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Show that there is a constant in an analytic function.

schwarz lemma questiona problem relate to analytic functionIf $|f|$ is constant, so is $f$ for $f$ analytic on a domain $D$.Show with maximum principle: $|f|$ constant, f analytic, then $f$ constant in open domain $D$Bound for analytic function on a disk given values at 0Existence of analytic function on a unit dsic (Converse of Schwaraz Pick Lemma )Conway, showing analytic function is constantAnalytic function satisfying given conditions is constantanalytic function on the upper half plane that maps to right half plane is constant.Analytic function $f$ either constant or surjective on unit disk.

1

$begingroup$

Suppose that $f(z)$ is analytic for $|z<1|$ and satisfies $|f(z)|<1, f(0)=0, $and $|f'(0)<1|.$ Let $r<1$. Show that there is a constant $c<1$ sucht that $|f(z)| leq c|z|$ for $|z|leq r$.

I'm struggling with this, but this is what I have right now:

Using the Schwarz Lemma, then we can assume $z$ is analytic thus we can factor $f(z)=z cdot h(z)$, where c is analytic. Given $r<1$. If $|z|=r$, then $|h(z)|=fracf(z)r leq frac1r$. Using the maximum principle then $|h(z)| leqfrac1r forall z$ satisfying $|z| leq r$.

By letting $r rightarrow 1$, then $|h(z)|<1 forall |z|<1$. This implies that $|f(z)|=|z||h(z)| leq |z|$. Since $|f(z_o)|=|z_o|$ for some $z_o neq 0$ then $|h(z_o)|=1$ for some $z_o neq 0$ and using the strict maximum principle h(z) is constant. We can say that $h(z)= lambda$. Then $f(z)= lambda z$.

What am I missing or did I miss the picture completely? Any hints would be appreciated. I tried using the Schwarz lemma since I feel it hits most of the points.

complex-analysis

share|cite|improve this question

edited Mar 22 at 14:51

Killercamin

asked Mar 22 at 14:39

KillercaminKillercamin

4541522

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you mean $|f(z)|le c|z|$?
  $endgroup$
  – Henning Makholm
  Mar 22 at 14:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes my bad, already corrected it.
  $endgroup$
  – Killercamin
  Mar 22 at 14:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It also seems like you're using $c$ in your attempt to mean a function whereas in the problem it is a constant. That's rather confusing at best.
  $endgroup$
  – Henning Makholm
  Mar 22 at 14:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Then I would need to change c into a function and then show that the function itself is constant?
  $endgroup$
  – Killercamin
  Mar 22 at 14:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Killercamin appologies for my (stupid) comment below. I misread the question totally.
  $endgroup$
  – Matematleta
  Mar 22 at 20:50
  

  
  
  
  

 | 
show 3 more comments

1

$begingroup$

Suppose that $f(z)$ is analytic for $|z<1|$ and satisfies $|f(z)|<1, f(0)=0, $and $|f'(0)<1|.$ Let $r<1$. Show that there is a constant $c<1$ sucht that $|f(z)| leq c|z|$ for $|z|leq r$.

I'm struggling with this, but this is what I have right now:

Using the Schwarz Lemma, then we can assume $z$ is analytic thus we can factor $f(z)=z cdot h(z)$, where c is analytic. Given $r<1$. If $|z|=r$, then $|h(z)|=fracf(z)r leq frac1r$. Using the maximum principle then $|h(z)| leqfrac1r forall z$ satisfying $|z| leq r$.

By letting $r rightarrow 1$, then $|h(z)|<1 forall |z|<1$. This implies that $|f(z)|=|z||h(z)| leq |z|$. Since $|f(z_o)|=|z_o|$ for some $z_o neq 0$ then $|h(z_o)|=1$ for some $z_o neq 0$ and using the strict maximum principle h(z) is constant. We can say that $h(z)= lambda$. Then $f(z)= lambda z$.

What am I missing or did I miss the picture completely? Any hints would be appreciated. I tried using the Schwarz lemma since I feel it hits most of the points.

complex-analysis

share|cite|improve this question

edited Mar 22 at 14:51

Killercamin

asked Mar 22 at 14:39

KillercaminKillercamin

4541522

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you mean $|f(z)|le c|z|$?
  $endgroup$
  – Henning Makholm
  Mar 22 at 14:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes my bad, already corrected it.
  $endgroup$
  – Killercamin
  Mar 22 at 14:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It also seems like you're using $c$ in your attempt to mean a function whereas in the problem it is a constant. That's rather confusing at best.
  $endgroup$
  – Henning Makholm
  Mar 22 at 14:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Then I would need to change c into a function and then show that the function itself is constant?
  $endgroup$
  – Killercamin
  Mar 22 at 14:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Killercamin appologies for my (stupid) comment below. I misread the question totally.
  $endgroup$
  – Matematleta
  Mar 22 at 20:50
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

Suppose that $f(z)$ is analytic for $|z<1|$ and satisfies $|f(z)|<1, f(0)=0, $and $|f'(0)<1|.$ Let $r<1$. Show that there is a constant $c<1$ sucht that $|f(z)| leq c|z|$ for $|z|leq r$.

I'm struggling with this, but this is what I have right now:

Using the Schwarz Lemma, then we can assume $z$ is analytic thus we can factor $f(z)=z cdot h(z)$, where c is analytic. Given $r<1$. If $|z|=r$, then $|h(z)|=fracf(z)r leq frac1r$. Using the maximum principle then $|h(z)| leqfrac1r forall z$ satisfying $|z| leq r$.

By letting $r rightarrow 1$, then $|h(z)|<1 forall |z|<1$. This implies that $|f(z)|=|z||h(z)| leq |z|$. Since $|f(z_o)|=|z_o|$ for some $z_o neq 0$ then $|h(z_o)|=1$ for some $z_o neq 0$ and using the strict maximum principle h(z) is constant. We can say that $h(z)= lambda$. Then $f(z)= lambda z$.

What am I missing or did I miss the picture completely? Any hints would be appreciated. I tried using the Schwarz lemma since I feel it hits most of the points.

complex-analysis

share|cite|improve this question

edited Mar 22 at 14:51

Killercamin

asked Mar 22 at 14:39

KillercaminKillercamin

4541522

$endgroup$

share|cite|improve this question

edited Mar 22 at 14:51

Killercamin

asked Mar 22 at 14:39

KillercaminKillercamin

4541522

share|cite|improve this question

edited Mar 22 at 14:51

Killercamin

asked Mar 22 at 14:39

KillercaminKillercamin

4541522

asked Mar 22 at 14:39

KillercaminKillercamin

4541522

asked Mar 22 at 14:39

KillercaminKillercamin

4541522

1 Answer
1

active

oldest

votes

1

$begingroup$

The function $h: Bbb D to Bbb C$ defined by
$$
h(z) = begincases
dfracf(z)z & text for z ne 0 \
f'(0) & text for z = 0
endcases
$$
is holomorphic in the unit disk $Bbb D$ with $|h(z)| le 1$ according to the Schwarz Lemma.

If $h$ is constant then we are done:
$$
|f(z)| = c |z|
$$
for all $zin Bbb D$, with $c = |f'(0)| < 1$.

Now assume that $h$ is not constant, and fix $0 < r < 1$. Then
$$
c = max :
$$
must satisfy $c < 1$, because $h$ can not have a maximum in the interior of the unit disk (the maximum modulus principle). If follows that
$$
|f(z)| le c |z|
$$
for $|z| le r$.

share|cite|improve this answer

edited Mar 22 at 19:44

answered Mar 22 at 19:38

Martin RMartin R

30.8k33561

$endgroup$

add a comment | 

Your Answer

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1

$begingroup$

The function $h: Bbb D to Bbb C$ defined by
$$
h(z) = begincases
dfracf(z)z & text for z ne 0 \
f'(0) & text for z = 0
endcases
$$
is holomorphic in the unit disk $Bbb D$ with $|h(z)| le 1$ according to the Schwarz Lemma.

If $h$ is constant then we are done:
$$
|f(z)| = c |z|
$$
for all $zin Bbb D$, with $c = |f'(0)| < 1$.

Now assume that $h$ is not constant, and fix $0 < r < 1$. Then
$$
c = max :
$$
must satisfy $c < 1$, because $h$ can not have a maximum in the interior of the unit disk (the maximum modulus principle). If follows that
$$
|f(z)| le c |z|
$$
for $|z| le r$.

share|cite|improve this answer

edited Mar 22 at 19:44

answered Mar 22 at 19:38

Martin RMartin R

30.8k33561

$endgroup$

add a comment | 

1

$begingroup$

The function $h: Bbb D to Bbb C$ defined by
$$
h(z) = begincases
dfracf(z)z & text for z ne 0 \
f'(0) & text for z = 0
endcases
$$
is holomorphic in the unit disk $Bbb D$ with $|h(z)| le 1$ according to the Schwarz Lemma.

If $h$ is constant then we are done:
$$
|f(z)| = c |z|
$$
for all $zin Bbb D$, with $c = |f'(0)| < 1$.

Now assume that $h$ is not constant, and fix $0 < r < 1$. Then
$$
c = max :
$$
must satisfy $c < 1$, because $h$ can not have a maximum in the interior of the unit disk (the maximum modulus principle). If follows that
$$
|f(z)| le c |z|
$$
for $|z| le r$.

share|cite|improve this answer

edited Mar 22 at 19:44

answered Mar 22 at 19:38

Martin RMartin R

30.8k33561

$endgroup$

add a comment | 

$begingroup$

The function $h: Bbb D to Bbb C$ defined by
$$
h(z) = begincases
dfracf(z)z & text for z ne 0 \
f'(0) & text for z = 0
endcases
$$
is holomorphic in the unit disk $Bbb D$ with $|h(z)| le 1$ according to the Schwarz Lemma.

If $h$ is constant then we are done:
$$
|f(z)| = c |z|
$$
for all $zin Bbb D$, with $c = |f'(0)| < 1$.

Now assume that $h$ is not constant, and fix $0 < r < 1$. Then
$$
c = max :
$$
must satisfy $c < 1$, because $h$ can not have a maximum in the interior of the unit disk (the maximum modulus principle). If follows that
$$
|f(z)| le c |z|
$$
for $|z| le r$.

share|cite|improve this answer

edited Mar 22 at 19:44

answered Mar 22 at 19:38

Martin RMartin R

30.8k33561

$endgroup$

share|cite|improve this answer

edited Mar 22 at 19:44

answered Mar 22 at 19:38

Martin RMartin R

30.8k33561

answered Mar 22 at 19:38

Martin RMartin R

30.8k33561

answered Mar 22 at 19:38

Martin RMartin R

30.8k33561