(4) x'fiXL'jfi), (3) L'_i(t)<x'(t), An application of a fixed point theorem

ROCZNIKI POLSKIEGO TOWARZYSTWA MATEMATYCZNEGO Séria I: PRACE MATEMATYCZNE XXV (1985)

L

P

(Krakôw)

An application of a fixed point theorem

Abstract. In this paper some conditions are given that ensure the equation Tx = x to have a solution with a given modulus of continuity (T operates on subsets of function spaces).

I. In [2 ] a theorem is given (Theorem 4) which in case of normed spaces can be expressed in the following form (see the proof).

T

. Let A be a com pact convex subset o f a normed space X.

Suppose that f : A -> X is a map such that fo r f (x) ф A, d ( f (x), x) > d ( f (x), A): = inf [ d ( f (x), y): y e ,4}. Then f has a f x e d point.

This theorem can be applied to cases when Int A = 0 and the Leray- Schauder theorem cannot be used.

Let С® (C/) denote the set of all real continuous (continuously differentiable) functions on the domain / = <0, 1 ). Let X := { x e C f : x(0)

= 0}, Y: = X n C } be two spaces equipped with the norm ||x||

= sup{|x(t)|: t e l } . Besides, let A a X consist of all functions x that satisfy the following conditions for tx, t2 e l , tl ^ t2:

(1) Ь м У - Ь - И Ь К х У - х ^ ) ,

(2) x(*2)“ * ( * i ) < M * 2 ) - - M * i ) ;

similarly let B c Y consist of all functions x such that they satisfy for t e l

(3) L '_ i(t)< x '(t),

(4) x'fiX L 'jfi),

w h ere L _ b L xe C f ( L _ l5 L 1e C } if th e d iffe re n tia tio n is m e n tio n e d ) an d fo r ti ^ f, t i , t e l , L _ i (t) L _ i ( t i ) ^ L i (t) L i ( t i ) (L _ i (t) ^ L± (t) in th e

“ d iffe re n tia b le ” case).

It can be seen that A, В are convex, the closure of В in X equals to A, A is compact and Int A = 0 .

L

1. Let x e A , zeX \ A be two functions satisfying fo r t b t2e l ,

316 L. P a s i c k i

(5) if (1) or (2) does not hold fo r z, then x satisfies (1) or (2), respectively, with the strong inequality.

Then d(z, A) < d(z, x).

P r o o f. Let us denote a — \\z — x\\ (a > 0). Let us consider the set Ж of all maximal intervals К for which \z — x\(K) = <a/2 , a >. The function z — x is uniformly continuous, which means that Ж is finite. Now we construct a special family of intervals. Let I 0 = <t0, t ^ e Ж’ contain the smallest number Po with \z-x\(p0) = a. If I k, pk are known, then I k+1 = ( t 2k+2, t2k+ i} e Ж contains the smallest number pk+l > pk for which (z —x)(pk)(z —x)(pk + 1) =

— a 2. Suppose we have obtained the family J — {V/k=o,i...,«- From the и

continuity of z — x it follows that sup (|z — x\ (t): t e l \ [ j I k] = : b < a. Let us fc

assume (z — x)(t0) > 0 and let for dk e ( 0 , 1 ) , к = 0, 1, . .. , n,

( 0 for t e <0, t2k),

ôk (t) = I L (_ 1)k(t2k) - ( x ( t ) ~ x ( t 2k)j)dk for t e ( t 2k, pk>,

[ ^k(Pk) for t e ( p k, 1>.

Suppose q i ^ t 0 ^ q ^ p 0, then for w := x + <50 we have w(q) — w (ql )

= w ( q ) - w ( t 0) + w (t0) - w ( q l ) = x { q ) ~ x (t0)AÔ0 (q) + x (t0) - x ( q l ) ^ L ^ q ) - - L 1(t0) + x {t0) - x ( q 1) ^ L l { q ) - L 1{ql ) and obviously, w { q ) - w ( q 1) ^ L_ t (q)

П

— L - i i q f ) . Now it is easily seen that у := x + ^ ô ke A. Besides, ^k(l) # 0, к k= о

= 0, . . . , n (see (5)) if x does not satisfy (1) or (2). The numbers dk can be i

established in such a manner that the sequence ((—l)1 £ <5k(l))j=o....*

k= 0

increases and is bounded by (a — b)/2. Now it can be seen that ||z —y|| < a.

C

1. I f T: A X is a continuous operator satisfying (5) fo r z : = Tx, the equation Tx = x has a solution.

L

2. Suppose x e A , z eY \ B are two functions such that

(6) i f (3) or (4) does not hold fo r z and a t e l , then (1) or (2) is satisfied with strong inequality fo r all tt < t < t2, tlf t2 e l , respectively.

Then d(z, A) < d(z, x).

P r o o f. Suppose z (t2) - z ( t i) > L x (t2) — L ^ r J . Then we have z'(t)

> L '1(f) on a subinterval of (tlf t2) and therefore x (t2) — х (^ ) < L l (t2)

— i.e., (5) holds.

C

2. I f T: A -> Y is a continuous operator satisfying (6) fo r z : = Tx, then the equation Tx = x has a solution.

L

3. Let x e B , zeX \ A be such that

(7) if (1) or (2) does not hold fo r z and tx, t2e l , t x ^ t2, then there exists t e ( t x, t2) such that (3) or (4) is satisfied with strong inequality, respectively.

Then d(z, В) = d(z, A) < d(z, x).

P ro o f. Suppose z (t2) — z (tl ) > L 1(t2) — L l (tl ) and x ' ( t ) < L l (t) for a

'2 l 2

t e ( t x, t2). Then х(г2) - х ( ^ ) = j x'(t)dt < j L x(t)dt = L x (t2) - L x (tx), i.e.,

n fi

(5) holds.

For an operator T: В -> X, let T: A -> X denote the continuous extension of T.

L

4. Let T. В -» X be a continuous operator satisfying (7) (z : = Tx).

Then (5) holds fo r Tx.

P ro o f. Suppose xeA \ B , (Tx) (t2) — (Tx) (f^ > L x (t2) — L x (tx) and x(r2)

— x ( t x) = L x(t2) — L x(tx) which means that x(t) = x (f1) + L 1 (t) — L x (гх) on

<tl 9 12) . It is seen that x can be approximated by the functions x ne B with x„ = x on <*!+£„, t2 — £n} , where

„ decreases to zero when n-> oo. So we have (Tx„)(r2)-(T x „) { t j > L l (t2) —L 1(t1), which implies х(г2) - х ( ^ )

f2 ~ £n *2

= lim J x'n(t)dt < j L x(t)dt = L x (t2) - L x (tx).

fl + £n * 1

C

3. I f T: В -> X is a continuous operator satisfying (7) (z : = Tx), then the equation Tx = x has a solution in A.

L

5. Let zeY \ B , x e B be such that

(8) if (3) or (4) does not hold fo r z and a t e I \ J (I\J = I), then (3) or (4) is satisfied with strong inequality, respectively.

Then d(z, A) < d(z, x).

P r o o f. Suppose z (t2) - z ( t x) > L 1(t2) - L 1(tl ). Then there exists a subinterval of (tx, t2) on which zf (t) > L x(t) and automatically, x'(t) < L x(t).

*2 *2

Hence x (t2) — x ( t x) = J x'(t)dt < | L x(t)dt = L x (t2) — L x (гД i.e., (5) holds.

*i *i

C

4. I f a continuous operator T: B -> Y satisfies (8) (z : = 7x), then the equation Tx = x has a solution in A.

C

5. L et T: A -» X be a continuous operator such that T\B satisfies (7) or (8). Then the equation Tx = x has a solution.

E

. Let — L _ x(t) = L x(t) = Lt, L > 0 and # f(x) = \P

= ( a , b } c z l \ x (t2) - x ( t t) = L i (t2 - t x) for tu t2e P , tx < t2), Qfix)

= U { P e ^ i(x )}, Q'i(x) = I\Qi(x) for i — — 1, 1. Assume f : I x R - > R is continuous and/ (0, 0) = 0. F o r / (t, x(t)), t e l , we define gx(t) = h (r, x(t)) as follows: gx: I -+ R is continuous, if (a , b ) is an element of ^ ,(x), then the

9 — Prace Matematyczne 25.2

318 L. P a s i c k i

graphs of Л (-,

( •)) and px are isometric on (a , b ) for

Px(t) = max { — L (t — a ) , f (t, x { t ) ) - f ( a , x(a))}

min \ L {t- a ), f ( t , x { t ) ) - f ( a , x(a))}

for i = — 1, for i = 1.

If (a, b) is any maximal open interval contained in QJ(x), then h (-, x (-)) and f (', x ( - ) ) k ( p x(')) + px(') have isometric graphs on (a, b) (к is continuous, k (0) = 0). From the continuity of / it follows that h: I x R -* R is continuous.

It is obvious that (7x)(t) : = h(t, x(tj), t e l , satisfies the assumptions of Corollary 1 and the assumptions of Corollary 2 hold for (Tx)(t):

t

= J h (t, x(t))dt. For x e B , there are obtained examples for Corollaries 3, 4.

о 1

Let us consider f ( t , x(t)) = — (x(t))1/3 = h(t, x(t)) {k = 0). We can see that / (Л) ф A as — (Lt)1/3 does not satisfy the Lipschitz condition and for x Ф 0 we obviously have ||x— 7x|| >%|| —x —7x|| which implies (1), (2).

Co r o l l a r y 5 .

L et T: A

X b e a continuous operator such that Т\в : В -> У and T\B satisfies (7) or (8). Then the equation Tx = x has a solution in A.

The previous considerations can be extended to the multi-dimensional case. Let x = (x 1, . . . , x ^ e A 1 x A 2 x ... x A k be a function, where A1, i

1, . .. , k, are analogs of A with the functions L'_ 1, l!1, and suppose T:

k k

P A1 -*■ P X is continuous. The domain of T may be the carthesian i = 1

= 1

product of the sets of type A1 or B l and similarly the range may contain X and У. Then if for every i = 1, . . . , k, ( Tx)' and x* satisfy one of conditions

( 5 ) - ( 8 )

(according to the case), then Tx — x has a solution.

II. Suppose now S>: 7 -^ 2 £" (E n euclidean n-space) to be an upper semicontinuous multi-valued mapping and D: = {] {(t, 3>(tj): t e l } . Let X

— {x e C £ : x (0 , •) = 0} and suppose А с X consists of all functions which satisfy the following conditions for all

^ ), (t2,

£

) ^ (

2> £2) (i.e. tj ^ t2, Й < f t for i = 1, n)

(9) x (t2, f t ) - x (t1, ft) ^ L 1(t2, f t ) - L 1(t1, ft), (10) L _ i( t 2, ft) ^ x (t2, f t ) - x ( t j , ft), (11) x ( t lt f t ) - x ( t i , f t ) < f t - f t ) , (12) v v .fttj, f t - f t ) ^ x {tu ft) x (f 1 , ft),

where L _ l5 L ls w _1? Wi are continuous on D, w ^ l (t1, 0) = wl (tl , 0) = 0, (9) holds for L _ i in place of x, and (11) — for w_i in place of x.

It is easy to verify that A consists of uniformly continuous functions as D is compact [1], Theorem 3, p. 116.

L

6. Let zeX \ A , x e A , be such that

(13) (11), (12) are satisfied with strong inequality fo r any t e { 0, 1), ^ £2 and (t, (t, £ 2) e A

(14) if fo r a fix ed £ and tl ^ t2, (fl5 £), (t2, <^)еД (9) or (10) does not hold fo r z, then (9) or (10) holds fo r x (at the same points) with strong

inequality.

Then d(z, A) < d(z, x).

P ro o f. Let a = \\z — x||. Let us consider a family Ж of cylinders {K x x A K} containing balls with the minimal diameters for which \x(KxAx)\

= <a/2, a } . The family Ж is finite. Let t' > 0 be the smallest number for which {f'} x ^ ( t ') n ( J [К x A K} Ф 0 and let L(q) denote m injlw ^t, Ç — —

- x ( t , Ç) + x (t, Ç J |, |w_i(t, Ç - Ç x) - x ( t , ç) + x (t, Çx)\: (t, Ç), ( t ,^ ) e D ,

^ <2;, d(Çx, Ç) ^ q, t ^ t'}. It is seen that Lis continuous, L(q) > 0 for q > 0 and L(0) = 0.

Let tK be the left end of К and let pK be the smallest number in К for which jz — x\ attains the value a in K x A K. We write

L ( A if (z — x )(K x A K) = <u/2, a ) , K ( L - j if (z — x )(K

A ff = < — a, —a/ 2 ), and for dKe ( 0, 1)

(0 for t < tK,

£) —

A ( L £) — L K(tK, £) — (x(L £) — x (tK, Ç)))dK for t e <(tK, Pjf),

lS (p K, { ) f o r t e ( p K, l ) .

Besides, let

/»*(», i) = » k (U {)2 L (d ((t, Q, D\I x AK))/diadK.

For pKQ = min {рк : К x A Ke Ж ] we can take any dKQe ( 0, 1 ). If pKl is a minimal number for which dKl is not defined yet, then it is possible to multiply all known dK by a small positive coefficient and choose the required dKi in such a manner that yKl = x + ^j8x e i (the sum is extended on all К with known pK) and yK approximates z “better”. It follows that there exists a function у = x + £ for which d(z, y) < d(z, x).

КеГ

Co r o l l a r y 6 .

L et T: A

X be a continuous operator such that ТхфА implies (13), (14) are satisfied ( z := 7 x ) . Then the equation Tx = x has a solution.

We can now generalize all considerations of part I in a natural way.

References

[1] C. B erge, Espaces topologiques, fonctions multivoques, Paris 1966, Dunod.

[2] L. P a sic k i, Retracts in metric spaces, Proc. Amer. Math. Soc. 78 (1980), 595-600.