied: $|h'(0)| le 1$.
Reversing the compositions we get
$$
f(z) = - left( frac{1+h(z)}{1-h(z)} right)^2 \
f'(z) = - 4 frac{1+h(z)}{(1-h(z))^3} h'(z) \
f'(0) = -4 h'(0)
$$
and the desired conclusion $| f'(0) | le 4$ follows. Equality holds if and only if
$$
f(z) = - left( frac{1+lambda z}{1-lambda z} right)^2 \
$$
for some $lambda in Bbb C$ with $|lambda| = 1$.
answered Dec 15 '18 at 17:21
Martin RMartin R
28.8k33356
$endgroup$
1
$begingroup$
non-negative real multiple
$endgroup$
– mathworker21
Dec 15 '18 at 17:28
$begingroup$
@mathworker21: Yes, thanks!
$endgroup$
– Martin R
Dec 15 '18 at 17:28
add a comment |
Your Answer
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1 Answer
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$begingroup$
$|1+f(z)| < 1+|f(z)|$ means that equality does not hold in the triangle inequality $$|1+f(z)| le 1+|f(z)|, ,$$ and that means that $f(z)$ is not a non-negative real multiple of $1$. In other words:
$$
f(z) in Bbb C setminus [0, infty) , .
$$
It is a bit more convenient to consider
$$
-f(z) in Bbb C setminus (-infty, 0] = U , .
$$
The “main branch” of the square root maps the slit domain $U$ conformally to the right-half plain, and that is mapped conformally to the unit disk with the Möbius transformation $T(w) = frac{w-1}{w+1}$.
Then
$$
h(z) = frac{sqrt{-f(z)}-1}{sqrt{-f(z)}+1}
$$
maps $Bbb D$ into $Bbb D$ with $h(0) = 0$, and the Schwarz Lemma can be applied: $|h'(0)| le 1$.
Reversing the compositions we get
$$
f(z) = - left( frac{1+h(z)}{1-h(z)} right)^2 \
f'(z) = - 4 frac{1+h(z)}{(1-h(z))^3} h'(z) \
f'(0) = -4 h'(0)
$$
and the desired conclusion $| f'(0) | le 4$ follows. Equality holds if and only if
$$
f(z) = - left( frac{1+lambda z}{1-lambda z} right)^2 \
$$
for some $lambda in Bbb C$ with $|lambda| = 1$.
answered Dec 15 '18 at 17:21
Martin RMartin R
28.8k33356
$endgroup$
1
$begingroup$
non-negative real multiple
$endgroup$
– mathworker21
Dec 15 '18 at 17:28
$begingroup$
@mathworker21: Yes, thanks!
$endgroup$
– Martin R
Dec 15 '18 at 17:28
add a comment |
$begingroup$
$|1+f(z)| < 1+|f(z)|$ means that equality does not hold in the triangle inequality $$|1+f(z)| le 1+|f(z)|, ,$$ and that means that $f(z)$ is not a non-negative real multiple of $1$. In other words:
$$
f(z) in Bbb C setminus [0, infty) , .
$$
It is a bit more convenient to consider
$$
-f(z) in Bbb C setminus (-infty, 0] = U , .
$$
The “main branch” of the square root maps the slit domain $U$ conformally to the right-half plain, and that is mapped conformally to the unit disk with the Möbius transformation $T(w) = frac{w-1}{w+1}$.
Then
$$
h(z) = frac{sqrt{-f(z)}-1}{sqrt{-f(z)}+1}
$$
maps $Bbb D$ into $Bbb D$ with $h(0) = 0$, and the Schwarz Lemma can be applied: $|h'(0)| le 1$.
Reversing the compositions we get
$$
f(z) = - left( frac{1+h(z)}{1-h(z)} right)^2 \
f'(z) = - 4 frac{1+h(z)}{(1-h(z))^3} h'(z) \
f'(0) = -4 h'(0)
$$
and the desired conclusion $| f'(0) | le 4$ follows. Equality holds if and only if
$$
f(z) = - left( frac{1+lambda z}{1-lambda z} right)^2 \
$$
for some $lambda in Bbb C$ with $|lambda| = 1$.
answered Dec 15 '18 at 17:21
Martin RMartin R
28.8k33356
$endgroup$
1
$begingroup$
non-negative real multiple
$endgroup$
– mathworker21
Dec 15 '18 at 17:28
$begingroup$
@mathworker21: Yes, thanks!
$endgroup$
– Martin R
Dec 15 '18 at 17:28
add a comment |
$begingroup$
$|1+f(z)| < 1+|f(z)|$ means that equality does not hold in the triangle inequality $$|1+f(z)| le 1+|f(z)|, ,$$ and that means that $f(z)$ is not a non-negative real multiple of $1$. In other words:
$$
f(z) in Bbb C setminus [0, infty) , .
$$
It is a bit more convenient to consider
$$
-f(z) in Bbb C setminus (-infty, 0] = U , .
$$
The “main branch” of the square root maps the slit domain $U$ conformally to the right-half plain, and that is mapped conformally to the unit disk with the Möbius transformation $T(w) = frac{w-1}{w+1}$.
Then
$$
h(z) = frac{sqrt{-f(z)}-1}{sqrt{-f(z)}+1}
$$
maps $Bbb D$ into $Bbb D$ with $h(0) = 0$, and the Schwarz Lemma can be applied: $|h'(0)| le 1$.
Reversing the compositions we get
$$
f(z) = - left( frac{1+h(z)}{1-h(z)} right)^2 \
f'(z) = - 4 frac{1+h(z)}{(1-h(z))^3} h'(z) \
f'(0) = -4 h'(0)
$$
and the desired conclusion $| f'(0) | le 4$ follows. Equality holds if and only if
$$
f(z) = - left( frac{1+lambda z}{1-lambda z} right)^2 \
$$
for some $lambda in Bbb C$ with $|lambda| = 1$.
answered Dec 15 '18 at 17:21
Martin RMartin R
28.8k33356
$endgroup$
$|1+f(z)| < 1+|f(z)|$ means that equality does not hold in the triangle inequality $$|1+f(z)| le 1+|f(z)|, ,$$ and that means that $f(z)$ is not a non-negative real multiple of $1$. In other words:
$$
f(z) in Bbb C setminus [0, infty) , .
$$
It is a bit more convenient to consider
$$
-f(z) in Bbb C setminus (-infty, 0] = U , .
$$
The “main branch” of the square root maps the slit domain $U$ conformally to the right-half plain, and that is mapped conformally to the unit disk with the Möbius transformation $T(w) = frac{w-1}{w+1}$.
Then
$$
h(z) = frac{sqrt{-f(z)}-1}{sqrt{-f(z)}+1}
$$
maps $Bbb D$ into $Bbb D$ with $h(0) = 0$, and the Schwarz Lemma can be applied: $|h'(0)| le 1$.
Reversing the compositions we get
$$
f(z) = - left( frac{1+h(z)}{1-h(z)} right)^2 \
f'(z) = - 4 frac{1+h(z)}{(1-h(z))^3} h'(z) \
f'(0) = -4 h'(0)
$$
and the desired conclusion $| f'(0) | le 4$ follows. Equality holds if and only if
$$
f(z) = - left( frac{1+lambda z}{1-lambda z} right)^2 \
$$
for some $lambda in Bbb C$ with $|lambda| = 1$.
answered Dec 15 '18 at 17:21
Martin RMartin R
28.8k33356
edited Dec 15 '18 at 18:21
answered Dec 15 '18 at 17:21
Martin RMartin R
28.8k33356
answered Dec 15 '18 at 17:21
Martin RMartin R
28.8k33356
answered Dec 15 '18 at 17:21
Martin RMartin R
28.8k33356
28.8k33356
1
$begingroup$
non-negative real multiple
$endgroup$
– mathworker21
Dec 15 '18 at 17:28
$begingroup$
@mathworker21: Yes, thanks!
$endgroup$
– Martin R
Dec 15 '18 at 17:28
add a comment |
1
$begingroup$
non-negative real multiple
$endgroup$
– mathworker21
Dec 15 '18 at 17:28
$begingroup$
@mathworker21: Yes, thanks!
$endgroup$
– Martin R
Dec 15 '18 at 17:28
1
1
$begingroup$
non-negative real multiple
$endgroup$
– mathworker21
Dec 15 '18 at 17:28
$begingroup$
non-negative real multiple
$endgroup$
– mathworker21
Dec 15 '18 at 17:28
$begingroup$
@mathworker21: Yes, thanks!
$endgroup$
– Martin R
Dec 15 '18 at 17:28
$begingroup$
@mathworker21: Yes, thanks!
$endgroup$
– Martin R
Dec 15 '18 at 17:28
add a comment |
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