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Lecture 1 Harmonic functions--

-

--

- -

/

Harmonic functions share many properties with holomorphicfunctions

. Cm legal formula , maximum modulus, convergence)

Riad G E G,re E 8%3

,re : G - R is harmonic if

I u = U×× t U = O

gy

RI D [ za@ A] If f : G → so holomorphic ⇒ Ref & Imf are harmonic.

- -

They satisfy CR agua teens n ~

Ux = V{ay = ,}

U 't are harmonic conjugates .

2).[ 22013] Conversely , if G is simply connected , any

harmonic function U arises this way . That is , F f : G→ a holomorphicwith U = Ref .

Remand If u is harmonic.u : G → IR

,⇒ re is to? ( Regularity) .

tooof This is a local statement. Tak a E G

,I ca

,r) EG

Suffices to show u is C-in ↳ Cair)

. By EzzoBT,U = Re f

,

f- holomorphic in D (a,-7 .

We have seen f is C't

. => UEC?

Properties of harmonic functions- - -

---

⑦ Mean Value Property Canute)

today / I② Maximum Modulus Principe Champ)

.

next time ) ③Poisson integral formula

④ Dirichlet problem ( for disc, maybe more ? )

1. Mean Value Property ( Mvp)- -

Theorem , Let an. G - IR harmonic

, D- (a. r) E G .Then- -

u satisfy- e

na. - ÷.si:c . . .⇒* .

G-

Proof j- -

Lety > m be chosen so I Cair) E D la

, g ) E G .In s cap

write u = Re f , f holomorphic in D. (9,9 ) .

By Cjauehy Integral Formula ( CIF).we have

f- Ca) = 2¥,. J f- CZ) ,

⑦ ④ I 2 = a + reit

2 D. Ca,r) I take Real part .

We have ?÷a,- = dt. then taking real part

-

2Tr

n ca) = ÷ ) nCat re 't ) dt, as needed

-

O

§

DI n : G → IR.

n e I (G). Satisfies have provided t a c- G

,

TI Ca,r) E G

.we have

u ca) = ¥ J! ucatreitsdt .

-2-

centeraverage of boundary

values

L Maximum (modulus) principle- - - -

- -

c

theorem u i s → its , u e ECG) satisfies MWE.-Assume 3- a EG such that

re Ca) Z re CZ) AZ EG.

Then re is constant.

Him , thmzRemark We have seen if u harmonic- -

=> U MV E.

⇒ u satisfies

tame.

T-

PII W LOG u (a) =o IF u ca) fo,then work instead with the function

anew= u - u ca)

. still satisfies KAVI.

WI re Eo. Let

SL = { 2- E G : vez) =o }. WINE or = G

.

a) oh =/ & .Since a E R

.

b) Is = closed in a.

because u continuous / G⇒"""EY g iz .

c) ropen in G

.

hoof of c ).Let 2- • ER . ⇒ F E (Zo ,r) EG .

- - -We show I Cfo , r) E SL

.

"t * ⇐ ⇐ "£"" £-t & = " " " ""

""s£

A = 2, tf eit

. for some to.

We know 2.E Sh =) U (2-0)=0 . By mean value

25

• = uC2o) = 2¥ f !" he 2. + g e 't ) dt . ⇒ So fat ) dt-

fct)antennaf : [ o , art] → its continuous,Fct) so 9 § fat,dt=o . Then

0

f- E. • . [ HWK, Rudin

.].

Apply this to f = UC2-• + g e 't)

.

E Uca) - o. By lemma fats Io

ht t . as fcto) = o =3 re (Zo + ye't) =o ⇒ Uca) =o => LE 02

.

T.

Remark If u has have ⇒ - n has have.

- -

→Minimum modulus property .

-

- - - -

.

.

Them n : G → IR continuous & have. .If u has a minimum in G =3 u constant '

Eine harmonische Function u kann nicht in einem Punkt im Innern einMinimum oder ein Maximum haben, wenn sie nicht uberall constant ist.

(A harmonic function u cannot have either a minimum or a maximum at aninterior point unless it is constant.)

"Grundlagen fur eine allgemeine Theorie der Functionen einer veranderlichen complexen Grosse," Dissertation (1851)

Georg Friedrich Bernhard Riemann17 September 1826 – 20 July 1866

HANNE IT Assume G E Q (bounded_= or unbounded).,u

,- : G → R

bounded continuous functions satisfying have Assume.

As a C- 2@ G.

limsup ulz) E hmm f v Cz)

.Then either

2 → a 2-→ a

• u < ~

everywhere on G

•U z re everywhere on G .

Promote Let y = u- re.

Then y satisfies have an 'd furthermore

Iimsup y Cz) = Iim sup (n - a) Cz) E din sup act) thmsupfvcz))

.

2 → a 2.→ a 2- →a 2 → a

= hmsup act) - limns f- UCHE o.

Rephrase 2 → a 2 -sa- - - limswp 9127 So Wish if Lo on a

2- → a•n else 4 I

.

Suffices to show 4 I o 'n as.( because Version I ) .- -

Assume that 3- a with Q Ca) > o .Let E - og Cx)> o .

Let k = { 2- n.GG : y (t) z s } . Since 2 E K => K =/ § .

Claim k is compact .

Then let Zo be the point where 9 achieves a maximum an K,.

'

=> 2-ois a maximum for Cf in G since in K

, Cf Z E and outside K,

y se.But the existence of a max . in G is

impossible by VersionI.

.

Proof of the claim Suffices to show k bounded d dosed in Q.

- --

nTor at an G.

- hm -up 4 Ct) ¥6 => F D Ca, ga) such that2- → a

Cf C E in DC a.ya ) .N G

. By Lebesgue coming Lemma for the covering

By Leb

S ca, ya) of 2 • G ,

we can find S > o s -t

-

a" discs D la,8) E D ca; Sj , )

.

Thus K E { 2- : d C 2,2•G) Z S ) . Then K is compact .

( You can also prove this directly w/o Lebesgue ) . pg

i Importantly.

If G is bounded, f : I → its continuous & have

.

e f zo on 28. Is f z o

.in G-

.

PIf this follows from MMR Version I. applied twice

.Let a C- 29

.

• u = f

{ O = limsup Fez ) s lrminfrrfz, " G

v =o.

= O e> U E re2- →a 2- →a VI

Es f so(ble f cent - on G-,f = o on 26

.)

•

gun g e.may, o , e.mm, µ, . .

. us ,

⇐ • .

2 → a 2- → aMens conf

⇒

-

f Zo.

Ty