Proof of Schwarz lemma 5
$begingroup$
I dont understand the following proof of Schwarz lemma:
Schwarz Lemma
In the last section it says, "Moreover, suppose that $|f(z)| = |z|$ for some non-zero $z$ in $D$, or $|f'(0)| = 1$. Then, $|g(z)| = 1$ at some point of $D$".
Why does $g(z)$ have a local maximum then?
maxima-minima
edited Dec 12 '18 at 11:28
amWhy
1
asked Dec 12 '18 at 10:51
Steven33Steven33
275
$endgroup$
add a comment |
$begingroup$
I dont understand the following proof of Schwarz lemma:
Schwarz Lemma
In the last section it says, "Moreover, suppose that $|f(z)| = |z|$ for some non-zero $z$ in $D$, or $|f'(0)| = 1$. Then, $|g(z)| = 1$ at some point of $D$".
Why does $g(z)$ have a local maximum then?
maxima-minima
edited Dec 12 '18 at 11:28
amWhy
1
asked Dec 12 '18 at 10:51
Steven33Steven33
275
$endgroup$
$begingroup$
There is a proof following it, what is the question?
$endgroup$
– Hawk
Dec 12 '18 at 11:52
add a comment |
$begingroup$
I dont understand the following proof of Schwarz lemma:
Schwarz Lemma
In the last section it says, "Moreover, suppose that $|f(z)| = |z|$ for some non-zero $z$ in $D$, or $|f'(0)| = 1$. Then, $|g(z)| = 1$ at some point of $D$".
Why does $g(z)$ have a local maximum then?
maxima-minima
edited Dec 12 '18 at 11:28
amWhy
1
asked Dec 12 '18 at 10:51
Steven33Steven33
275
$endgroup$
I dont understand the following proof of Schwarz lemma:
Schwarz Lemma
In the last section it says, "Moreover, suppose that $|f(z)| = |z|$ for some non-zero $z$ in $D$, or $|f'(0)| = 1$. Then, $|g(z)| = 1$ at some point of $D$".
Why does $g(z)$ have a local maximum then?
maxima-minima
maxima-minima
edited Dec 12 '18 at 11:28
amWhy
1
asked Dec 12 '18 at 10:51
Steven33Steven33
275
edited Dec 12 '18 at 11:28
amWhy
1
asked Dec 12 '18 at 10:51
Steven33Steven33
275
edited Dec 12 '18 at 11:28
amWhy
1
edited Dec 12 '18 at 11:28
amWhy
1
edited Dec 12 '18 at 11:28
amWhy
1
1
asked Dec 12 '18 at 10:51
Steven33Steven33
275
asked Dec 12 '18 at 10:51
Steven33Steven33
275
asked Dec 12 '18 at 10:51
Steven33Steven33
275
275
$begingroup$
There is a proof following it, what is the question?
$endgroup$
– Hawk
Dec 12 '18 at 11:52
add a comment |
$begingroup$
There is a proof following it, what is the question?
$endgroup$
– Hawk
Dec 12 '18 at 11:52
$begingroup$
There is a proof following it, what is the question?
$endgroup$
– Hawk
Dec 12 '18 at 11:52
$begingroup$
There is a proof following it, what is the question?
$endgroup$
– Hawk
Dec 12 '18 at 11:52
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Why does $g$ have a local maximum? We cannot even ask that question because $g$ is complex valued. Maximum Modulus Theorem says if $|g|$ attains its maximum at an interior point of the domain then $g$ is a constant.
It is already proved that $|g(z)|leq 1$. So $|g(z)|=1$ for some $z$ implies that $|g|$ attains its maximum at that point. This implies that $g$ is a constant $c$ so $f(z)=cz$.
answered Dec 12 '18 at 11:44
Kavi Rama MurthyKavi Rama Murthy
59.5k42161
$endgroup$
$begingroup$
I wanted to make sure that that you fullfil all requirements to apply the Maximum Modulus Theorem. |g| =1 is a local maximum, because we proved $|g(z)| leq 1$. Thank you very much :)
$endgroup$
– Steven33
Dec 12 '18 at 12:10
$begingroup$
Sorry. I have another question. Why does |c| have to be 1?
$endgroup$
– Steven33
Dec 12 '18 at 15:31
$begingroup$
@Steven33 Just apply the hypothesis: either $|f(z)|=|z|$ (with $ zneq 0$) or $|f'(0)|=1$. In either case you get $|c|=1$.
$endgroup$
– Kavi Rama Murthy
Dec 12 '18 at 23:12
$begingroup$
It is $ g(z)=c$ Then we get $ f'(0) =c $ and$ f(z)=cz. $. When you compare this expressions c must be 1?
$endgroup$
– Steven33
Dec 13 '18 at 8:59
add a comment |
$begingroup$
On that page, $D$ is defined to be the open complex unit disc: $D={zin {mathbb C}: |z|<1}$ and $g$ is defined as
$$left{ g = begin{eqnarray} frac{f(z)}{z} quad mbox{ if } f(z) not= 0 \ f'(0) quad mbox{ otherwise } end{eqnarray} right. $$
So by definition, either $|g(z)| = |f(z)/z| = |f(z)|/|z| = 1$ (since $|f(z)| = |z|$) or $|g(z)| = |f'(0)| = 1$.
Note that this isn't a local maximum really -- by the maximum modulus principle $g$ is now constant on $D$.
answered Dec 12 '18 at 11:43
postmortespostmortes
1,93621117
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Why does $g$ have a local maximum? We cannot even ask that question because $g$ is complex valued. Maximum Modulus Theorem says if $|g|$ attains its maximum at an interior point of the domain then $g$ is a constant.
It is already proved that $|g(z)|leq 1$. So $|g(z)|=1$ for some $z$ implies that $|g|$ attains its maximum at that point. This implies that $g$ is a constant $c$ so $f(z)=cz$.
answered Dec 12 '18 at 11:44
Kavi Rama MurthyKavi Rama Murthy
59.5k42161
$endgroup$
$begingroup$
I wanted to make sure that that you fullfil all requirements to apply the Maximum Modulus Theorem. |g| =1 is a local maximum, because we proved $|g(z)| leq 1$. Thank you very much :)
$endgroup$
– Steven33
Dec 12 '18 at 12:10
$begingroup$
Sorry. I have another question. Why does |c| have to be 1?
$endgroup$
– Steven33
Dec 12 '18 at 15:31
$begingroup$
@Steven33 Just apply the hypothesis: either $|f(z)|=|z|$ (with $ zneq 0$) or $|f'(0)|=1$. In either case you get $|c|=1$.
$endgroup$
– Kavi Rama Murthy
Dec 12 '18 at 23:12
$begingroup$
It is $ g(z)=c$ Then we get $ f'(0) =c $ and$ f(z)=cz. $. When you compare this expressions c must be 1?
$endgroup$
– Steven33
Dec 13 '18 at 8:59
add a comment |
$begingroup$
Why does $g$ have a local maximum? We cannot even ask that question because $g$ is complex valued. Maximum Modulus Theorem says if $|g|$ attains its maximum at an interior point of the domain then $g$ is a constant.
It is already proved that $|g(z)|leq 1$. So $|g(z)|=1$ for some $z$ implies that $|g|$ attains its maximum at that point. This implies that $g$ is a constant $c$ so $f(z)=cz$.
answered Dec 12 '18 at 11:44
Kavi Rama MurthyKavi Rama Murthy
59.5k42161
$endgroup$
$begingroup$
I wanted to make sure that that you fullfil all requirements to apply the Maximum Modulus Theorem. |g| =1 is a local maximum, because we proved $|g(z)| leq 1$. Thank you very much :)
$endgroup$
– Steven33
Dec 12 '18 at 12:10
$begingroup$
Sorry. I have another question. Why does |c| have to be 1?
$endgroup$
– Steven33
Dec 12 '18 at 15:31
$begingroup$
@Steven33 Just apply the hypothesis: either $|f(z)|=|z|$ (with $ zneq 0$) or $|f'(0)|=1$. In either case you get $|c|=1$.
$endgroup$
– Kavi Rama Murthy
Dec 12 '18 at 23:12
$begingroup$
It is $ g(z)=c$ Then we get $ f'(0) =c $ and$ f(z)=cz. $. When you compare this expressions c must be 1?
$endgroup$
– Steven33
Dec 13 '18 at 8:59
add a comment |
$begingroup$
Why does $g$ have a local maximum? We cannot even ask that question because $g$ is complex valued. Maximum Modulus Theorem says if $|g|$ attains its maximum at an interior point of the domain then $g$ is a constant.
It is already proved that $|g(z)|leq 1$. So $|g(z)|=1$ for some $z$ implies that $|g|$ attains its maximum at that point. This implies that $g$ is a constant $c$ so $f(z)=cz$.
answered Dec 12 '18 at 11:44
Kavi Rama MurthyKavi Rama Murthy
59.5k42161
$endgroup$
Why does $g$ have a local maximum? We cannot even ask that question because $g$ is complex valued. Maximum Modulus Theorem says if $|g|$ attains its maximum at an interior point of the domain then $g$ is a constant.
It is already proved that $|g(z)|leq 1$. So $|g(z)|=1$ for some $z$ implies that $|g|$ attains its maximum at that point. This implies that $g$ is a constant $c$ so $f(z)=cz$.
answered Dec 12 '18 at 11:44
Kavi Rama MurthyKavi Rama Murthy
59.5k42161
edited Dec 12 '18 at 11:50
answered Dec 12 '18 at 11:44
Kavi Rama MurthyKavi Rama Murthy
59.5k42161
answered Dec 12 '18 at 11:44
Kavi Rama MurthyKavi Rama Murthy
59.5k42161
answered Dec 12 '18 at 11:44
Kavi Rama MurthyKavi Rama Murthy
59.5k42161
59.5k42161
$begingroup$
I wanted to make sure that that you fullfil all requirements to apply the Maximum Modulus Theorem. |g| =1 is a local maximum, because we proved $|g(z)| leq 1$. Thank you very much :)
$endgroup$
– Steven33
Dec 12 '18 at 12:10
$begingroup$
Sorry. I have another question. Why does |c| have to be 1?
$endgroup$
– Steven33
Dec 12 '18 at 15:31
$begingroup$
@Steven33 Just apply the hypothesis: either $|f(z)|=|z|$ (with $ zneq 0$) or $|f'(0)|=1$. In either case you get $|c|=1$.
$endgroup$
– Kavi Rama Murthy
Dec 12 '18 at 23:12
$begingroup$
It is $ g(z)=c$ Then we get $ f'(0) =c $ and$ f(z)=cz. $. When you compare this expressions c must be 1?
$endgroup$
– Steven33
Dec 13 '18 at 8:59
add a comment |
$begingroup$
I wanted to make sure that that you fullfil all requirements to apply the Maximum Modulus Theorem. |g| =1 is a local maximum, because we proved $|g(z)| leq 1$. Thank you very much :)
$endgroup$
– Steven33
Dec 12 '18 at 12:10
$begingroup$
Sorry. I have another question. Why does |c| have to be 1?
$endgroup$
– Steven33
Dec 12 '18 at 15:31
$begingroup$
@Steven33 Just apply the hypothesis: either $|f(z)|=|z|$ (with $ zneq 0$) or $|f'(0)|=1$. In either case you get $|c|=1$.
$endgroup$
– Kavi Rama Murthy
Dec 12 '18 at 23:12
$begingroup$
It is $ g(z)=c$ Then we get $ f'(0) =c $ and$ f(z)=cz. $. When you compare this expressions c must be 1?
$endgroup$
– Steven33
Dec 13 '18 at 8:59
$begingroup$
I wanted to make sure that that you fullfil all requirements to apply the Maximum Modulus Theorem. |g| =1 is a local maximum, because we proved $|g(z)| leq 1$. Thank you very much :)
$endgroup$
– Steven33
Dec 12 '18 at 12:10
$begingroup$
I wanted to make sure that that you fullfil all requirements to apply the Maximum Modulus Theorem. |g| =1 is a local maximum, because we proved $|g(z)| leq 1$. Thank you very much :)
$endgroup$
– Steven33
Dec 12 '18 at 12:10
$begingroup$
Sorry. I have another question. Why does |c| have to be 1?
$endgroup$
– Steven33
Dec 12 '18 at 15:31
$begingroup$
Sorry. I have another question. Why does |c| have to be 1?
$endgroup$
– Steven33
Dec 12 '18 at 15:31
$begingroup$
@Steven33 Just apply the hypothesis: either $|f(z)|=|z|$ (with $ zneq 0$) or $|f'(0)|=1$. In either case you get $|c|=1$.
$endgroup$
– Kavi Rama Murthy
Dec 12 '18 at 23:12
$begingroup$
@Steven33 Just apply the hypothesis: either $|f(z)|=|z|$ (with $ zneq 0$) or $|f'(0)|=1$. In either case you get $|c|=1$.
$endgroup$
– Kavi Rama Murthy
Dec 12 '18 at 23:12
$begingroup$
It is $ g(z)=c$ Then we get $ f'(0) =c $ and$ f(z)=cz. $. When you compare this expressions c must be 1?
$endgroup$
– Steven33
Dec 13 '18 at 8:59
$begingroup$
It is $ g(z)=c$ Then we get $ f'(0) =c $ and$ f(z)=cz. $. When you compare this expressions c must be 1?
$endgroup$
– Steven33
Dec 13 '18 at 8:59
add a comment |
$begingroup$
On that page, $D$ is defined to be the open complex unit disc: $D={zin {mathbb C}: |z|<1}$ and $g$ is defined as
$$left{ g = begin{eqnarray} frac{f(z)}{z} quad mbox{ if } f(z) not= 0 \ f'(0) quad mbox{ otherwise } end{eqnarray} right. $$
So by definition, either $|g(z)| = |f(z)/z| = |f(z)|/|z| = 1$ (since $|f(z)| = |z|$) or $|g(z)| = |f'(0)| = 1$.
Note that this isn't a local maximum really -- by the maximum modulus principle $g$ is now constant on $D$.
answered Dec 12 '18 at 11:43
postmortespostmortes
1,93621117
$endgroup$
add a comment |
$begingroup$
On that page, $D$ is defined to be the open complex unit disc: $D={zin {mathbb C}: |z|<1}$ and $g$ is defined as
$$left{ g = begin{eqnarray} frac{f(z)}{z} quad mbox{ if } f(z) not= 0 \ f'(0) quad mbox{ otherwise } end{eqnarray} right. $$
So by definition, either $|g(z)| = |f(z)/z| = |f(z)|/|z| = 1$ (since $|f(z)| = |z|$) or $|g(z)| = |f'(0)| = 1$.
Note that this isn't a local maximum really -- by the maximum modulus principle $g$ is now constant on $D$.
answered Dec 12 '18 at 11:43
postmortespostmortes
1,93621117
$endgroup$
add a comment |
$begingroup$
On that page, $D$ is defined to be the open complex unit disc: $D={zin {mathbb C}: |z|<1}$ and $g$ is defined as
$$left{ g = begin{eqnarray} frac{f(z)}{z} quad mbox{ if } f(z) not= 0 \ f'(0) quad mbox{ otherwise } end{eqnarray} right. $$
So by definition, either $|g(z)| = |f(z)/z| = |f(z)|/|z| = 1$ (since $|f(z)| = |z|$) or $|g(z)| = |f'(0)| = 1$.
Note that this isn't a local maximum really -- by the maximum modulus principle $g$ is now constant on $D$.
answered Dec 12 '18 at 11:43
postmortespostmortes
1,93621117
$endgroup$
On that page, $D$ is defined to be the open complex unit disc: $D={zin {mathbb C}: |z|<1}$ and $g$ is defined as
$$left{ g = begin{eqnarray} frac{f(z)}{z} quad mbox{ if } f(z) not= 0 \ f'(0) quad mbox{ otherwise } end{eqnarray} right. $$
So by definition, either $|g(z)| = |f(z)/z| = |f(z)|/|z| = 1$ (since $|f(z)| = |z|$) or $|g(z)| = |f'(0)| = 1$.
Note that this isn't a local maximum really -- by the maximum modulus principle $g$ is now constant on $D$.
answered Dec 12 '18 at 11:43
postmortespostmortes
1,93621117
answered Dec 12 '18 at 11:43
postmortespostmortes
1,93621117
answered Dec 12 '18 at 11:43
postmortespostmortes
1,93621117
answered Dec 12 '18 at 11:43
postmortespostmortes
1,93621117
1,93621117
add a comment |
add a comment |
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$begingroup$
There is a proof following it, what is the question?
$endgroup$
– Hawk
Dec 12 '18 at 11:52