1. MEAN VALUE PROPERTY AND THE MAXIMUM PRINCIPLE AND THEIR CONNECTION, SCHWARZ LEMMA

1.1           Mean Value Property

Motivation.  If is an analytic function,  then it can be represented as a  power series on a disk of convergence D(0,r). In addition, we know by Cauchy’s Integral Formula that, in particular, or  by parametrization which states that the value of  f  at 0 is the mean (average) value of  f  on a circle centered at 0 with radius r.  More generally, it follows that if S is a closed disk contained in an open set D where f is holomorphic, the value of  f at the center of S is the mean of the values of  f on the circumference of S (considering the more general power series ).

Definition.  A complex-valued function defined and continuous on an open subset D of C is said to have the Mean Value Property (MVP) if for each compact disk S contained in D, the value of f at the center of S is the mean value of f on the circumference of S.

Remark.  If f has the MVP, then (where and are complex constants) have the MVP.

1.2          Maximum Modulus Property

Definition.  Let D be an open subset of C and let   Let f  be a real-valued function defined and continuous on D.  We say that f  has a relative maximum at if such that and .

Definition.  .  Let D be an open subset of C and let f be a complex-valued function defined and continuous on D.  We say that f has the Maximum Modulus Property (MMP) on D if and we have

is attained at a point of is constant on (I)

Equivalently (exercise), f has the MMP on D if we have:

 

 |f| has a relative maximum at ais constant in a neighborhood of a                        (II)

1.3  Connection between MVP and MMP

Theorem 1.  Suppose that  f  is defined, continuous, and has the MVP on an open subset D of C.  Then f has the MMP on D.

Proof.  We apply (II).  First, if f is constant in a neighbourhood of , then has a relative maximum at .  Then there exists such that and   We need to show that f is constant on   If , then on and in particular on Suppose then that .  Define a function g on D by

Note that and g has the MVP (since f does).  Now consider the function on D. Then h has the MVP on D.  Note that and .  So    Thus   Let and .  Since f  has the MVP on D, we get 

But .  It follows that or

So   Now or .  Thus and therefore ;  that is,   Consequently, .

Theorem 2.  Let D be a bounded domain in C and let f be defined and continuous on .  Suppose that f has the MVP on D and let .

If M is attained at a point in D, then f is constant on D (and hence on since f is continuous on D).

Proof.  Suppose that M is attained at a point in D; that is, for .

Put .  Since .  Moreover, if , then with and so for , .  Thus has a relative maximum at w.  Since f  has the MVP on D, f has the MMP on D.  It follows that f must be a constant in a neighbourhood V of w.  Thus . So   Therefore E is open in D.  Since D is connected, E=D.  Consequently, constant on D.

1.3          Maximum Principle

Theorem 3. (Maximum Principle). Suppose is a non-constant  analytic function in an open subset of . Then the function has no maximum in .

Proof. Suppose Then there is a disk contained in the image of But in this disk there are points whose moduli are Consequently, is not the maximum of

The Maximum Principle is sometimes stated as:

Theorem 4. (Another formulation of the Maximum Principle). Suppose that is analytic on a closed and bounded set . Then the maximum of is attained on the boundary of

Proof. Since is a closed and bounded subset of is compact. Thus attains its maximum on If is constant, there is nothing to prove. We can now assume that   is not a constant function. Suppose, to get a contradiction, that the maximum is attained at a point inside Then there is inside such that is maximum. Thus there is a disk in such that is the maximum among all the values , with  in this disk. But this is only possible if is constant. This is the desired contradiction.

1.4          Minimum Principle

Theorem 5.  (Minimum Principle)  Let f(z)  be analytic inside and on a simple closed curve C.  If inside C, then |f(z)| assumes its minimum value on C.

Proof.  Since f(z) is analytic inside and on C and since , it follows that is analytic inside C.  By the Maximum Principle cannot assume its maximum inside C and so |f(z)| cannot assume its minimum value inside C.  Since |f(z)| has a minimum, this minimum must be attained on C.

1.5          Schwarz Lemma

Theorem 6. (Schwarz Lemma). Suppose   is analytic for , and Then where equality holds only when for some constant C whose absolute value is 1.

Proof. It is natural to define the function

if  z≠ 0 and f’(0) if z=0. 

Since is analytic for . Let On the circle,   By the maximum Principle, for Letting for all

Suppose that equality holds; that is suppose that there is a such that Then Let By the first part, for all that is,  for all   which means that the maximum is attained at By the Maximum Principle must be constant; that is, there exists a constant C such that which means that Moreover, since