On the Indices of Fixed Points of Mappings in Cones and Applications

On the Indices of Fixed Points of Mappings in Cones and Applications

E. N. DANCER

Department of Mathematics, University of New England, Armidale, New South Wales. Australia

Submitted by C. L. Dolph

Assume that K is a cone in a Banach space and A: K-t K is completely continuous. We obtain a formula for the index in K of a fixed point of A under the assumption that a linearization exists and satisfies an invertibility condition. We then use this formula to generalize some results of Amann on the number of fixed points of A to the case where K has empty interior.

The fixed point index in cones has been used a great deal to study the fixed points of positive mappings in cones. (See, for example, the survey article [2], where further references can be found.) There seems to be no known formula, however, for the fixed point index of a solution 4’ of x = A(x) in a cone K when A: E + E is completely continuous, A(K) E K, A is Frechet differentiable at y and I-A’(y) is invertible. (Here E denotes a Banach space and K is a cone in E.) This contrasts with the situation for mappings on all of E. The main result of this paper rectifies this situation. We obtain a simple formula for the fixed point index under the above assumptions. In fact, our methods could still be used if K is replaced by an arbitrary closed convex set.

We then use this result to extend some important results of Amann 111 on the number of fixed points of mappings in cones. In doing this, we discuss the index of minima1 solutions. Amann had to assume that his cone had nonempty interior. We do not need this assumption. This improvement is important because many of the most used cones have empty interior. Amann applies his results to elliptic partial differential equations. To do this. he has to construct an unusual Banach space. Our results enable one to much more easily derive his results by using natural subspaces of continuous functions.

Our method has the advantage that it naturally applies to elliptic partial differential equations on much more irregular domains and it can be used to treat mixed boundary conditions under reasonable hypotheses. Moreover, our results are much easier to use in application than Amann’s. In generalizing Amann’s results, we at the same time make some minor improvements in them.

131

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We also use our index calculations to prove some results on the continuation of solution branches in cones. In particular, these show that, under reasonable hypotheses, a simple arc of solutions in E which intersects the cone K must be entirely in K.

In Section 1, we obtain some preliminary results; in Section 2, we prove the result on the index of a solution in K; in Section 3, we study the continuation of branches of solutions and in Section 4, we generalize Amann’s results and also discuss the index of minimal solutions.

Note that, in Section 2, we prove our results for wedges rather than cones.

This is of some interest in itself because wedges occur naturally for systems of equations (cf [7]). Moreover, even if one tried to prove the results for cones, our proof requires the study of wedges.

1. PRELIMINARIES ON WEDGES

We assume that W is a wedge in a real Banach space E such that E, = W - W is dense in E. Recall that a wedge W is a closed convex subset of E such that a WC W for a > 0. A wedge is said to be a cone if Wn - W = (0). The reader is warned that the terminology in the literature varies widely. As usual, we write y > x if y -x E W.

IfyEW,let W,=(xEE:y+yxEWforsomey>O}.Itiseasytosee

that WY is convex, f y E WY, W g WY, and a WY c WY for a > 0. Thus W, is a wedge containing W, y, and -y. Let S, = {x E I?‘,,: -x E W,,}. Since W:,, is a wedge it is easy to see that S, is a closed subspace of E and y E S,.

Let W*= (fEE*:f(x)>O on W} and y’={fE W*:f(y)=O}. Here E* denotes the dual of E.

LEMMA 1. @,, = ‘(y’) E {u E E: f(u) > 0 for all f E y’} and S, = {x E E:f(x) = Ofor aflf E y’}.

Remark. Note that our annihilators and preannihilators are not the usual ones.

ProoJ: (i) We first show that WY c ‘(y’). If x E W,,, y + yx E W for some y > 0. Thus, if f E yi, 0 <f (y + yx) = rf (x) (since f(y) = 0). Hence, since y > 0, f(x) > 0. Thus x E ‘(y’) and hence WY 5 ‘(y’). Since ‘(y’) is closed, it follows that W,, cl ‘(y’).

To complete the proof, it s&ices to show that ‘(y’) G Wy. Suppose by way of contradiction that z E ‘(y’) - W,,. Then, by the Hahn-Banach theorem, there is an fin E * and a c in R such that f (z) < c and f (x) > c for all x in WY. Since 0 E WY, c < 0. Then, because a WY z-WY for a > 0, it follows that c = 0. Thus f(x) > 0 in WY. Since W c WY, f E W* and,

because ky E WY, we see thatf( y) = 0. Thusf E y-. Sincef(z) < c = 0 and f E y’, z 6Z ‘(y’). Thus we have a contradiction and (i) is proved.

(ii) If u E S, and

f

E yL, then u E WY and thusf(u) > 0. Similarly, since --u E WY, f(-u) > 0, that is, f(u) ,< 0. Hence f(u) = 0. Thus SY c (U E

E:

f(u) = 0 for all f in y’). Conversely, if x E {u E

E: f(u) = 0

for all f in J’-}.

.Y E @L by part (i). Since f(-x) = 0 for all f in ~3~. we see similarly that --x E

W,.

Thus x E S,, as required. 1

This lemma provides convenient formulae for WY and SY. Note that, in general, SY is larger than EY, where

E,. = {u E E: -;y < u < 15~

for some y > 0) (cf. [ 18, p. 1361). On the other hand, EY = S,. if (i) S?, =

E

and

W

has nonempty interior, or if (ii)

W

is the natural cone of a Banach lattice (cf.

( 19, Theorem 2.6.3 1).

An element y in

W

is said to be a

demi-interior point

if S,. = E (equivalently if WY =

E

or, by Lemma 1, if y = {O)). Note that this is weaker than assuming that y is a quasi-interior point (that is, EL=

E).

Let & denote the quotient space

E/S,

and pY the image of

W,.

under the quotient mapping. Then %‘Y is a closed cone in $,. It is closed because w,. + S,. s WV. It is a cone because the coset u + S, is in gV if and only if u E I@,‘and because u and --u belong to I?$ implies that u E S,. Moreover.

F& - py is

dense in ,!?,, since

W - W

is dense in

E.

Now assume that 77

E

+

E

is a compact linear operator such that T( W,.) E Iv,,. If U E s,,, fu E ws and thus

Tu

E WY and

-Tu = T(-u)

E @,. Hence

Tu

E EV. Thus

T

maps SY into itself. Hence

T

induces a compact linear mapping

T

of gY =

E/S,

into inself. Since

T( w,) G w,..

?;($)z cy..

We say that

T

has

Property

a if there is a t in (0, 1) and a ~$3 in @‘,.\S, such that )L’ -

tTw

E S,. Let

r(T)

denote the spectral radius of

T

and

o(T)

the spectrum of

T.

More generally, if

W

is a wedge in

E.

let T,, (7J denote the spectral radius of

T

on

W

(cf. ( 18, p. 2661).

LEMMA

2. Assume that I - T is invertible.

(a)

T has Property CL t> r(F) >

1.

(b)

If T has an eigenvalue in

(1,

co) M’ith a corresponding eigenvector in ii;‘,\S,.. then T has Property a.

(c) rf r(T) <

1,

or if a(T)n

(1,

00) =

0,

then 7‘ does not have Property a.

(d)

If T has Property a, there is a z in I?? such that the equation x ~ TX = : has no solution in my.

Proof:

(a)

T

has Property

a

if and only if there is a t in (0, 1) and an x in w,.\S,. such that x -

tTx E S,.

Thus

T

has Property a if and only if there

is a t E (0, 1) and an x’ in qY such that z-- ti%= 0 (where x’= x + SL).

Hence T has Property a if and only if F has an eigenvector in W, corresponding to an eigenvalue in (1, co). Since “i;; - RY is dense in g,,., (a) now follows from the Krein-Rutman theorem ([ 18, Appendix, 2.4 Corollary]) applied to i;.

(b) This is trivial.

(c) Since T is compact and T(S,) G S,, [ 16, Theorem 1.5.31 implies that (Z - AT))’ (S,) c S, if I E R\cr(T). Thus T has Property a implies that a(T) n (1, oo) # 0 and hence that r(T) > 1.

(d) If T has Property a, then thercexist t E (0, 1) and w in m,\S, such that w - tTw = m E S,. Then -w @ W,, and (-w) - T(-w) = -m + (1 - t) Tw E WY. Since the equation x - TX = z has a unique solution for each z in E (because Z - T is invertible), the result follows. m

Remarks. (1) The result is useful because Property a is of some impor- tance in the next section. Note that, since T maps w, into itself, it is not difftcult to show that r(n E o(T). A variant of the result is still true if we only assume that TIFY is completely continuous. (We must replace r(T) by rw,(T’) and so on and some of the proofs need to be changed.)

(2) If a(0 f? a(T Is,)

n

(1, co) = 0, then T has Property a if and only if T has an eigenvalue in (1, co) with a corresponding eigenvector in Ws\SY.

(3) If S,, has a closed complement M in E and P is the corresponding projection onto M, then T has Property a if and only if r(PTI,4,) > 1. (This is simply a special case of Lemma 2(a).) This remark is useful in applications.

(4) By similar arguments, T has Property a if and only if there is a t in (0, 1) and a w in Z&S), such that w - tTw E S,,.

Finally, we shall need some approximation and extension results.

LEMMA 3. (i) Zf M is a closed subspace of E and T: M+ M is completely continuous and positive homogeneous (that is, T(ax) = aT(x) if a 2 0), then T can be extended to a completely continuous positive homogeneous map of E into M.

(ii) Suppose that Z is a wedge in E, that T, is a dense convex subset of Z such that aT, c T, for a > 0, that L: Z + Z is completely continuous and positive homogeneous and that E > 0. Then there is a continuous positive homogeneous map L, : Z + Z such that L, {x E Z: 11 x )I = 1 } is contained in the convex hull of aftnite number of points in T, and IJL,x- Lx/l <E llxlj on Z.

ProoJ: (a) Let M, and E, denote the unit spheres in M and E, respec- tively. By the Dugundji extension theorem [20, Proposition 3.5.81, TI,, can

be extended to a continuous mapping of

E,

into the convex hull of T(M,) in M. Thus

T(E,)

c

M

and

T(E,)

is compact (since

T(M,)

is). We then extend

T

to all of

E

by positive homogeneity.

(b) Since L(z,) is a compact subset of Z (where Z, = (X E Z: //x(/ = 1).

it has a finite is-net {xi}:=, in Z. Since T, is dense in Z, we can choose yi E Ti such that // yi - xi/( < SE. Then ( yi}f_ , is a finite F net for L(Z,). If we examine the construction of the corresponding Schauder projection

P

(cf.

[ 12, p. 146 I), we see that I/

Pz

- z/l < E on L(Z,) and the range of

P

is contained in the convex hull of (yi}:,,. We define L, on Z, to be

PL

and then extend by positive homogeneity. It is easy to see that L, has the required properties. fl

Remark,

As a consequence of Lemma 3, we see that, if

L: E ---) E

is compact and linear, L(w?) G WY -and E > 0, then there is a positive homogeneous continuous map

L,

:

WY

+ WY such that /IL,

x - Lxll < c (/x :’

on WY and

L,(?Y)

is contained in the convex hull C of a

Jnite

number of points ( yi}f=, in

W,.

Here P” denotes the unit sphere in gY. Then there is a y > 0 such that y + tyyi E

W

if 0 < t < y and i = l,...,

k.

(Recall the definition of

W,.)

Since

W

is convex it follows that y +

tz

E

W

if z E C and 0 < t < ;I.

Hence y +

L,x

E

W

if (Ix/( <y and x E WY (since

L,xE //x/I C).

We shall use this in Section 2.

2. CALCULATION OF THE FIXED POINT INDEX

We assume that A : W -+ W is completely continuous,

where

W

is a wedge in the Banach space

E.

For simplicity, we suppose that

W - W

is dense in

E. We also assume that y E W is a fixed point of A and that A is Frechet differentiable at

y in

W,

that is, there is a continuous linear operator

L

on

E

such that

IjhJj-‘jJA(y+h)-A(y)-Lh+O as h+O

and

y+hE W.

Finally, we assume that L is compact on E.

It is easy to see (cf. 112, Lemma 3.11) that

L

maps bounded subsets of

W

to precompact sets and thus this last assumption automatically holds if

W

is reproducing (that is, if

W

-

W

= E). Simple examples, however, show that, in general, the other assumptions do not force

L

to be compact on w,. (even for quasi-interior points).

The following is our main result.

THEOREM 1.

L maps W.,, into itself. If I - L is inoertible. then:

(i) index..(A,

y) = 0 if L has Property a on W,. ; and

(ii) index,(A, y) = index,,(L, 0) = index,(L. 0) if

L does not hate

Property a on WY.

Note that index,@, y) denotes the index in

W

of the isolated solution y and that index,@, 0) and index&, 0) are the indices of linear maps on a linear space and thus they can be evaluated by a simple well-known formula (cf. [ 11, Theorem 2.4.5.1).

Remarks.

(1) In particular, we see that index,@, JJ) = 1 if r(L) < 1 and that index&l, y) = index,(A,y) if y is demi-interior to

E.

This second special case is of particular importance. We conjecture that this second special case is still true if we do not assume that Z - A ‘(u) is invertible but assume that y is an isolated solution of x = A(x) in

E.

In the general case, Lemma 2 of Section 1 is of use in deciding whether

L

has Property ~1 on WY.

(2) For most of Theorem 1, it would suffice to assume that

L

is compact and continuous on WY and h #

Lh

if

h E qy\{O}.

(3) It is easy to check that our assumptions imply that y is an isolated solution of x = A(x) and hence the index must be defined.

(4) Theorem 1 is still true if

W

is replaced by an arbitrary closed convex set C. Two slight changes need to be made. First, we now define y1 to be (f E

E*:

f(x) > f(u) for all x in C}. Second, instead of assuming that

W- W

is dense in

E,

we assume that C is not contained in a closed hyper- plane in

E.

Otherwise, the theory is as before.

We prove the result by a series of lemmas.

LEMMA

1.

L maps & into itself:

Proof:

IfhEW,,,thereisay>Osuchthaty+thEWifO<t,<y.

Thus, if 0 ,<

t

< y, we see from the differentiability of A at y in

W

that

A (y + th) = Ay + tLh + tr(t),

where r(t) -+ 0 as t--t 0. Since A(W) c

W

and Ay = y, we have that y +

t(Lh

+

r(t))

E

W

if 0 < t < y. Thus, by the definition of W),,

Lh

+ r(t) E

WY

if 0 <

t

< y. Since

r(t)

+ 0 as t -+ 0, it follows that

Lh E W,, .

Thus

L( W,,)

G WY. Since

L

is continuous, the result follows.

Since Z-L is invertible, there is a

K

> 0 such that

Ilx - Lxll > K I/XII (‘1

for x E WY. Let E =

bK.

By Lemma 3(ii) of Section 2 and the remark after it, there is a y > 0 and a compact positive homogeneous map

L, : @, -+ @,

such thaty+L,hE Wif

hE

W,, and

l(h(l<y

and such that

J(Lh-L,hll<

E((hl( on WY. I

Let

W,(z)

= {U E

W:

I) u - z]j <

r}

and

WY,(z) = {u E WY: ]Iu -z )I < r}.

LEMMA

2.

index,,.(L, 0) = deg,(L,, W,(JJ))

if 0 < r < ‘J, where ,?, u = y + L,(u .- y).

ProoJ

Since

L

is positive homogeneous,

index,,(L, 0) = degw,(L. W,.,(O)) (2) if r > 0. Since ((L,x -

Lxl(

<E /(x/( on pY and /(x -

Lx11 2-4~ //x//

on @\-. we see that x#tL,x+(l

-t)Lx

if O<t< 1 and xE W,.\(O\. Thus. by homotopy invariance,

deggv(L W,.,(O)) = deg,x(L,. W,.,(O))

if Y > 0. Moreover,

deg,,(i:, , Y + W,,(O)) = de,->@, , W,.,(O)),

(31

(4)

where V = y + %‘? and z, is defined in the statement of the lemma. This last equation follows from the commutativity theorem for the degree. (In 113.

Theorem D3 I. take g,(x) = x + y and gz(x) =

L ,(x - y).

If 0 < I < y, L, maps W,,(y) into a subset of

W.

(See the discussion preceding the statement of Lemma 2.) Thus by the restriction property of the degree (cf. 113, Section C. Property 4’]),

de&@, , Y + W,,,(O)) = da,@, . W,(Y)) (5) if 0 < r ( y. (Note that (-v + W,,(O)) n

W

= W,.(V), as is easily seen from the definition of I?$.) The result follows from (2~(5). 1

LEMMA

3. If r is sufjciently small and positive.

deg,,(A W,(Y)) = deg..(~, , W,.(y)).

Prooj:

Since

L

is the derivative of

A

at .r. there is a p > 0 such that p K ;’

and

J(AX

- Ay - L(X - y)j\ < F

Ii-Y - .I/1 (6)

if x E W,(J). (Recall that y occurs just prior to the statement of Lemma 2.) In addition.

if +r + h E W,( ?I) c y + W,,@(O). Thus, since J’=

Ay.

j/Ax - 4’ - L,(x - y))! < 2& 11-y - ?‘/I (8)

if xE W,(y). By (1) and (7),

ll~-L,~ll~3clI~ll (9)

if y + h E W,(y). Thus, by (8), (9), and a simple estimation, we see that x #

&4x + (1 - t)(y + L,(x - y)) if 0 <t < 1 and x E W,,(y)\{O}. Hence, by homotopy invariance,

if 0 <

r

< ,u. This proves the lemma. I

By Lemmas 2 and 3, we only need to evaluate indexwV(L, 0) in order to prove the result.

LEMMA 4. (i)

IfL has Property

a, indexw,(L, 0) = 0.

(ii)

IfL does not have Property a,

indexw,(L, 0) = indexSY(L, 0) = index,(L, 0).

ProoJ

(i) Suppose by way of contradiction that indexpV(L, 0) # 0.

Thus, since

L

is linear,

deg@,,(L,

W,,,(O)) # 0. If

h

E WY, we see by using the homotopy (x, t) -+

Lx

+

th

that deg,-(L +

th,

W,,(O)) # 0 if t is sufficiently small and positive. Thus, if t is small and positive, the equation x =

Lx + th

has a solution in KY. It follows that the equation x =

Lx

+

h

has a solution in p,, for all

h

in WY. Since this result contradicts Lemma 2(d) of Section 1, we have completed th proof of (i).

(ii) By Lemma 3(i) of Section 1, there is a positive homogeneous completely continuous mapping C: WY--f S,, such that Cl,) =

LlsY.

If

hE

W,\{O} and

tE [O, l), hftLh+(l-t)Ch.

(Otherwise,

h-tLh=

(1 - t)

Ch

E S,. Thus, since

L

does not have Property

a, h E S,.

Since Clsy = LISP) it follows that

h

=

Lh,

which contradicts our assumptions.)

Thus, by homotopy invariance, indexw,(L, 0) = indexw,.(C, 0)

= indexs,,(C, 0)

= index& Is,,, 0)

(since C( WY) G S,V) (since Clsy = L Is,>.

That index,(L, 0) = indexSy(LISyr 0) can be proved by almost exactly the same argument. (Replace

WY

by

E

and use Remark 4 after Lemma 2 of Section 1.) This completes the proof. I

Theorem 1 follows immediately from Lemmas l-4 of this section.

Finally, note that, for the case of wedges, Theorem 1 is new even if y = 0.

Indeed, it could be used to generalize the global bifurcation results for a particular sort of wedge in [7] to arbitrary wedges.

3. STABILITY OF SOLUTIONS IN CONES

In this section, we discuss rather briefly the application of theorems on the calculation of the index to the continuation of solution branches in wedges.

These results are of some interest in themselves and, in addition, we use one of the results in Section 4.

The main use of knowing that the index in W of an isolated solution .Y of x = A(x) (where A(W) c W) is nonzero is that it implies that nearby mappings which map W to itself (and are completely continuous) also have fixed points in W near y. Note that this is nontrivial even when A: E x ]O, 1 ] -+ E is completely continuous, A( W x ]O, I]) c W, A is Frechet differentiable and I - A;(y, 0) is invertible. (Here A’, denotes the partial derivative with respect to the first variable.) It is nontrivial in this case because, while x = A(x, t) has a unique solution x(t) near y for 1 near 0 (say 0 < t < a), we do not know whether this solution lies in W. (Of course. it does if 4’ is interior to W.) Indeed, x(t) does not belong to W in some cases where ind,,,(A( , 0), y) = 0. For a more precise result, assume that y is an isolated solution in W of x = A(x, 0) such that index,.(A( , 0), J) # 0, where A: W X [0, 1) + W is completely continuous. Choose a bounded neighbourhood V of .Y in W such that x # A(x, 0) on aP’ (where al’ means the boundary of V in W) and then choose 6 > 0 such that x # A(x, t) if x E aV and t E ]O, S]. (This is certainly true if 6 is sufficiently small.) Then, by homotopy invariance, deg,,(A( . t), V) = deg,,.(A( , 0), V) = index,+(A( , 0), y) # 0 if 0 < t < 6. Thus, for each t in 10, S] the equation s = A(x, t) has a solution in I’. In fact, a simple modification of the proof 01

[ 14, Lemma A5] shows that there is a connected subset of ((x. t) E 7~

(0.6 ]: x = A(x, t)} intersecting both t = 0 and t = 6. It follows from our comments above that in the case where index&A, J’) # 0 and the assumptions of the implicit function theorem hold (as above), then x(t) E W for 0 ,< t < influ, 6). Here x(t) denotes the unique solution of x = A(x, t) in E near y.

In many cases a more global result is needed. For instance. we need one in Section 4. Results of this type are well known but it is difIicult to find a reference. For completeness, we sketch a proof.

PROPOSITION 1. Assume that A: Wx (O,oo)+ W is completeij~

continuous, that ,a > 0, and that y is an isolated solution in W of x = A(x. u) with indW(A( , p), y) # 0. Suppose E, 6 > 0 are such that x # A(x, A) if either (i) XE W, 0 < lix-yll < E and L =,u or (ii) xE W. I/x --l/I = E amf

11 -,u( < 6. Let T denote the component of {(x, 1) E WX (0, co): x=

A(x, l)}\((x, I): I/x - ~11 < E, p - 6 < I < ,uu) containing (y,,u). Then (i) T is unbounded or (ii) inf{A: (A, x) E T} = 0 or (iii) T contains (z, A) with 2. = /I - 6.

Remarks. (1) Such an E and 6 can always be found. (First choose E and then choose 8.)

(2) Intuitively that (iii) holds for all small positive 6 means T contains a loop.

Proof (sketch). We only sketch the proof because it is a simple modification of standard proofs (cf. [ 151 or [6, Theorem 2)) and is, in fact, easier than the two mentioned above. Let g = ((x, A) E W X [O, co): x = A(x,A)} and let g, =g\((x,A): ]lx-~]i<e, p-6 <A <,u}. Suppose the result is false. Then, by a standard separation argument, a, = T, U T,, where T, is compact, T, n T, = 0, T E T,, A > 0 if (A, x) E T, and

{(x, A) E G: /Ix - y ]I < E, A= p - S} E T, . It is easy to see that we can choose a bounded open subset 6’ of W x (0, co) such that T, c 19, sn T, = 0 and 0 does not contain any points (x, A) with either (i) I/x - y]J < E and A = p - 6 or (ii) ((x - y]I = E and ,U - 6 < 1 < ,u. Hence, if (x, A) E 80 n .@, then Ilx-yl(<~and~--6~<~1.Thusx#A(x,~)if(x,1)~aeandeither1~

,u - 6 or A >,p. Hence, by homotopy invariance,

deg,(A ( ,A>, 8,) = 0 (9)

if A <,u - 6, where 0, = (x E W: (x, A) E t!?}. (Note that 8, is empty.) Similarly deg,(A( , A), e,) = 0 if A 2~. Define 8’ = 8u ((x, A):

llx - YII < ~1. BY our construction of e, ae; n G is empty if P - 6 < A ,<,K Hence, by homotopy invariance,

deg,(A( ~~-4~~:-,)=deg,(4 4J:). (10)

Since 0; and t9,, contain exactly the same solutions of x = A (x, ,u) and since deg,(A( , y), 8,) = 0, the right-hand side of (10) must be zero. Since, by our construction of e, ewe, n W,(y) is empty, it follows that (10) becomes

deg,(A( ,iu - 4, e& + deg,(A( .P - 4, W,(Y)) = 0.

Thus, by (9), deg,(A( ,,u - S), W,(y)) = 0. By homotopy invariance, the left hand side is deg,(A( ,,u), W,(y)) = ind,(A( ,p), y) # 0. Hence we have a contradiction. This completes the proof. m

With a little care, a slightly more precise result could be proved (as in [6, Section 1 I). Many variants of this result can be proved.

COROLLARY.

Assume that A satisfies the assumptions of Proposition 1

and that there is a neighborhood V of (y,,u) in E X (0, co) such that

A

extends to a map of (W

x

[0, (lo))

U

V into E and such that V, = ((x, /1) E V:

x = A(x, J.)} is homeomorphic to an arc or a circle (that is, a connected

l-

manifold). Then V, c W

x

(0, 00).

Remark.

This result makes it rather difficult for part of an arc of solutions to leave a wedge except near a bifurcation point.

Proof:

We assume that

V,

is homeomorphic to an arc. (The other case is similar.) By Proposition 1,

V,

is necessarily an open arc and thus we may assume that

V,

= ((x(s), L(s)): s E (-1, l)), where x(0) = 4’ and /1(O) =,u.

Since the component of Wn ((x, 1): L > ,u) containing (y, ,u) is not contained in a small neighbourhood of 01, p) in

W

X

10, co)

(bq Proposition l), we must have that L(s) > iu for small positive s or that 1(s) > ,U for small negative s. (Note that both cannot occur or we would contradict the analogue of Proposition I with ,u - 6 < L < ,U replaced by ,u < 1 < ,D + S.) We consider only the case where L(s) > p for small positive s since the other case is similar. We shall show that s(s) E

W

for 0 < s < 1.

(The case where s < 0 follows by similar arguments.) If not let s!. = inf(s E (0, 1): x(s) 65

W).

Thus 0 6 se < 1. It is then easy to see that, if c. ti are small and positive, then the component of

containing (y, ,u) is ((x(s), n(s)): s E (0, se I}. Since this contradicts Proposition 1, we have completed the proof. 1

Note that the solutions are locally an arc near (J,P) if

A

is continuously differentiable and the map

(h, t) + h - A;( y, p) h - A;( y, ,u) t

is onto (where A { and

A;

denote the respective partial derivatives).

Finally, we mention one final result on perturbations. Recall that, if

K

is a normal cone, if

A: K

+

K

is completely continuous and increasing (that is, A(x) >

A(y)

if x > y > 0) and if

A

has a fixed point in

K,

then

A

has a minimal fixed point 2 in

K.

If X is isolated and has nonzero index in

K.

then it is easy to see that any completely continuous increasing map close to A has its minimal fixed point near 2. In other words the minimal solution varies continuously with perturbations if X is isolated and has nonzero index.

(If A and the permitted perturbations all map each order interval to ;I bounded set, the normality condition can be avoided.)

4. ON SOME RESULTS OF AMANN

In [ 11, Amann obtained some pretty results on the solutions of equations x =

A(x,

A) in cones. He assumed that the cone

K

has nonempty interior and

he placed strong assumptions on A. We prove similar results for arbitrary cones. Moreover, our assumptions on A weaken his even for cones with interior. Many, but not all, of our proofs are based on his.

We assume that K is a normal cone in E with K - K dense in E, that A:

K X [0, a) -+ K is completely continuous, that A(x, 0) = 0 for x E K and that A is increasing (that is,if x > y > 0 and 12 ,u > 0, then A(x, A) > A(y, p)).

In addition, we assume (i) there is an open set V in E x [0, co) with V 2 K x [0, a) such that A is continuously differentiable on V and Ai(x, I) is compact for (x, I) E K X [0, co); and (ii) for each solution (z, ,a) of x = A(x, 1) in K x (0, 00) with r(A ;(z, ,a)) > 0, r(A ;(z, ,u) I -A ;(z, ,a)) has a one- dimensional kernel and A;(z,,u) is not in its range.

A few remarks need to be made on these assumptions. First, the assumption that K-K is dense in E can be avoided by replacing the spectral radius on E by the spectral radius restricted to K-K. Second, the normality of K could be avoided if we assumed that A mapped order intervals to bounded sets. Finally, with a good deal of care, it would suffice to consider derivatives at points in the cone and in directions in the cone (as in Section 2). Condition (ii) on the derivative is only required at solutions with 3, > 0. This is often an advantage because these solutions are often demi-interior points of K x [0, co) and the derivative there may have properties which are not true at all points of K x [0, co). (For example, all solutions are demi-interior if A(0, A) is demi-interior for 1 > 0.) Many results are known which ensure that, under appropriate assumptions on A{(z, ,u), N(r(A ;(z, p)) Z - A;(z, p)) is one dimensional, where N(B) denotes the kernel of B. See, for example, [ 11, 18, 19). Another result of this type is as follows.

If a compact positive linear mapping B with r(B) > 0 has the property that (,U - B) ’ maps nonzero elements of K to demi-interior points for some 1 > r(B), then N(r(B) I - B) is one dimensional. (We call an operator with this property a demi-irreducible operator.) This follows by an easy modification of the proof of Theorem 3.2 [ 18, Appendix]. It is easy to show that the above condition is equivalent to assuming that, for each x E K\(O}

and each

f

E K*\{O}, there is a nonnegative integer n such that f(B”x) > 0.

(This follows simply from the Neumann series for (U - B) -‘.) Finally, the second condition in (ii) is equivalent to assuming that f(A;(z. ,a)) > 0, where f E K* and f spans N(r(AI(z,,a)) 1- A:(z,,a)*). (Note that f exists by the Krein-Rutman theorem.) Thus the second condition in (ii) holds if either (a) A;(z,,n) is demi-interior to K or (b) A;(z,p) # 0 and f is strictly positive.

(The latter condition in (b) holds if A;(z,,u) is demi-irreducible.) It is not difficult to show that the second part of condition (ii) is a stronger condition than requiring z(=A(z,,a)) does not belong to the range of r(A;(z,,u) Z - A;(z,P).

If, for some II > 0, the equation x = A(x, A) has a solution, it follows (cf.

( 11) that it has a minimal solution z(n) in K. (In other words, x > z(L) for

every solution x in K of .x =,4(x, A).) Moreover, as in 111, (A: x =A(x. A) has a solution in K} is an interval T in [0, co) with 0 E T, z(k) is increasing on T and z(n) is left continuous on T. (It is for these results that normality of K is used.) Let A* = sup T.

PROPOSITION 2. Zf 2 E T and z(A) # 0, then r(A i(z(A), A)) < 1.

ProoJ Suppose not. It is convenient to write z instead of z(n). If 0 < t < 2, then t E T, z(t) <z and z(t) = A(z(t), t). Thus, by a simple calculation, h(t) = z - z(t) E K and is a solution of

h(t) = C,h(t) + (A - t) R,. (11) where C, = :iA A;(sz + (1 - s) z(t), A) ds and R, = J’i Ai(z(t), st $- (1 - s) 1) ds. Smce z(t) is left continuous at A, C, + A ;(z, 1) and R, -+ A ;(z, 1) as f--f 1. Thus, since the spectral radius is continuous for compact operators

(by

near ll”fll

NOM

10, Theorem 4.3.1 and Section 4.3.5 I), it follows that Y(C,) > -1 if t is A. By the Krein-Rutman theorem, there is an f, in K* such that

= 1 and CTfl = r(C,)f,. Thus, by (1 1 ), and a simple calculation (1 - r(C,))f,(W)) = (A - t)./i(R,).

r(C,) > 1, S,(h(t)) > 0, and f,(R,) > 0 (since f, E K*. h(t) E K and R, E K). Hence we must have that &(/z(t)) =f;(R,) = 0. Suppose that f, --f as t--t 1, where f E K*, lifll = 1 and (A’(z, n))*f= v(A I(z, A))$ (This will be proved in a moment.) Since f, + f and R, --) Ai(z, A) as t + 2 and since f,(R,) = 0, it follows that f(A;(z, A)) = 0. Because this contradicts the second

part of condition (ii), we have the required contradiction.

It remains to show that f,- f as f + 1. Since C, -A{(z, 2) and r(C,) -+

r(A {(z, A)) as t ---t 1, we see from a simple calculation that

as

t

--t A. Thus, by a simple compactness argument, a subsequence off, must converge to an eigenvector in K* of norm 1 of (A ;(z, A))* corresponding to the eigenvalue r(A{(z, A)). Since, by our condition (ii), this limit must be J the whole sequence

f,

must tend to

f

as required. I

Remarks. (1) The argument above can be used to establish the following result which is of some interest in its own right. Assume that A:

K --) K is completely continuous, continuously differentiable and increasing on the normal cone K, that A’(x) is compact for x E K and that A has a fixed point W. Thus, as before, A has a minimal fixed point z. Then r@‘(z)) < 1 if T(z) > 0 for every nonzero 7 E K* n N(r(A’(z)) I -.4’(z)*).

(For example, this condition holds if z is demi-interior to K.) This result is

proved by applying the argument in the proof of Proposition 2 to the map (x, t) + tA(x). Note that, in the case where A depends on A and r(A ;(z, A)) Z - A;(z, 1) has a one-dimensional kernel, the assumption that f(z) > 0 is weaker than assuming thatf(Ai(z, A)) > 0 (since z = A(z, 1)). It is possible to prove many variants of the above results by using known results on the spectra of linear operators in [ 12 or 181. For example, if A ‘(z) does not have an eigenvector in K corresponding to the eigenvalue 1, it s&ices to assume that?(z) > 0 for some nonzeroJE K* f7 N(r(A’(z)) Z-A’(z)*). We conjecture that r(A’(z)(,.) < 1 always. (Our notation is that of Section 1.) The above argument can- be modified to prove that this is true if A’(z) does not have an eigenvector in K corresponding to the eigenvalue 1 and either (a) S, = ,!?1 (for example, in the case of a Banach lattice) or (b) more generally, z is demi-interior to Sz and (K f7 S;) - (K f’ S,) is dense in S:.

Note that there are examples of afline positive maps on R” (with the usual order structure) for which the minimal solution z satisfies r(A’(z)) > 1. Thus some extra condition is needed. Finally, note that, if A’(z) is a demi- irreducible operator and if z # 0, then it is not difficult to show that z must be demi-interior to K (since A’(z) maps S, into itself).

Let & = {(x, 2) E V x [0, co): x = A(x, A)} and 3 = G n (K X [0, 03)).

LEMMA 1.

Assume that (z,,u) is a solution of x = A(x, A) in K

X

(0, co) and that r(A i(z, ,u)) = 1. Then there is a neighbourhood 8 of (z, ,u) in V x (0, co), an E > 0 and continuously dtferentiable functions 4: (-e, E) + R and w: (--E, E) -+ E, 3 (x E E: y(x) = 0) such that q+(O) = ,u, w(O) = 0 and a n 19 = {(z + ah + ~(a), @(a)): a E (--E, E)}. Here h E K spans N(Z -A ;(z, p)) and f’E K* is such that T(h) = 1. Moreover, I- A;(w(a), #(a)) is invertible if and onZy if o’(a) # 0 (where w(a) = z + ah + w(a)).

ProoJ This is simply a minor variant of a result of Amann [ 1, Theorem 2.11. It follows by applying the implicit function theorem to the map ,(x, 1) -+ x - A(x, A). Our assumptions imply that the derivative T(z, p) of this map is onto at (z, p) and that {(x, 0): x E E,?(x) = 0) @ ((0) x R) is a complement to N(T(z, p)). For the last part, we simply note that (w’(a), #‘(a)) is in the kernel of T(w(a), $(a)) and the kernel of Z(w(a), #(a)) is one dimensional. (A similar argument appears in the proof of (4, Theorem 1.171.) I

The corollary to Proposition 1 of Section 3 implies that, under weak hypotheses, 9 n 8 s K x (0, co).

PROPOSITION

3. Assume that I. < 1*, that z(2) is continuous at ,u, that

z(p) # 0 and that z(u) is an isolated solution of x = A(x,p) in K. Then

index,(A( , p), z@)) = 1.

Proof:

We write z for Z(U). By Proposition 2, r(Ai(z, ,u)) < 1. if r(A I(z, ,u)) < 1, the result follows from Theorem 1. Thus we may assume that r(A ;(z, ,u)) = 1. Hence Lemma 1 applies and all the solutions near (z, ,u) are of the form (z + ah + ~(a), g(a)), where our notation follows that of Lemma 1. Since z(A) is continuous at ,u, (z(A), A.) E 0 for II near ,U and thus z(A) = w(a) for some a in (--E, E).

Choose a neighbourhood V, of z in

K

and then a 6 > 0 such that v, X@-&y+6)Gf? and such that x#A(x,A) if xE V, and k=p or if x E aV, and /A - ,D 1 < 6. Since 4 is continuously differentiable. we can choose a r in @,p + S) which is a regular value of 4. Then z(r) = w(q) for some a, in (--E, E). Since #‘(a,) # 0, Z-A

I(w(a,),

z) is invertible by the second part of Lemma 1. Since

r(A;(w(a,),

r)) < 1 by Proposition 2 and since

r(Ai(w(a,),

5)) is in the spectrum of

Ai(w(a,),

r)) by the Krein- Rutman theorem, it follows that r(A;(w(a,), T)) < 1. Thus, by Theorem I.

index,,@ ( , r),

w(a,))

= 1. The corollary to Proposition 1 of Section 3 now implies that (w(a), 4(a)) E

K

X

[0, a~)

for

a E (--E, E).

If /1> r, z(A) > z(t) and thus T(z(A)) > r(z(r)). Since

f(w(a)) = a + f(z)

(by the formula for

w(a)),

it follows that, if 1 > p and is close to ,u, then

z(A) = w(a),

where

a

is

positive.

Moreover, since z(k) is the minimal solution, we see by a similar argument that, for 1 > ,U and near 1, T(z(A)) is the smallest positive

a

for which

#(a)

= ,k Since

d(a)

#,u for

a

small and nonzero (by our isolatedness assumption), it follows that

a(#(a) -u) > 0

for small nonzero a.

Since z(r) is the minimal solution and

w(ao)

= z(r), a0 is the smallest positive

a

for which g(a) = r. Let a, = inf(a > a,: $(a) < t}. (If $(a) > r for a > a,,, choose a, in

(a,, E).)

Let V, = {x E V, : T(x) > J’(z) + ;(a,, +

a,)}.

By our construction, the set of solutions of x = A(x, z) in Vz are those in V, . excepting

w(a,,).

(Remember that, if w(a) is a solution,

a

> a,,.) Moreover.

x = A(x, A) has no solution on aV, if p < A< r (since, if

$(a)

< r. then a < a0 or a 2

a,).

Thus, by homotopy invariance,

since the only solution of x = ,4(x, p) in V, is z which is not in Vz. Now index,(A( ,pu>, z) = deg,(A( ,,a>, v,)

= de&AC , r), v,)

= index&( , r),

w(a")) +

deg,(A( , r), Vz)

=1+0=1.

Here, for the second line, we have used homotopy invariance and, for the third, we have used that the solutions in V, are

w(a,)

and the ones in V,. 1

Remarks. (1) We have shown that, under the assumptions of Proposition 3, w(a) E K for a E (--E, E). This is still true even if z(A) is not continuous at p or if ,u = i*.

(2) It is easy to see that index,(A( ,p), Z(D)) = 0 if Z(J) is not continuous at p and z@) is an isolated solution in K of x = A(x, p). (This holds much more generally.) We shall discuss the calculation of the index of an isolated minimal solution in more detail at the end of this section.

(3) A slight modification of our proof shows that index&, z@)) = 1 as well.

(4) Suppose that p < A* and z@) is discontinuous at ,u. Let z(u ’ ) be the right limit of z(1) as L tends to p. (Since z(d) is decreasing as i decreases, a simple compactness argument shows that z@‘) exists.) Then p is a point of discontinuity of z@) if and only if Z(D) # z& ‘). If Z(,LI ‘) is an isolated solution of x = A(x,,D), then index,(tl( ,,D), z@ ‘)) = 1. To see this, we argue as follows: By Proposition 1, r(A;(z@), A)) < 1. Thus, by passing to the limit as 2 decreases p, r(A;(z@‘),~)) < 1. (Recall that the spectral radius is continuous on the set of compact linear maps.) As before, it suffices to prove the result in the more diffkult case where r(A i(z@ ’ ), cl)) = 1. Once again, we can apply Lemma 1 and find that the solutions of x = A (x, A) near (z(u+),,u) are of the form (w(a), $(a)), where w(a) = z@ ‘) + ail + ~(a).

Now z(A) > z(,u’) for ,J > p. Thus we see, as in the proof of Proposition 3, that @(a) > p if a is small positive and that, if 1 > p, then z(n) = w(a), where a > 0. Moreover, #(a) < p if a < 0 and a is small. (Otherwise, there would exist a, < 0 and a, > 0 such that #(a,) = #(az) and w(aJ is the minimal solution of x =,4(x, $(a*)). Since w(al) is also a solution of this equation, w(a,) > w(a,). Thus, since %E K*, T(w(a,)) >r(w(a,)). Hence, by the formula for w(a), a, > a*. This is impossible because a, < 0 < a, .) The result can now be completed by the same argument as in the proof of Proposition 3.

(5) If A is real analytic (as a function of (x, A)), then z@) and z@ +) are isolated solutions of x = A(x, p). We only prove this for the case of Z(D) and when r(A ;(z@), p) = 1. (The other cases are similar or much easier.) By the proof of Lemma 1 and the real analyticity of A, we easily see that 4 is real analytic in a. (Here our notation follows that of Lemma 1.) Thus Q is constant or #(a) #p for small nonzero a. In the former case, it follows that all solutions of x = A(x, A) near (z@), p) satisfy I = p. This is impossible since z(n) -+ Z(D) as A increases to p. Hence 4(a) #p for small nonzero a and the result follows.

(6) Under Amann’s original assumptions, Proposition 3 has a much simpler proof. It would be of interest to see if some variant could be used to give a simpler proof of Proposition 3. The idea is that, if ;1 > p, A ( , ,D) maps

the closed convex set Y1 = {x E K: 0 <x < z(n)) into itself. Thus, provided that A( , p) has no fixed points on aY,, index,(A( , y), int Y,i) = 1. Since z@) is an isolated solution of x = A(x, p), a simple continuity argument shows that A ( , p) has z@) as its only fixed point in Y.k if A is near ,u (and ,? > p). Because z(n) is strongly increasing, z@) E int Y,\ and the result follows. This proof has another advantage in that it is valid under much weaker smoothness assumptions.

We now prove the main result of this section.

THEOREM 2. Assume that 0 < ‘J c A* < co and p > 0 such that x # A(x, 1) if y < b < A*, x E K and jl xi1 3 p. Then the equation x = A (x, ,I) has at least 2 solutions in K for each ,I in [ y, I* ).

Proof: Suppose that y < ,u < L*. If z(u) is a nonisolated solution of I = A(x, p) in K or if z&) # z@’ ), the result is trivial. Otherwise, p is a point of continuity of z(n). Thus, by Proposition 3. index,(A( ,p), z@)) = 1.

On the other hand, since x # A(x, A) if x E K, //x// = p and A > y- we see by homotopy invariance that

deg,(A( ,,a), K,) = deg,(A( , ,I* i- 1). K,) = 0,

where K, = (x E K: (1x1( < p}. (The last equality follows since, by the definition of A*, x = A(x, 1) has no solution in K if k > A*.) Since z(,u) E K, and z(u) has index 1, it cannot be the only solution in K,, as required. fl

Remarks. (1) If y <cl < A* and p is a point of discontinuity of z(n), it is not difficult to use Remark 4 after Proposition 3 and Lemma 3 to show that there are at least 3 solutions in K for 1 =,u. Moreover, if z($) and z(,u ’ ) are both isolated solutions of x = A(x, p) and if y < ,u, then there are at least 4 solutions for each A in (J - E, p) for some positive E. (Note that it can be shown that there are at most a countable number of points in T, where z(J) is discontinuous.)

(2) A* is easily seen to be finite if there is a compact linear map L. a g E K and a ,J > 0 such that (i) A(x, A) > Lx + g on K, (ii) r(L) > I, and (iii)f(g) > 0 for some f E K* n N(r(L) I - L *). (If f is strictly positive, we could assume instead that g = 0 and r(L) > 1.) Other results of this type could be obtained from Proposition 1 in [Sl.

The simplest case where the assumptions of Theorem 2 are satisfied is the following one. Assume that A is asymptotically linear, that is, there exist linear maps B(A): E--t E such that llxll_ ’ IIA(x, A) - B(k) xl/ + 0 as 11x/j --+ cc and x E K uniformly for 2 in bounded subsets of [O, co). It is easy to show that B(A) is increasing in A, that B(a) is positive and that the map (x, A) -+

B(J) x is continuous and maps bounded sets in K x [O, 00) to precompact sets. For simplicity, we strengthen this by requiring that the map

(x, A) -+ B(A) x maps bounded sets in E

x

[0, co) to precompact sets and that the map A -+ B(A) is continuous in the uniform topology. Let us also assume that r@(O)) < 1 and r(B(A,)) > 1 f or some A1 > 0. Let A, = sup@ >O:

@(A)) < I}. Note that since there is a A, for which r(B(&)) > 1, there must be a A for which r@(A)) = 1 (by an easy continuity argument). We assume that I, < 00 and that B(1) h # h if h EK\(O} and 1 > 1,. (Since r(B(;1)) > 1 for A > A,, this certainly holds if B(A) is always a demi- irreducible operator.) If A* < co, then, as in [ 11, we easily find that, for each E > 0, there is an M> 0 such that (IxJ] <M if (x, A) E g n (K

x

[A, + E, A*]>. Finally, let us assume that Am is the only A with r(B(1)) = 1 and that there is a 6 > 0 and a nonzerof E K* n N(Z - B(A,) *) such thatf(F(x,A))<O ifxEK, IIxIj>%‘, /A--l,/,<6 and d[JJxll-‘x, NZ - W,))l < 6. 0-I ere F(x, A) = A (x, A) - B(A) x and d[ y, C] denotes the distance of y from the set C.) Then 1, < A* and Theorem 2 applies (with y > A,). To see this, we argue as follows: If A, + 1, if (Ix,]] + co and if (x,, A,) E 5?, we easily see (cf. [I]) that I= A,. By a simple calculation, y, - B(;l,)y, -+ 0 as n + co, where y, = I~x,])-’ x,. Since B(A,) is compact, it follows easily that d[ y,, N(Z - B(n,))] + 0 as n -+ co. Now, because x, =

~kl~ 4J~

f(ml~ u> = m, - mm) 5) - f@@,) x, - wco) 4 > 0 if 1, < 1, (since B(&,) ( B(A,) and f annihilates the range of Z - B(1,)).

Since this contradicts our assumption on F, it follows that A,, > 1, for large n. Thus A* > 1, and Theorem 2 applies. Note that, since A * < co, a global bifurcation result (cf. [S, Theorem 11) implies that there must exist (x,, A,) in g and 2 E (0, co) such that ]]x,(] --f co and 1, + 3 as n -+ 00.

Remarks. (1) The assumptions above may look complicated that they can be easily verified in many cases. The key assumption that f(F(x, A)) < 0 for certain f, x, ;1 can often be verified by using the dominated convergence theorem or a similar estimation to that in [8, Lemma 41. It is a simpler condition than the one in [ 11. The assumption that B(A) h # h if I > A, and h E K\(O) can be avoided to some extent. We still have 2 solutions for each A in (A,, A*) such that B(A) h = h has no solution in K\(O) provided that we assume that, whenever (x,, , A,) 5 97, J]x, ]I -+ 00 as n -+ co, A, -+ 1 as n -+ co, and r(A i(x, , A,,)) < 1, then r@(A)) < 1. This holds if there is a compact set Z in-E such that A ;(x,, , A,)(E,) c Z for all n and such that f(A ;(x, , 1,)~) + f(B(l)y) for all y in K and for all f in a dense subset of K*. (Here E,

denotes the unit ball in E.)

Our methods could also be used to improve some of the results in [3] for the convex case.

We want to return and consider the index of a minimal solution a little

further. The main question is as follows: Assume that

K

is as above and A:

K

+

K

is completely continuous and increasing and A(0) # 0. Assume that z is the minimal solution of x=,4(x) in

K,

that A is continuously differentiable in

K

near z and A’(x) is completely continuous on

E

for x in K near z and A’(x) is completely continuous on

E

for x in

K

near z and that z is an isolated solution in

K

of x = A(x). The question is as follows: Is index,(A,

z) always 0 or

1

and, if so, is there a simple formula for deciding when it is zero and when it is

l? We mention some partial results.

First,

if

r(A’(.z))

> 1, the index is zero. (We

sketch

a proof. If the index is not zero.

we reduce to the case where

E

is separable, then choose u demi-interior to

E

and finally show that the minimal solution z(6) of x =

A(x)

+ 6~ is near z if 6 is small. Moreover, z(6) > z. We then use a similar argument to that in the proof of Proposition 2 to obtain a contradiction.)

Second,

if

r(A’(z)) <

1.

Theorem 1 implies that the index is 1. Thus the only case where there is a difficulty is when

r(A’(z))

= 1. In this case, one considers the map (x, t) +

tAx.

Let ?- = (t: x =

tA(x)

has a solution in K} and let

z(t)

denote the minimal solution. Thus z(1) = z. If ?= [0, 1 1 or 1 E int F and if

z(t)

is not continuous at 1, then it is easy to show that the index is zero. If 1 E int

T.

z(t)

is continuous at 1, N(Z -

A’(z))

is one dimensional and z is not in the range of I -A’(z) (for example, if z is demi-interior), then the index is 1.

This is a variant of Proposition 3. Note that we only need to assume that

A

is defined on

K.

This needs some care. It is possible, however, to construct an example where only the last of the above conditions fails but the index is zero. (This follows from the next result.) If there is a compact positive linear mapping I/ such that

r(V)

> 1 and

A(x)

> VX on

K,

then any isolated nonzero fixed point of

A

has index zero. This follows because there are arbitrary small perturbations of

A

which have no nonzero fixed points (cf.

the proof of the second assertion of 15, Theorem 11, The index can be calculated in many other cases by considering the local bifurcation structure of solutions of x =

tA(x)

and applying Theorem 1 to each of the branches.

Finally, note that our methods for evaluating the index of the minimal solution could also be used to evaluate the index of the minimal nonzero solution in many cases, where 0 is a solution and a minimal nonzero solution exists. (These assumptions are satisfied in a number of bifurcation problems.)

We consider one last case where our methods are useful. We assume:

(i)

K

is a normal cone in

E

with

K - K

dense in

E,

(ii) uO, co E E and r+, < u,, , (iii) A :

E

--)

E

is completely continuous and continuously differen tiable,

A

is increasing on the order interval [uo, tl,,],

A(u,)

> u,, and

A(u,)

< L’(,, (iv)

A’(x)

is demi-irreducible for x E (u,,, c,,], and (v) there is a nonzero z in

K

such that z <

A(u,)

- u,, and z < c0 -

A(r,).

It can then be shown: that (a) there are at least two solutions of

A(x) = x

in [u,,, L’,,) or (b) there is a solution of

A(x)

= x in IuO, uO] which is a nonisolated solution

in

E

or (c) there is a solution z of A(x) = x with index,@, z) = 1. This result is proved by applying analogues of Propositions 2 and 3 and Lemma 1 of this section for the map (t, x) *Ax +

tz

on [uO, u,]. (Remark 4 after Theorem 1 is used.) Rather more general results could be proved but the above suffices in many applications. For example, it could be used to simplify the proof of 19, Theorem 4(2)] and indeed it can be used to greatly weaken the regularity assumptions in 191, especially those on the underlying domain. If (i)-(iii) above hold, if (vi) uO - uO is quasi-interior to

K,

if (vii) each fixed point of A in [u,, u,,] is demi-interior to [u,, v,, ] and if (viii) I-A’(z) is invertible for each fixed point z in [u,,, v,], then CF= i index&, zi) = 1, where zi, i = l,..., n, are the fixed points of A in [u,, u,,]. (This is proved by showing that x7=, index,,0,,.0,(,4, zi) = 1 and then using Remark 4 after Theorem 1 to show that the indices in

E

and [uO, u,,] are equal.) It is likely that this result can be considerably generalized. Many conditions can be given which ensure that (vii) holds.

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