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  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.

1.4Minimum 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.

1.5Schwarz 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