THE ONE THEOEM JOEL A. TOPP Abstact. We pove a fixed point theoem fo functions which ae positive with espect to a cone in a Banach space. 1. Definitions Definition 1. Let X be a eal Banach space. A subset of X is called a cone if the following ae tue: 1. is nonempty and nontivial (i.e. contains a nonzeo point); 2. λ fo any nonnegative λ; 3. is convex; 4. is closed; and 5. ( ) = {0}. These conditions have faily intuitive meanings. The second one shows that a cone is a collection of ays emanating fom the oigin, while the fist foces the cone to contain at least one ay. The thid equiement ensues that the cone contains no holes. The fouth guaantees that the cone contains its boundaies. And the last condition makes sue that the cone is not too big; it must not take up moe than half the space. Examples of cones ae easy to come by. In 3, the definition of a cone pecisely matches ou geometical intuition, with the stipulation that the vetex must coincide with the oigin. In, the nonnegative numbes fom a cone. In 2, any wedge which extends to infinity fom the oigin is a cone (see Figue 1). We can also find moe abstact examples. In any L p space (including L ), the set = {f : f 0} is a cone. Similaly, in l p spaces, the set of nonnegative sequences fom a cone. Next, we will define a patial odeing of the Banach space with espect to a cone. Definition 2. We say that y x if and only if x y. Poposition 1. The elation defines a patial odeing on X. Date: 5 Octobe 2000. 1
THE ONE THEOEM 2 3 2 Figue 1. Examples of cones. Poof. We simply check that the elation satisfies the necessay citeia. eflexivity: Since 0, we see that x x fo any point x. Antisymmety: If x y and y x, then both y x and (y x). It follows that y x = 0, and so x = y. Tansitivity: If x y and y z, then y x and z y. By convexity and the ay popety, 2{ 1(y x) + 1 (z y)}. 2 2 We see that z x, and we may conclude that x z. So is a patial odeing. The patial odeing also satisfies seveal othe popeties, which follow immediately fom the definition of a cone. 1. Multiplication by a nonnegative scala peseves the odeing. Togethe, x y and λ 0 imply that λx λy. 2. Addition of a fixed vecto peseves the odeing. The inequality x y implies that x + z y + z fo any vecto z. 3. The odeing peseves limits. If y n y and each y n x, then y x. It tuns out that any patial odeing in a Banach space which satisfies these thee additional popeties will define a cone. We make one final definition befoe we tun to ou main theoem. Definition 3. Let be a cone contained in a Banach space X. We say that an opeato A : X X is positive with espect to the cone if A(). 2. The one Theoem Theoem 4 (Kasnoselskii). Let be a cone in a Banach space X. Let A : X be continuous and compact. Suppose that fo distinct, positive eal numbes and, the following conditions hold fo x : 1. When x =, we have Ax x. 2. When x =, we have x Ax.
THE ONE THEOEM 3 Then we may conclude that A has a fixed point ϕ within the cone which, moeove, satisfies ϕ (, ). To undestand the content of this theoem bette, let us conside the case when X = and is the set of nonnegative numbes. Then, the cone theoem is a slightly specialized vesion of the intemediate value theoem (see Figue 2). f(x) f() f(ϕ) = ϕ f() Figue 2. The one Theoem in. It is moe difficult to undestand how this theoem applies to 2. The eade may wish to contemplate this situation. Moe o less, the theoem demonstates that a function which always points outwad (o inwad) at the bounday of an annula secto has a fixed point within that secto (see Figue 3). x Aϕ = ϕ ϕ Aϕ = ϕ ϕ 0 0 Figue 3. The one Theoem in 2. That being said, we poceed to the poof of the theoem, which is an easy application of degee theoy. Poof. Fist, we simplify the poblem by esticting ou consideations to the topological space endowed with the elative topology. In the sequel, it must be undestood that any set which extends beyond the cone should be intepeted as the intesection of that set with the cone. This does not affect the degee theoy, except insofa as it changes the definition of a bounday point.
THE ONE THEOEM 4 Fo simplicity, we shall also assume that <. We can handle the opposite case easily once we have completed the cuent demonstation. Using the homotopy popety, we shall detemine the degee of zeo within B (0) and within B (0) with espect to the function I A. Then, we shall apply the additivity popety to detemine the degee of zeo within the annula secto = B (0) \ B (0). This degee, being odd, will imply the existence of a fixed point within the annula secto (see Figue 4). B (0) 0 B (0) Figue 4. Dividing the cone to detemine the degee of zeo inside the annula secto. We define a function Φ = I A, which we shall examine using the degee theoy. Fist, we wish to detemine the degee of the egula value zeo within the set B = B (0). To this end, we define the homotopy Φ t (x) = (I ta)(x) fo 0 t 1. We need to ensue that the homotopy is valid. It is clealy a continuous function of t. Now, we fix a bounded set E. Since A is compact, the set [0, 1] A(E) is pecompact. Scala multiplication is continuous, so the image of this poduct space unde scala multiplication is also pecompact, which is what we needed to check. We must also check that the homotopy is admissible fo calculating the degee of zeo within B. That is, no zeos may ente the set though the bounday duing the tansition fom t = 0 to t = 1. To the contay, suppose that Φ t (x) = 0 fo some t and some point x on the bounday of B. Since we ae woking in the elative topology, the bounday contains only the points on S (0); we do not need to conside the fontie between the cone and the emainde of the space. Now, we may ewite ou supposition as x = tax. Since tax and t 1, we have tax Ax. onsequently x Ax fo some x S, which contadicts the lowe cone condition.
THE ONE THEOEM 5 Now, we apply the homotopy popety to see that deg(φ 0, B, 0) = deg(φ 1, B, 0). In othe wods, deg(i, B, 0) = deg(i A, B, 0). It is obvious that the lefthand membe equals one. And we conclude that deg(i A, B, 0) = 1. Next, we wish to detemine the degee of zeo within the set B = B (0). Fist, we note that the set (I A)(B ) is bounded. Since the cone is unbounded, we may select a point w fom the potion of the cone which is not contained in (I A)(B ). Now, we define a new homotopy Φ t (x) = (I A tw)(x). The fixed pat (I A) is invetible, and the set t {tw} is evidently compact. Thus, the homotopy is valid. We must check that the new homotopy is admissible. To the contay, we assume that thee exists a t and a bounday point x fo which Φ t (x) = 0. Then, we have x Ax = tw, which implies that x Ax. Due to the topology, the only bounday points of B lie on the sphee S (0). But then we each a contadiction to the uppe cone condition. Theefoe, deg(i A, B, 0) = deg(i A w, B, 0). We have chosen w in such a way that the function (I A w) neve vanishes on B, so the ighthand membe is zeo. We conclude that deg(i A, B, 0) = 0. Using the additivity popety, we see that deg(i A,, 0) = deg(i A, B, 0) deg(i A, B, 0) = 1. In consequence, thee exists at least one egula point ϕ at which (I A)(ϕ) = 0. So A has a fixed point in. If instead we epeat the poof with the assumption that <, we each deg(i A,, 0) = deg(i A, B, 0) deg(i A, B, 0) = +1. Once again, we discove that A has a fixed point with nom between and.