Show that $Lgeq2pi|f'(0)|$
$begingroup$
Suppose $f$ is analytic on and inside the unit circle $partialmathbb{D}$. Let $L$ be the length of the curve $f(partialmathbb{D})$. Show that $Lgeq2pi|f'(0)|$.
I have tried different approaches from Schwarz Lemma, and maximum modulus principle, but got nowhere. Any guidance is appreciated.
Edit:
I think I may have found it (please confirm):
By maximum modulus principle, $f$ attains it's max value on $partialmathbb{D}$. By Cauchy Estimates:
$$begin{align*}
|f'(0)|leq|sup f|
end{align*}$$
Note that since $L=2pi |sup f|$, we conclude that
$$begin{align*}|f'(0)|&leq frac{L}{2pi}\
2pi|f'(0)|&leq L
end{align*}$$
complex-analysis
asked Dec 27 '18 at 18:28
Ya GYa G
536211
$endgroup$
|
show 1 more comment
$begingroup$
Suppose $f$ is analytic on and inside the unit circle $partialmathbb{D}$. Let $L$ be the length of the curve $f(partialmathbb{D})$. Show that $Lgeq2pi|f'(0)|$.
I have tried different approaches from Schwarz Lemma, and maximum modulus principle, but got nowhere. Any guidance is appreciated.
Edit:
I think I may have found it (please confirm):
By maximum modulus principle, $f$ attains it's max value on $partialmathbb{D}$. By Cauchy Estimates:
$$begin{align*}
|f'(0)|leq|sup f|
end{align*}$$
Note that since $L=2pi |sup f|$, we conclude that
$$begin{align*}|f'(0)|&leq frac{L}{2pi}\
2pi|f'(0)|&leq L
end{align*}$$
complex-analysis
asked Dec 27 '18 at 18:28
Ya GYa G
536211
$endgroup$
1
$begingroup$
Why does $L=2pilvertsup frvert$? What does it mean to take sup of (strictly) complex numbers?
$endgroup$
– user10354138
Dec 27 '18 at 19:17
$begingroup$
@user10354138 Circumference of a circle is defined by $2picdot r$. Here, radius is $f(z)$ where $|z|=1$. Since, by maximum modulus principle, that's where $f$ attains its maximum, i.e. $|sup f|$. I may be totally wrong.
$endgroup$
– Ya G
Dec 27 '18 at 19:20
$begingroup$
@YaG But $f(partial mathbb{D})$ need not be a circle
$endgroup$
– angryavian
Dec 27 '18 at 19:24
1
$begingroup$
@YaG Also, I think user10354138 is trying to get you notice that you probably meant $sup |f|$ rather than $|sup f|$.
$endgroup$
– angryavian
Dec 27 '18 at 19:30
1
$begingroup$
That is one aspect of it. The other part is indeed $f(partialmathbb{D})$ need not be a circle and $f$ need not be univalent, so it doesn't make sense to multiply by $2pi$ and say it is $L$.
$endgroup$
– user10354138
Dec 27 '18 at 19:33
|
show 1 more comment
$begingroup$
Suppose $f$ is analytic on and inside the unit circle $partialmathbb{D}$. Let $L$ be the length of the curve $f(partialmathbb{D})$. Show that $Lgeq2pi|f'(0)|$.
I have tried different approaches from Schwarz Lemma, and maximum modulus principle, but got nowhere. Any guidance is appreciated.
Edit:
I think I may have found it (please confirm):
By maximum modulus principle, $f$ attains it's max value on $partialmathbb{D}$. By Cauchy Estimates:
$$begin{align*}
|f'(0)|leq|sup f|
end{align*}$$
Note that since $L=2pi |sup f|$, we conclude that
$$begin{align*}|f'(0)|&leq frac{L}{2pi}\
2pi|f'(0)|&leq L
end{align*}$$
complex-analysis
asked Dec 27 '18 at 18:28
Ya GYa G
536211
$endgroup$
Suppose $f$ is analytic on and inside the unit circle $partialmathbb{D}$. Let $L$ be the length of the curve $f(partialmathbb{D})$. Show that $Lgeq2pi|f'(0)|$.
I have tried different approaches from Schwarz Lemma, and maximum modulus principle, but got nowhere. Any guidance is appreciated.
Edit:
I think I may have found it (please confirm):
By maximum modulus principle, $f$ attains it's max value on $partialmathbb{D}$. By Cauchy Estimates:
$$begin{align*}
|f'(0)|leq|sup f|
end{align*}$$
Note that since $L=2pi |sup f|$, we conclude that
$$begin{align*}|f'(0)|&leq frac{L}{2pi}\
2pi|f'(0)|&leq L
end{align*}$$
complex-analysis
complex-analysis
asked Dec 27 '18 at 18:28
Ya GYa G
536211
asked Dec 27 '18 at 18:28
Ya GYa G
536211
edited Dec 27 '18 at 18:57
Ya G
asked Dec 27 '18 at 18:28
Ya GYa G
536211
asked Dec 27 '18 at 18:28
Ya GYa G
536211
asked Dec 27 '18 at 18:28
Ya GYa G
536211
536211
1
$begingroup$
Why does $L=2pilvertsup frvert$? What does it mean to take sup of (strictly) complex numbers?
$endgroup$
– user10354138
Dec 27 '18 at 19:17
$begingroup$
@user10354138 Circumference of a circle is defined by $2picdot r$. Here, radius is $f(z)$ where $|z|=1$. Since, by maximum modulus principle, that's where $f$ attains its maximum, i.e. $|sup f|$. I may be totally wrong.
$endgroup$
– Ya G
Dec 27 '18 at 19:20
$begingroup$
@YaG But $f(partial mathbb{D})$ need not be a circle
$endgroup$
– angryavian
Dec 27 '18 at 19:24
1
$begingroup$
@YaG Also, I think user10354138 is trying to get you notice that you probably meant $sup |f|$ rather than $|sup f|$.
$endgroup$
– angryavian
Dec 27 '18 at 19:30
1
$begingroup$
That is one aspect of it. The other part is indeed $f(partialmathbb{D})$ need not be a circle and $f$ need not be univalent, so it doesn't make sense to multiply by $2pi$ and say it is $L$.
$endgroup$
– user10354138
Dec 27 '18 at 19:33
|
show 1 more comment
1
$begingroup$
Why does $L=2pilvertsup frvert$? What does it mean to take sup of (strictly) complex numbers?
$endgroup$
– user10354138
Dec 27 '18 at 19:17
$begingroup$
@user10354138 Circumference of a circle is defined by $2picdot r$. Here, radius is $f(z)$ where $|z|=1$. Since, by maximum modulus principle, that's where $f$ attains its maximum, i.e. $|sup f|$. I may be totally wrong.
$endgroup$
– Ya G
Dec 27 '18 at 19:20
$begingroup$
@YaG But $f(partial mathbb{D})$ need not be a circle
$endgroup$
– angryavian
Dec 27 '18 at 19:24
1
$begingroup$
@YaG Also, I think user10354138 is trying to get you notice that you probably meant $sup |f|$ rather than $|sup f|$.
$endgroup$
– angryavian
Dec 27 '18 at 19:30
1
$begingroup$
That is one aspect of it. The other part is indeed $f(partialmathbb{D})$ need not be a circle and $f$ need not be univalent, so it doesn't make sense to multiply by $2pi$ and say it is $L$.
$endgroup$
– user10354138
Dec 27 '18 at 19:33
1
1
$begingroup$
Why does $L=2pilvertsup frvert$? What does it mean to take sup of (strictly) complex numbers?
$endgroup$
– user10354138
Dec 27 '18 at 19:17
$begingroup$
Why does $L=2pilvertsup frvert$? What does it mean to take sup of (strictly) complex numbers?
$endgroup$
– user10354138
Dec 27 '18 at 19:17
$begingroup$
@user10354138 Circumference of a circle is defined by $2picdot r$. Here, radius is $f(z)$ where $|z|=1$. Since, by maximum modulus principle, that's where $f$ attains its maximum, i.e. $|sup f|$. I may be totally wrong.
$endgroup$
– Ya G
Dec 27 '18 at 19:20
$begingroup$
@user10354138 Circumference of a circle is defined by $2picdot r$. Here, radius is $f(z)$ where $|z|=1$. Since, by maximum modulus principle, that's where $f$ attains its maximum, i.e. $|sup f|$. I may be totally wrong.
$endgroup$
– Ya G
Dec 27 '18 at 19:20
$begingroup$
@YaG But $f(partial mathbb{D})$ need not be a circle
$endgroup$
– angryavian
Dec 27 '18 at 19:24
$begingroup$
@YaG But $f(partial mathbb{D})$ need not be a circle
$endgroup$
– angryavian
Dec 27 '18 at 19:24
1
1
$begingroup$
@YaG Also, I think user10354138 is trying to get you notice that you probably meant $sup |f|$ rather than $|sup f|$.
$endgroup$
– angryavian
Dec 27 '18 at 19:30
$begingroup$
@YaG Also, I think user10354138 is trying to get you notice that you probably meant $sup |f|$ rather than $|sup f|$.
$endgroup$
– angryavian
Dec 27 '18 at 19:30
1
1
$begingroup$
That is one aspect of it. The other part is indeed $f(partialmathbb{D})$ need not be a circle and $f$ need not be univalent, so it doesn't make sense to multiply by $2pi$ and say it is $L$.
$endgroup$
– user10354138
Dec 27 '18 at 19:33
$begingroup$
That is one aspect of it. The other part is indeed $f(partialmathbb{D})$ need not be a circle and $f$ need not be univalent, so it doesn't make sense to multiply by $2pi$ and say it is $L$.
$endgroup$
– user10354138
Dec 27 '18 at 19:33
|
show 1 more comment
1 Answer
1
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oldest
votes
$begingroup$
From definition of length of a curve and mean value property:
$$L = int_0^{2pi} |(f(e^{it}))'| dt = int_0^{2pi} |f'(e^{it})|dt geqleft|int_0^{2pi}f'(e^{it})dt right| = 2pi |f'(0)| $$
answered Dec 27 '18 at 19:34
JakobianJakobian
2,733721
$endgroup$
add a comment |
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1 Answer
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$begingroup$
From definition of length of a curve and mean value property:
$$L = int_0^{2pi} |(f(e^{it}))'| dt = int_0^{2pi} |f'(e^{it})|dt geqleft|int_0^{2pi}f'(e^{it})dt right| = 2pi |f'(0)| $$
answered Dec 27 '18 at 19:34
JakobianJakobian
2,733721
$endgroup$
add a comment |
$begingroup$
From definition of length of a curve and mean value property:
$$L = int_0^{2pi} |(f(e^{it}))'| dt = int_0^{2pi} |f'(e^{it})|dt geqleft|int_0^{2pi}f'(e^{it})dt right| = 2pi |f'(0)| $$
answered Dec 27 '18 at 19:34
JakobianJakobian
2,733721
$endgroup$
add a comment |
$begingroup$
From definition of length of a curve and mean value property:
$$L = int_0^{2pi} |(f(e^{it}))'| dt = int_0^{2pi} |f'(e^{it})|dt geqleft|int_0^{2pi}f'(e^{it})dt right| = 2pi |f'(0)| $$
answered Dec 27 '18 at 19:34
JakobianJakobian
2,733721
$endgroup$
From definition of length of a curve and mean value property:
$$L = int_0^{2pi} |(f(e^{it}))'| dt = int_0^{2pi} |f'(e^{it})|dt geqleft|int_0^{2pi}f'(e^{it})dt right| = 2pi |f'(0)| $$
answered Dec 27 '18 at 19:34
JakobianJakobian
2,733721
answered Dec 27 '18 at 19:34
JakobianJakobian
2,733721
answered Dec 27 '18 at 19:34
JakobianJakobian
2,733721
answered Dec 27 '18 at 19:34
JakobianJakobian
2,733721
2,733721
add a comment |
add a comment |
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1
$begingroup$
Why does $L=2pilvertsup frvert$? What does it mean to take sup of (strictly) complex numbers?
$endgroup$
– user10354138
Dec 27 '18 at 19:17
$begingroup$
@user10354138 Circumference of a circle is defined by $2picdot r$. Here, radius is $f(z)$ where $|z|=1$. Since, by maximum modulus principle, that's where $f$ attains its maximum, i.e. $|sup f|$. I may be totally wrong.
$endgroup$
– Ya G
Dec 27 '18 at 19:20
$begingroup$
@YaG But $f(partial mathbb{D})$ need not be a circle
$endgroup$
– angryavian
Dec 27 '18 at 19:24
1
$begingroup$
@YaG Also, I think user10354138 is trying to get you notice that you probably meant $sup |f|$ rather than $|sup f|$.
$endgroup$
– angryavian
Dec 27 '18 at 19:30
1
$begingroup$
That is one aspect of it. The other part is indeed $f(partialmathbb{D})$ need not be a circle and $f$ need not be univalent, so it doesn't make sense to multiply by $2pi$ and say it is $L$.
$endgroup$
– user10354138
Dec 27 '18 at 19:33