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
complexvalued 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 realvalued 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 complexvalued 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 nonconstant 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
