Schwarz lemma - Revision history

Vipul: 6 revisions

2008-05-18T19:18:32Z

6 revisions

← Older revision	Revision as of 19:18, 18 May 2008
(No difference)

Vipul at 19:17, 26 April 2008

2008-04-26T19:17:09Z

← Older revision		Revision as of 19:17, 26 April 2008
Line 1:		Line 1:
	{{basic fact}}		{{basic fact}}
			{{application of\|maximum modulus principle}}
	==Statement==		==Statement==

Vipul: /* Proof */

2008-04-26T18:56:06Z

Proof

← Older revision		Revision as of 18:56, 26 April 2008
Line 26:		Line 26:
	Consider the function:		Consider the function:

	<math>g(z) := \frac{f(z)}{z} (z \ne 0), \qquad, f'(0), (z = 0)</math>		<math>g(z) := \frac{f(z)}{z} (z \ne 0) \qquad, \qquad f'(0), (z = 0)</math>

	Clearly, <math>g</math> is a [[holomorphic function]] on <math>D</math>.		Clearly, <math>g</math> is a [[holomorphic function]] on <math>D</math>.

Vipul: /* Applications */

2008-04-26T18:55:37Z

Applications

← Older revision		Revision as of 18:55, 26 April 2008
Line 20:		Line 20:
	==Applications==		==Applications==

	* [[Schwarz-Pick lemma]]		* [[Schwarz-Pick lemma for disk]]

	==Proof==		==Proof==

Vipul at 18:55, 26 April 2008

2008-04-26T18:55:22Z

← Older revision		Revision as of 18:55, 26 April 2008
Line 1:		Line 1:
			{{basic fact}}
	==Statement==		==Statement==

Line 9:		Line 10:
	<math>\|f'(0)\| \le 1</math>		<math>\|f'(0)\| \le 1</math>

	Moreover, if there is any point <math>z \ne 0</math> such that <math>\|f(z)\| = \|z\|</math> or if <math>\|f'(0)\| = 1</math>, then <math>f</math> is a rotation about zero, i.e. there exists <math>\alpha \in \mathbb{C}</math> with <math>\|\alpha\| = 1</math>, such that:		Moreover, if there is any point <math>z \ne 0</math> such that <math>\|f(z)\| = \|z\|</math> OR if <math>\|f'(0)\| = 1</math>, then <math>f</math> is a rotation about zero, i.e. there exists <math>\alpha \in \mathbb{C}</math> with <math>\|\alpha\| = 1</math>, such that:

	<math>f(z) = \alpha z \ \forall \ z \in D</math>		<math>f(z) = \alpha z \ \forall \ z \in D</math>

Vipul: /* Statement */

2008-04-21T16:11:30Z

Statement

← Older revision		Revision as of 16:11, 21 April 2008
Line 9:		Line 9:
	<math>\|f'(0)\| \le 1</math>		<math>\|f'(0)\| \le 1</math>

	Moreover, if there is any point <math>z \ne 0</math> such that <math>\|f(z)\| = \|z\|</math>, then <math>f</math> is a rotation about zero, i.e. there exists <math>\alpha \in \mathbb{C}</math> with <math>\|\alpha\| = 1</math>, such that:		Moreover, if there is any point <math>z \ne 0</math> such that <math>\|f(z)\| = \|z\|</math> or if <math>\|f'(0)\| = 1</math>, then <math>f</math> is a rotation about zero, i.e. there exists <math>\alpha \in \mathbb{C}</math> with <math>\|\alpha\| = 1</math>, such that:

	<math>f(z) = \alpha z \ \forall \ z \in D</math>		<math>f(z) = \alpha z \ \forall \ z \in D</math>

Vipul: New page: ==Statement== Let $D$ denote the open unit disc. Any holomorphic map $f:D \to D$ with $f(0) = 0$ satisfies: $|f(z)| \le |z| \ \forall \ z \in D<...$

2008-04-20T23:02:06Z

New page: ==Statement== Let <math>D</math> denote the open unit disc. Any holomorphic map <math>f:D \to D</math> with <math>f(0) = 0</math> satisfies: <math>|f(z)| \le |z| \ \forall \ z \in D<...

New page

==Statement==

Let <math>D</math> denote the open unit disc. Any [[holomorphic map]] <math>f:D \to D</math> with <math>f(0) = 0</math> satisfies:

<math>|f(z)| \le |z| \ \forall \ z \in D</math>

and:

<math>|f'(0)| \le 1</math>

Moreover, if there is any point <math>z \ne 0</math> such that <math>|f(z)| = |z|</math>, then <math>f</math> is a rotation about zero, i.e. there exists <math>\alpha \in \mathbb{C}</math> with <math>|\alpha| = 1</math>, such that:

<math>f(z) = \alpha z \ \forall \ z \in D</math>

==Facts used==

* [[Maximum modulus principle]]

==Applications==

* [[Schwarz-Pick lemma]]
==Proof==

Consider the function:

<math>g(z) := \frac{f(z)}{z} (z \ne 0), \qquad, f'(0), (z = 0)</math>

Clearly, <math>g</math> is a [[holomorphic function]] on <math>D</math>.

Now, for any <math>r \in (0,1)</math>, we have:

<math>\max_{|z| = r} |g(z)| \le \frac{1}{r}</math>

Thus, by the [[maximum modulus principle]], we get:

<math>\max_{|z| \le r} |g(z)| \le \frac{1}{r}</math>

Taking the limit as <math>r \to 1</math>, we get:

<math>\max_{z \in D} |g(z)| \le 1</math>

which yields that <math>|f(z)| \le |z|</math> for all <math>z</math> and <math>|f'(0)| = 1</math>. Moreover, if <math>|g(z)| = 1</math> for any <math>z \in D</math>, then the maximum modulus principle forces <math>g</math> to be a constant function with modulus <math>1</math>.