Full Text
N. V. MARCHENKO
ON THE EXTENSION OF AN OPERATOR AND THE EXISTENCE OF FIXED POINTS
(Presented by Academician I. G. Petrovskii, 21 VI 1962)
Let \(E\) be a Banach space. We shall denote by \(\theta \in E\) the zero element. We consider operators acting from \(E\) into \(E\).
The following is known \((^1)\).
Theorem 1. Let \(F\) be a closed set; let \(A\) be a completely continuous operator defined on \(F\); let \(T\) be the closure of the convex hull \(A(F)\), and let \(\varepsilon>0\). Then there exists a completely continuous operator \(A^\varepsilon\), defined on the whole space \(E\), for which
\[
A^\varepsilon x = Ax \qquad \text{for } x \in F;
\tag{1}
\]
\[
\rho(A^\varepsilon x, T)<\varepsilon \qquad \text{for } x \in E.
\tag{2}
\]
In \((^1)\) the question is raised of the possibility of constructing an operator \(\widetilde A\) such that
\[
\widetilde A x = Ax \qquad \text{for } x \in F;
\tag{3}
\]
\[
\widetilde A(E) \subset T.
\tag{4}
\]
A positive answer is given by
Theorem 2. Under the conditions of Theorem 1 there exists a completely continuous operator \(\widetilde A\), defined on the whole space \(E\) and satisfying conditions (3), (4).
Proof. Take a number \(\varepsilon>0\). Let the operator \(A^\varepsilon\) satisfy the requirements of Theorem 1. Consider the closed ball \(\overline O_R\) of radius \(R>0\) with center at the point \(\theta\). Let \(\eta_1,\eta_2,\ldots,\eta_p\) be an \(\varepsilon_1\)-net for \(A^\varepsilon(\overline O_R)\), with some \(\varepsilon_1>0\). For each point \(\eta_i\) choose a point \(y_i \in T\) so that
\[
\rho(\eta_i,y_i)<\rho(\eta_i,T)+\varepsilon_1.
\]
Define on \(\overline O_R\) the operator by the equality
\[
A_1x=\sum_{i=1}^{p}\mu_i(x)y_i \bigg/ \sum_{i=1}^{p}\mu_i(x),
\]
where
\[
\mu_i(x)=
\begin{cases}
2\varepsilon_1-\|A^\varepsilon x-\eta_i\|, & \text{for } \|A^\varepsilon x-\eta_i\|\le 2\varepsilon_1,\\
0, & \text{for } \|A^\varepsilon x-\eta_i\|>2\varepsilon_1.
\end{cases}
\]
Next, let on the number line a function be defined for \(z\ge 0\) by
\[
f(z)=
\begin{cases}
z/\varepsilon_1, & \text{for } 0\le z\le \varepsilon_1,\\
1, & \text{for } z>\varepsilon_1.
\end{cases}
\]
Define the operator \(B_1\) on the set \(P_1=F+\overline{O}_R\) in the following way:
\[ B_1x= \begin{cases} A^\varepsilon x+f[\rho(A^\varepsilon x,T)](A_1x-A^\varepsilon x), & \text{for } x\in \overline{O}_R-F,\\ Ax, & \text{for } x\in F. \end{cases} \]
The operator \(B_1\) is completely continuous on \(P_1\), and \(B_1x=Ax\) for \(x\in F\). It turns out that for \(\varepsilon_1<\varepsilon/14\) one has
\[ \rho(B_1x,T)<\frac{\varepsilon}{2}\quad \text{for } x\in P_1, \]
\[ \|B_1x-A^\varepsilon x\|<2\varepsilon \quad \text{for } x\in P_1. \]
Applying our construction to the operator \(B_1\), we construct the operator \(B_2\), and so on. We obtain a sequence of completely continuous operators \(\{B_n\}\), defined on \(P_1\) and satisfying, for \(n=1,2,\ldots,\infty\), the conditions
\[ B_nx=Ax \quad \text{for } x\in F, \]
\[ \rho(B_nx,T)<\varepsilon/2^n \quad \text{for } x\in P_1, \]
\[ \|B_nx-B_{n-1}x\|<\varepsilon/2^{\,n-2} \quad \text{for } x\in P_1. \]
It is easy to show that \(\{B_n\}\) converges on \(P_1\) to a completely continuous operator \(C_1\), for which
\[ C_1x=Ax \quad \text{for } x\in F, \]
\[ \rho(C_1x,T)=0 \quad \text{for } x\in P_1. \]
Starting from the operator \(C_1\), construct an operator \(C_2\), defined on the set \(P_2=F+\overline{O}_{2R}\), where \(\overline{O}_{2R}\) is the ball of radius \(2R\) with center at the point \(\theta\), and so on. We obtain a sequence of completely continuous operators \(\{C_n\}\) on the sets \(P_n=F+\overline{O}_{2^nR}\) such that
\[ C_nx=C_{n-1}x \quad \text{for } x\in P_{n-1}, \]
\[ \rho(C_nx,T)=0 \quad \text{for } x\in P_n. \]
The operator \(\widetilde A\) is defined on \(E\) by the equality
\[ \widetilde Ax=C_{n_0}x, \]
where \(n_0\) is the least \(n\) among those satisfying the inequality \(2^nR\ge \|x\|\). \(\widetilde A\) satisfies all the requirements of Theorem 2.
Remark. Theorems 1 and 2 remain valid if the domain of definition of the operator \(A\) belongs to an arbitrary metric (not necessarily Banach) space. The essential point is that the values of the operator \(A\) belong to a Banach space.
Let \(\Gamma\) be the boundary of an open bounded set \(G\), and let the operator \(A\) be completely continuous on \(\overline G\) and have no fixed points on the boundary \(\Gamma\). Then the topological degree \({}^{(3)}\) \(d(\Phi,G,\theta)\) is defined for the mapping \(\Phi x=x-Ax\) at the point \(\theta\) on the set \(G\). It is known \({}^{(3)}\) that if \(d(\Phi,G,\theta)\ne0\), then the operator \(A\) has a fixed point.
Another criterion for the existence of a fixed point is due to Schauder \({}^{(4)}\):
If a continuous operator \(A\) maps a convex closed set \(F\) into a compact set \(\Delta\subset F\), then the operator \(A\) has a fixed point in \(F\).
Theorem 2 makes it possible to prove this assertion by means of the notion of topological degree. Let \(T\) be the closure of the convex hull of the set \(A(\Delta)\). Let the operator \(\widetilde A\) satisfy all the conditions of Theorem 2
and \(G\) is a bounded open set containing \(\Delta\). Then for \(\widetilde{\Phi}x=x-\widetilde{A}x\) we have \(d(\widetilde{\Phi},G,\theta)=1\), which proves the required assertion. Denote by \(K\) the cone in the space \(E\) [5].
Theorem 3. Let \(\Gamma\) be the boundary of an open bounded set \(G\), \(\theta\in G\), and let the operator \(A\) be completely continuous on the set \(F=K\Gamma\), with \(A(F)\subset K\), \(\rho(\theta,A(F))>0\). Then there exists an element \(x_0\in F\) such that \(Ax_0=\lambda x_0\) and \(\lambda>0\).
Theorem 3 was proved in [1]. Theorem 2 makes it possible to obtain a simple proof of it.
Let \(u_0\in K\) and \(u_0\ne\theta\). Consider the completely continuous operator
\[ Bx=\widetilde{A}x+\rho[\widetilde{A}x,A(F)]u_0, \]
where the operator \(\widetilde{A}\) satisfies the requirements of Theorem 2. It turns out that
\[ \inf_{x\in G} Bx>0,\qquad Bx=Ax\quad \text{for } x\in F. \]
Then there exists an element \(x_0\in\Gamma\) [1] such that \(Bx_0=\lambda x_0\) and \(\lambda>0\), whence Theorem 3 follows.
Theorem 4. Let \(\Gamma\) be the boundary of an open bounded set \(G\), \(\theta\in G\). Suppose, further, that the operator \(A\) is completely continuous on the set \(F=K\cap\overline{G}\), and that for \(x\in K\cap\Gamma\) the equality \(\varphi=\lambda A\varphi\) \((\lambda>0)\) implies \(\lambda\ge 1\). Then the operator \(A\) has a fixed point on \(F\).
Theorem 5. Let \(\Gamma_1,\Gamma_2\) be, respectively, the boundaries of open bounded sets \(G_1,G_2\); \(\theta\in G_1\), \(\theta\in G_2\), \(G_1\ne G_2\). Suppose, further, that the operator \(A\) is completely continuous on the set \(F=(\overline{G}_1+\overline{G}_2-G_1G_2)K\), \(A(F)\in K\), and that for \(x\in K\Gamma_1\), from the condition \(\varphi=\lambda A\varphi\) \((\lambda>0)\) it follows that \(\lambda\ge 1\), while for the set \(F_2=K\Gamma_2\) there is an element \(u_0\in K\) \((u_0\ne\theta)\) such that \(x-Ax\ne tu_0\) for \(t>0\), \(x\in F_2\). Then the operator \(A\) has a fixed point on \(F\).
Theorems 4 and 5 are proved by means of Theorem 2 by a topological method and are generalizations of Theorems 1 and 2 of [2].
Received
12 V 1962
CITED LITERATURE
- M. A. Krasnosel’skii, Topological Methods in the Theory of Nonlinear Integral Equations, Moscow, 1956.
- M. A. Krasnosel’skii, DAN, 135, No. 3 (1960).
- J. Leray, J. Schauder, UMN, 1, 3–4 (13–14) (1946).
- V. V. Nemytzkii, Mat. sbornik, 41, 4 (1934).
- M. G. Krein, M. A. Rutman, UMN, 3, 23 (1948).