A Fredholm mapping of index zero

J. Korean Math. Soc. 45 (2008), No. 2, pp. 527–535 A FREDHOLM MAPPING OF INDEX ZERO Jos´ e M. Soriano Arbizu Reprinted from the Journal of the Korean Mathematical Society Vol. 45, No. 2, March 2008 c °2008 The Korean Mathematical Society J. Korean Math. Soc. 45 (2008), No. 2, pp. 527–535 A FREDHOLM MAPPING OF INDEX ZERO Jos´ e M. Soriano Arbizu Abstract. Sufficient conditions are given to assert that between any two Banach spaces over KFredholm mappings share exactly Nvalues in a specific open ball. The proof of the result is constructive and is based upon continuation methods. 1. Preliminaries Let Xand Ybe two Banach spaces. If F:X→Yis a continuous mapping, then one way of solving the equation (1) F(x) = 0 is to embed (1) in a continuum of problems (2) H(x, t) = 0 (0 ≤t≤1), which is resolved when t= 0.When t= 1,the problem (2) becomes (1). In the case when it is possible to continue the solution for all tin [0,1] then (1) is solved. This method is called continuation with respect to a parameter [1]-[23]. In this paper, sufficient conditions are given in order to prove that two differentiable mappings share exactly Nvalues in a specific open ball. Other conditions, sufficient to guarantee the existence of zero points in finite and infinite dimensional settings, have been given by the author in several other papers [10]-[23]. In this paper we use continuation methods. The proof supplies the existence of implicitly defined continuous mappings whose ranges reach zero points [5]-[7]. The key is the use of the Continuous Dependence theorem on a parameter in Banach spaces [25] , properties of Fredholm C1-mappings [25, 26], the Weierstrass theorem relative to extremum points [26], and a consequence of the properties of the algebra of Banach whose elements are the linear continuous mappings from a Banach space into itself. We briefly recall some theorems and concepts to be used. Received September 15, 2006. 2000 Mathematics Subject Classification. Primary 58C30; Secondary 65H10. Key words and phrases. zero point, continuation methods, continuous dependence theorem, C1-homotopy, proper mapping, compact mapping, regular value, Fredholm mapping, isomorphism. This work is partially supported by D.G.E.S. Pb 96-1338-CO 2-01 and the Junta de Andalucia. c °2008 The Korean Mathematical Society 527 528 JOS´ E M. SORIANO ARBIZU Theorem 1 ([25, pp. 17-19] Continuous Dependence Theorem).Let the following conditions be satisfied: (i) Pis a metric space, called the parameter space. (ii) For each p∈P, the mapping Tpsatisfies the following hypotheses: (1) Tp:M⊆X→M, i.e., Mis mapped into itself by Tp; (2) Mis a closed non-empty set in a complete metric space (X, d); (3) Tpis k-contractive for fixed k∈[0,1). (iii) For a fixed p0∈P, and for all x∈M, limp→p0Tp(x) = Tp0(x). Thus, for each p∈P, the equation xp=Tpxphas exactly one solution, where xp∈Mand limp→p0xp=xp0. Definition ([25, 26]).We will assume Xand Yare Banach spaces over K, where K=Ror K=C. Mapping F:D(F)⊆X→Y, is said to be compact whenever it is continuous and the image F(B) is relatively compact (i.e., its closure F(B) is compact in Y) for every bounded subset B⊂D(F). Mapping Fis said to be proper whenever the pre-image F−1(K) of every compact subset K⊂Yis also a compact subset of D(F). If D(F) is open, then mapping Fis said to be a Fredholm mapping if and only if both Fis a C1-mapping and F0(x) : X→Yis a Fredholm linear mapping for all x∈D(F).That L:X→Yis a linear Fredholm mapping means that Lis linear and continuous and both the numbers dim(ker(L)) and codim(R(L)) are finite, and therefore ker(L) = X1is a Banach space and has topological complement X2,since dim(X1) is finite. The integer number ind(L) = dim(ker (L)) −codim(R(L)) is called the index of L, where dim signifies dimension, codim codimension, ker kernel and R(L) stands for the range of mapping L. Let F(X, Y ) denote the set of all linear Fredholm mappings A:X→Y. Let L(X, Y ) denote the set of all linear continuous mappings L:X→Y. Let Isom(X, Y ) denote the set of all the isomorphisms L:X→Y. Let B(x0, ρ) denote the open ball of centre x0and radius ρ, and S(x0, ρ) the sphere of centre x0and radius ρ. If u:X→Yis a linear continuous bijective operator, the inverse linear continuous operator to uwill be denoted by u−1. Theorem 2 ([27, pp. 23-24]).(a)The set IsomL(X, Y )is open in L(X, Y ). (b)The mapping β: Isom(X, Y )→ L(Y, X), β(u) := u−1is continuous. Theorem 3 ([26, p. 296]).Let g:D(g)⊂X→Ybe a compact mapping, where a∈D(g).If the derivative g0(a)exists, then g0(a)∈ L(X, Y )is also a compact mapping. Theorem 4 ([26, p. 366]).Let S∈ F(X, Y ).The perturbed mapping S+C verifies S+C∈ F(X, Y )and ind(S+C) = ind(S)if C∈ L(X, Y )and Cis a compact mapping. Definition ([26, p.318]).Let F:X→Ybe a C1-mapping. A FREDHOLM MAPPING 529 The point u∈Xis called a regular point of Fif and only if F0(u)∈ L(X, Y ) maps onto Y, and ker(F0(x)) splits Xinto a topological direct sum. The point v∈Yis called a regular value of Aif and only if the pre-image F−1(v) is empty or consists solely of regular points. 2. A Fredholm mapping If we can say u:= f−g, then uhas a zero if and only if fand gshare a value, that is, there is x∈Xwith f(x) = g(x). We thereby establish our result in terms of f, g. Theorem 5. Let f, g :X→Ybe two C1-mappings,where Xand Yare two Banach spaces over K=Ror K=C. (i) fis a proper and Fredholm mapping of index zero and gis a compact mapping. (ii) Mapping fhas Nzeros, xi, i = 1, . . . , N in B(x0, ρ). (iii) Zero is a regular value of the mapping f(·)−tg(·) : X→Yfor each parameter t∈[0,1]. (iv) If (x, t)∈S(x0, ρ)×[0,1] then f(x)6=tg(x). Hence the following statement holds true: (a)fand gshare exactly Nvalues in the open ball B(x0, ρ). Proof. (a) Henceforth X×Ris provided by the topology, given by the product norm. L(X, Y ),L(Y, X) are provided by the topologies given by their respective operator norm. (a1) Let us construct the following homotopy H:X×[0,1] →Y,H(x, t) := f(x)−tg(x), which is a C1-homotopy between the mappings fand f−g. Henceforth partial derivatives will generally be denoted by writing initial spaces as subindices of mappings. We will see here for any (x, t)∈X×[0,1] that HX(x, t) = f0(x)−tg0(x) verifies HX(x, t)∈ F(X, Y ), and indHX(x, t) = 0. Since gis a compact mapping and the derivative g0(x) exists for any fixed x∈X, Theorem 3 implies that g0(x)∈ L(X, Y ) is a compact mapping and therefore for any (x, t)∈X×[0,1], tg0(x)∈ L(X, Y ), is also a compact mapping. Since fis a Fredholm mapping of index zero, then f0(x)∈ F(X, Y ),∀x∈X and ind(f0(x)) = 0,∀x∈X. These results together with Theorem 4 imply that HX(x, t)∈ F(X, Y ),and indHX(x, t) = 0,∀(x, t)∈X×[0,1]. (a2) We will now prove that, if H(x, t) = 0,(x, t)∈B(x0, ρ)×[0,1],then HX(x, t)∈isom(X, Y ). Let (x, t)∈X×[0,1], H(x, t) = 0.Since zero is a regular value of f(x)−tg(x), therefore HX(x, t) maps onto Y, therefore codim(R(HX(x, t))) = dim(Y/Y ) = 0 and hence ind(HX(x, t)) = dim(ker(HX(x, t))). 530 JOS´ E M. SORIANO ARBIZU Furthermore, since ind(HX(x, t)) = 0,therefore dim(ker(HX(x, t))) = 0, and hence HX(x, t) is also injective. Thus, HX(x, t) is a bijective linear continuous mapping, and since Yis a Banach space, the linear inverse mapping HX(x, t)−1∈ L(Y, X) is also continuous. Hence, HX(x, t)∈isom(X, Y ). (a3) We will prove the existence of a compact set V00 which contains all x∈Xsuch that f(x)−tg(x) = 0,when (x, t)∈B(x0, ρ)×[0,1].Let us define the set V:= g(D),where D:= {x∈B(x0, ρ) : ∃t∈[0,1], t =t(x),such that f(x) = tg(x)}. Since f(xi) = 0 = f(xi)−0g(xi), i= 1, . . . , N, therefore xi∈D,i= 1, . . . , N, hence Dis not empty. Owing to V⊂g(B(x0, ρ)), we know that Vis a bounded set, and with gas a compact mapping, then Vis a relatively compact set. We now construct the set V0:= {ty :t∈[0,1], y ∈V}. V 0is a compact set in Ydue to the fact that it can be written in the following way V0=v(V×[0,1]), where vis the continuous mapping v:V×[0,1] ⊂Y×[0,1] →Y, v(y, t) = ty, and V×[0,1] is a compact set in the topological product space Y×R. Since fis a proper mapping and V0is a compact set on Y, the pre-image of V0under f, V 00 := f−1(V0) is a compact set on X, which contains all xwhich verify the following f(x)−tg(x) = 0,(x, t)∈B(x0, ρ)×[0,1]. (a4) We will prove that there is a real number C > 0 such that if (x, t)∈(H−1{0})∩(B(x0, ρ)×[0,1]), where H−1{0}is the pre-image of zero under H, then kHX(x, t)−1k ≤ C, where HX(x, t)−1is the inverse mapping of HX(x, t). Since His a C1-mapping, the mapping HX:X×R→ L(X, Y ),(x, t)7→ HX(x, t) is continuous. From (a2) if (x, t) belongs to (H−1{0})∩(B(x0, ρ)× [0,1]),then HX(x, t)∈Isom(X, Y ).From Theorem 2, the mapping inverse formation β: Isom(X, Y )⊂ L(X, Y )→ L(Y, X), β(u) = u−1,is a continuous mapping. Consequently, by composition of continuous mappings, the mapping k·k◦β◦HX:H−1{0} ∩ (V00 ×[0,1]) ⊂X×R→R,(x, t)7→ kHX(x, t)−1k, is continuous. Since H:X×[0,1] is a continuous mapping, H−1{0} ⊂ X×[0,1] is a closed set, and as V00 ×[0,1] ⊂X×Ris a compact set, therefore H−1{0} ∩ (V00 ×[0,1]) ⊂X×Ris a compact set. Weierstrass Theorem implies that there is maximum of kHX(x, t)−1kwhen (x, t)∈H−1{0} ∩ (V00 ×[0,1]),and hence, there is a real number C > 0,such that kHX(x, t)−1k ≤ C, ∀(x, t)∈ (H−1{0})∩(V00 ×[0,1]). (a5) Let suppose that (xa, ta)∈B(x0, ρ)×[0,1] and that H(xa, ta) = 0. Therefore: From (a3), (xa, ta)∈V00 ×[0,1]. From (a2), HX(xa, ta)∈Isom(X, Y ). A FREDHOLM MAPPING 531 We will now prove the existence of r0>0, r > 0 and the existence a continuous mapping x(·) : [ta−r0, ta+r0]∩[0,1] →X, which verifies kx(t)k< r, H(xa+x(t), t) = 0,∀t∈[ta−r0, ta+r0]∩[0,1]. To this end, we define G(x, t) := H(xa+x, t),∀x∈X, and we solve the equation (3) G(x, t) = 0 for x. Obviously, we have G(0, ta) = H(xa, ta) = 0,and furthermore, GX(0, ta) verifies GX(0, ta) = HX(xa, ta). We transform Equation (3) into the following equivalent equation: (4) HX(xa, ta)−1[HX(xa, ta)(x)−G(x, t)] = x. Equation (4) leads us to define the two following mappings h(x, t) := HX(xa, ta)(x)−G(x, t), and Tt(x) := HX(xa, ta)−1((h(x, t)), where his a C1-mapping, and (5) h(0, ta) = 0. Equation (4) is equivalent to the following “key equation” (6) Tt(x) = x. Let us observe that tin the definition of Ttis an index and not a partial derivative as is usually written. Equation (3) is equivalent to the Fixed Point Equation (6), which will be studied below. Let x, x0∈B(x0, ρ); t, ta∈[0,1] such that |t−ta|,kxk,kx0k< r,|t−ta|< r0, where r, r0will be fixed at a later stage. Since hX(x, t) = HX(xa, ta)−GX(x, t),hence (7) hX(0, ta) = 0. From Equation (7) and since hX:X×[0,1] → L(X, Y ),(x, t)7→ hX(x, t) is continuous, the Taylor theorem implies that (8) kh(x, t)−h(x0, t)k ≤ sup{khX((x0+θ(x−x0), t)k:θ∈[0,1]}kx−x0k =o(1)kx−x0k, o(1) →0 as r→0. Due to Equations (5) and (8), and since his a continuous mapping, therefore kh(x, t)k ≤ kh(x, t)−h(0, t)k+kh(0, t)k=o(1)kxk+o0(1), o(1) →0 as r→0, o0(1) →0 as r0 0→0. Hence kTt(x)k ≤ kHX(xa, ta)−1kkh(x, t)k ≤ kHX(xa, ta)−1k(o(1)kxk+o0(1)), (9) o(1) →0 as r→0, o0(1) →0 as r0 0→0. 532 JOS´ E M. SORIANO ARBIZU Now ris fixed so that o(1) ≤1 2C,and then the closed and non-empty set M:= {x∈X:kxk ≤ r}is constructed. We are now able to fix r0 0,so that o0(1) <r 2C,and the set M0:= {t∈[0,1] : |t−ta| ≤ min{r, r0 0}=r0}is constructed. We prove below that the hypotheses of Theorem 1 are verified by the spaces and mappings, we have just defined. The Metric Space (M0,|·|) will be considered as the parameter space of the hypothesis (i) of Continuous Dependence Theorem 1. Mwill be considered as the closed and non-empty set and (X, k·k) as the complete metric space considered in hypothesis (ii) of Theorem 1, which is verified as we will see in the two following paragraphs. Owing to Equation (9), for any fixed t∈M0,and for all x∈M, kTt(x)k ≤ kHX(xa, ta)−1k(o(1)kxk+o0(1)) ≤C(1 2Cr+1 2Cr)≤r, therefore Tt(x)∈M, and hence Tt:M→M. That is, Ttmaps the closed non-empty set Mof the Banach space Xinto itself. Due to Equation (8), for any x, x0∈Mand all fixed t∈M0 kTt(x)−Tt(x0)k ≤ kHX(xa, ta)−1(h(x, t)−h(x0, t))k ≤ kHX(xa, ta)−1k1 2Ckx−x0k ≤ 1 2kx−x0k, therefore Ttis half-contractive for any t∈M0which has been fixed. Hence hypothesis (ii) of Theorem 1 is verified. For any fixed t0∈M0and for all x∈M, limt→t0,t∈M0Tt(x) = limt→t0,t∈M0HX(xa, ta)−1(HX(xa, ta)(x)−G(x, t)) =HX(xa, ta)−1(HX(xa, ta)(x)−G(x, t0)) = Tt0(x), and hence hypothesis (iii) of Theorem 1 is also verified. Thus, Theorem 1 implies, for any t∈M0,that Tthas a unique fixed point Tt(x) = x:= x(t),and it is verified that x(t)→x(t0) as t→t0,t, t0∈M0, that is, x(·) is a continuous mapping. Thus for each t∈M0there is only one x(t)∈M⊂Xsuch that G(x(t), t) = 0,and hence (10) H(xa+x(t), t) = 0. Let us also observe that Tta(0) = 0, x(ta) = 0.Equation (10) can be written in the following way: H(α(t), t) = 0,∀t∈M0,where αis the following continuous mapping α:M0→Y, α(t) := xa+x(t). (a6) We will now prove that fand gshare exactly Nvalues in the open ball B(x0, ρ),using iteratively the process of the previous section a finite number of times, to find each shared value. To this end we have to prove that it is possible to select the same r0for each iteration of the process of the previous section. Let us define the mapping ϕ:V00 ×[0,1] ×V00 ×[0,1] ⊂X×[0,1] ×X×[0,1] →Y, A FREDHOLM MAPPING 533 ϕ(xa, ta;x, t) := HX(xa, ta)x−H(xa+x, t), which, as a composition of continuous mappings, is uniformly continuous on the compact set V00 ×[0,1] ×V00 ×[0,1] of the product topological space X× R×X×R.Therefore for any fixed r > 0, there is δ(r 2C)>0 such that, if (xa, ta;x, t),(xa0, ta0;x0, t0)∈V00 ×[0,1] ×V00 ×[0,1], with k(xa, ta;x, t)− (xa0, ta0;x0, t0)k< δ(r 2C), then kϕ(xa, ta;x, t)−ϕ(xa0, ta0;x0, t0)k<r 2C. If we restrict the domain of mapping ϕby fixing any (xa, ta)∈V00 ×[0,1] such that H(xa, ta)=0,we obtain mapping hconsidered in the previous section, that is h: (H−1{0})∩(V00 ×[0,1]) ⊂X×R→Y, h(x, t) = ϕ(xa, ta;x, t) = HX(xa, ta)(x)−H(xa+x, t). We are now able to fix r0 0considered in the previous section by taking r0 0= δ(r 2C),where rwill be established later in this section. On the other hand the mapping ϕX:V00 ×[0,1] ×V00 ×[0,1] → L(X, Y ), ϕX(xa, ta;x, t) = HX(xa, ta)−HX(xa+x, t),is uniformly continuous on the compact set V00 ×[0,1]×V00 ×[0,1],and therefore there is δ(1 2C)>0 such that, ∀(xa, ta;x, t),(xa0, ta0;x0, t0)∈V00 ×[0,1] ×V00 ×[0,1], k(xa, ta;x, t)−(xa0, ta0;x0, t0)k< δ(1 2C) ⇒ kϕX(xa, ta;x, t)−ϕX(xa0, ta0;x0, t0)k<1 2C. Let us observe that the mapping hXconsidered in the previous section is the mapping ϕX, when (xa, ta) is fixed: hX:V00 ×[0,1] → L(X, Y ), hX(x, t) = ϕX(xa, ta;x, t) = HX(xa, ta)−HX(xa+x, t). At this point we determine the previously mentioned rby taking r=δ(1 2C). Since rand r0 0have been fixed, we are now able to fix r0in the same way as in the previous section, that is, r0= min{r, r0 0}. The previous section implies that if H(xa, ta) = 0,(xa, ta)∈B(x0, ρ)×[0,1] then there is a continuous mapping x(·):[ta−r0, ta+r0]∩[0,1] →X, which verifies H(xa+x(t), t) = 0,∀t∈[ta−r0, ta+r0]∩[0,1]. This lets us construct the continuous mapping α: [ta, ta+r0]→Y,α(t) = xa+x(t) with H(α(t), t) = 0, ∀t∈[ta, ta+r0], α(ta) = xa. We repeat this process of (a5) by taking (α(ta+r0), ta+r0) as an initial point in each iteration, where (xa, ta) is the previous initial point, and (xi,0) ∈ B(x0, ρ)×[0,1], i ∈1, . . . , N as the initial point of the first iteration with α: [0, r0]→Y, α(t) = xi+x(t), α(0) = xito be extended in successive iterations of the process. A point (x0 i,1) ∈B(x0, ρ)×[0,1] which verifies H(x0 i,1) = 0, i ∈ 1, . . . , N is reached in a finite number of iterations, since [0,1] is a compact set, and from the frontier condition established in hypothesis (iv) of the theorem. In an identical way, but starting at in a value shared by fand g, and by the same process but taking initial successive points conveniently, we reach a zero 534 JOS´ E M. SORIANO ARBIZU of fon the ball B(x0, ρ).Therefore fhas the same number of zeros that shared values by fand g. ¤ References [1] E. L. 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