ma'¥

Today we cover the PflMax modulus theorem

-

Iemma:_ Suppose f isanalytic on

a region A andthat If # I

is constant onA

.

Then ftz )

is constant on A .

proof's Suppose that

f- ( xti ,) = UH, y ) tivcx ,y ) .

We are assumingthat on

A we have

Ifl'

-_ ( Tutu )! u 'tvZ=c

-

for someconstant c .

so:c:#" tax.n¥÷÷÷÷:f- =D on A ,So nowassumec

Differentiating we

get④2u¥×t2v¥=O ( * ,

Zu 't2V = O

on A .

Since f- is analytic onA

,by Cauchy -

Riemann we know f÷=Ey and Fx=

-

IF

Subbing these into( * ) and dividing Ct I

by 2 we get:

uff , - v¥y=o ( * * )v ¥×+u¥y=0

onA

.

pg( HH ) becomes ↳

Kill

:YY=Kl***For any fixed input

( x ,y ) theabove

is a linear systemwith two

equations andtwo

unknowns .

Since det ( Y-

I )=u7f= CFO .

Thus , foreach ( x. y )

there is

a unique solutionto C

Htt )

Which is f exist= }Jlx,y)=O .

Thus,

f-'

(xtiyl-IICx.yltiffcx.gl=

Oti 0=0

xtiy EA .SinceHainaut

' -0

for all domainAstonish.

pg

theorem: Suppose that f is analytic 4Lin a neighborhood D ( Zo ; E )where Zo E IC and E ? O .

If I f ( z ) ) E Ifczo ) I forall

ZED ( Zo ,-

E ),

then f is

constant on DCZOJE)

.

aDlzo

,

-

E )proof: Suppose IHZHEIHZOH ,

-

-

¥÷¥¥¥÷:c'Let p =/ Zo - Zit . - ,Let 8 , be the #circle centered

at Zo

with radius p

,oriented counter - clockwise .

By the Cauchy- integral theorem

f Hot ziti ! dz

Parameterize 8,

as pg

8pct ) = Zo t p eit

,OETEZIT

&

and then Y'ltl = iceit

.

So we get

f- Hot = ÷i §,

III. dz

* ,= S

.

"

ieeitdt

= ¥ Jj"

fczotpeit ) d t

[ This result is called Gauss's mean

value theorem ]

From CHI we set pg

Iffzosl =/ It ! Czoteeitdtl⑥

E ¥ ! IfCzotpeitldt

Since lftetpe.it//EIfCzo ) ) for all

t, by assumption , we

get

Iftzolpffzoteeitldt

Ifczo,ldt=¥GHfHolD

= If C Zo ) ) .

Thus, ¥ Iflzoteeit )/dt= If Holl .

pg⇐"Iflzoteeittldt

-1¥HzoHdt=o

-

Iffto ) I

Thus,

÷toffeedt = O70

because Ifczoll >Ifczotpeit ) )

We are integratinga

continuousfunction

that is 70and the

integral

equals O .

The only waythis can

happen is

if Iffzo ) ) - Iffzotpeit) 1=0

for all t .

So, PglIf I = Ifczo ) )

for all Z on Tp .

We can vary Z ,to get all curves

Jp inside on D ( Zo ,-

E ) .

So,

If # I = Ifczoll for

all Z E D ( Zo ,-

E ).

So,

Iflztl is constant

on D ( Zois ) .

By the lemma ,

f- is constanton

Dfzoje ).

( Mjth ) pgtheorem: Suppose that f 9L

is analytic on a domain A

and f is not constant on A .

Then f does nothave a

max

value on A .

That is , theredoes not exist

ZOE A where If# I > If # I

for all 2- EA .

ti:÷:D-

proof in Churchill / Brown

probably also in Hoffman / Maiden

pg

Defy: Let SEE with ↳S # 0 . We say that EEG

is a boundarypointof#if Zo is

not aninterior point

of Sand Zo

is notan

interiorpointoflc-S.is

,

a boundary pointis a point where

all of

boundary points

⇐ The boundaryof

ots

5={2-112-14} I

is { zlizki}h¢*

Def : Let SEE , Sto . Ps

The cofS is'll

al ( S ) = S U (boundary of S ) .

-

Theorem(Max-ModThm)

Let Abe an open ,

connected ,

boundedset in

Cl . Suppose

f- : al ( A)→ I

is analytic

onA and

continuouson

al ( A ) .

Then Hfz )I has a

maximum

Value whichlies on

theboundary

of A . That is ,I Zo

DCA )

on theboundary of A

Z°•w.ie#izit:e:o"If ( Z , )l=If(

Zo ) I where

z,

is in the interiorof A

,then Eonisfaitn!