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(1)

e

o

### 4 (1969), p. 711-744

<http://www.numdam.org/item?id=ASNSP_1969_3_23_4_711_0>

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http://www.numdam.org/

(2)

### OF MINIMAL SURFACES by

DAVID KINDERLEHRER

that the

### boundary

of a mininal surface contains an open

arc, r whose

is (n

### - I).times

differentiable with

to arc

### length (I’E C’z),

1. In this paper we determine the

of the conformal

### representation

of the surface restricted to that are V in the

of the

domain whose

is r. The

recent

of this

date from M.

### Tsnji [10]

who extended

to minimal surfaces the theorem of F. and M. Riesz

p.

In

II.

that if ~’ is

### analytic,

then the conformal

tion is

### analytically

extensible across y. To solve this

em-

his method of

the conformal

to its ana-

### lytic

extension with an

### auxiliary

function constructed as the solution to

an

differential

### eqnation. Although

this method does not

### apply

in

our case, the differential

itself does

an

role.

Criteria

of the conformal

have been

### given by

S. ITildehrandt in

### [2], [3].

In his latest paper, he has

### proved

that if I’E C" and tlie

### (n -1)

81 derivative of’ the

### tangent

vector satisfies

a Holder condition with

a, 0 a

for

then

the conformal

### representation

is of class on y. Hildebrandt’s

### proof proceeds by « straightening &#x3E;&#x3E;

the arc 7~ to a line

a suitable

### ’diffeomorphism

of Euclidean space,

the

to a

with certain

### boundary

conditions. This new

### problem

is then solved with the assistance of methods in the

of

this

seems to

that 1’E

The

### primary object

of this paper is to prove that if then the conformal

is also of class

In an

### elementary

Pervenuto alla Redazione il 14 Agosto 1969.

(3)

manner, it follows that the conformal

### representation

is of class when The

### precise

statements of these results are Theorems 4 and 5.

Motivated

### by

the conformal character of minimal

the

### proof given

here utilizes methods of function

For each

of

we

a

of a

### portion

of the surface where oue of the harmonic functions is Re = x

and 1n an odd

The

### points

of r for which 1n

### ~ 1, corresponding

in a sense to interior

branch

for a minimal

### surface,

pose serious difficulties.

such «

### boundary singular points &#x3E;&#x3E; comprise

a discrete subset of r. These considerations

### only require

that I’ has a continuous

The use of

a

to describe a

### particular problem

in the behavior of minimal surfaces has been found useful

### by

many authors. We mention

as

L. Bers

J. C. C. Nitscle

and Al. Shiffman

The

is proven

the

### special

conformal repre- sentation and

Theorem

or

### [12]). Higher

deri-

vatives are discussed

### in §

6.

Each theorem we prove has an

### analogue

for a minimal surface whose

### boundary

contains a cusp. As an

### example,

we state Theorem 4’. The me-

thods used here

### boundary

behavior when r is assumed to be differentiable at a

or when the

### tangent

to

A

r has a modulus of

m

t-1 oi

dt oo, for a 6

### &#x3E;

0.

o

We do not discuss these

### questions

here. The local character of the results insure that the conclusions are valid for minimal surfaces bounded

### by

several Jordan curves which may even have self intersections.

1.

Denote

Q’ the upper

### half ( --- ~ -+- iq plane.

Let S be a minimal

surface in three dimensional

### (x~ , x2 , x3)

space which is the conformal

of (~ under the

### triple

of harmonic functions

which are continuous in

the closure of

= ± oo,

### Suppose

that the curve as ; r = r

- oo

oo, the boun-

### dary of ~~,

is rectifiable and that r = r

is a

monotone map of

oo,

onto

In the conformal

### representation, r (C)

satisfies the isother-

(4)

mal relation

### Here,

we have

the notations a. b =

a.

=

= a. a for =

The

### subacripta \$, *

denote

differentation with

Let

= Xj

the harmonic

of Xj

and

to the results

the

are bounded

functions in (~

continuous boun-

values

on

### Im ~ = 0.

Their derivatives

to the

class

### H’

of the upper half This

there exists

an

### M,

0 oo, so that for 1

### ~ j S 3,

The isothermal relations

hold almost

on

=

null sets on

= 0

to null sets on

### 6S)

and null sets on as cor-

to null sets on

= 0. In

The

### assumption

that as be rectifiable is a local one in this sense: if the

C of -1

1 under

is

then there is a

(~ whose

under r =

is bounded

a

curve of finite

for

Lemma 1.1 in

2.

### Geometric Considerations.

Let 8 be a minimal surface with rectifiable

as.

### Suppose

that 8S contains the arc

(5)

where

are

continuous

a. e.

in

=

=

3. Here a)

is a modulus of

at t = 0

Since t = c

is a monotone

### mapping,

we may assume that 1’ is the

of -1 1

### (- l ) and xi (1 )

= B.

It follows where is the arc

### length

of I in S measured

-

In this

### section,

the behavior of S on I’ will be utilized to

conformal

### representation

for a subset of S. The

### proofs

of

Theorem 1 and the lemmas which

it are

### only

technical modifications of those

2 of

are

here for

### completeness.

LEMMA 2.1 : There is an

### absolutely

continuous function such that

=

- B

and

(1(0

a. e. - B x1

### B,

where c is a constant

which

on

### 1’,

PROOF: With the notations introduced

a.e. in

### (I by (3),

so that

It follows that the inverse to Xi

oil

is

### absolutely

continuous. Since

is also

H

=

is

continuous in

From

and

where c

on r,

### Using

the isothernal relations once more,

By modulus of continuity at t == 0 we understand a non-decreasing continuous

function co (t) with w (0) .= 0.

(6)

Statement

follows upon the

### integration

of H’ from 0 to a

and

### applying

the estimate above.

LEMMA 2.2 : Recall that

= xi

= 0. Then

lim

= 0

lim

_ ~c, where

denotes the

number in

x

=

2 2

## /

Arg

is

for -

0 and

in-

for 0

where

0 is

small.

PROOF : For - 1

i H

Hence lim

the

lemma.

B XI / -o-

The same

in the

--&#x3E; 0+.

For -1

set R

=

H

an

is

in

is

for

is

for

0

some

### \$0 .

LEMMA. 2.3 : The function

### Fi (C)

has an isolated zero

= 0.

The idea of this

### proof

is due to Donald Sarason. It derives from a well known theorem. If a function in the

space H 1 of the

### disc

1

is real valued almost

on an arc y

then it can be

### analytically

continued across y,

PROOF OF LEMMA.. Let o = z

### (C)

map f~ onto the unit z = x

and set g

=

Lemma

for some

0 and a-,

(7)

Here

has the same

as in Lemma 2.2.

denote the

### bounded,

harmonic function

1 with

values

Lemma 2.2

### (i),

is continuous at z = l. Let

### cp. (z)

de

note the harmonic

The real

of the ana-

function h

= is

1.

Hence

to the

### Hardy

class H I so that the function g

h

to H I.

for almost every

lim g

It

=

, ,

r - I

real number.

the

### analytic

extension theorem for H’ functions

may be

continued

Since h

is zero

### free, z

= 1 is an isolated zero

### Therefore (

= 0 is an isolated zero of

### F1 (C).

THEOREM 1 : Let S be a minimal surface with rectifiable

as.

### Suppose

that aS contains the arc

where

are

continuous

ro

a. e in

vi :- co

x,

### 1),

a. e in

Then there is a subdomain U of G bounded

of

-1

### C ~

1 and a Jordan arc a

### joining b

to a in G such that

### log F1

is a

univalent map of U onto a

of bounded

### argument

in the Riemann surface of

### log (xi -~- ixt ).

The notations of the theorem and its

are

1. The

function co

is a modulus of

### continuity

at t = 0.

(8)

PROOF. We may assume that x,

is

in -1

0 so

= 0 is the

zero of

in

and so

is

in

and

in

This is

Lemmas

### 2.2,

2.3.

Choose a semicircle 0 0 ~ ;r, with r such that

For 0

### 1 L,

it will be shown that the set

is a

arc,

in G.

and strict

of

there are

a 0 and b

0 on a.

of

a n G

### ~ 0. Since (~) ~ I

is not constant in

a n G is

### locally

one or

several branches of

### analytic

curves. No more than one branch passes

each

of a.

that

at a

of a and

### traversing

a

in one direction necessitates

a

### point

of a more than once.

Then a would

### sepa-rate G

into at least two

On the

### boundary

of at least one of these

### components,

the harmonic function

### log ~ (f)) I

would be constant. Hence

### Pt

would be constant in

=

### 0,

a does not contain the

L

### ] ~,,

a does not intersect the semicircle

0

Hence a c

at the

and

### point

a. Hence a is a Jordan arc from b to a.

E a is a

= 0. Then in a

of

### this (,

It follows that more than one branch of a passes

But this can-

occur.

a is an

arc in G.

For a

0

### 2"’

2 choose ro so small that

and ro r 1’1. Let a = 0153¡ ,,"here A = mill

Â.

. Put ro =

### b,

let a ~ - ro be the terminus of and denote

U the

### 8imply

connected subdomaill of C bounded and the

### segment

12. Ânnali delta Scuola Norm. Sup.. · PiBa.

(9)

Since

### F1 (~)

does not vanish in

a

valued branch of

U may be defined

that

=

i

g to the maximum

p 1’ dit

E ai n

### G,

where it denotes the outward normal derivative on

with s as

### arc-length

on measured from its

hand end

d

the

Y

q

pY

arg

0

E aA. Hence ds g

arg

is

### strictly increasing

on (Xl.

We now show that arg

is finite for

E

For a fixed

E

0 so that the

### segment

Since the total variation of

on

to

axes

is bounded

and since

is not zero in

oo. Hence

for a

E

there is a

odd

1n, in view of

### (6),

such

that

It also follows that

is continuous on

0. Given

6

choose an

-

### ~’ C

e and a box B witli

e, for

0 so that

Since

is in the

space

(10)

-

o,

This

that

holds for

E

with m

dent

E. D.

the

### assumptions

and notations of the

determine

a branch of

The

of U under the

=

### F,

is a domain in the

z = x

### + iy plane.

COROLLARY 1: With the

of Theorem

### 1,

there is a subdo-

main U of C and a

odd

such that z =

### Fi

is a uni.

tn

-

valent map onto S =

denote the

of

n U under

this

is a

### simple

rectifiable arc, null sets on

### correspond

to null sets on y, null sets on 7

to null sets on

and

PROOF :

we have the

for suitable

### b’,

b". Hence the strict

in

n U in-

sures that y is

4

which

the

### correspondence

between null sets.

### Rectifiability of y

is

shown with Lemma 2.1

and

### (7)

is shown with Lemma 2.2

### (i).

COROLLARY 2: With the

of Theorem

The arc

### a~

is rectifiable for almost

and

(11)

For almost

0

, there is a

( oo such that

PROOF: For

### (i), let 99 (z)

denote the inverse to

For each

### ~01~ _ ~ (b) 111m,

let ce- denote the arc in lie Then the

of

_ ~

A =

### e- ,

satisfies

Hence

where A denotes the area of T7 c G. Since the

### integrand I§ / o

is finite al- most

is finite almost

0

R.

To prove

### (ii)

we utilize two well known theorems. Let E be the set of e, 0 ~ e ~

such that

### j = 1; 2, 3,

and the limits are not zero

= 1.

Since y is

the set of

g not

a. has measure

zero.

### By

a Theorem of Lindelof

p.

preserves

at each

where y has a

### tangent.

It follows that the set

n

for a suitable 6

0

### depending

on 11, is contained in a sector with vertex at ae for almost every 0 s Lo

### R.

The same is true for almost

### every bg , According

to a well known Theorem

p.

a function h of

### Hardy

class satisfies lim h

= h

### (E) ( 0, oo) uniformly

in any sector with vertex

for almost every

Hence the

### complement

of E has measure zero.

(12)

, since

### Fl’ (C)

is known not to vanish in U.

3.

### Estimations.

Limitations for the conformal

determined

### in §

2 are pro- vided in this

In wha,t

### follows,

let S be a minimal surface with

a rectifiable

### boundary

which contains the arc r described

For con-

### venience,

we suppose that the modulus of

c~

### (t)

satisfies a) 2- 5.

With m the

### positive integer

and U the subdomain of ( determined in Theo- 5n

rem

we set z = x

=

S~ =

y the

of

and

=

3. Since 1n is

### positive, Q

does not inter- sect the

### negative imaginary

axis. We assume that D is bounded

and

an arc

= R

with R~

### satisfying

the conclusions of

2.

Note that

= zm .

LEMMA 3.1 : Set

= 4

4w

with w

~ 2-5.

= x

### (o) + iy (o),

where a denotes arc

on y, Then

There is a

function It

### (x)

so that y : z = x

ih

and

For t

and fixed

### z E y,

determine a branch of the

so

that t - z =

y 2

2 for a fixed t

Re

0.

Then cos 20

and

where c

### depends only

Then COB 29 &#x3E;

### 2 and I

dr 1 2 for t where 1

on r.

For

E Q there is a

### path

C from z to z’ in j such that

(13)

and there is a

### path

Z from 0 to z in S~ such that

This Lemma is proven in a,n

manner. To

use the

LEMMA 3.2 :

be

in S and

### satisfy

Then there is a subdomain !J’ - D n

and a

### &#x3E;

0 such that

PROOF. Ijet z = x

and choose p

### &#x3E;

0 so small that z

E ~.

Then z

### Q ; hence,

4

Choose a branch of

-

so that t -- z =

### reiÐ,

t E y, as in LEMMA 3.1

### (iii).

Then

Hence

(14)

It is easy to see that there is an

0

so that

of (p,

R. Set c’ = max

LEMMA 3.3.

be

in S~ and

Then there is a

### subdomain Q’

such that

PRooi’ : In view of the

### previous lemma,

it suffices to show that

The domain Q’ and constant c’ will then be those of the

lem-

ma.

in

fractions and

### integrating,

Note that all the

in the

above are

### convergent.

We

4!how that those on the

### right

hand side vanish. For t = E

sin

2

Lemma 3.1

Hence

sin I

For t E

we also

2

Hence for

(15)

the functions

a. e. as 6 --~

and

since

### - 18 Y Q,

LEMMA 3.4. For

PROOF. With s as arc

### length

of S on T and a as arc

### length

on y,

Since the isothermal relations hold a. e. on y

and

a. e. on 7

a. e, on y.

### Hence,

a, e. on y

a~, e. on y

a~. e, on y.

LEMMA 3.5. There is a subdomain such that

E =

where 1~I

### &#x3E;

0 is a constant.

PROOF. We first show that is

### (essentially)

bounded on For

(16)

Since we have assnmed that «i satisfies the conclusions of

oo

E =

In view of Lemma

is

bounded on

are bounded

functions in

continuous in

Hence

Since

the above

may be in-

to

### yield

that

The Lemma follows from Lemma 3.3.

4.

### Boundary Singular Points.

Let S be a minimal surface with a rectifiable

aS. In this

### section,

we show that when the arc r c as

### (cf. (4))

has a continuous tan-

certain

defined

### below,

form a discrete subset of F.

The

### assumption

that r be smooth is

as

illustrate.

### Suppose, then,

that as contains the arc

.

where the and Let

denote the modulus of

of

in

### [- B, B],

To I’ we associate a

handed

=

ak2,

### ak3), Ic _-_ 1, 2, 3,

where is the unit

### tangent

vector to I" at P and the functions = akj

### (P)

are continuous. Note that

=

### ~k~ ,

the

Kronecker delta. For any figed P with the distance OP

### sufficiently small,

there is a subarc

### Fp C

I’ which contains the

### origin

and C1 1 functions

=

so that

(17)

The Inodulus of

### continuity

of is for a constant

of 11 for Oll

small.

for OP

the arcs

### 1’j~, P E r,

contain a coinmon subarc

with

the

matrix

=

### (P))

we determine ; conformal

=

y2

Y3

that P = r

Let =

=

where

is the harmonic

to

and

Then

The

### hypotheses

of Theorem I are fulfilled for the conformal

### representation )2 (~) at I == ~l, . Hence,

there is an open subset

of

bounded

a

of the = 0 axis

### containing ~~,

and a Jordan arc in G

### ,joining b

to c~, and an odd

### integer &#x3E;

0 such that 1t’ =

### 6~~

is a

conformal map of onto a domain

in the if, = 1£

iv

We

refer to the

### parameter If

defined in the domain

### Qp

obtained in this way

as the local

### parameter

at P. We call P a «

of S on

or

a

### singular point, »,

if 1.

In the case M.liere the

### tangent

to I’ satisfies a Holder condition at a

the notion of

### singular point

coincides with that of branch

### point

for a minimal surface

p.

and related

not

germane to the

will be

elsewhere.

In our

we shall

snppress the

of

the

### )

and other functions on P. We assume that p

### )

32 4t1’HEOREM 2. Let S be a minimal surface with rectifiable

that

### ~S

contains the C’ arc

Then the set of

### singular points

of S on r is discrete.

As an immediate

COROLLARY 3. l,et S be a minimal surface with a C’

### boundary

aS.

If S has N interior branch

and M

on

then

### M+N00

(18)

PROOF OF THEOREM. We

=

----

3. Let z

be the local

### parameter

at 0 E 1’ and in = ma be the

determined

### by

Theorem 1. Let D =

be the

domain

### of z,

7 = Yo the arc of

whose

### image

is a subarc of V, and

the

func-

tions whose real

form a conformal

### representation

of S at 0. We shall

show that for non-zero z

.~

r

is not a

### singular point.

Let w be the local

at

With

### 1np = n,

there is

a subset - which is

onto a subset

V =

c

.

The

?? = 1{’

is a

map of

onto

to

### for,j=2,3;

First we establish a limitation on

and which does

not

on P E I’ for or

small.

Lemma

there is an

0 so

for z

Since the may

trix is

### orthogonal

and lim we may choose

0 so small

_

P- o 0

that

p

with

### where c, -t&#x3E;

1 is the constant from Lemma 3.1 which

on r.

for 1’0

and z E

(19)

and

The

above and

arse valid

### except

for a set of measure

zero in a V.

Lemma 3.1

### (iv),

The remainder of the

is a

### simple

examination of the

small and z

zrn - 0. Hence a

valued

branch of

### log (z-

- may be defined for 0

with r

small. Given 8

choose

iv’ = ic

- w

in

with 0

r so that

Let C be a

### path

from a,’ to 1(’" in

### jo 1 r).

Then the varia- tion of the

of z- - on

(20)

When

i

1 Hence

arg

n.°

Lemma

the

### complement

of Q

4 8

contains a sector with vertex at

" and central

### angle q 2013

n.

4 Therefore

7 2. Therefore n = 1.

### Q.

E. D.

COROLLARY 4. Let 8 be a minimal surface whose

contains

the Ci arc

JT:

= 1f’2

X3 =

- 13

1pj

=

= =

3.

There is

0 such that

### if 0

for then P is not

a

of S on r.

Let

### .t) (z)

and be the conformal

for S at 0

### and P, OP p, respectively. Suppose

that is a conformal map

(21)

of

outo V c

### Qp.

Then for iv E ~’ and almost all w E

### 8 V,

This follows from

and

5.

the

The Holder

is shown

### initially

in the domain of the local

### parameter

with a diminished

Theorem

p.

the diminished

is

the

one.

### Using Kellogg’s

Theorem once more, we prove Holder

of the

conformal

### representation.

The main result is Theorem 4.

### Throughout

we consider a minimal surface b’ whose

contains the arc

v,here yyj

E 01,

0

and y)j

=

=

=

### 2,

3.

We assume that the Holder constant of the functions

=

### 2, 3,

the

derivatives of the functions

the arc

### rP (cf. § 4),

is the same as

the Holder constant for =

3.

As a

we refer to the

functions

gj

=

whose real

### parts

form the conformal

for S at

### (or P E r)

as « the conformal

of S at 0

### (or P)

».

Let z denote tlie local

at S~ the domain

the

### preimage

of the arc of I’ in and

### fj (z), j = 1, 2, 3,

the conformal re-

near 0. Let »1

0 be the

determined

### by

Theorem 1.

4

LEMMA 5.1. Let o denote the arc

of y. Then

PROOF. For

there is an

### absolutely

continuous such that

= x,

iH

Lemma 2.1,

=

iH

(22)

b’ x,

for suitable

b". Therefore

and the

to Theorem

the arc

s of 8 on r

satisfies

### dq 2+

0 a. e. on . It follows that

### # 0

a. e. on y. From the

.

ag

isothermal

we obtain

For this

refer to

With =

Hence

e 1

Lemma

MO 1

### 1 1

2 a. e. on y ; so there is a branch of the square root such that

LEMMA 5.2. There are e

0

### and Ao &#x3E;

0 such that when Of e and

?v

=

is a

conformal map of

onto a subset

of where r is

### independent

of P.

(23)

PROOF. We remark that for r

0 and A

For z E

r

0 so small that

Hence

### (iii)

is satisfied. Now

E

where C is the

### path

from x’ to z determined in Lemma 3.1

(24)

’fherefore 1£’

### (z)

is univalent in In

An

### analogous

conclusion holds for a set

that r = min

and let Q’ = S~

### n [ rl.

The estimate below is a well known fact in

We pre-

, ,

I

sent a

### proof’

because conclusion occurs in an unusual form. For

### efficiency,

"7e introduce the notations

LEMMA 5.3. Let ~p

iv

be bounded and

in

v

that for a fixed

and 0

1

Then

The constant c

### depends 6,

q, gj - e2 .

PROOF. With « chosen so that 0 1

0

for 0

### 0

S 7l. Therefore the harmonic function ic

6 = is

_

### ’0. By hypothesis,

13. Anllali della Scuola Norm. Sup. - Pi8a.

(25)

this

with

### respect

to the Poisson kernal of D

we find that

E J

D

### (et , q)

contains the disc

Since

Green’s

### Theorem,

we have that

The same estimate holds for

and

C3

--

there is a

method for

that

99 E

pp.

### 51-53), since C

is restricted to a sector a

direct

is

J

e2 ,

let L : t

-

0

be the line

### segment

between them. Since t E arg

-

=-

### = 9 s a -

6. We define a branch of

### (t -

Holder continuous in

-

for a fixed t.

4

(26)

### Here, Op

is the Holder constant of

-

### ).

The lemma follows.

IJEMMA 5.4. Let

map

onto lii’ =

with

the

of

### [-1,1 J.

Then F

and F-I are Holder continuous.

PROOF.

### Ilere, r

is the radius from Lemma 5.2. This lemma is a con-

sequence of Warschawski

and

Lemma

### 1,

once a certain chord-arc

### length

ratio for aQ" is established. We must show that there is a constant

~ &#x3E; 0

### satisfyng’

where Jo denotes the

### length

of the shorter arc of OQ’

and x’.

at

0 E r,

For two

on the

### arc I z

= rt the condition

is obvious.

let z E

### z’ !

= r, and a be the end

### point

of y n ii’ closest to z’. Then the

### length

.L of zaz’ in satisfies

Hence

LEMMA 5.5. Let

=

E

and

= F

Let v = w

= v’

### (e)

denote the Holder

of

### F I D(/,

the restriction of

(27)

F to and Then for 1 there is such that

PROOF. We note first that the existence of a

to y at z = 0

that F

is conformal

= 0

### by

a theorem of Lindelof. Conse-

the

of

= F-’

### (z)

contains a sector

An

### analogous

statement holds for the set =

A arg z

### ~ -f - 1,

We derive an estimate for the local

of S.

With o as arc

on y, .

, dx

Since

0 a. e, on r,

### by

Lemmas 3.1 and 3.4.

### Therefore,

with d as the Holder constant of the

to

### F,

(28)

Hence

This estimate also holds if a =1.

Let w denote the local

at P = r

be the conformal

of 8 at P. Since

that is

ro

does not

on

### P,

the conclusion of the

above may

be

At this

the

### analysis

is transformed to D. Set

and

a

=

### 2,

3. This estimate is

### easily

extended to the semi-

e1 ,

0 since

is bounded.

for a

So

Lemma

E

~02, ~ -

=

for

0

Qi ’ y and any

z

a. We

g2 . e1.

= F-I

where r

= P.

In view of

=

a. e. for 0

1.

and Lemma

5.2

### (ii),

we obtain that for 0 S

### ep s o,

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