**ARKIV FOR MATEMATIK Band 6 nr 22 **

1.65 12 1 Communicated 13 October 1965 b y L. GXI~DING and L. CARLESO~

**M i n i m i z a t i o n p r o b l e m s f o r t h e f u n c t i o n a l **

*supxF(x,f (x),f'(x)). * ^{(II) }

### By Gu~NAR AnoNsso~

**Introduction **

The present paper is a continuation of the paper [1]. This means t h a t we shall t r y to minimize the functional

### /t(/) =sup~/~(z,/(x),/'(x))

over the class :~ of all absolutely continuous functions/(x) which satisfy the boundary conditions/@1)

*=Yl *

and/(Xz) =Y2-
I n Chapter 1 we consider the question of the existence of a minimizing function, which was left open in [1].

I n Chapter 2 some of the previous results are generalized to a wider class of func- tions

*F(x,y,z): *

the previous condition
~F

~z

**> 0 ** **for z > 0 **

**[!o orz> x **

**[!o orz> x**

= 0 for z = 0 is changed into ~-z 0 for z = ~o(x, y),

< 0 for z < 0 , 0 for

*z<o:,(x,y). *

An existence proof for absolutely minimizing functions is given in the general case.

I n Chapter 3 we examine further the properties of a.s. minimals. We also consider uniqueness questions for a.s. minimals and minimizing functions. For reasons of simplicity, this chapter is restricted to the case

*o(x,y)=-0. *

I n Chapter 4, finally, we give some examples in order to illustrate the previous exposition.

We shall now introduce notations for some conditions on

*F(x,y, z). *

They are as-
sumed to hold for x E J and for all real y and z. Here J is an interval which will be
stated explicitly in most cases. If it is not explicitly given, then the choice of J
will be clear from the context.
*1. F(x,y,z) *

is continuous and *~F/~z *

^{exists. }

*~F(x, y, z) *

is
2A. ~z

> 0 if

= 0 if

< 0 if z > 0 , Z ~ 0 , z < 0 .

409

c. AROrCSSO~,

*Minimization problems. I I *

2 B . There is a continuous f u n c t i o n

*co(x, y) *

such t h a t
*OF(x, y, z) *

Oz

**f! **

^{O if }

*z>o~(x,y), *

is 0 if z = o ~ ( x , y ) , 0 if

*z<co(x,y). *

3. liml~l_,~r

*F(x, *

y, z ) = + oo if x a n d y are fixed.
T h e conditions 1 a n d 3 will be assumed to hold t h r o u g h o u t this paper, t o g e t h e r with 2 A or 2 B . 1 2 A is a s s u m e d to hold in C h a p t e r 1 a n d Chapter 3; in Chapter 2 we shall use t h e m o r e general condition 2 B. Xt is easy t o see (using t h e H e i n e - B o r e l covering theorem) t h a t

lim ( inf

*F(x, y, z))= + *

I z l ~ ~x~<~

*l y K K *

for every c o m p a c t interval [:r c J a n d for every K > 0.

**Chapter 1. The question of the existence of a minimizing function **
T h e assumptions on

*F(x,y,z) *

in this c h a p t e r are m a i n l y t h e same as in [1]. W e
shall assume t h a t t h e following conditions are satisfied t h r o u g h o u t t h e c h a p t e r :
1, 2 A a n d 3. T h e interval J will be chosen as *Xl<~X<~X ~. *

1. W e shall n o w give an example which shows t h a t t h e existence of a minimizing function does n o t follow f r o m these properties of

*F(x, y, z). *

*A counterexample. *

^{Choose }

*F(x, y, z) =z2/(y 2 + *

1) 2 - cosx - x ( 2 ~ - x ) c o s x - ~ ( y ) (2~ - ~ ( y ) ) ,
*where~(y)=arctgy+~/2; *

x l = 0 a n d

*I x2=2~' *

Yl = 0 [ Y2 = 0.

L e t /E :~. W e h a v e

*H(I) *

>I F ( ~ , / ( ~ ) , 0) = 1 + ~ 2 - ~ ( / ( ~ ) ) [ 2 ~ - ~ ( / ( ~ ) ) ] > 1,
since 0 < ~ ( / ( ~ ) ) < z . H e n c e H ( / ) > I for all /E:~. Suppose n o w t h a t 0 < ~ < g / 2 a n d consider t h e function

1 T h e r e is one exception: E x a m p l e 5 in C h a p t e r 4.

ARKIV FOR MATEMATIK. Bd 6 nr 22 tg x

tg (2 :~ **- ** **x ) **

I t is easy to verify t h a t

for 0 ~ < x - ~ - ~, for 2 - - (~ ~ < x ~ + (5, for 3 - ~ + ~ ~<x~<27~.

*H(g~) = F ( z , g~(ze), O) = 1 + ~2. *

This proves t h a t infr~ ~ H ( / ) = 1 and t h a t there is no minimizing function. Compare the r e m a r k to Theorem 1.1.

2. I t follows from this example that, in order to ensure the existence of a mini-
mizing function, we m u s t impose suitable extra conditions on *F(x,y,z). * We shall
describe a few such conditions (they are separately sufficient), but we are not aiming
at a complete discussion of the question.

I n each existence proof, we shall need

**Lemma 1.1. ***Let *{/~(x)}~ be a sequence o//unctions in :~ such that supl.<~<~ *H(/~) < *

*and suppose that ]~(x) -+/(x) uni/ormly /or x I <<. x <~ x 2 . Then / 6 :~ and H(/) <~ *lim~_~ ~ H(/r).l
*Proo/. *Since ]/r(x)] ~<K1 a n d *H(/~)<~K 2 for all v, it follows from the properties of *
F t h a t ]/:(x)] ~<K a. Hence all/~(x) a n d / ( x ) satisfy a uniform Lipschitz condition.

Clearly /6:~, and the rest of the proof follows easily. (Use the fact t h a t if */'(xo) *
exists, then we have

*~(xo, /(xo), z) >~ F(Xo, /(xo), /'(xo)) *

for all *z >~/'(xo) *or for all *z <<./'(xo), together with a continuity argument.) *

L e t us now return to the minimization problem and introduce the notation
M 0 =inf~E~ *H(/). *Suppose t h a t {/~}F is a minimizing sequence, i.e. lim~_~ ~ *H ( / , ) = M 0. *

I f the f u n c t i o n s / , ( x ) are *uni/ormly bounded, *then it follows (as above) t h a t t h e y are
equicontinuous, and then there is a uniformly convergent subsequence. L e t the limit
function b e / ( x ) . I t follows from the l e m m a t h a t *H ( / ) = M *0.

Hence, in order to prove t h a t there is a minimizing function, it is sufficient to prove t h a t there is a minimizing sequence of uniformly bounded functions.

I n m a n y cases, it can be seen from the function *F(x, y, 0) t h a t there is a minimizing *
function. This possibility is illustrated b y the following theorem.

Theorem 1.1. *Suppose that there is a number Y1 >~max (Yl, Y2) such that *
m a x *F(x, Yl, O) = m a x (F(~, Yl, 0), F(/~, Y1, 0)) *

*/or all :r and fi satis/ying xl <~<fl<~x ~. Suppose also that there is a *
Y2 ~< rain (Yl, Y2) with the same property. Then there is a minimizing/unction.

*number *
1 In other words, the functional H(]) is lower semi-continuous.

411

G. AROrr

*Minimization problems. I I *

*Proo/. *

Y1 if / ( x ) > Y1, L e t / ( x ) e ~ and form

*g(x) *

= /(x) if *Y2 <~/(x) <~ Y1, *

[Y2 if

*/(x)<Y2. *

Clearly

*H(g)<.H(/) *

and the rest of the proof is obvious.
I t is not difficult to find different conditions on

*F(x, *

y, 0) which lead to the same
result. (For instance, *F(x, *

y, 0) increasing in y for y >/]71 and all x, decreasing in y for
Y ~< Ye and all x, are also sufficient conditions for the existence of a minimizing
function.)
3. Suppose that there is a number

*M > M o *

(M0=infrE~ *H(/)) *

such t h a t *H ( / ) < M *

implies t h a t max *I/(x) l<~g, *

where K depends on M but not on/(x). Then, clearly,
every minimizing sequence is bounded and there is a minimizing function.
I t is evident t h a t every number M > M 0 has this property if limlzl_~

*F(x, y, z) = + *

uniformly for all x and y. But there m a y exist numbers M with this property even if
the limit is not uniform. We can describe the result of this section thus: if the roots
zl and z2 of the equation *F(x, y, z ) = M *

do not increase too fast in absolute value
when [Yl -+c~, then the attainable set *E(M) *

is bounded, and then there is a mini-
mizing function. For instance, it will follow from Theorem 1.2' t h a t there is a
minimizing function if liml~l_~r *F(x, *

y, y) = + ~ uniformly in x.
Let us denote b y

*El(M) *

the set of points (x, y) such that
**a) ** **xl<~x<~x~, **

**xl<~x<~x~,**

b) there is an absolutely continuous function

*g(t), *

joining the points @1, Yl) a n d
*(x,y), *

such t h a t
sup

*F(t, g(t), g'(t)) <~ M. *

*t *

The set

*Ee(M ) *

is defined analogously, but with *(x2,y~) *

instead of *(xl, Yl). *

^{We are }

interested in the set

*E(M) = EI(M ) N E2(M ). *

I t is clear that

*H(/) <~ M *

implies t h a t max *I/(x) I <~ K *

if and only if *E(M) *

is bounded,
and in t h a t case the constant K is determined by *E(M). *

We shall first examine

*E(M) *

in the case *F = F(y, z). *

Later, we shall t r y to carry
over some of the results to the general case.
Using some results from [1], we can easily prove the following theorem (concerning

*dPM(t), ~pM(t) *

and the integrals below, see section 2 of [1]):
Theorem 1.2.

*Assume that F is independent o / x and that M > Mo. *

A)

*E(M) is unbounded above (i.e. it contains points with arbitrarily large ordinates) * *i/and only i/the integrals *

*f ~ dt * *and 12= f~ * *dt *

*11 *

**J~, O ~ ( t )**

**J ~ -****~,(t) **

**~,(t)**

*are well-de/ined, convergent and *

*I i + I ~ < x 2 - x 1. *

(1)
**A R K I V FOR M A T E M A T I K . B d 6 n r 2 2 **

B)

*E(M) is unbounded below i / a n d only i/the integrals *

*i3=f * *_ - V,~(t) * ^{dt } ^{and } ^{i4=fu } *_ opt(t) * ^{dt } *are well-de/ined, convergent and *

^{dt }

^{and }

^{i4=fu }

^{dt }

*I3 + I4 < x 2 - x 1. *

(2)
*Proo]. *

We restrict the proof to the first p a r t of the theorem. Assume first t h a t *B(M) *

is unbounded above. L e t (x0, Yo) C *E(M) *

and Yo > m a x *(Yl, *

Y2). I t follows from Theorem 2
in [1] t h a t
*f yV~ dt *

1 ~ ~< x ~ xl a n d *f y . dt *

~<x 2 - x 0.
**, - ~ , ~ ( t ) **

I f we add these inequalities and let Y0 tend to infinity, then we have proved one p a r t of the assertion.

Next, suppose t h a t

*I i + I 2 < x 2 - x 1. *

I t follows from the discussion of " t h e at-
tainable cone" in section 2B of [1] t h a t *(x, y ) E E I ( M ) *

if x 1 + I l < x < x 2 and *Y>~Yl. *

Similarly, it follows t h a t

*(x, y)EE2(M ) *

if *x 1 <x<~x 2 - I 3 *

and *Y>~Y2. *

Clearly, all points (Xl + I1, y), where y ~> m a x (Yl, Y2), belong to

*E(M). *

This completes
the proof of Theorem 1.2, since the reasoning in the case B is analogous.
A consequence of the theorem is t h a t

*E(M) *

is bounded if a n d only if neither (1)
nor (2) is satisfied.
L e t us now consider the general case

*F = F ( x , y, z). *

I t can be *"reduced" *

to the
previous case simply b y neglecting the dependence on x. This m a y be considered
to be a crude method, but, nevertheless, it works in m a n y cases.
The functions

*~gM(y) *

and *~fM(Y) *

^{w e r e }defined in [1]. Clearly, we can introduce here the corresponding functions

*O(x,y,M) *

and ~(x, y, M). Now fix y and M and
suppose t h a t U = *{x IF(x, *

y, 0) ~<M} is non-empty.
*De/inition: * *I ~(y' *

M) = max~ *~ v O(x, y, M), *

**[ ** *fl(y, *

M) = minx ~ v ~f(x, y, M).
**[**

I t is easy to verify t h a t these functions are continuous and t h a t they are monotonic in M, as are (I)~(y) and ~v~(y). Finally, an argumentation, similar to the one used in the special case F =

*F(y,z), *

leads to the following result:
Theorem 1.2'.

*Assume that M > M o. *

A)

*I / E ( M ) is unbounded above, then the integrals * *dt *

*I1= * *, a(t,M) and *

*dt * *12= *

**~ - f l ( t , M )***are well.de/ined, convergent and I 1 + 12 <<-x2 -x l . *

B) *I / E ( M ) is unbounded below, then the integrals *

413

c. ARO~SSOr~, *Minimization problems. I I *

**i3=f dt **

**i3=f dt**

**_ ****- f l ( t , M ) ***and * *Y~ * *dt *

*I ~ = _ * *a(t, M ) *
*are well-defined, convergent and I a + I a <~ x 2 - x 1, *

An application of this theorem is given in Example 1 in Chapter 4. Compare also the counterexample.

**Chapter **2. A **more general case. ** P r o o f of **the existence of absolutely **minimizing
**functions **

]. I n this section we shall generalize some of the previous results to a more general class of functions F(x, y, z).

We are not going to change the condition limN_~cF(x, y, z ) = + c o , which has
turned out to be very useful and which is a rather natural condition. However, the
assumption t h a t F(Xo, Yo, z) has a minimum at z = 0 for all (x0, Y0), is more restrictive,
and we shall assume instead t h a t F(x0, Yo, z) has a minimum at z=eo(x0,yo) , where
*oo(x,y) *is a continuous function. Compare Example 5 in Chapter 4.

Thus, let us assume t h a t the following conditions hold throughout this chapter:

F(x, y, z) E C1; 2B and 3.

The minimization problem is meaningful, since we have
*H(/) >~ *m a x [F(Xl, Yl, O~(Xl, Yl)), *F(x2, Y2, *c~

for every admissible function/(x).

We shall now carry over some of the results of the paper [1] and of the previous chapter to this more general case.

The analogues of the Theorems 5 and 6 in [1] are
**Theorem **5'. *L e t / ( x ) be an admissible/unction such ** t h a t : *
a)

*/'(x) is continuous on I (x 1 <~x <~x2),*

### b)

*F z ( x , / ( x ) , / ' ( x ) ) > ~ O on I , o r < O on I ,*

### c) F(x,/(x),/'(x))

*= M on I .*

*T h e n / ( x ) is a m i n i m i z i n g / u n c t i o n 1 (not necessarily unique). *

**Theorem **6'. *Suppose that/(x) E ~ satisfies: *

a) */'(x) is continuous on I (x 1 ~x<~x~), *
b) *F z ( x , / ( x ) , / ' ( x ) ) # 0 on I , *

### c) F(x,/(x),/'(x))

*= M on I .*

*T h e n / ( x ) is a unique m i n i m i z i n g / u n c t i o n . 1 *

The proofs are analogous to those of the case *~o(x, y)=~0 *and they are omitted.

Lemma 4 in [1] holds in this case without changes. The proof is simple.

1 It also follows that *](x) *is an a.s. minimal on I.

**ARKIV FOR MATEMATIK, B d 6 n r 22 **
A c o n s e q u e n c e of t h i s l e m m a is t h a t we can n e g l e c t sets of m e a s u r e zero. This
m e a n s t h a t we can j o i n t w o f u n c t i o n s a t a f i n i t e n u m b e r of p o i n t s w h e r e t h e y a r e
e q u a l b y t a k i n g one or t h e o t h e r on successive i n t e r v a l s . T h e n t h e s e p o i n t s will n o t
affect t h e v a l u e of t h e f u n c t i o n a l for t h e r e s u l t i n g f u n c t i o n . (This s i t u a t i o n will
o c c u r in t h e n e x t section.)

:Next, we m a k e a j u m p t o T h e o r e m 9, w h i c h is c a r r i e d o v e r w i t h o u t changes:

T h e o r e m 9'. *I / ] ( x ) is an a.s. minimal on the interval 1 (x 1 ~ x <~x2), then: *

1) */(x)EC 1 on I, *

2) *the differential equation *

### dF(~,/(x),/'(~))

*Fz(X, / ( x ) , / ' ( x ) ) = 0*

*dx*

*is satisfied on I in the sense that i / F ~ ( x , / ( x ) , / ' ( x ) ) =#0 on a sub-interval I 1 c I, then *
F ( x , /(x), /'(x)) *is constant on 11. *

W e shall p r o v e t h e a s s e r t i o n s 1) a n d 2) u n d e r t h e following, w e a k e r a s s u m p t i o n s

**on/(x): **

I) */(x) *satisfies a L i p s c h i t z c o n d i t i o n on [Xl, x2],

I I ) t h e r e is a c o n s t a n t K > 0 such t h a t if [tl, t2] is a n a r b i t r a r y s u b - i n t e r v a l of I ,
t h e n t h e r e a r e two (not n e c e s s a r i l y different) m i n i m i z i n g f u n c t i o n s *u(x) *a n d *v(x) *be-
t w e e n ( t i , / ( t i ) ) a n d *(t2,/(t2) ) *such t h a t [u(x)[ *~ K , Iv(x)] ~ K * a n d *u(x) <~/(x) <~v(x) *
for t 1 ~< x ~< t 2.

(These a s s u m p t i o n s a r e s a t i s f i e d if */(x) *is a n a.s. m i n i m a l , since we c a n choose
K = m a x z ~ *]/(x) l *a n d *u ( x ) = v ( x ) = / ( x ) *in t h a t case.)

T h e r e a s o n w h y we shall p r o v e t h i s m o r e g e n e r a l r e s u l t is t h a t i t will be useful in t h e n e x t section.

*Proo]. 1 *Consider a n a r b i t r a r y p o i n t x o on t h e i n t e r v a l x 1 ~ x < x 2 a n d s u p p o s e t h a t
t h e u p p e r a n d lower r i g h t - h a n d d e r i v a t i v e s of */(x), :r *a n d /~, a r e d i f f e r e n t a t x 0.

(Clearly, t h e y a r e finite.) Choose a n u m b e r 7 ~:o)(Xo,/(Xo)) such t h a t g > ~ >ft. T h e
i n i t i a l - v a l u e p r o b l e m *F(x, g(x), g'(x))=constant, g(xo)=/(x0), g'(xo)=~, * h a s a solu-
t i o n ~(x) in a n e i g h b o u r h o o d of x 0. Since ~ > ~ >fi, t h e r e a r e p o i n t s x > x o, a r b i t r a r i l y
close t o x0, w h e r e */(x)=~0(x). *B u t since ~, ~=eo(x0,/(xo)), i t follows f r o m t h e condi-
t i o n I I a n d f r o m T h e o r e m 6' t h a t / ( x ) *=q~(x) *on [Xo, x0+~t ] for s o m e ~ > 0 . T h e con-
t r a d i c t i o n shows t h a t / ( x ) h a s a r i g h t - h a n d d e r i v a t i v e . S i m i l a r l y , i t follows t h a t / ( x )
h a s a l e f t - h a n d d e r i v a t i v e for x 1 < x ~<x 2.

O u r n e x t s t e p is to p r o v e t h a t t h e o n e - s i d e d d e r i v a t i v e s a r e equal. Consider a
p o i n t x0, a n d a s s u m e t h a t *o:=/'(xo)+4fi=/'(xo)-. * Choose a n u m b e r y b e t w e e n
a n d ft. C l e a r l y

F(x0,/(Xo), *y) < *m a x [F(xo,/(xo) *, o:), F ( x o, ](Xo), *fl)].

S e l e c t a sequence of i n t e r v a l s *[Pn, q~] *such t h a t
*p~ <xo <q~, q~-p~-->O *

a n d */(q~) - / ( P ~ ) *

q~ - p~

1 The reasoning will be similar to the one used in [1], but not identically the same.

2s: 4 415

**G. ARONSSO~, **

**Minimization problems. II **

**Minimization problems. II**

I t is obvious t h a t

lim

*M(pn, qn; /(P,), /(q,)) < F(xo,/(xo), ~). *

*n ---> o v *

(The n o t a t i o n M(...) was i n t r o d u c e d in [1].) L e t us assume, for example, t h a t g > y > f l a n d

*F(xo,/(xo) , *

~) > F(x0,/(xo) *, *

y). L e t the functions *Un(X) *

correspond to t h e intervals
[p~, q~] according to I I . I t follows t h a t lim~_~r (SUpx *u'~(x))>~ ~z. *

Thus
Consequently

lira

**H(un) ~ F(xo, /(xo), ** *~). *

**H(un) ~ F(xo, /(xo),**

*n ---> ~ r *

lim

*H(u~) *

> lira *M(pn, qn; /(Pn), /(qn)), *

*n .--> o r * *n ---> ~ *

which is a contradiction.

The reasoning is analogous in the other cases. This proves t h a t / ' ( x ) exists.

Suppose n o w t h a t

*Fz(xo,/(xo) ,/'(xo) ) #0. *

T h e n we assert t h a t / ( x ) *E C 2 *

in a neigh-
b o u r h o o d U of x 0 a n d t h a t *F(x, /(x), /'(x)) *

is c o n s t a n t in U. This is p r o v e d in t h e
same w a y as T h e o r e m 8 in [1]. W e omit the details.
N o w t h e only thing left to prove is t h a t / ' ( x ) is continuous a t those points where

*Fz(X, /(x), /'(x)) *

= 0 . Assume t h e n t h a t *]'(Xo)=o~(x *

o,/(x0) ). Assume f u r t h e r t h a t there
are n u m b e r s ~n->x0, ~ > x 0 such t h a t */'(~)>~/'(Xo)+~. *

I f the functions *Un(X)cor- *

respond to the intervals x 0 ~ x ~ according to I I , t h e n we h a v e sup~ *u'~(x) >~/'(xo) § ~. *

H e n c e l i m n ~

*H(Un) ~ F(Xo,/(Xo) ,/'(Xo) *

+~). B u t this contradicts the obvious relation
lim *M(xo, ~n; /(xo), /(~n)) <~ F(xo, /(Xo), /'(Xo)). *

*n - - > o ~ *

T h e other cases can be t r e a t e d similarly. This completes the proof.

W e h a v e t h u s established some i m p o r t a n t properties of a.s. minimals. B u t we h a v e n o t e x a m i n e d the properties of minimizing functions in general. T h e reason for this is evident f r o m E x a m p l e 2 in Chapter 4.

I n t h e previous c h a p t e r we t r e a t e d the question of the existence of a minimizing function for the case

*co(x,y)~O. *

I t is easy to verify t h a t L e m m a 1.1 holds in t h e
present case. This means t h a t if there is a minimizing sequence of u n i f o r m l y b o u n d e d
functions, t h e n there is a minimizing function. I n particular, this is t r u e if *E(M) *

is
b o u n d e d for some *M > M o . 1 *

W e shall n o t discuss generalizations of the other criteria in Chapter 1.

2. A problem which we h a v e n o t y e t discussed, is t h a t of t h e existence of absolutely minimizing functions. W e shall give an existence proof for the general case.

Theorem 2.1.

*Assume that F(x, y, z) E C 1 and satisfies the conditions 2 B and 3 / o r * *X I <~x<~X 2 and all y, z. (Thus J=[X1,X2]. ) *

*Assume/urther that, /or every choice o/(~1,~1) and (~2,~2), there is a minimizing/unc- * *tion between these points, and, i/ there are several such/unctions, that they are uni/ormly * *bounded. (The bounds will depend on ~1, ~h, ~ and ~2") *

1 A sufficient condition for this is that limlz[-~r

*F(x,y, *

z) = + c~ uniformly for all x and y.
**ARKIV FSR MATEMATIK. B d 6 n r 22 **

*Then there is an absolutely m i n i m i z i n g / u n c t i o n / o r every minimization problem in * *the strip X I <~ x ~ X2, - ~ < y < ~ . *

*Proo/. *

I n order to m a k e the proof more lucid, we shall divide it into several parts.
l) Consider the minimization problem between

### (Xl,

Yl) and (x2, Y2). L e t ~ be the class of minimizing functions, which, b y assumption, is not empty.P u t

**[ **

u(x) ~ inf [(x)
a n d
**lv (x) = sup/(x). **

f e T n

I t is clear t h a t

*u(x) *

and *v(x) *

belong to ~ .
Consequently there are always a smallest minimizing function

*u(x) *

and a greatest
minimizing function *v(x). *

2) We shall introduce two notations in order to simplify the expressions later on.

L e t

*p(x) *

be a function on an interval I . Let [Xl,X2] be a sub-interval of I and
consider the minimization problem between (Xl, *p(xl) ) *

and *(x2, p(x2)). *

L e t *u(x) *

be
the smallest and *v(x) *

the greatest minimizing function, as above. We shall agree to
say t h a t *p(x) *

has the p r o p e r t y A on I if *p(x)>~u(x) *

on Ix1, xe]/or *every *

sub-interval
[xi, x2] of I and t h a t *p(x) *

has the p r o p e r t y B on I if *p(x) <~v(x) *

on *[xl, xeJ/or every *

sub-interval Ix1, x2] of I . (Compare subharmonic and superharmonic functions.)
3) Consider our minimization problem on an interval I with some b o u n d a r y values. I t is obvious t h a t the greatest minimizing function

*v(x) *

is not less t h a n the
*greatest *

minimizing function ### Vi(X )

between a n y two points (tl,### v(ti) )

and*(t2, v(t2) ). *

Similarly, the smallest minimizing function

*u(x) *

is not greater t h a n the *smallest *

minimizing function *u1(x ) *

between a n y two points *(tl, u(tl) ) *

and *(t2, u(t2) ). *

4) We shall now construct an a.s. minimal.

*From now on, we consider a /ixed * *minimization problem, namely between *

(Xl, y~) and (x2, Y2). L e t G be the class of those
minimizing functions which have the p r o p e r t y A on *[xl,x2]. G *

is not empty, since
the greatest minimizing function belongs to G.
*We introduce the/unction *

*h(x) *

= inf *g(x) *

*geG *

*and we assert that h(x) is an a.s. minimal. *

I t is clear t h a t

*h(x) *

is a minimizing function.
5) Now let us prove t h a t

*h(x) *

has the properties A and B.
I) Choose an a r b i t r a r y interval it1, t2] a n d let

*u(x) *

be the smallest minimizing func-
tion between *(tl, h(tl) ) *

a n d *(t2, h(t2) ). *

L e t *g(x)CG. *

We assert t h a t *g(x)>~u(x) *

on
t 1 ~<x ~ t 2. I f this were not true, then there would be an interval s 1 ~<x ~<s 2 such t h a t
*U(81)=g(81) , U(S2)=g(82)*and

*u(x)>g(x) *

for *81<x<82. *

Let *ul(x ) *

be the smallest
minimizing function between (Sl, *u(sl) ) *

and *(se, u(s2) ). *

I t follows from the definition
of G t h a t *g(x)~>Ul(X *

). B u t we also have *u ( x ) ~ u l ( x ). *

(See p a r t 3 of the proof.) This
gives *g(x)>~ul(x ) ~ u ( x ) *

and we have a contradiction.
Thus

*u(x) <~g(x), *

and the inequality *u(x) <~h(x) *

follows from the definition of *h(x). *

This proves t h a t

*h(x) *

has the p r o p e r t y A.
I I ) Assume t h a t

*h(x) *

has not the p r o p e r t y B. This means t h a t there is an interval
It1, t2] such t h a t the corresponding greatest minimizing function *v(x) *

does not satisfy
417
**c . AROr~SSOrr ** **Minimization problems. II **

t h e inequality

*v(x)>~h(x). *

I t follows f r o m p a r t 3 of the proof t h a t we m a y assume
t h a t *h(x)>v(x) *

for t I < X < t 2. L e t us f o r m the function
h(x) for

*x<~q, * *p(x)=]v(x) *

for *tl<x<t2, *

**[h(x) for **

**x ~ t 2.**I f we can prove t h a t

*p(x)E G, *

t h e n this will contradict the definition of *h(x), *

since
we h a v e *p(x) < h(x) *

on (tl, t2). I t is clear t h a t *p(x) *

is a minimizing function on [xl, x~],
a n d it remains to prove t h a t *p(x) *

has t h e p r o p e r t y A. So let us consider a n interval
Is1, s2] a n d let the corresponding smallest minimizing function be *ul(x ). *

W e h a v e
*ul(sl)=p(sl)<h(sO *

^{a n d }

*ul(s2)=p(%)<~h(% ). *

^{Since }

*h(x) *

has t h e p r o p e r t y *A, *

^{this }

implies t h a t

*h(x)~>Ul(X *

). If p ( ~ ) < u l ( ~ ) for some ~E(Sl, S2), t h e n we m u s t h a v e
*p(~)=v(~)<ul(~ ). *

Consequently there m u s t be an interval [rl, r2] such t h a t
*v(rl) *

= u l ( r 0 , *v(r2) -ul(r2) *

a n d *v(x) < ul(x ) *

for r 1 < x < r 2. L e t *u2(x ) *

a n d *v2(x ) *

corre-
spond to this interval a n d these b o u n d a r y values in the usual manner. T h e n we h a v e
*udx) *

<u2(x) <v2(x) < v(x).
B u t this contradicts t h e inequality

*v(x)<ul(x *

). This proves t h a t *h(x) *

has the pro-
p e r t y B.
6) I t is obvious t h a t

*h(x) *

satisfies the conditions I a n d I I which, as we h a v e
seen, g u a r a n t e e t h a t t h e assertions in T h e o r e m 9' are true. Thus *h(x)E C 1 *

a n d the
differential equation
*dE(x, h(x), h'(x)) *

*dx * *9 Fz(x, h(x), h'(x)) = 0 *

is satisfied in t h e sense described there.

7) W e are n o w in a position to prove t h a t

*h(x) *

is an a.s. minimal. Consider an
a r b i t r a r y interval t 1 < x < t 2. L e t M 0 be t h e m i n i m u m value of the functional, as
usual, a n d p u t
M = m a x

*F(x, h(x), h'(x)). *

*t x ~ x ~ t ~ *

W e w a n t to prove t h a t M = M 0.

I f

*F(t 1, *

h(tl), h'(t 0) = M, t h e n t h e result follows at once, since *h(x) *

has the properties
A a n d B, a n d t h e same is true for t 2. So let us assume t h a t *F(t 1, *

h(tl), *h'(tl))<M *

a n d t h a t

*F(t~, h(t~), h'(t2))<M. *

I t follows f r o m p a r t 6 t h a t there is a n u m b e r
*2, tl<~<t2, *

such t h a t h'(~)=r h(~)) a n d F(~, h(~), h'(~)) = M . L e t us assume t h a t
M > M o
This means t h a t there is no minimizing function co(x) for which 0)(2)=h(~). (I.e.

*(2, h(~)) CE(Mo). ) *

L e t K be the class of those minimizing functions a)(x), for which 0)(2)<h(~). (K is n o t e m p t y , see p a r t 5.) P u t k ( x ) = s u p ~ K co(x). I t is clear t h a t k ( x ) E K a n d t h a t k(~) < h(~).

P u t

/ 81 = sup {x ix < ~,

*h(x) *

= k(x)}
a n d )

[ s2 = inf {x ] x > ~,

*h(x) = k(x)}. *

**ARKIV FOR MATEMATIK. B d 6 n r 22 **
W e introduce the function

h(x) for

*x < s 1, * *q(x)= * jk(x)

^{for }

*s l < x < s 2, *

[h(x) for x ~ s 2.

If we can prove t h a t

*q(x)E G, *

t h e n this will c o n t r a d i c t the definition of h(x), since
*q(x) *

<h(x) on (s 1, s2). (Compare p a r t 4.) *So all we have to prove is that q(x) has the * *property A. *

Consider t h e n a n a r b i t r a r y interval [rl, r2] a n d let *u(x) *

be t h e smallest
minimizing f u n c t i o n between ### (rl, *q(rl) ) *

a n d *(r2, q(r2) ). *

W e h a v e *u(rl)=q(ri)<~h(rl) *

a n d

*u(r2) *

=q(r2) *<~h(r2). *

Since *h(x) *

has the p r o p e r t y A, it follows easily t h a t *h(x) >~u(x) *

for r~ ~< x ~< r e.
So if we assume t h a t

*u(xo)>q(x0) *

for some x0, t h e n we m u s t h a v e *s 1 <xo<s 2 *

a n d
*q(xo) * =k(x0).

Consequently there m u s t be a n interval (~h, ~2) such t h a t s 1~<~1 <~2 ~<s2, u(~h) = k(~l), u(~2) = k(~2) a n d

*u(x) > k(x) *

for ~1 < x <V2.
If

*H(k; ~ , ~2) <~H(u; ~ , ~2), *

t h e n we get a contradiction to the definition of *u(x). *

Suppose t h e n t h a t L e t us form the function

*H(k; ~1, *

~ 2 ) > H ( u ; ~1, ### ~2)"

**[ **

~(X) for Z ~ l ,
*kl(x ) = j u ( x ) *

for ~ 1 < x ~ 2 ,
[k(x) for x ~ 2 .

### Clearly, kl(X )

is a minimizing function between (tl,*h(tl) ) *

a n d (t~, *h(t2) ). *

F u r t h e r , we
h a v e kl(~)~<h(~ ) (for, as we m e n t i o n e d above, *u(x)~h(x) *

for all x, where *u(x) *

is
defined). H e n c e kl(x ) EK. :But this contradicts the definition of k(x). Hence we get
a contradiction in a n y case.
This completes the proof of T h e o r e m 2.1.

3. The m e t h o d of a p p r o x i m a t i n g a m a x i m u m b y a sequence of integral m e a n values is well known. I n our case it means t h a t we should consider t h e functional

*H(/) *

as the limit of the sequence of functionals
H n ( l ) = 1 ,

*[F(x,/(x),f(x))]ndx] 11~, *

n = 1 , 2 , 3 , . . . .
W e have used this a p p r o a c h in [1] to derive t h e differential e q u a t i o n

*(dF/dx). F z = O. *

I t can also be used to give a different existence proof for a.s. minimals. This is v e r y n a t u r a l since a minimizing function in the calculus of variations a u t o m a t i c a l l y is minimizing on e v e r y sub-interval. However, we shall n o t c a r r y t h r o u g h this proof since it is more complicated t h a n the one already given, a n d since it requires stronger conditions on

*F(x, y, z). *

A n o t h e r result which can be p r o v e d with this "integral m e t h o d " is the following:

if/o(X) is t h e only a.s. minimal for

*H(/) *

a n d if/~(x) minimizes *Hn(/) *

for n = 1, 2, 3, ...,
t h e n limn_~ ~/~(x) =/o(X) uniformly.
419

**c. ARONSSON, ****Minimization problems. II **

**Chapter 3. Further examination of absolutely minimizing functions. **

**Uniqueness questions **

For reasons of simplicity, this chapter will treat only the case

*(o(x,y)=-O. *

N o
doubt, the results can be modified for the general case.
1. The following theorem gives us further information on the structure of absolutely minimizing functions:

**Theorem **3.1.

*Assume that F(x, y, z) satis/ies these conditions: *

*F(x, y, z) is an analytic/unction o/ x, y and z in a complex neighbourhood o/the set * *o/ values *

**{ **

*X l ~ X ~ X*

*2*

y, z real;

*2 A and 3. *

*Assume/urther that/(x) is an absolutely minimizing/unction on the compact interval * *xl ~ x ~ x 2. *

*Then: *

a)

*/(x) EC 1 on *

[xl, x2] *(already proved); *

b)

*the set {x I F~(x, /(x), /'(x)) # 0 } consists o / a f i n i t e number o/intervals (to be proved * *now); *

c)

*the di//erential equation *

*dF(x, /(x), /'(x)) Fz(x, /(x), /'(x)) = 0 * *dx *

*is satis/ied in the sense that F(x, /(x), /'(x)) is constant on each o/these intervals (al- * *ready proved). *

*Proo/. *

I n order to facilitate the further references, we divide the proof into several
parts.
1) Assume t h a t

*] ' ( x ) # 0 *

for *p < x < q *

and t h a t / ' ( p ) *=/'(q)=0. *

*Fx(p,/(p), o) < o, *

Then ^{/ }

**[ Fx(q,/(q), O) >10. **

**[ Fx(q,/(q), O) >10.**

To see this, we recall t h a t / ( x )

*EC 2 *

on *(p,q) *

and t h a t *F(x, /(x), /'(x)) *

is constant there
(Theorem 8 in [1]). So we have
*Fx(X, /(x), /'(x)) + F~(...) /'(x) + Fz(...)/"(x) = 0 *

for

*p < x < q . *

Clearly, there m u s t be points arbitrarily close to p, where */'(x) *

a n d
*/"(x) *

have the same sign. At such a point *Fz(...)/"(x)>0, *

which gives
*Fx(x, l(x), l'(x)) + ~'y(...) /'(x) *

< 0 .
420
**ARKIV FOR MATEMATIK. B d 6 n r 2 2 **
Since l i m x - ~ v / ' ( x ) = O , t h i s p r o v e s t h e first i n e q u a l i t y , a n d t h e s e c o n d is p r o v e d
s i m i l a r l y (there a r e p o i n t s a r b i t r a r i l y close t o q w h e r e

*Fz(...)/"(x ) <0, *

etc.).
N o w s u p p o s e t h a t t h e r e is a n u m b e r

*t>q *

such t h a t / ' ( t ) *4=0. *

T h e n i t follows f r o m
t h e a b o v e r e s u l t t h a t we can define
*r=inf{xlx>~q, *

F~(z,/(z), 0)=0},
a n d i t follows t h a t

*q<r<t. *

I f *q<r, *

t h e n we o b v i o u s l y h a v e */'(x)=O *

for *q~x<~r *

( a n d if q = r , t h e n t h i s is t r i v i a l l y true).
I f t h e r e is a n u m b e r

*u < p *

such t h a t *]'(u)4=0, *

t h e n we define
**s = s u p {x; x < p , F ~ ( x , / ( x ) , 0) = 0 } .**W e h a v e / ' ( x ) = 0 for

*s ~ x ~ p . *

F i n a l l y , we o b s e r v e t h a t

**{ ** *Fx(x,/(x),O)~O *

for *s<~x~p, * *Fx(x,/(x), *

0)~>0 for *q<~x<~r. *

2) W e shall give a n i n d i r e c t p r o o f of t h e t h e o r e m . So we a s s u m e t h a t t h e set {x[ x 1 < x < x2,

*/'(x) *

4 0 } is t h e u n i o n of a n i n f i n i t e n u m b e r of d i s j o i n t o p e n i n t e r -
vals. These i n t e r v a l s m u s t h a v e a l i m i t - p o i n t a n d we m a y a s s u m e t h a t i t is *x=O. *

W e m a y also a s s u m e t h a t / ( 0 ) = 0 a n d t h a t t h e r e a r e i n f i n i t e l y m a n y such i n t e r v a l s i n e v e r y i n t e r v a l 0 < x < e. C l e a r l y / ' ( 0 ) = 0 a n d Fx(0, 0, 0) = 0.

T h e f u n c t i o n

*Fx(x, *

y, 0) will b e i m p o r t a n t for t h e r e s t of t h e proof. I t m a y be
i d e n t i c a l l y zero or not. T h i s gives us t w o cases a n d we shall s t a r t w i t h t h e s i m p l e s t
of t h e m .
3) A s s u m e t h a t Fx(x, y, 0) ~ 0. T h i s m e a n s t h a t we h a v e

*F(x, y, O) = F(O, y, O) =of(y), *

w h e r e ~c(y) is a n a l y t i c in a (complex) n e i g h b o u r h o o d of y = 0 .
A) ~c(y)--- c o n s t a n t , i.e.

*F(x, *

y, 0) --- c o n s t a n t .
Consider a n i n t e r v a l *(p, q) *

as in 1). P u t *t=89 +q). *

T h e n we h a v e
F(t, /(t), /'(t)) > F(t,/(t), 0) =

*F(p, tip), *

0),
w h i c h gives a c o n t r a d i c t i o n .

B) ~c(y) ~ c o n s t a n t .

I t follows f r o m T h e o r e m 10 in [1] 1 t h a t / ( x ) is m o n o t o n i c , for i n s t a n c e n o n - d e c r e a s - ing.

B u t ~ c ' ( y ) # 0 for 0 < y < ~. H e n c e ~c(y) is s t r i c t l y m o n o t o n i c for 0 ~ y ~ . Consider a n i n t e r v a l

*(p,q) *

such t h a t *O</(p)</(q)<~. *

T h e n we m u s t h a v e *cf(/(p))=cf(/(q)) *

w h i c h gives a c o n t r a d i c t i o n .
1 I t is c l e a r t h a t t h e c o n d i t i o n o n F z i n T h e o r e m 10 n e e d o n l y b e a s s u m e d t o h o l d for z = 0.

I n f a c t , t h e c o n d i t i o n of T h e o r e m 1.1 i n t h i s p a p e r is s u f f i c i e n t if i t h o l d s f o r a l l y.

**4 2 1 **

**G. AROr~SSOr~, ****Minimization problems. II **

4) ~qow let us assume t h a t

*F~(x, *

y, 0) ~ 0. Clearly *F~(x, y, *

0) is a n analytic function
a n d we can a p p l y Weierstrass' p r e p a r a t i o n t h e o r e m (see [2], p. 89). I n our case it
says t h a t
*F~(x, y, O) =x~o~(x, y)W'(x, y) *

in a n e i g h b o u r h o o d U of the origin. Here # is an integer ~>0;

*~o(x,y) *

is analytic in U
a n d # 0 ; ~F(x, y) is *either =- 1 or *

of t h e f o r m
*vt~(x ' y) =ym + A~(x)y,~-i + A2(x)ym-2 + ... + A,~(x), *

where t h e functions

*Ak(x) *

are analytic in a (sufficiently small) n e i g h b o u r h o o d of
*x=O *

a n d A k ( 0 ) = 0 for all k.
W e k n o w f r o m 2) t h a t

*F:~(x, *

y, 0) - 0 at t h e points (r,/(r)) a n d *(s, ](s)). *

^{Therefore }we can exclude t h e case ~F ~= 1 and, instead, we inquire a b o u t the zeros of

*uz.(x,y ) =ym + A~(x)yr~-I + ... +Am(x). *

L e t us, for the present, replace x a n d y b y the complex variables ~ a n d ~]. W e shall t r y to determine ~ as a function of $ f r o m t h e relation ~F(~,~) = 0 . W e are only in- terested in the solutions of this e q u a t i o n in a n e i g h b o u r h o o d of ~ = 9 = 0 .

Consider a complex n e i g h b o u r h o o d V of $ = 0 a n d c u t it along t h e negative real axis. T h e n every root of ~ can be defined as a regular function in V, a n d the roots

~k of ~F(~, ~]) = 0 m a y be written

*= ~ C * *i~l/n~ *

for k = l , 2 , . . , m .
,~,o, j , k ~ ,
Concerning this expansion, see [3], p. 50; [4], pp. 98-103 a n d [5], Chapter X I I I . (The roots can be divided into cyclic systems, a n d the n u m b e r n can be chosen as t h e p r o d u c t of the n u m b e r of roots in each system.)

W e m a y assume t h a t ~l/n is real for ~ > 0, which gives the expressions

~ *~ - ~ J . k [ x */ ^{1/n~i }*) ~ *k = 1, 2, *. . . , * m.

j = l

Clearly, we need only consider those series in which Cj.g are real for all ?'. (If there are no such series, t h e n there is n o t h i n g left to prove.) Therefore, let us suppose t h a t

*~]k(x), *

k = 1,2 ... *M, *

are real for 0 < x < 0 , a n d t h a t ~k(x) are complex (not real) for k > M a n d 0 < x < 0 . W e k n o w f r o m t h e first p a r t of t h e proof t h a t

*Fx(r,/(r), *

0)= Fx(8,/(s), 0)=0.
H e n c e t h e points (r,/(r)) a n d

*(s,/(s)) *

m u s t lie on the curves *y =~k(x) *

if r a n d s are
sufficiently close to x = 0.
ARKIV FOll MATEMATIK. B d 6 n r 22 5) W e shall n o w s t u d y in some d e t a i l t h e f u n c t i o n s

**~k(x) -= ~ ~].k **

**~k(x) -= ~ ~].k**

**( x l / n ) ~ =**

*g k ( X l / n ) 9*

j = l

H e r e t h e f u n c t i o n s

*g~(z) *

a r e a n a l y t i c for ]z[ < 8. F u r t h e r , e v e r y *g~(z) *

is r e a l if z is real.
D i f f e r e n t i a t i o n gives

. _ _ - 1 - - 1

*,(X)=gk(xl/n ) 1 "X n *

for *O < x < 8 n. *

*n *

P u t t =

*x ~/n, *

w h i c h g i v e s
**~k (x) = ** **g; (0 [ ~ = ~ ( t ) . **

T h e f u n c t i o n ~ ( z ) is a n a l y t i c in 0 < I z] < 8 a n d t h e s i n g u l a r i t y a t z = 0 is e i t h e r re- m o v a b l e or a pole. W e h a v e

l i m ~ (x) = l i m F~(t),

x --.-~ + 0 t ' - - > + O

a n d t h i s l i m i t m u s t be real. I t can b e + o% - 0% f i n i t e b u t ~ 0 a n d 0. Since ]'(0) = 0 , we can e x c l u d e f r o m c o n s i d e r a t i o n t h o s e v a l u e s of/c for which we do

*not *

h a v e ~ ( 0 ) = 0 .
T h e r e f o r e we c a n a s s u m e t h a t ~ ( z ) is a n a l y t i c for l zl < ~ .
A c o n s e q u e n c e of t h i s is t h a t t h e r e is a 8 1 > 0 such t h a t ~vk(z ) 4=0 for 0 < ]z] <81, unless ~vk(z)~0. F r o m t h e r e l a t i o n ~ ( x ) =~k(xl/~), we can infer a c o r r e s p o n d i n g r e s u l t for ~ ( x ) .

N o w let us c o m p a r e t h e f u n c t i o n s ~ ( x ) =gk(t) a n d ~ l ( x ) - g l ( t ) . T h e f u n c t i o n s

*gk(z) *

a n d

*gz(z) *

a r e a n a l y t i c a n d n o t i d e n t i c a l . H e n c e t h e r e m u s t e x i s t a 82 > 0 such t h a t
*9k(z)#gl(z) *

for 0 < l z ] <82.
T h i s gives a c o r r e s p o n d i n g r e s u l t for ~k(x) a n d ~z(x).

W e h a v e f o u n d t h a t ~/k(x)=gk(x

*1/'~) *

a n d ~ ( x ) = ( v k ( x l / ~ ) , w h e r e *gk(z) *

a n d ~k(z) a r e
a n a l y t i c for Iz[ < 8 . L e t us w r i t e *xl/n=t, *

as a b o v e . This gives
**~'(x, ~(x), o)= ~'(t ~, g~(t), **

**~'(x, ~(x), o)= ~'(t ~, g~(t),**

**o ) = u ( t )**a n d

*F(x, ~k(x), ~'k(x) )= F(t ~, gk(t), qJk(t) )=v(t). *

Clearly, t h e f u n c t i o n s

*u(z) *

a n d *v(z) *

are a n a l y t i c in a n e i g h b o u r h o o d of z = 0 . H e n c e
t h e r e is a 8~ > 0 such t h a t e a c h of t h e f u n c t i o n s *F(x, *

~k(x), 0) a n d *F(x, ~(x), ~'k(x)) *

is e i t h e r increasing, c o n s t a n t or d e c r e a s i n g on t h e whole i n t e r v a l 0 ~< x ~< 83.

W e h a v e e x c l u d e d t h o s e f u n c t i o n s

*~k(x), *

for w h i c h we d i d n o t h a v e ~]~(0) = 0 . A f t e r
a r e n u m b e r i n g , we h a v e ~k(x), k = 1, 2 . . . P , left.
I f we collect o u r results, t h e n we see t h a t t h e r e e x i s t n u m b e r s ~ > 0 a n d % such t h a t : a)

*~k(X):4:~l(X) *

if 0 < X ~ : r a n d *lc:4:l. *

This m e a n s t h a t one of t h e f u n c t i o n s ~k(x)
is s m a l l e s t a n d one is g r e a t e s t on t h e whole i n t e r v a l 0 < x ~< ~.
b) ~]k(x) C C 1 on 0 ~<x ~ cr a n d ~k(0) =~/~(0) = 0 . I f ~k(x) ~: 0, t h e n ~ ( x ) 4=0 for 0 < x ~< ~.

c) I f we h a v e p , q, r a n d s as in 1) a n d 0 < s < r ~ < c ~ , t h e n

*](r)=~k(r ) *

for s o m e *k<~p *

a n d

*](s)=~(s) *

for s o m e k 1 ~<P.
423

c. xRo~ssorr

*Minimization problems. I I *

d) E v e r y function

*F(x, ~k(x), *

0) is either c o n s t a n t or strictly m o n o t o n i c on
0~<x~<~. The same is true for *F(x, *

~k(x), ~ ( x ) ) (1 *<~k<~P). *

6) W e are interested only in those functions ~k(x) for which one of t h e relations u n d e r c) above really occurs on t h e interval 0 < x ~< ~. I f we exclude t h e others, t h e n we get (after a renumbering) t h e functions ~k(x), k = 1, 2 . . . . , N, left. This does n o t affect t h e validity of t h e s u m m a r y a)-d).

I f N = 1 a n d ~ l ( x ) ~ 0 , t h e n we get a contradiction at once a n d there is n o t h i n g left t o prove.

I f this is n o t t h e case, t h e n t h e greatest of t h e functions ~ ( x ) is > 0 (for 0 < x ~< r162 o r t h e smallest is < 0.

L e t us choose t h e first case a n d let ~N(X) be the function in question. T h e n

*~N(X) > 0 *

a n d ~]N(X) > 0 on 0<x~<r162
I t follows from our choice of

*~N(X) *

t h a t there is an interval (P0, %) such t h a t / ( % ) *= *

*~N(ro). *

W e have t h u s
*/(qo)--/(ro)=~N(ro); *

/'(qo) = 0
a n d ~N(X) > 0 .
H e n c e

*/ ( x ) > ~ ( x ) *

for q0 - ~ < x < % . B u t we also h a v e / ( P o ) =/(So)<~N(So) (owing
t o our choice of ~ ( x ) ) a n d */'(po)-O. *

H e n c e */ ( x ) < ~ ( x ) *

for *p o < x < p o § *

Conse-
q u e n t l y there m u s t be a ~, p 0 < ~ < q 0 , such t h a t / ( ~ ) = ~ ( ~ ) a n d / ' ( ~ ) > ~ N ( ~ ) > 0 .
N o w assume first t h a t

*F(x, *

~?N(X), 0) is c o n s t a n t or decreasing. T h e n we have
*F(~, ~N(~), O) >~ F(r o, ~]N(rO), O) >~ F(q o, ](qo), 0). *

(See t h e first p a r t of the proof.) This gives

*F(8, /(8), /'(8)) *

> F(q0,/(qo), *0), *

which is a contradiction.
Assume t h e n t h a t

*F(x, ~N(X), *

0) is increasing. A n obvious consequence of this is
t h a t *F(x, ~N(X), ~N(X)) *

is increasing. L e t us n o w consider t h e minimization problem
between t h e points x l = y l = 0 a n d x2=~, Y2=/(~)=~N(~). Since */(x) *

is an absolute
minimal, we h a v e
M o = H ( / ) >~F(~,/(~), 1'(~)).

:But

*F(x, ~TN, ~7"N) *

is increasing, which gives
*M o <~ H(~tN ) = F(~, ~IN(~), ~'N(~)) <~ H(/) = M o. *

T h u s

*H(~tN ) = M o *

a n d *~N(X) *

is also a minimizing function.
Choose a function r C 1 on 0 ~<x ~<~ such t h a t ~ ( 0 ) = r a n d r 0. F o r m t h e function

*g(x)-~N(x)+),~b(x), *

where A > 0 is a parameter. W e h a v e
*.F(x, g(x), g'(x) ) = F(x, ~N(x), V]'N(X) ) § ~.(a(x)r * *d-b(x)r * *) § R(x, 2.). *

H e r e

*a(x) = Fy(x, ~N(x), ~7'N(X)) *

a n d *b(x) = F~(x, *

~N(x), ~v(x)) a n d [ *R(x, 4) 1 <~ C~ 2 *

where
C is i n d e p e n d e n t of x a n d 2, if 0 < ~ < 1.
Now, recalling the fact t h a t

*F(x, ~TN, ~'N) *

takes its m a x i m u m o n l y at x = ~ a n d
t h a t F~(~, ~N(~), ~TN(~)) > 0 , it is easy to verify t h a t *H(g) *

< H ( ~ ) , if A is small enough.
This gives

*H(g) < M o *

which is a contradiction. This completes the proof.
**ARKIV FOR MATEMATIK. B d 6 n r 2 2 **
2. This section will t r e a t b r i e f l y t h e q u e s t i o n of u n i q u e n e s s for a.s. m i n i m a l s . I t
will be s h o w n b y e x a m p l e s t h a t t h e r e m a y be s e v e r a l a.s. m i n i m a l s for a g i v e n mini-
m i z a t i o n p r o b l e m . A few sufficient c o n d i t i o n s for u n i q u e n e s s will also be given.

Choose *F(x, y, *z) = (1 *+ x ~ + y2)2 § z2, *2

*x 1 = - 2 , yl=O, * x 2 = 2 a n d y ~ = 0 .

I t follows f r o m T h e o r e m 2.1 t h a t t h e r e is a n a.s. m i n i m a l / ( x ) . Clearly, *g(x) = - / ( - x ) *
is also a n a.s. m i n i m a l . Now, if t h e r e is a u n i q u e a.s. m i n i m a l , t h e n we g e t - ] ( - x )
*](x), *i.e. /(0) = 0. C o n s e q u e n t l y M o >~ F ( 0 , 0, 0) = 2.

P u t *h ( x ) = 1 2 - x * for x~>0,

**[2+ **

x for x ~ 0 .
C l e a r l y *H(h)*= 2 / 9 § 1 < 2.

T h e c o n t r a d i c t i o n shows t h a t t h e r e is no u n i q u e a.s. m i n i m a l . (This s i t u a t i o n c a n b e d e s c r i b e d thus: e v e r y m i n i m i z i n g f u n c t i o n m u s t e v a d e t h e m a x i m u m a t x = y = 0 , w h i c h , owing t o t h e s y m m e t r y , l e a d s t o n o n - u n i q u e n e s s . )

P u t *F(x, y, z ) = y 2 + z 2 *a n d let */o(X) *be d e f i n e d as in E x a m p l e 3 in [1]. W e k n o w
t h a t all f u n c t i o n s *p/o(x+q)(p, q *a r e c o n s t a n t s ) a r e a.s. m i n i m a l s . I t is e v i d e n t t h a t
t h e r e a r e i n f i n i t e l y m a n y a.s. m i n i m a l s if Y2 = - Y l 44=0 a n d x 2 - x 1 >~r.

I t follows f r o m t h i s e x a m p l e t h a t t h e r e m a y b e s e v e r a l a.s. m i n i m a l s e v e n if 2'(x, y, z) is c o n v e x in y a n d z. I n t h e calculus of v a r i a t i o n s , t h i s c o n d i t i o n l e a d s t o u n i q u e n e s s (for all b o u n d a r y values). I t follows f r o m t h i s e x a m p l e t h a t t h e c o n d i t i o n (Yl - C) (Y2 - C) ~> 0 in T h e o r e m 3.3 c a n n o t b e o m i t t e d .

These e x a m p l e s s h o w t h a t e x t r a c o n d i t i o n s on *F(x, y, z) *o r on t h e b o u n d a r y v a l u e s
m u s t be a d d e d to o u r u s u a l ones in o r d e r to secure uniqueness. Two w a y s to do t h i s
a r e s h o w n b y t h e following t h e o r e m s :

T h e o r e m 3.2. *Assume that F(x,y,z) satisfies the conditions o/ Theorem 2.1, with *
*o)(x, y ) ~ 0 . Then, as we know ]rom that theorem, there is at least one a.s. m i n i m a l / o r *

*every choice o / ( x D Yl) and (x2, y~). *

*Assume now that *

*~Y(x, y, O) *

*0 for all x and y. *

*~x *

*Then there is a V•IQVE a.s. minimal/or every choice o/(xl, Yl) and *(x2, Y2).

*Proo/. *W e a s s u m e t h a t *Fx(x,y,O ) *> 0 . W e m a y also a s s u m e t h a t Yl <Y2 (the case
*yl>y2 *is a n a l o g o u s a n d t h e case *Yl=Y2 *is t r i v i a l ) . L e t */(x) *b e a n a.s. m i n i m a l . I f
*]'(x) *> 0 for *x 1 ~ x <~x2, *t h e n t h e a s s e r t i o n follows f r o m t h e T h e o r e m s 6 a n d 9 in [1].

I f / ' ( x ) *=0 *for some x, t h e n i t follows f r o m p a r t 1 of t h e p r o o f of T h e o r e m 3.1 t h a t
t h e r e is a n u m b e r ~, x 1 < ~ ~x2, such t h a t

> 0 for *xl<~x<~, *
/'(x) is = 0 for ~ < x ~ < x ~ .

425

a. ARO~SSO~, *Minimization problems. I I *

I f *g(x) *is a different a.s. minimal, then the same reasoning holds for *g(x). *Let ~'
correspond to *g(x). Assume t h a t ~' <~. This gives g'(xl) >]'(xl) and *

B u t we also have

### M1: F(Xl,

*Yl, /' (xl) ) < F(xl, Yl, g' (x) ) = M e.*

M 1 = F ( ~ , Y2, 0) > F(~', Ye, 0) = M e.

The contradiction proves the theorem.

Theorem 3.3. *Assume that F(x,y,z) satis/ies these conditions: *

*F(x,y,z) is analytic, as in Theorem 3.1; *

*2 A and 3; *

*the condition concerning existence and boundedness o/minimizing/unctions, which *
*was given in Theorem 2.1; *

*t > : * *i / y > C ' *

*~F(x,y,Z) is = * *i/ y = C , *

@ [ < 0 *i/ y < C , *
*(C is a constant); *

*F(x, ~Yl + ( 1 - 2 ) y~, 4z 1 + (1-4)z2) ~<~F(x, y~, *Zl) + ( 1 - 4 ) *F(x, y~, z~) *

*/or all x, Yl, Y2, zl and z e and/or all 4 such that *0 < 4 < 1 , *equality holds i/ and only i/ *

*Yl =Y~ and z I =z2. *

*We consider the minimization problem between (x 1, Yl) and *(x 2, Ye). We assume that
*the boundary values satis/y the inequality *

*(Yl - C) (Ye - C) >~ O. *

*Then there is a VNIQV]~ a.s. minimal between (x 1, Yl) and *(xe, Ye).

The proof is omitted, since it is rather laborious and since the result is not used in this paper.

3. I n the previous section, we discussed the uniqueness question for a.s. minimals.

We shall now briefly consider the same question for minimizing functions. The
following theorem shows t h a t those functions *F(x,y,z), for which every minimization *
problem has a unique solution, constitute a v e r y "small" class.

Theorem 3.4. Let F(x, y, z) satis/y the/ollowing conditions/or X 1 <~x <~X2 and all y,z:

*F(x, y, z) E Ce; *

*2 A and 3; *

*the condition concerning existence and boundedness o/ minimizing /unctions which *
*was given in Theorem 2.1. *

*We consider the minimization problem between (x~, y)and (xe, Y2) (X1 <~xl <x2 <<-X2) 9 *
*Then there is a VNTQV~ minimizing /unction /or every choice o/ xl, Yl, x2 and Y2 *
*i/ and only i/ *

426

**ARKIV FOR MATEMATIK. B d 6 n r 2 2 **

*_F(x, y, O) =-constant. *

*Proof. *

1) S u p p o s e t h a t *F(x, *

y, 0) - c o n s t a n t . Consider t h e m i n i m i z a t i o n p r o b l e m
b e t w e e n (xl, Yl) a n d (x2, Y2). W e k n o w t h a t t h e r e is a n a.s. m i n i m a l / ( x ) a n d we k n o w
t h a t / ( x ) *E C 1. *

*If ] ' ( x ) 4 0 *

for x 1 ~<x ~<x~ t h e n i t follows f r o m T h e o r e m 8 a n d T h e o r e m 6 in [1] t h a t
*](x) *

is a u n i q u e m i n i m i z i n g f u n c t i o n .
I f t h e r e is a n ~ such t h a t / ' ( r 1 6 2 = 0 , t h e n

*] ' ( x ) - 0 . *

I n o r d e r to see t h a t , s u p p o s e t h a t
*/'(fl) # 0 *

for s o m e fi > ~. :Put
**= sup {x Ix < ~ , / ' ( x ) = o}. **

T h e n we g e t

F(/~, / @ , / ' @ ) > P ( ~ , 1 @ , 0) = F ( ~ , / ( r ) , 0).

B u t this c o n t r a d i c t s t h e r e l a t i o n

*F ( x , / , / ' ) *

= c o n s t a n t w h i c h h o l d s o n e v e r y i n t e r v a l
w h e r e *]'(x) #0. *

So we h a v e

*](x)-yl(=y2). *

I f *g(x) *

is a n a d m i s s i b l e f u n c t i o n a n d g ' ( ~ ) # 0 for s o m e
~, t h e n we g e t

*H(g) *

>~ F ( $ , g(~), g'($)) > F(~, g(~), 0) = H ( / ) ,
a n d we h a v e p r o v e d one h a l f of t h e t h e o r e m .
2) S u p p o s e n o w t h a t t h e r e is a u n i q u e m i n i m i z i n g f u n c t i o n for e v e r y choice of xl, x2, Yl a n d Y2.

I f F=(Xl, Yl, 0) =4=0, t h e n we choose Y2 =Yl a n d x 2 such t h a t Fz(X , Yl, 0) # 0 for

*xl<~x<<.x ~. *

Clearly, */ ( x ) - Y l *

is a m i n i m i z i n g f u n c t i o n , b u t n o t t h e o n l y one. T h i s
p r o v e s t h a t *Fx(x , y, O) - O. *

I f

*Fy(Xl, *

Yx, 0) # 0 , t h e n we choose *Yz=Yl *

a n d x 2 such t h a t *Fy(x, *

Yl, 0) # 0 for
x 1 ~<x ~<x z. W e shall p r o v e t h a t / ( x ) *=Yl *

is n o t t h e o n l y m i n i m i z i n g f u n c t i o n .
I t is no r e s t r i c t i o n to a s s u m e t h a t x 1 =Yl = 0 a n d Fy(0, 0, 0 ) < 0. Consider t h e func- t i o n y = ax2 w h e r e ~ > 0 is a c o n s t a n t . T a y l o r s f o r m u l a gives

F ( x , ~x 2, 2~x) =

*F(x, O, O) + ax~F~(x, O, O) * *+ l(~2xaF~(x, O. ax2, O. *

2ax)
*+ 2. ax~ .2~XF~z(...) + 4~x~F~z(...)).*

I f we consider a s u i t a b l e n e i g h b o u r h o o d of t h e o r i g i n a n d a s s u m e t h a t 0 < ~ < 1, t h e n F ~ ( x , 0 , 0 ) < - K < 0 , a n d t h e m o d u l u s of t h e e x p r e s s i o n in b r a c k e t s is n o t g r e a t e r t h a n K 1 9 ~2. x 2, w h e r e K a n d K 1 a r e i n d e p e n d e n t of x a n d ~.

This m e a n s t h a t

*F(x, ax e, *

2ax) ~< *F(x, *

0, 0) - K . ~x 2 + K 1 ~2x2 = *F(x, *

0, 0) - *ax2(K - *

~K1).
I f we choose ~ so s m a l l t h a t K - ~ K 1 > 0, t h e n we h a v e

*F(x, ax2, *

2~r ~< *F(x, *

0, 0) for
0 ~ x < ~ , for s o m e 8 > 0 .
A s i m i l a r c o n s t r u c t i o n c a n be c a r r i e d t h r o u g h for @2, Y2), a n d t h e r e s t of t h e p r o o f is obvious.

Therefore, if F ( x , y, 0) is n o t c o n s t a n t , t h e n we m u s t i m p o s e c o n d i t i o n s on t h e b o u n d a r y v a l u e s in o r d e r t o secure u n i q u e n e s s . T h e following t h e o r e m i l l u s t r a t e s t h i s p o s s i b i l i t y .

427

G. ARO~SSOrr *Minimization problems. I I *

T h e o r e m 3.5. *Assume that F(x, y, z) satisfies these conditions: *

*F(x, y, z) E C1; *

*2 A and 3; *

*the condition concerning existence and boundedness o/ minimizing /unctions which *
*was given in Theorem 2.1; *

*either ~F(x, y, O) >~ 0 / o r all (x, y) or < 0 / o r all (x, y). *

*~x *

*Let us denote the admissible linear/unction by l(x). Finally, we assume that *
m i n *F(x, l(x), l'(x)) *> m a x *F(x, y, 0), *

*X I ~ X ~ X ~ * *( X , y ) e K *

*where K is the set * *I xl ~ x <~ x 2 *

**[ **

**[**

*Yl <~ Y <~ Y2 (Yl >t Y >1 Y~)"*

*Then there is a unique minimizing ]unction/(x). Moreover, ]'(x) # 0 /or x 1 ~x<~x2. *

*Proo/. *W e k n o w t h a t t h e r e is a n a.s. m i n i m a l / ( x ) *EC 1. *I f / ' ( x ) = 0 for some x, t h e n
i t follows f r o m t h e T h e o r e m s 9 a n d 10 (with a t r i v i a l change) in [1] t h a t

M 0 ~ m a x *F(x, y, 0). *

*(X,y) E K *

B u t i t was p r o v e d in L e m m a 5 in [1] t h a t M0>~minxl<x<x ~ *F(x, l(x), l'(x)). *So t h e as-
s u m p t i o n t h a t *]'(x)~0 *for s o m e x l e a d s t o a c o n t r a d i c t i o n . F i n a l l y , i t follows f r o m
T h e o r e m 6 in [1] t h a t *](x) *is a u n i q u e m i n i m i z i n g f u n c t i o n .

*Remark. *I t is e a s y t o f i n d n e w r e s u l t s of t h e s a m e t y p e b y using n e w e s t i m a t e s .
I t s h o u l d be m e n t i o n e d h e r e t h a t t h e i n e q u a l i t y

M 0~> inf *F(x,g(x),g'(x)) *

*x1~x<~x2 *

h o l d s for e v e r y a d m i s s i b l e m o n o t o n i c f u n c t i o n *g(x)EC 1 *( c o m p a r e L e m m a 5 in [1]).

This gives

M 0 >~ sup [ inf *F(x, g(x), *g'(x))],

*g(x) * *xl<~x~x~ *

w h e r e t h e s u p r e m u m is t a k e n o v e r all such f u n c t i o n s .

I t is also clear t h a t t h e c o n d i t i o n on F x in t h e t h e o r e m c a n b e r e p l a c e d b y t h ~ c o n d i t i o n in T h e o r e m 1.1 (in t h i s case a s s u m e d t o h o l d for all y).

**Chapter 4. Comments and examples **

I n t h i s c h a p t e r , we shall i l l u s t r a t e some of t h e t h e o r e m s b y m e a n s of e x a m p l e s .
*Example 1. *S u p p o s e t h a t

*F(x, y, z) = G(x, y) + A y 2 + By + z 2, *

w h e r e *G(x,y) *is c o n t i n u o u s a n d b o u n d e d f r o m b e l o w a n d A , B are c o n s t a n t s .

**ARKIV FOR MATEMATIK, B d 6 n r 2 2 **
Is there a minimizing function? W e shall a p p l y T h e o r e m 1.2'. F o r m a l calculation
gives

*l O(x, y , M ) = § (M - G ( x , y) - A y e - B y ) ~/2, *
*9 (x, y, M) = - ( M - G(x, y) - A y 2 - By) 1/2. *

I f :r is defined for y

### ~Yl,

t h e n we see f r o m the expression for O(x,y, M) t h a t*a(y,M)*=O(y), when y--> + ~ . H e n c e t h e integral 11 in T h e o r e m 1.2' c a n n o t exist finite. I t follows in the same w a y t h a t none of t h e integrals Ie, /3 a n d 14 can exist finite.

So we can conclude t h a t *E ( M ) *is b o u n d e d for all M. Consequently there is a mini-
mizing f u n c t i o n for e v e r y choice of *xl, x2, Yl *a n d Y2-

*Example 2. *W e h a v e n o t studied t h e properties of minimizing functions in general,
we h a v e o n l y studied the special cases of a.s. minimals a n d unique minimizing func-
tions. The reason for this is clear f r o m t h e following example:

Choose *F(x, y, y') = x + y,2, *xi = Yl = 0, x 2 = 1 a n d Y2 = 0. This gives *M o >~ F(x~, y~, O) = 1, *
a n d if *g(x)~-0, *t h e n *H(g)= 1. *

Thus we h a v e M 0 = 1. I t follows t h a t i f / ( x ) is an admissible function, t h e n it is a
minimizing f u n c t i o n if a n d only if *I/'(x) l ~ V i - x * a.e.

Clearly, the fact t h a t / ( x ) is a minimizing function implies v e r y little a b o u t / ( x ) . This m o t i v a t e s the i n t r o d u c t i o n of a.s. minimals.

On the other hand, suppose t h a t we h a v e a minimizing function *g(x) which belongs *
*to C 1. *T h e n the local v a r i a t i o n m e t h o d , which was used in the end of t h e proof of
T h e o r e m 3.1, can be applied to *g(x). *This m e t h o d works also in t h e general case of
variable *~o(x,y). *F o r instance, it can be used to derive t h e following result: if

*Fz(X, g(x), g'(x)).o *

for all x, t h e n *F(x, g(x), g'(x)) = M o *for all x. (This is m e n t i o n e d in [6] with a s k e t c h
of a proof.) This result leads us once again to the differential equation *(dF/dx)" Fz = 0 *
for a.s. minimals.

*Example 3. *W e h a v e p r o v e d in Chapter 2 t h a t if *F(x, y, z) *satisfies certain condi-
tions, t h e n there is an a.s. minimal for e v e r y choice of the b o u n d a r y values, a n d we
h a v e also p r o v e d t h a t every a.s. minimal belongs to C 1.

*Consequently, there is a minimizing/unction in C 1. *

However, there need n o t exist a minimizing function in C 2, as can be seen f r o m the following example:

P u t *F(x, y, y') =y,2_cos2x, xl =Yl =0, x 2 =n7~ *a n d Y2 = 2 n . W e assert t h a t

### /(x)= [cos t l dt

is the only minimizing function. Clearly,/'(x) *>~0 *a n d *F(x, /(x), /'(x)) - 0 . If H(g) <~0, *
t h e n it follows t h a t *g'(x)~/'(x) *a.c. This implies t h a t g(x)~/(x), which proves o u r
assertion.

Finally, it is easy to verify t h a t / " ( x ) has a j u m p at x =ze/2 § k.~, k = 0,1,2 ... n - 1.

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