Complex Analysis Inequality.
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Im studying for an exam and this is a practice problem. Not sure where to start or what theorems to even start with.
if $f$ is analytic on $|z|<1$ and $|f(z)|<1$ for $|z|<1$, prove that $$left|frac{f(z)-f(0)}{z}right|leq left|1-overline{f(0)}f(z)right|,$$
for all $0<|z|<1$.
thanks!
complex-analysis
edited Jan 8 at 6:20
Riemann
3,3281321
asked Jan 8 at 4:50
Alex SampsonAlex Sampson
1
New contributor
Alex Sampson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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add a comment |
$begingroup$
Im studying for an exam and this is a practice problem. Not sure where to start or what theorems to even start with.
if $f$ is analytic on $|z|<1$ and $|f(z)|<1$ for $|z|<1$, prove that $$left|frac{f(z)-f(0)}{z}right|leq left|1-overline{f(0)}f(z)right|,$$
for all $0<|z|<1$.
thanks!
complex-analysis
edited Jan 8 at 6:20
Riemann
3,3281321
asked Jan 8 at 4:50
Alex SampsonAlex Sampson
1
New contributor
Alex Sampson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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Have you seen the Schwarz-Pick theorem?
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– lEm
Jan 8 at 5:03
$begingroup$
I have not. I just looked it up and I think the proof is clear from there. Thank you.
$endgroup$
– Alex Sampson
Jan 8 at 5:32
add a comment |
$begingroup$
Im studying for an exam and this is a practice problem. Not sure where to start or what theorems to even start with.
if $f$ is analytic on $|z|<1$ and $|f(z)|<1$ for $|z|<1$, prove that $$left|frac{f(z)-f(0)}{z}right|leq left|1-overline{f(0)}f(z)right|,$$
for all $0<|z|<1$.
thanks!
complex-analysis
edited Jan 8 at 6:20
Riemann
3,3281321
asked Jan 8 at 4:50
Alex SampsonAlex Sampson
1
New contributor
Alex Sampson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Im studying for an exam and this is a practice problem. Not sure where to start or what theorems to even start with.
if $f$ is analytic on $|z|<1$ and $|f(z)|<1$ for $|z|<1$, prove that $$left|frac{f(z)-f(0)}{z}right|leq left|1-overline{f(0)}f(z)right|,$$
for all $0<|z|<1$.
thanks!
complex-analysis
complex-analysis
edited Jan 8 at 6:20
Riemann
3,3281321
asked Jan 8 at 4:50
Alex SampsonAlex Sampson
1
New contributor
Alex Sampson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jan 8 at 6:20
Riemann
3,3281321
asked Jan 8 at 4:50
Alex SampsonAlex Sampson
1
New contributor
Alex Sampson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jan 8 at 6:20
Riemann
3,3281321
edited Jan 8 at 6:20
Riemann
3,3281321
edited Jan 8 at 6:20
Riemann
3,3281321
3,3281321
asked Jan 8 at 4:50
Alex SampsonAlex Sampson
1
New contributor
Alex Sampson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 8 at 4:50
Alex SampsonAlex Sampson
1
asked Jan 8 at 4:50
Alex SampsonAlex Sampson
1
1
New contributor
Alex Sampson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Alex Sampson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Alex Sampson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Have you seen the Schwarz-Pick theorem?
$endgroup$
– lEm
Jan 8 at 5:03
$begingroup$
I have not. I just looked it up and I think the proof is clear from there. Thank you.
$endgroup$
– Alex Sampson
Jan 8 at 5:32
add a comment |
$begingroup$
Have you seen the Schwarz-Pick theorem?
$endgroup$
– lEm
Jan 8 at 5:03
$begingroup$
I have not. I just looked it up and I think the proof is clear from there. Thank you.
$endgroup$
– Alex Sampson
Jan 8 at 5:32
$begingroup$
Have you seen the Schwarz-Pick theorem?
$endgroup$
– lEm
Jan 8 at 5:03
$begingroup$
Have you seen the Schwarz-Pick theorem?
$endgroup$
– lEm
Jan 8 at 5:03
$begingroup$
I have not. I just looked it up and I think the proof is clear from there. Thank you.
$endgroup$
– Alex Sampson
Jan 8 at 5:32
$begingroup$
I have not. I just looked it up and I think the proof is clear from there. Thank you.
$endgroup$
– Alex Sampson
Jan 8 at 5:32
add a comment |
1 Answer
1
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oldest
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Let $phi (z)=frac {z-f(0)} {1-overline {f(0)} z}$. Then $phi$ is holomorphic, maps the open unit disk into itself and $phi(f(0))=0$. Hence $phi circ f$ maps the open unit disk into itself, maps $0$ to $0$. By Schwarz Lemma $|phi circ f(z)| leq |z|$ which is precisely what you are trying to prove.
answered Jan 8 at 5:33
Kavi Rama MurthyKavi Rama Murthy
53.2k32055
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add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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$begingroup$
Let $phi (z)=frac {z-f(0)} {1-overline {f(0)} z}$. Then $phi$ is holomorphic, maps the open unit disk into itself and $phi(f(0))=0$. Hence $phi circ f$ maps the open unit disk into itself, maps $0$ to $0$. By Schwarz Lemma $|phi circ f(z)| leq |z|$ which is precisely what you are trying to prove.
answered Jan 8 at 5:33
Kavi Rama MurthyKavi Rama Murthy
53.2k32055
$endgroup$
add a comment |
$begingroup$
Let $phi (z)=frac {z-f(0)} {1-overline {f(0)} z}$. Then $phi$ is holomorphic, maps the open unit disk into itself and $phi(f(0))=0$. Hence $phi circ f$ maps the open unit disk into itself, maps $0$ to $0$. By Schwarz Lemma $|phi circ f(z)| leq |z|$ which is precisely what you are trying to prove.
answered Jan 8 at 5:33
Kavi Rama MurthyKavi Rama Murthy
53.2k32055
$endgroup$
add a comment |
$begingroup$
Let $phi (z)=frac {z-f(0)} {1-overline {f(0)} z}$. Then $phi$ is holomorphic, maps the open unit disk into itself and $phi(f(0))=0$. Hence $phi circ f$ maps the open unit disk into itself, maps $0$ to $0$. By Schwarz Lemma $|phi circ f(z)| leq |z|$ which is precisely what you are trying to prove.
answered Jan 8 at 5:33
Kavi Rama MurthyKavi Rama Murthy
53.2k32055
$endgroup$
Let $phi (z)=frac {z-f(0)} {1-overline {f(0)} z}$. Then $phi$ is holomorphic, maps the open unit disk into itself and $phi(f(0))=0$. Hence $phi circ f$ maps the open unit disk into itself, maps $0$ to $0$. By Schwarz Lemma $|phi circ f(z)| leq |z|$ which is precisely what you are trying to prove.
answered Jan 8 at 5:33
Kavi Rama MurthyKavi Rama Murthy
53.2k32055
answered Jan 8 at 5:33
Kavi Rama MurthyKavi Rama Murthy
53.2k32055
answered Jan 8 at 5:33
Kavi Rama MurthyKavi Rama Murthy
53.2k32055
answered Jan 8 at 5:33
Kavi Rama MurthyKavi Rama Murthy
53.2k32055
53.2k32055
add a comment |
add a comment |
Alex Sampson is a new contributor. Be nice, and check out our Code of Conduct.
Alex Sampson is a new contributor. Be nice, and check out our Code of Conduct.
Alex Sampson is a new contributor. Be nice, and check out our Code of Conduct.
Alex Sampson is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Have you seen the Schwarz-Pick theorem?
$endgroup$
– lEm
Jan 8 at 5:03
$begingroup$
I have not. I just looked it up and I think the proof is clear from there. Thank you.
$endgroup$
– Alex Sampson
Jan 8 at 5:32