UDC 513.88

MATHEMATICS

L. P. VLASOV

ON CHEBYSHEV SETS*

(Presented by Academician P. S. Novikov on 21 V 1966)

A Chebyshev set, according to (¹), is a set \(M\) in a normed linear space \(X\) such that for every \(x \in X\) there exists a unique element \(x' \in M\) for which \(\rho(x,M)=\rho(x,x')\) (abbreviated \(xM=xx'\)). The point \(x'\) is called the projection of \(x\) onto \(M\). A Chebyshev set \(M\) is called a sun (¹) if, for every \(x \notin M\) and for each point of the ray issuing from \(x'\) and passing through \(x\), the projection is the point \(x'\). \(M\) is called approximatively compact (²) if, for every \(x \in X\), every sequence \(y_n \in M\) with \(xy_n \to xM\) has a limit point \(y \in M\). The space \(X\) is called locally uniformly convex (³) if the relations \(\|x\|=\|x_n\|=1\), \(\|x+x_n\|\to 2\) imply \(x_n\to x\). \(X\) is called uniformly smooth in the direction \(h \in X\) (⁴) if the limit \(\lim_{t\to0}(\|x+th\|-1)/t\) exists and is uniform over all \(x\) with \(\|x\|=1\).

Lemma 1. Let \(M\) be an approximatively compact Chebyshev set in a normed linear space \(X\). Let \(x \notin M\). Then

\[ \lim_{\lambda\to+0} (vM-xM)/vx=1, \tag{1} \]

where \(v=\lambda(x-x')+x\).

Proof. We first show that from \(v\to x\) it follows that \(v'\to x'\). Assuming the contrary, we find \(\varepsilon>0\) and \(v_n\) with \(v_n'x'\ge \varepsilon\) for all \(n\). But \(xv_n'\to xM\), so that, in view of the approximative compactness of \(M\), \(v_n'\) has a limit point \(v_0\in M\). Passing to the limit gives \(v_0x'\ge \varepsilon\), \(xv_0=xM=xx'\), which contradicts the fact that \(M\) is a Chebyshev set.

Since \(xv'\ge xx'\) and one may assume \(xv\le xx'\), on the segment \([v,v']\) there is a point \(\bar v\) with \(x\bar v=xx'\). Note the obvious inequalities

\[ v\bar v \le vv' \le vx'. \tag{2} \]

Put \(z=\bar v+\dfrac{vx}{xx'}(\bar v-x')\). Hence

\[ z\bar v/\bar v x' = vx/xx', \tag{3} \]

and since \(v=x+\dfrac{vx}{xx'}(x-x')\), we have

\[ z-v=(\bar v-x)+\frac{vx}{xx'}(\bar v-x), \]

\[ vz=(1+vx/xx')xx'=xx'+vx=vx'. \tag{4} \]

Taking (2), (3), (4) into account, we obtain:

\[ 0\le 1-(vM-xM)/vx =1-(vv'-xx')/vx \le 1-(v\bar v-xx')/vx \]

\[ =(vx+xx'-v\bar v)/vx =(vx'-v\bar v)/vx =(vz-v\bar v)/vx \le z\bar v/vx =\bar v x'/xx'. \]

* Dedicated to the blessed memory of my friend Mikhail Panteleevich Ulanov.

It remains to show that \(\bar v x'\to 0\). Since \(xx'-vx\leq v\bar v\leq xx'+vx\), we have \(v\bar v\to xx'\), and, in view of \(vv'\to xx'\), we obtain \(\bar v v'\to 0\). The required assertion now follows from the inequality \(\bar v x'\leq v'x'+\bar v v'\) and the assertion at the beginning of the proof. The lemma is proved.

Lemma 2. Let \(M\) be an approximately compact Chebyshev set in a Banach space \(X\). Denote by \(K(\sigma,x)\) the set
\[ \{z\in X:\ zx\leq \sigma(zM-xM)\}. \]
Then for any \(\sigma>1\) and \(x\notin M\) the set \(K(\sigma,x)\) intersects every sphere with center at the point \(x\).

Proof. Let \(R>0\) be arbitrary and let \(V\) be the ball of radius \(R\) with center at \(x\). In the set \(V\) one can introduce an order as was done in (5), p. 401. Namely, put \(x<x'\) if \(x'x\leq \sigma(x'M-xM)\). Antisymmetry and transitivity are obvious. We shall show that \(V\) is inductive (see (6), p. 300). Let \(x_\alpha\) be a chain in \(V\). For \(\alpha<\alpha'\) we have \(x_{\alpha'}x_\alpha\leq \sigma(x_{\alpha'}M-x_\alpha M)\). The numerical chain \(x_\alpha M\) converges, since it is bounded. Therefore \(x_\alpha\) converges in itself, and since \(X\) is Banach, \(x_\alpha\to x\in V\). It is clear that \(xx_\alpha\leq \sigma(xM-x_\alpha M)\), so that \(x_\alpha<x\) for every \(\alpha\), and the chain \(x_\alpha\) has the upper bound \(x\). This proves that \(V\) is inductive. The set \(K(\sigma,x)\cap V=\{z\in V: z<x\}\) is also inductive and, by Zorn’s lemma, has a maximal element \(x_0\). Suppose that \(x_0\) lies inside \(V\). Then, by Lemma 1, there is a point \(v=\lambda(x_0-x_0')+x_0\in V\) with \((vM-x_0M)/vx_0\geq 1/\sigma\), which contradicts the maximality of the element \(x_0\). If, however, \(x_0\) lies on the boundary of the ball, then the lemma is proved.

Theorem 1. In a Banach space \(X\) there do not exist approximately compact Chebyshev sets of the form \(M=X\setminus T\), where \(T\) is a bounded set.

This is an immediate consequence of Lemma 2.

Theorem 2. In a locally uniformly convex Banach space \(X\), every approximately compact Chebyshev set \(M\) is a sun.

Proof. If \(M\) is not a sun, then there is a point \(x\notin M\) such that not all points of the ray issuing from \(x'\) and passing through \(x\) have projection \(x'\). It is easy to see that on the ray there exists a point farthest from \(x'\) whose projection is \(x'\). Let this point be \(x\). By Lemma 2, for \(\sigma_n>1\), \(\sigma_n\to 1\), there is a point \(x_n\) with \(x_nx=xM\), \(x_nx\leq \sigma_n(x_nM-xM)\). Hence \((1+1/\sigma_n)xM\leq x_nM\), and since \(x_nM\leq x_nx'\leq 2\cdot xM\), we have \(x_nx'\to 2\cdot xM\). Setting \(y=(x-x')/xM\), \(y_n=(x_n-x)/xM\), we have
\[ \|y\|=\|y_n\|=1,\qquad \|y+y_n\|=x_nx'/xM\to 2. \]
Since \(X\) is locally uniformly convex, we obtain \(y_n\to y\), i.e. \(x_n\to 2x-x'\). Obviously, \(\rho(2x-x',M)=2\cdot xM=\rho(2x-x',x')\). This contradicts the fact that \(x\) is the point farthest from \(x'\) on the ray with projection \(x'\). The theorem is proved.

Theorem 3. In a smooth locally uniformly convex Banach space, every approximately compact Chebyshev set is convex.

It follows from Theorem 2 and the fact that in a smooth normed space every sun is convex (see, for example, (7)).

The assertion of Theorem 3 was given without proof in (11).

Lemma 3. Let the space \(X\) be uniformly smooth in every direction \(h\in X\). Then for any \(a,b\in X\), \(a\ne b\), and \(\delta>0\) there exists \(R>0\) such that for every \(z\) with \(zc=R\) (where \(c=(a+b)/2\)) we have
\[ zc+\delta\geq \min\{za,zb\}. \]

Proof. Suppose the contrary, that for some \(a,b\in X\), \(a\ne b\), and \(\delta>0\), and for every \(R>0\), there is a \(z\) with \(zc=R\) such that
\[ zc+\delta\leq za,\qquad zc+\delta\leq zb. \tag{5} \]

Make a similarity transformation \(U\) of the ball of radius \(R+\delta\) with center at the point \(z\) into the unit ball \(V\): \(Ux=(x-z)/(R+\delta)\). If \(\mathring V=\{x\in V:\|x\|<1\}\), then, by (5),

\[ a'=(a-z)/(R+\delta)\notin \mathring V,\qquad b'=(b-z)/(R+\delta)\notin \mathring V, \]
\[ c'=(c-z)/(R+\delta)\in \mathring V. \]

Then on the segment \([a',b']\) we find points \(a'', b'', a''\ne b''\), of intersection of the segment \([a',b']\) with the unit sphere. On the segment \([a'',b'']\) some point \(c''\) has the least norm \(1-\varepsilon\), whence we have \(1-\varepsilon\le \|c'\|=R/(R+\delta)\), \(\varepsilon\ge \delta/(R+\delta)\). Obviously, as \(R\to\infty\), \(a''b''\to0\). Put

\[ x=c''/\|c''\|=c''/(1-\varepsilon),\qquad t=b''c''/(1-\varepsilon). \]

Then \(t>0\), \(t\to0\). Let \(h=(b-a)/\|b-a\|\). The point \(c''\) was chosen by us so that \(\|c''+\lambda h\|\ge 1-\varepsilon\) for all \(\lambda\), \(-\infty<\lambda<+\infty\); therefore we have \(\|x+th\|\ge1\) and \(\|x-th\|\ge1\). By virtue of uniform smoothness in the direction \(h\), we obtain (\(x\) and \(t\) depend on \(R\) and \(z\)):

\[ (\|x+th\|-1)/t+(\|x-th\|-1)/t\to0. \]

Since each summand is positive, \((\|x+th\|-1)/t\to0\), which is equivalent to the following: \(\varepsilon/b''c''\to0\). But

\[ \delta/ab=\delta/a'b'(R+\delta)\le \delta/b''c''(R+\delta)\le \varepsilon/b''c''\to0. \]

This is impossible, since \(a\), \(b\), \(\delta\) do not depend on \(R\). The lemma is proved.

Theorem 4. In a Banach space that is uniformly smooth in each direction \(h\in X\), every approximately compact Chebyshev set is convex.

Proof. Suppose, to the contrary, that an approximately compact Chebyshev set \(M\) is not convex. Then there exist \(a,b\in M\), \(c=(a+b)/2\notin M\). Take \(0<\delta<cM\). By Lemma 3 there exists \(R>0\) such that for any \(z\) with \(zc=R\) we have \(zc+\delta\ge \min\{za,zb\}\ge zM\). Take arbitrary \(\sigma>1\). By Lemma 2 there exists \(z_0\) with \(R=z_0c\le \sigma(z_0M-cM)\). We have

\[ z_0M-\delta\le z_0c\le \sigma(z_0M-cM), \]

whence

\[ \sigma\cdot cM\le(\sigma-1)\cdot z_0M+\delta\le (R+\delta)(\sigma-1)+\delta. \]

In view of the arbitrariness of the number \(\sigma>1\), we obtain \(cM\le\delta\), which contradicts the choice of \(\delta\). The theorem is proved.

Let us use the occasion to give a series of theorems generalizing the results of note \({}^{(8)}\) on approximately convex sets. The proofs remain the same.

We shall call a set \(M\) approximately acyclic if for any \(x\in X\) the set \(\{y\in M:xy=xM\}\) is acyclic (see \({}^{(10)}\)).

Theorem 5. In a Banach space, every boundedly compact and approximately acyclic set is a sun (in the sense of note \({}^{(8)}\)).

Theorem 6. In a smooth Banach space, every boundedly compact and approximately acyclic set is convex.

A special case of this theorem was noted in \({}^{(3)}\).

Theorem 7. In an \(n\)-dimensional Banach space every approximately acyclic set is convex if and only if the space is smooth.

The author expresses his gratitude to Yu. A. Shashkin for valuable comments.

Ural State University
named after A. M. Gorky

Received
4 V 1966

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