Let \(D\) be an open subset of \(\mathbf R^N\) and \(f: \overline D\to \mathbf R^N\) a continuous function. The classical topological degree for \(f\) demands that \(D\) be bounded. The boundedness of domains is also assumed for the topological degrees for compact displacements of the identity and for operators of monotone type in Banach spaces. In this work, we follow the methodology introduced by Nagumo for constructing topological degrees for functions on unbounded domains in finite dimensions and define the degrees for Leray-Schauder operators and \((S_+)\)-operators on unbounded domains in infinite dimensions.
The theory of topological degrees has progressed significantly in recent years because of its applicability to the analysis of ordinary and partial differential equations and continuation methods in nonlinear analysis in general (e.g. see [1, 2, 3, 4, 5, 6, 7, 8, 9]). The classical topological degree theory developed by Brouwer [10] in 1912 for continuous functions on finite-dimensional spaces and the Leray-Schauder degree [11] in 1934 for compact displacements of the identity in Banach spaces both assume the boundedness of the domains over which the degrees are defined. There are numerous generalizations and extensions of these degree theories all of which are expressed, one way or the other, in terms of the Brouwer degree or the Leray-Schauder degree, and therefore the boundedness of domains becomes an essential consideration. Nagumo [12] defined the Brouwer degree based on infinitesimal analysis and indicated the possibility for developing degree theories over unbounded domains. In this paper, we construct topological degrees on unbounded domains both in finite-dimensional and infinite dimensional spaces and discuss their properties. For the development of degree theories for operators of monotone type that involve \((S_+)\)-operators and their generalized and/or multivalued versions, the reader is referred to Kartsatos and Skrypnik [1, 2], Berkovits [13], Berkovits and Mustonen [14], Kartsatos and the first author [5, 7], Kartsatos and Kerr [8], Hu and Papageorgiou [15], Kittilä [16] and the references therein. For the coincidence degree developed by Mawhin for nonlinear perturbations of certain Fredholm operators in normed spaces, the reader is referred to [17].
In Section 2, we elaborate on the Nagumo’s definition of the Brouwer degree on unbounded domains by verifying the assertions made in [12]. We then prove a version of the Leray-Schauder lemma for this new degree. In Section 3, we construct the Leray-Schauder degree on unbounded domains anddiscuss its desirable properties. Section 4 deals with the degree theory on unbounded domains for operators of type \(\alpha_0(S_+)\) introduced by Skrypnik [18].
Lemma 2.1. Let \(f: [0, 1] \times \overline D \to \mathbf R^N\) be continuous, \(t_0\in [0, 1]\) be fixed, and \(p\notin f(t;t_0) \{D\} \). Then there exist a neighborhood \(U\) of \(p\), a number \(\delta>0\) and a bounded open set \(D_0\) such that \(\overline D_0 \subset D\) and \(\)U\cap f(t, D\setminus D_0) = \emptyset\(\) for all \(t\in (t_0-\delta, t_0+\delta).\)
Proof. Assume that the conclusion does not hold. For each \(n\in\mathbb N\), define $$D_n = \{ x\in D: \|x\| 1/n\}.$$ Then each \(D_n\) is open and bounded. It is also clear that \(\overline{D}_n\subset D\) for all \(n\). By the assumption, for each \(n\in\mathbb N\), there exist \(y_n\in\mathbf R^N\) and \(t_n\in [0, 1]\) such that \(t_n\to t_0\) and $$y_n\in B(p; 1/n)\cap f(t_n, D\setminus D_n).$$ Then there exists \(x_n\in D\setminus D_n\) such that \(f(t_n, x_n) = y_n \to p.\) Since \(x_n\not\in D_n\), we have either \(\|x_n\|\ge n\) or \(\mbox{dist}(x_n, \partial D) \le 1/n\). If \(\|x_n\|\ge n\) for infinitely many values of \(n\), then we may assume that \(\{x_n\}\in \mathscr D\), and so \(p\notin f(t;t_0) \{D\} \), which is a contradiction. Therefore \(\mbox{dist}(x_n, \partial D) \le 1/n\) for infinitely many values of \(n\). If \(\{x_n\}\) has a subsequence \(\{x_{n_k}\}\) such that \(x_{n_k}\to x_0\) for some \(x_0\in\mathbf R^N\), then \(\mbox{dist} (x_{n_k}, \partial D) \to \mbox{dist}(x_0, \partial D) = 0,\) which implies \(x_0\in \partial D\). Since \(f\) is continuous, we have \(f(t_0, x_0) = p\) so that \(p\in f(t_0, \partial D)\) which contradicts \(p\notin f(t;t_0) \{D\} \) because \(f(t_0, \partial D)\subset f(t;t_0) \{D\} \) . This means that \(\{x_n\}\) has no convergent subsequence, and therefore \(\{x_n\}\in \mathscr D\). However, this implies \(p\notin f(t;t_0) \{D\} \), which is again a contradiction. This completes the proof.
Proposition 2.2. Let \(f: [0, 1] \times \overline D \to \mathbf R^N\) be continuous, and assume that, for a fixed \(t_0\in[0, 1]\), \(p(t)\notin f(t;t_0) \{D\} \) for all \(t\in [0, 1]\). Then there exist number \(\delta >0 \), an open set \(U\) and a bounded open set \(D_0\) with \(\overline D_0 \subset D\) such that $$p(t) \in U\quad\mbox{ and }\quad U\cap f(t, D\setminus D_0) = \emptyset $$ for all \( t\in (t_0-\delta, t_0+\delta).\)
Proof. Since \(p(t_0) \notin f(t;t_0) \{D\} \), by Lemma 2.1 there exist a neighborhood \(U\) of \(p(t_0)\), a number \(\delta>0\) and a bounded open set \(D_0\) such that $$\overline D_0\subset D\quad \mbox{and} \quad U\cap f(t, D\setminus D_0) =\emptyset $$ for all \(t\in (t_0-\delta_1, t_0+\delta)\). The continuity of \(p\) ensures that we can shrink \(\delta\), if necessary, so that \(p(t) \in U\) for all \(t\in (t_0-\delta, t_0+\delta)\).
Theorem 2.3 (Homotopy Invariance). Let \(f: [0, 1] \times \overline D \to \mathbf R^N\) be continuous, and let \(p:[0, 1]\to \mathbf R^N\) be continuous such that, for every \(t_0\in [0, 1]\), \(p(t)\notin f(t;t_0) \{D\} \) for all \(t\in [0, 1]\). Then the degree \({\rm d} (f(t, \cdot) , D, p(t))\) is a constant for all \(t\in [0, 1]\).
Proof. In the settings of Proposition 2.2, we have \(\partial D_0= \overline D_0\setminus D_0 \subset D\setminus D_0\) and \( U\cap f(t, D\setminus D_0) =\emptyset\) for all \(t\in (t_0-\delta, t_0+\delta)\), and therefore \(p(t)\notin f(t, \partial D_0)\). This implies $${\rm d} (f(t, \cdot) , D, p(t))= {\rm d}_{\rm B} (f(t, \cdot) , D_0, p(t))$$ for all \(t\in (t_0-\delta, t_0+\delta)\). Since \([0, 1]\) is compact, by applying a standard covering argument we find that the degree \({\rm d} (f(t, \cdot) , D, p(t))\) is constant for all \(t\in [0, 1]\).
We next give a version of the Leray-Schauder lemma [18, Lemma 1.1] for the Brouwer degree on unbounded domains. If \(x= (x_1, \dots, x_{N-1}, x_N)\in\mathbf R^N\), we write \(x = (x’, x_N),\) where \(x’= (x_1, \dots, x_{N-1})\in\mathbf R^{N-1}\).Theorem 2.4. Let \(D\subset \mathbf R^N\) be an unbounded open set and \(0\in D\). Let \(f: \overline{D}\to\mathbf R^N\) be continuous with \(f= (f_1, \dots, f_N)\) and satisfy $$f_N(x) \equiv x_N \mbox{ for } x= (x’, x_N)\in D.$$ Suppose that \(0\not\in f\{D\}\) and \(D’= \{x’: (x’,0)\in D\}\) is nonempty. Then $${\rm d} (f, D, 0) = {\rm d} (f’, D’, 0),$$ where \(f’: \overline {D’}\to\mathbb R^{N-1}\) defined by $$ f'(x’) = (f_1(x’,0), \dots, f_{N-1}(x’, 0)).$$
Proof. We assert that \(0\notin f’\{D’\}.\) Otherwise, there would exist a sequence \(\{u’_n\}\in \mathscr D’\) such that \(f'(u’_n)\to 0 \). This implies that \(f_i(u’_n, 0) \to 0\) for each \(i= 1, \dots, N-1\). Let \(v_n = (u’_n, 0) \). Then \(f_N(v_n) = 0\) and \(\{v_n\}\in\mathscr D\) and \(f(v_n) \to 0\). This implies \(0\in f\{D\}\), a contradiction. Next, let \(D_0\subset \mathbf R^{N}\) be a bounded open set containing \(R= \{x\in D : f(x) = 0\}\) such that \(\overline {D}_0\subset D\). Since \(x\in R\) implies \(x_N= 0\), we define \(R’ = \{x’\in\mathbf R^{N-1}: x\in R\}\). Then \(R’= \{x’\in D’: f'(x’) = 0\}\), and therefore \(D’_0 := \{x’: x\in D_0\}\) is a bounded open set in \(\mathbf R^{N-1}\) satisfying \(R’\subset D’_0\) and \(\overline {D’}_0 \subset D’.\) Applying the Leray-Schauder lemma [18, Lemma 1.1], we see that $${\rm d}_{B}(f, D_0, 0) = {\rm d}_{B}(f’, D’_0, 0).$$ Since these degrees are independent of the choice of \(D_0\), by the definition of the degree in (1), we have $${\rm d} (f, D, 0) = {\rm d} (f’, D’, 0).$$
Lemma 3.1. Let \(p\notin (I-T) \{D\} \). Then there exist a neighborhood \(U\) of \(p\) and a bounded open set \(D_0\) such that \(\overline D_0 \subset D\) and $$U\cap (I-T)(D\setminus D_0) = \emptyset.$$
Proof. Assume that the conclusion does not hold. For each \(n\in\mathbb N\), define \(D_n\) as in Lemma 2.1. Then \(D_n\) is open and bounded. It is also clear that \(\overline{D}_n\subset D\) for all \(n\). By the assumption, for each \(n\in\mathbb N\), there exists $$y_n\in B(p; 1/n)\cap (I-T)(D\setminus D_n). $$ Then, for each \(n\in\mathbb N\), there exists \(x_n\in D\setminus D_n\) such that \(x_n-Tx_n = y_n \to p.\) Since \(x_n\not\in D_n\), either \(\|x_n\|\ge n\) or \(\mbox{dist}(x_n, \partial D) \le 1/n\). If \(\|x_n\|\ge n\) for infinitely many values of \(n\), then we may assume that \(\{x_n\}\in \mathscr D\), and so \(p\in (I-T)\{D\}\), which is a contradiction. Therefore \(\mbox{dist}(x_n, \partial D) \le 1/n\) for infinitely many values of \(n\). If \(\{x_n\}\) has a subsequence \(\{x_{n_k}\}\) such that \(x_{n_k}\to x_0\) for some \(x_0\in X\), then \(\mbox{dist} (x_{n_k}, \partial D) \to \mbox{dist}(x_0, \partial D) = 0,\) which implies \(x_0\in \partial D\). Since \(T\) is compact, we have \(x_0- Tx_0 = p\) so that \(p\in (I-T)(\partial D)\), a contradiction. This means that \(\{x_n\}\) has no convergent subsequence, and therefore \(\{x_n\}\in \mathscr D\). However, this also implies \(p\in (I-T)\{D\}\), a contradiction.
Let \(T: [0, 1] \times \overline D \to X\) be a compact mapping, and let \((I-T)(t;t_0) \{D\}\) denote the set of all limit points of \(\{x_n -T(t_n, x_n)\}\), where \(\{t_n\}\subset [0, 1]\), \(t_n\to t_0\), and \(\{x_n\}\in \mathscr{D}.\) As before, if \(D\) is bounded, then \((I-T)(t; t_0) \{D\} = (I-T)(\{t_0\}\times \partial D).\) In view of Lemma 3.1, we can prove an analog of Lemma 2.1 which we only state as follows.Lemma 3.2. Let \(T: [0, 1] \times \overline D \to X\) be compact and \(t_0\in [0, 1]\) be fixed. Let \(p\notin (I-T)(t;t_0) \{D\} \). Then there exist a neighborhood \(U\) of \(p\), a number \(\delta>0\) and a bounded open set \(D_0\) such that \(\overline D_0 \subset D\) and $$ U\cap (I-T)(t, D\setminus D_0) = \emptyset $$ for all \(t\in (t_0-\delta, t_0+\delta).\)
Theorem 3.3 (Homotopy Invariance). Let \(T: [0, 1] \times \overline D \to X\) be compact, and let \(p:[0, 1]\to \mathbf R^N\) be continuous such that, for all \(t_0\in [0, 1]\), \(p(t)\notin (I-T)(t;t_0) \{D\} \) for all \(t\in [0, 1]\). Then the degree \({\rm d} (f(t, \cdot) , D, p(t))\) is constant for all \(t\in [0, 1]\).
Proof. For each fixed \(t_0\in [0, 1]\) as in Lemma 3.2, we have \(\partial D_0= \overline D_0\setminus D_0 \subset D\setminus D_0\) and \( U\cap (I-T)(t, D\setminus D_0) =\emptyset\) for all \(t\in (t_0-\delta, t_0+\delta)\), and therefore \(p(t)\notin (I-T)(t, \partial D_0)\). This implies $${\rm d} (I-T(t, \cdot) , D, p(t))= {\rm d}_{\rm LS} (I-T(t, \cdot) , D_0, p(t))$$ for all \(t\in (t_0-\delta, t_0+\delta)\). Since \([0, 1]\) is compact, by applying the Borel’s covering argument we find that the degree \({\rm d} (I-T(t, \cdot) , D, p(t))\) is constant for all \(t\in [0, 1]\).
Definition 4.1. Let \(D\subset X\) be open. An operator \(T:\overline{D}\to X^*\) is said to be of class \(\alpha_0(S_+)\) if each sequence \(\{u_n\}\) in \(\overline{D}\) with \(u_n\rightharpoonup u_0\) in \(X\), \(Tu_n\rightharpoonup 0\) in \(X^*\) and $$\limsup_{n\to\infty}\langle Tu_n, u_n – u_0\rangle \le 0$$ is in fact strongly convergent to \(u_0\).
The operators of class \(\alpha_0(S_+)\) were first studied by Skrypnik [18] and are more general than \((S_+)\)-operators considered by Browder [19], Berkovits [13], and several other authors. Let \(T:\overline{D}\to X^*\) be a bounded demicontinuous of type \(\alpha_0(S_+)\), and let \(T\{D\}\) be the set of all weak limits of \(\{Tu_n\}\) where \(\{u_n\}\in \mathscr D\), where \(\mathscr D\), as in previous sections, denotes the set of all sequences in \(D\) that have no limit points in \(D\). We want to make sure that the set \(T\{D\}\) contains \(T(\partial D)\). In fact, let \(p\in T(\partial D)\). Then \( p = Tu_0\) for some \(u_0\in \partial D\), and therefore there exists a sequence \(\{u_n\}\in\mathscr D\) such that \(u_n\to u_0\). By the demicontinuity of \(T\), we get \(Tu_n\rightharpoonup Tu_0 =p\), which implies \(p\in T\{D\}\). Thus, \(T(\partial D)\subset T\{D\}\). One can verify that when \(D\) is bounded and \(T\) maps bounded sets to relatively compact sets, then \(T(\partial D)= T\{D\}\). We proceed to define the degree, \({\rm d} (T, D, 0)\), of a bounded demicontinuous mapping \(T\) of type \(\alpha_0 (S_+)\) under the condition \(0\notin T\{D\}\) when \(D\) is unbounded. In this setting, let \(R=\{x\in D: Tx=0\}\). As in the previous sections, \(R\) is bounded and closed in \(X\) here as well. In fact, let \(x_0\in \overline{ R}\). Then there exists a sequence \(\{x_n\}\subset D\) with \(Tx_n = 0\) such that \(x_n\to x_0\). Since \(T\) is demicontinuous on \(\overline{ D}\), \(Tx_n \rightharpoonup Tx_0 = 0\). If \(x_0\in D\), we are done. Otherwise, \(x_0\in\partial D\). Then \(x_0\in T(\partial D)\subset T\{D\}\), a contradiction. Thus, \(R\) is closed in \(X\). To show the boundedness of \(R\), suppose that there exists a sequence \(\{x_n\}\) in \(R\) such that \(\|x_n\|\to \infty\). Clearly, \(\{x_n\}\in\mathscr D\) and \(Tx_n = 0\) for all \(n\). This contradicts \(0\notin T\{D\}\), and therefore \(R\) must be bounded. Choose a bounded open set \(U\) such that \(R\subset U\subset D\).Theorem 4.2. Let \(D\subset X\) be an unbounded open set and \(T:\overline D\to X^*\) a bounded demicontinuous mapping of type \(\alpha_0 (S_+)\). Assume \(0\notin T\{D\}\). Then there exists \(n_0\in\mathbb N\) such that \(0\notin T_n(\partial U_n)\) for all \(n\ge n_0\) and the degree \({\rm d}_{\rm B}(T_n, U_n, 0)\) is defined. Moreover, the degree \({\rm d}_{\rm B}(T_n, U_n, 0)\) is independent of \(n\).
Proof. Since \(0\notin T\{D\}\), there exists a bounded open subset \(U\) of \(X\) such that \(R\subset U\subset D\), where \(R = \{x\in D: Tx= 0\}\). It now follows that \( 0\notin T(\partial U)\). Let \(\{v_i\}, i = 1, 2, \dots,\) be a complete system of \(X\). Suppose that \(\{v_1, \dots, v_n\}\) is linearly independent for every \(n\), and let \(F_n =\mbox{span}\{ v_1, \dots, v_n\}\). We now define the finite-dimensional approximation \(T_n\) of \(T\) as
Remark 4.3. It would be interesting to determine whether a version of Theorem 4.2 may be given by directly using the approximations \(T_n\) defined on \(\overline{ D}_n\), where \(D_n = D\cap F_n\). The set \(D_n\) is open in \(F_n\), but it may be unbounded. In this case, the degree theory introduced in Section 2 may be used to compute \({\rm d} (T_n, D_n, 0)\). A suitable condition replacing \(0\notin T\{D\}\) in Theorem 4.2 is needed. Such a condition must be contradicted by the condition that \(0\in T_{n_k}\{\mathscr D_{n_k}\}\) for a sequence \(\{n_k\}\) of positive integers with \(n_k\to\infty\) as \(k\to\infty\). Here, \(\mathscr D_{n_k}\) has the same meaning as that of \(\mathscr D\) in Section 2. A version of the Leray-Schuader lemma given in Theorem 2.4 may be useful to establish the independence of \({\rm d} (T_n, D_n, 0)\) for sufficiently large \(n\).