L. P. VLASOV

APPROXIMATIVELY CONVEX SETS IN BANACH SPACES

(Presented by Academician P. S. Novikov on 31 XII 1964)

In the present article the concept of a Chebyshev set, introduced in \((^1)\), is generalized, and several theorems are proved on the connection of generalized Chebyshev (otherwise: approximately convex) sets with convex sets. These theorems are an extension of the author’s results concerning Chebyshev sets \((^6)\) to approximately convex sets.

Let \(X\) be a linear normed space, and let \(T\) be a multivalued mapping of \(X\) into itself, more precisely, a single-valued mapping of \(X\) into the set of subsets of \(X\). We shall call such a mapping convex-valued if, for every \(x \in X\), the set \(Tx\) is nonempty and convex. \(T\) is called continuous here if from \(x_n \to x\), \(y_n \to y\), \(y_n \in Tx_n\), it follows that \(y \in Tx\). Let \(M\) be a set in \(X\). In what follows we shall denote by \(T\) the operator of metric projection (metric projection) of the space \(X\) onto the set \(M\), i.e., such a mapping which assigns to each point \(x \in X\) the set \(Tx\) of all points \(y \in M\) satisfying the condition \(\rho(x,y)=\rho(x,M)\) (abbreviated: \(xy=xM\)). If \(y \in Tx\), then one says that \(x\) is projected into \(y\). \(T\) will be continuous if and only if \(M\) is closed. In the case when \(T\) is a convex-valued operator, we call the set \(M\) approximately convex. It is clear that an approximately convex set is always closed, and the operator \(T\) for such a set is continuous. If, in particular, \(T\) is single-valued, then \(M\) is called a Chebyshev set \((^1)\). Recall that the segment \([x,y]\) (respectively, the ray \(\overrightarrow{xy}\)) is the set of points of the form \((1-\lambda)x+\lambda y\), where \(\lambda\) runs through the interval \([0,1]\) (respectively, the ray \([0,\infty)\)). Generalizing the notion introduced in \((^1)\), we shall call a set \(M\) a sun if for every \(x \in X\) there exists a point \(y \in Tx\) such that all points of the ray \(\overrightarrow{yx}\) are projected into \(y\). Let us note that the points of the segment \([y,x]\) are always projected into \(y\), if \(y \in Tx\), as the following proposition shows:

\(1^\circ\). If \(xy=xM\), \(y \in M\), \(z \in [x,y]\), then \(zy=zM\).

We shall say that a point \(x'\) majorizes a point \(x\), and write \(x' \succ x\), if there exists \(y \in Tx'\) such that \(x \in [x',y]\). For example, if \(y \in Tx\), then \(x \succ y\).

\(2^\circ\). If \(x'' \succ x\), \(x' \in [x'',x]\), then \(x'' \succ x'\), \(x' \succ x\).

\(3^\circ\). If \(x' \succ x\), \(x' \ne x\), then \(x'M > xM\).

\(4^\circ\). If \(M\) is approximately convex, then the relation \(\succ\) is transitive.

Proof. Let \(x'' \succ x'\), \(x' \succ x\); then
\(\alpha)\ \ x'y=x'M;\)
\(\beta)\ \ x \in [y,x'];\)
\(\gamma)\ \ x''y'=x''M;\)
\(\delta)\ \ x' \in [y',x''].\)

From \(\alpha\), \(\beta\), and \(1^\circ\) it follows that
\(\varepsilon)\ \ xy=xM.\)

From \(\gamma\), \(\delta\), and \(1^\circ\) it follows that
\(\zeta)\ \ x'y'=x'M.\)

Extend the line \(x''x\) to its intersection with the line \(yy'\) at the point \(y''\). (It is easy to see that the points \(x,x',x'',y,y'\) lie in one plane and that \(y'' \in [y,y']\).) We have:

\[ x''M \le x''y \le x''x' + x'y \stackrel{(\alpha)}{=} x''x' + x'M \stackrel{(\zeta)}{=} x''x' + x'y' \stackrel{(8)}{=} x''y' \stackrel{(\gamma)}{=} x''M . \]

Hence \(x''y=x''M=x''y'\). This gives \(y\in Tx''\), \(y'\in Tx''\), and, by the approximative convexity of \(M\), \([y,y']\subset Tx''\) and \(y''\in Tx''\). Taking into account that, by construction, \(x\in [x'',y'']\), we infer \(x''\succ x\), and \(4^0\) is proved.

Let \(x\in X\). Denote by \(K_x\) the set of all points majorizing \(x\). Recall that a set is called boundedly compact \((^2)\) if its intersection with every closed ball is compact.

\(5^0\). For every \(x\in X\setminus M\), the set \(K_x\) is boundedly compact, provided the set \(Tx\) is compact.

Proof. \(K_x\) is contained in the set \(D\) of all points of the form \((1-\lambda)y+\lambda x\), where \(\lambda\) runs through the ray \([1,\infty)\), and \(y\) through the set \(Tx\). \(D\) is obviously boundedly compact; therefore it suffices to show that \(K_x\) is closed. Let \(x_n'\in K_x\), \(x_n'\to x'\). There exists \(y_n\in Tx_n'\) such that
\[ x=(1-\lambda_n)x_n'+\lambda_n y_n,\quad 0\leq \lambda_n\leq 1. \]
Choosing subsequences \(\lambda_{n_k}\to\lambda\), \(y_{n_k}\to y\in M\), we obtain: \(x_{n_k}\to x'\), \(y_{n_k}\to y\), \(y_{n_k}\in Tx_{n_k}'\). Hence, by the continuity of \(T\), we have \(y\in Tx'\). Passing to the limit in the equality \(x=(1-\lambda_n)x_n'+\lambda_n y_n\), we obtain \(x\in[x',y]\), and \(5^0\) is established.

For what follows we shall need the following assertion:

\(6^0\). In a Banach space, every convex-valued continuous mapping of a convex compact set into itself has a fixed point.

By a fixed point of a multivalued mapping \(R\) is meant a point \(x_0\) such that \(x_0\in Rx_0\).

The validity of this theorem follows from the following, more general proposition (taking into account \(8^0\) and \(9^0\), and the obvious fact that every compact convex set is contractible to a point; for the substance of the terms occurring in the references, see \((^3,^5)\)).

\(7^0\). Theorem \((^3)\). Let \(M\) be a compact metric space which is an acyclic absolute neighborhood retract, and let \(T:M\to M\) be a continuous multivalued function such that, for every \(x\in M\), the set \(Tx\) is acyclic. Then \(T\) has a fixed point.

\(8^0\). Theorem \((^4)\). A separable and convex subset of a linear normed space is an absolute retract (and hence an absolute neighborhood retract).

\(9^0\). Theorem \((^3)\). Every space contractible to a point is homologically trivial (i.e., acyclic).

\(10^0\). In a Banach space, every convex-valued continuous mapping \(R\) of a convex closed set \(V\) into its compact part has a fixed point.

For the proof it suffices to apply \(6^0\) to the compact set \(\widetilde V\), the closed convex hull of the set \(RV\), and to the mapping \(\widetilde R\), the restriction of \(R\) to \(\widetilde V\).

\(11^0\) Theorem. Every approximatively convex and boundedly compact set \(M\) of a Banach space \(X\) is a sun.

First we establish the following property of the set \(M\):

\(12^0\). For every \(x\in X\setminus M\), the set \(K_x\setminus\{x\}\) is nonempty.

Since \(M\) is closed, there exists a closed ball \(V\) with center at the point \(x\) and radius \(r>0\) such that the following relation holds: for every \(z\in V\),
\[ zM\geq \rho(V,M)=\inf_{x\in V,\;y\in M}\rho(x,y)>r. \]
Put, for \(z\in V\),
\[ Rz=\frac{r}{zM}(z-Tz)+x. \]
Since the functional \(zM\) is continuous and nonzero for \(z\in V\), and \(T\) is continuous, it is not hard to see that \(R\) will also be continuous. \(R\) maps \(V\) into itself, since \(\|Rz-x\|=r\). The set \(RV\) is relatively compact, because the set \(TV\subset M\) is such. Finally, since \(Rz\) is linearly related to \(Tz\), the convex-

sets \(Tz\) pass into convex sets \(Rz\). Thus, \(R\) satisfies all the conditions of Theorem \(10^\circ\). Consequently, there is a point \(x_0 \in V\) for which \(x_0 \in Rx_0\), i.e., there exists \(y \in Tx_0\) such that

\[ x_0=x+\frac{r}{x_0M}(x_0-y). \]

Hence

\[ x_0=\left(-\frac{r}{x_0M}y+x\right)\frac{1}{1-r/x_0M}=(1-\lambda)y+\lambda x=x_\lambda, \]

where \(\lambda=\dfrac{1}{1-r/x_0M}>1\), i.e. \(x_\lambda>x,\ x_\lambda\ne x\). This proves property \(12^\circ\).

Let us pass to the proof of \(11^\circ\). Let \(x\) be an arbitrary point of the set \(X\setminus M\). If \(K_x\) were bounded, then it would be compact (by \(5^\circ\); here the compactness of \(Tx\) follows from the bounded compactness of \(M\)); therefore \(K_x\) would contain a point \(x'\), farthest from \(M\). By \(12^\circ\) there exists \(x''>x'\), \(x''\ne x'\), such that, by \(3^\circ\), \(x''M>x'M\) and, by \(4^\circ\), \(x''\in K_x\), which contradicts the choice of the point \(x'\). Consequently, \(K_x\) is unbounded. The theorem will evidently be proved if we show that \(K_x\) contains a ray emanating from the point \(x\). Let \(x_n'\in K_x\) and \(\|x_n'\|\to\infty\). Put

\[ z_n=x+\frac{x_n'-x}{\|x_n'-x\|}. \]

Then \(\|z_n-x\|=1,\ z_n\in[x,x_n']\), so that, by \(2^\circ\), \(z_n\in K_x\), and, by \(5^\circ\), it has a limit point \(z\in K_x\). We shall assume that already \(z_n\to z\). We shall show that the ray \(\overrightarrow{xz}\) is contained in \(K_x\). Let \(u\in\overrightarrow{xz}\), i.e. \(u=x+A(z-x)\), where \(A\ge 0\). Put

\[ u_n=x+A\frac{x_n'-x}{\|x_n'-x\|}=x+A(z_n-x). \]

Then \(u_n\to u\), and

\[ u_n=\left(1-\frac{A}{\|x_n'-x\|}\right)x+\frac{A}{\|x_n'-x\|}x_n', \]

therefore, for sufficiently large \(n\), \(u_n\in[x,x_n']\) and \(u_n\in K_x\). Hence \(u\in K_x\), as was required.

\(13^\circ\). Theorem. In a smooth linear normed space every sun is a convex set.

Proof. Suppose the contrary, that the sun \(M\) is not convex. Then there exist points \(a\in M,\ b\in M,\ x\in[a,b]\setminus M\). Take \(y\in Tx\). Suppose that the segments \([a,y]\) and \([b,y]\) do not intersect the interior of the ball \(V\) with center at \(x\) and radius \(\|x-y\|\). Then the corresponding straight lines \(ay\) and \(by\) also do not intersect the interior of \(V\). If, for example, we had \(\|d\|<\|x-y\|,\ y\in[d,b]\), then the point of intersection of the segments \([a,y]\) and \([x,d]\) (obviously, they lie in one plane) would be interior to \(V\). By the Hahn–Banach theorem, there exist at the point \(y\) supporting hyperplanes to the ball \(V\), passing respectively through \(ay\) and \(by\), and therefore distinct. This contradicts the assumed smoothness of the space. Thus, for every \(y\in Tx\) one of the segments \([a,y]\), \([b,y]\) intersects the interior of the ball \(V\). By the definition of a sun, there is such a \(y\in Tx\) that \((\overrightarrow{yx}\setminus[y,x])\subset K_x\). Suppose, for example, that the segment \([a,y]\) intersects the interior of the ball \(V\). Then for some ball \(V'\), similar to the ball \(V\) with center of similarity at the point \(y\), the point \(a\) is interior. It is clear that the center \(x'\) of the ball \(V'\) majorizes \(x\), but, on the other hand, the point \(a\in M\) is closer to \(x'\) than \(y\). This is impossible. The theorem is proved.

From 11° and 13° it follows that

14°. Theorem. In a smooth Banach space every approximately convex and boundedly compact set is convex.

15°. Theorem. In an \(n\)-dimensional Banach space every approximately convex set is convex if and only if the space is smooth.

Sufficiency is given by Theorem 14°.

Necessity. If \(y\) is a point of nonsmoothness of the unit sphere, then as the desired approximately convex but nonconvex set one may take the union of two distinct closed supporting half-spaces at the point \(y\), not containing the unit ball and intersecting with it in a common set (obviously convex).

The author expresses gratitude to Yu. A. Shashkin, who pointed out the possibility of posing and solving the problem considered here.

Ural State University
named after A. M. Gorky

Received
23 XII 1964

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