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The convex hull and the Carathéodory number of a set in terms of the metric projection operator
K. S. Shklyaevab a Faculty of Mechanics and Mathematics, Lomonosov Moscow State University, Moscow, Russia
b Moscow Center of Fundamental and Applied Mathematics, Moscow, Russia
Abstract:
We prove that each point of the convex hull of a compact set M in a smooth Banach space X can be approximated arbitrarily well by convex combinations of best approximants from M to x (values of the metric projection operator PM(x)), where x∈X. As a corollary, we show that the Carathéodory number of a compact set M⊂X with at most k-valued metric projection PM is majorized by k, that is, each point in the convex hull of M lies in the convex hull of at most k points of M.
Bibliography: 26 titles.
Keywords:
metric projection, convex hull, Banach space, smoothness, Minkowski functional, Carathéodory number.
Received: 23.02.2022 and 11.05.2022
§ 1. Introduction Let (X,‖⋅‖) be a real Banach space. The distance of a point x∈X to a set M⊂X is defined by The closed and open balls, and the sphere in X with centre x∈X and radius r>0 are denoted by ¯B(x,r), B(x,r) and S(x,r), respectively. We let riM denote the relative interior of a set M⊂X, that is, the interior of M in its affine hull
affM:={k∑i=1λixi:k∈N,xi∈M,λi∈R,k∑i=1λi=1}.
The metric projection of a point x onto a set M is defined by
PM(x)={y∈M:‖x−y‖=d(x,M)}.
We say that M is a Chebyshev set if PM(x) is a singleton for each x∈X. The convex hull of a set M is denoted by convM. The problem of the convexity of Chebyshev sets in concrete and abstract normed spaces has been studied extensively (see, for example, Efimov and Stechkin [1], Klee [2], [3], Berdyshev [4], Brø ndsted [5], Brown [6], [7], Vlasov [8], Balaganskii and Vlasov [9], Tsar’kov [10], [11], and Alimov [12], [13]). The investigations of the geometry of Chebyshev sets in finite-dimensional normed linear spaces date back to Bunt [14], Mann [15] and Motzkin [16], [17]. In his thesis [14] (1934) Bunt proved that, in strictly convex finite-dimensional Banach spaces with second-order modulus of smoothness (and, in particular, in finite-dimensional Euclidean spaces), each Chebyshev set is convex; he also showed that in two-dimensional spaces the condition of strict convexity can be dropped. In particular, this result implies that, in finite-dimensional Euclidean spaces Rn, the class of Chebyshev sets coincides with the class of convex closed sets. A bit later, Motzkin [16] showed that any Chebyshev set in a two-dimensional normed plane is convex if and only if the space is smooth (a space is smooth if the unit sphere there has a unique support hyperplane at each point). Klee extended Bunt’s and Motzkin’s results in his early papers [2] and [3]: he showed that in any finite-dimensional smooth space every Chebyshev set is convex. Efimov and Stechkin were among the first authors to investigate Chebyshev sets in infinite-dimensional Banach spaces. In particular, they stated the problem of the convexity of Chebyshev sets in (infinite-dimensional) Hilbert spaces. Open Problem A. Is every Chebyshev set in an infinite-dimensional Hilbert space convex? For a survey of results for infinite-dimensional spaces, see [9]. As a natural generalization of Chebyshev sets, we consider sets with at most k-valued, nonempty metric projection. Definition 1. Given a set M⊂X and a natural number k, set
convkM:={k∑i=1λixi:xi∈M,λi∈[0,1],k∑i=1λi=1}.
Definition 2. The Carathéodory number of a set M is defined as the smallest natural number k such that convkM=convM. The following Carathéodory convex hull theorem is well known (see [18]): if M⊂Rd, then convd+1M=convM, that is, the Carathéodory number of M is at most d+1. Under some additional conditions on a set M the upper estimate of its Carathéodory number can be improved. Results of this type are due to Fenchel (see [10]), Bárány and Karasev (see [20]), and other authors. A detailed account of the available results on the Carathéodory number can be found in [20]. In the present paper we examine the Carathéodory numbers of sets M with at most k-valued metric projection PM. In 2010 Borodin stated the following problem. Problem A. Prove that in any finite-dimensional smooth space X the Carathéodory number of each closed set M⊂X with at most k-valued metric projection PM is at most k. Note that for k=1 this claim is valid because each Chebyshev set in a smooth finite-dimensional space is convex. Also note that for k⩾ it holds automatically by Carathéodory’s theorem. For k= d=2 Problem A was solved by Flerov [21]. In our paper Problem A is solved under certain constraints on M, for spaces of arbitrary dimension (in particular, for boundedly compact sets M in infinite-dimensional Banach spaces X). In Theorems 1–5 we present more general results on representations of a point in the convex hull of a set M \subset X as a convex combination of points in the metric projection P_M(x) for some x \in X. Using these results, in Corollaries 1–4 we derive analogues and extensions of Problem A. Moreover, with the help of Corollaries 1–4 we readily obtain sufficient conditions for the existence of a point with at most k-valued metric projection. The corresponding sufficient conditions are as follows: \operatorname{conv}{M} \neq \operatorname{conv}_k{M} in Corollaries 1 and 4, and \overline{\operatorname{conv}{M}} \neq \overline{\operatorname{conv}_k{M}} in Corollaries 2 and 3. Note that the infinite-dimensional analogue of Problem A for a Hilbert space X, with no constraints on M, includes Open Problem A as a particular case (for k=1).
§ 2. Auxiliary lemmas In what follows X_d denotes a normed space of dimension d. Lemma 1. Let a, b \in X_d, r > 0, let a \in B(b,r), and let f\colon \overline{B}(b,r) \to X_d be a continuous mapping such that a \notin [x, f(x)] for all x \in S(b,r). Then there exists a point x_* \in B(b,r) such that f(x_*)=a. Proof. It can be assumed without loss of generality that b=0 and r=1. Assume that f(x) \neq a for all x \in B(0,1). Since a \notin [x, f(x)] for all x \in S(0,1), and since f is continuous, there exists \tau \in (0,1) such that a \notin [x, f(t x)] for all x \in S(0,1) and all t \in [1-\tau, 1]. Hence
\begin{equation}
g(x,t) := \frac{t+\tau-1}{\tau}x+\frac{1-t}{\tau}f(tx) \neq a
\end{equation}
\tag{1}
for all x \in S(0,1) and all t \in [1-\tau, 1]. Consider the automorphism of the sphere S(0,1):
\begin{equation*}
u\colon x \mapsto \frac{x-a}{\| x-a\|}.
\end{equation*}
\notag
The map \widetilde{f}\colon \overline{B}(0,1) \to S(0,1) defined by
\begin{equation*}
\widetilde{f}(tx) := \begin{cases} u^{-1}\biggl( \dfrac{f(tx)-a}{\| f(tx)-a \| }\biggr), &x \in S(0,1),\ t \in [0, 1-\tau), \\ u^{-1}\biggl(\dfrac{g(x,t)-a}{\| g(x,t)-a \| }\biggr), &x \in S(0,1),\ t \in [1-\tau,1], \end{cases}
\end{equation*}
\notag
is well defined in view of (1). Moreover, for all x \in S(0,1),
\begin{equation*}
\frac{g(x,1-\tau)-a}{\| g(x,1-\tau)-a \|}=\frac{f((1-\tau)x)-a}{\| f((1-\tau)x)-a \|}, \qquad \frac{g(x,1)-a}{\| g(x,1)-a\|}=\frac{x-a}{\| x-a\|}.
\end{equation*}
\notag
Hence \widetilde{f} is continuous and \widetilde{f}|_{S(0,1)} \equiv \mathrm{id}, that is, \widetilde{f} retracts the ball \overline{B}(0,1) onto its boundary S(0,1), which is impossible (see [22], § 38).
Lemma 1 is proved. Let (X, d_X) and (Y, d_Y) be metric spaces. A set-valued map F\colon X \to 2^Y \setminus \{\varnothing\} is said to be upper semicontinuous at a point x_0 \in X if for each \varepsilon > 0 there exists \delta > 0 such that, for all x \in B(x_0,\delta),
\begin{equation*}
F(x) \subset B_\varepsilon(F(x_0)) := \Bigl\{ y \in Y\colon d_Y(y, F(x_0)) := \inf_{z \in F(x_0)}d_Y(y, z) < \varepsilon \Bigr\}.
\end{equation*}
\notag
The graph of a set-valued mapping F\colon X \to 2^Y \setminus \{\varnothing\} is defined by
\begin{equation*}
\Gamma_F := \bigl\{ (x,y)\colon x \in X, \, y \in F(x) \bigr\};
\end{equation*}
\notag
it is equipped with the distance d( (x_1, y_1), (x_2, y_2) ) \,{:=} \max (d_X(x_1, x_2), d_Y(y_1, y_2) ). Given a linear space Y, we denote the class of nonempty closed convex subsets of Y by \mathrm{Conv}(Y). Lemma 2. Let a, b \in X_d, r > 0, let a \in B(b,r), and let F\colon \overline{B}(b,r) \to \mathrm{Conv}(X_d) be an upper semicontinuous mapping such that a \notin \operatorname{conv}(\{x\} \cup F(x)) for all x \in S(b,r). Then a \in F(x_*) for some x_* \in B(b,r). Proof. It can be assumed without loss of generality that b=0 and r=1. Suppose that a \notin F(x) for all x \in \overline{B}(0,1). Since F is upper semicontinuous, there exists \varepsilon \in (0, (1-\| a \|)/2) such that, for all x \in \overline{B}(0,1) and all y \in \overline{B}(0,1) \setminus B(0, 1-\varepsilon),
\begin{equation}
B(a,2\varepsilon) \cap F(x)=\varnothing\quad\text{and} \quad B(a,\varepsilon) \cap \operatorname{conv}(\{y\} \cup F(y))=\varnothing.
\end{equation}
\tag{2}
Next we require the following result.
Theorem B (see [23], Theorem 1). Let S be a compact metric space, Y be a normed linear space and \Phi\colon S \to \mathrm{Conv}(Y) be an upper semicontinuous mapping. Then for each \varepsilon > 0 there exists a single-valued \varepsilon-approximation of \Phi, that is, a continuous single-valued mapping \varphi\colon S \to \operatorname{conv} \Phi(S) such that the graphs \Gamma_\Phi and \Gamma_\varphi of \Phi and \varphi, respectively, satisfy d^*(\Gamma_\varphi, \Gamma_\Phi):= \sup_{y \in \Gamma_\varphi} d(y, \Gamma_\Phi) < \varepsilon.
By Theorem B there exists a single-valued \varepsilon-approximation f\colon \overline{B}(0,1) \to X of the set-valued mapping F. Since d^* (\Gamma_{f},\Gamma_F) < \varepsilon, by (2) we have
\begin{equation}
B(a,\varepsilon) \cap f(\overline{B}(0,1))=\varnothing.
\end{equation}
\tag{3}
Next we claim that a \notin [y,f(y)] for y \in S(0,1). Indeed, otherwise there exists \lambda \in [0,1] such that
\begin{equation}
a=\lambda y+(1-\lambda)f(y).
\end{equation}
\tag{4}
Since d^* (\Gamma_{f},\Gamma_F) < \varepsilon, there would exist z \in B(y,\varepsilon) \cap \overline{B}(0,1) and w \in B(f(y),\varepsilon) such that w \in F(z). Hence by (4)
\begin{equation*}
\bigl\| \lambda z+(1-\lambda)w-a\bigr\| \leqslant \|\lambda(z-y\|+\bigl\| (1-\lambda)(w-f(y)) \bigr\| < \varepsilon,
\end{equation*}
\notag
that is, B(a,\varepsilon) \cap \operatorname{conv}(\{z\} \cup F(z)) \neq \varnothing, which contradicts (2). So f satisfies the conditions of Lemma 1, and therefore there exists x_* \in B(0,1) such that f(x_*)=a. However, this is impossible by (3).
Lemma 2 is proved.
§ 3. Results for starlike domains The results in this section are related to combinatorial geometry; they are close to [24] in spirit. Recall that a domain U \subset \mathbb{R}^d is starlike with respect to 0 if \lambda U \subset U for all \lambda \in (0,1) and 0 \in U. Let U be a domain starlike with respect to 0, and let p_u be its Minkowski functional, namely,
\begin{equation*}
p_u\colon \mathbb{R}^d \to \mathbb{R}_+, \qquad x \mapsto \sup\bigl\{ \lambda \geqslant 0\colon \lambda x \in U\bigr\}.
\end{equation*}
\notag
Traces of the following definition can be found in [25], Remark 2. Definition 3. Let U \subset \mathbb{R}^d be a bounded starlike domain with respect to 0. Given a point x \in \mathbb{R}^d and a closed set M \subset \mathbb{R}^d, we set
\begin{equation*}
\begin{gathered} \, U(x, r):= x+rU, \qquad \overline{U}(x,r) := \overline{U(x,r)}, \qquad d_u(x,M)= \inf\bigl\{ p_u(y-x)\colon y \in M \bigr\} \\ \text{and}\quad P^u_M(x) := M \cap \overline{U}(x, d_u(x,M)). \end{gathered}
\end{equation*}
\notag
Note that
\begin{equation*}
P^u_M(x) \neq \varnothing\quad\text{and} \quad M \cap U(x, d_u(x,M))= \varnothing,
\end{equation*}
\notag
since M is closed and U is open. However, in general, the equality M \cap \overline{U}(x,\lambda)=\varnothing need not hold for all \lambda \in (0, d_u(x,M)) . Recall that a function f\colon \mathbb{R}^d \to \mathbb{R} is said to be upper semicontinuous on \mathbb{R}^d if \varlimsup_{x \to x_0} f(x) \leqslant f(x_0) for all x_0 \in \mathbb{R}^d. Lemma 3. Let U \subset \mathbb{R}^d be a bounded starlike domain with respect to 0 and let M be a nonempty closed subset of \mathbb{R}^d. Then 1) the function d_u(\,\cdot\,, M) is upper semicontinuous on \mathbb{R}^d; 2) the set-valued map P^u_M is upper semicontinuous on \mathbb{R}^d. Proof. 1) Let x \in \mathbb{R}^d. We claim that the function d_u(\,\cdot\, , M) is upper semicontinuous at x. In other words, we need to show that, for each sequence x_n \to x, n\to\infty,
\begin{equation*}
d_u(x,M) \geqslant \varlimsup_{n \to \infty} d_u(x_n,M) =: r.
\end{equation*}
\notag
Assume on the contrary that d_u(x,M) < r. Hence M \cap U(x,r) \neq \varnothing. We choose y \in M \cap U(x,r) and set z := (y-x)/r \in U. It is clear that y=x+r z. Let \varepsilon > 0 be such that
\begin{equation*}
B(x+rz, 4r\varepsilon) \subset U(x,r)=x+r U.
\end{equation*}
\notag
We also set r_n := d_u(x_n,M). The above inclusion is invariant under homothetic transformations and translations, so
\begin{equation}
B(x_n+r_n z, 4 r_n \varepsilon ) \subset (x_n+r_n U )=U(x_n,r_n).
\end{equation}
\tag{5}
We choose n so that
\begin{equation*}
\| x-x_n\| < r \varepsilon\quad\text{and} \quad |r-r_n| < \min\biggl(\frac{r \varepsilon}{\| z \|+1}, \frac{r}{4}\biggr).
\end{equation*}
\notag
In this case we have the estimate
\begin{equation*}
\| x+r z -(x_n+r_n z)\| \leqslant \| x-x_n \|+\| (r-r_n)z \| < 2 r \varepsilon, \qquad r_n > \frac{3r}{4}.
\end{equation*}
\notag
Using the last two inequalities and (5) we obtain
\begin{equation*}
B(y, r \varepsilon)=B(x+r z, r \varepsilon) \subset B(x_n+r_n z, 3 r \varepsilon) \subset B(x_n+r_n z, 4 r_n \varepsilon) \subset U(x_n, r_n ).
\end{equation*}
\notag
Hence
\begin{equation*}
M \cap B(y, r\varepsilon) \subset M \cap U(x_n, r_n )=\varnothing;
\end{equation*}
\notag
however, y \in M \cap B(y, r \varepsilon) because y has been chosen in M \cap U(x,r). Therefore, M \cap U(x,r)=\varnothing, and so d_u(x, M) \geqslant r. This proves the first assertion of the lemma.
2) We claim that the set-valued mapping P^u_M is upper semicontinuous at x. To prove this it suffices to show that if x_n \to x as n\to\infty, y_n \in P^u_M(x_n), and y_n \to y as n\to\infty, then y \in P^u_M(x). Passing to subsequences we can assume without loss of generality that r_{n} := d_u(x_{n}, M) \to r. The sets \overline{U}(x_{n},r_{n}) converge to \overline{U}(x,r) in the Hausdorff metric. Hence y \in \overline{U}(x,r). In addition, y \in M because y_n \in M and since M is closed. On the other hand, since d_u(\cdot,M) is upper semicontinuous, we have R := d_u(x,M) \geqslant r. Therefore, y \in \overline{U}(x,R) \cap M, that is, y \in P^u_M(x).
Lemma 3 is proved. As a direct corollary to Lemmas 2 and 3 we have the following. Theorem 1. Let U \subset \mathbb{R}^d be a bounded starlike domain with respect to 0, let {M \subset \mathbb{R}^d} be a closed set, and let a, b \in \mathbb{R}^d and r > 0 be such that a \in B(b,r) and a \notin \operatorname{conv}(\{x\} \cup P^u_M(x)) for all x \in S(b,r). Then there exists x_* \in \mathbb{R}^d such that a \in \operatorname{conv}(P^u_M(x_*)). Under additional constraints on a set M and a domain U, the above result can be refined as follows. Theorem 2. Let U \subset \mathbb{R}^d be a bounded starlike domain with respect to 0, let \partial U be a C^1-smooth closed (d-1)-dimensional manifold, and let M \subset \mathbb{R}^d be compact. Then \operatorname{ri}(\operatorname{conv} M) \subset \bigcup_{x \in \mathbb{R}^d} \operatorname{conv}(P^u_M(x)) and \operatorname{conv} M=\overline{\bigcup_{x \in \mathbb{R}^d} \operatorname{conv}(P^u_M(x))}. Proof. 1) First we show that \operatorname{ri}(\operatorname{conv} M)\,{\subset} \bigcup_{x \in \mathbb{R}^d} \operatorname{conv}(P^u_M(x)).
Consider the case when \operatorname{aff} M= \mathbb{R}^d. Then
\begin{equation*}
\operatorname{ri} (\operatorname{conv} M)=\operatorname{int} (\operatorname{conv} M).
\end{equation*}
\notag
Let a \in \operatorname{ri} (\operatorname{conv} M) \setminus M. We can find a_1, \dots, a_\nu \in M and \delta \in (0,1) such that
\begin{equation}
B(a, 3\delta) \subset \operatorname{conv}(\{a_1, \dots, a_\nu\}), \qquad B(a, 3\delta) \cap M=\varnothing.
\end{equation}
\tag{6}
We claim that there exists a point x_* \in \mathbb{R}^d such that a \in \operatorname{conv}(P^u_M(x_*)). By Theorem 1, to prove this it suffices to find R > 0 such that
\begin{equation*}
a \notin \operatorname{conv}(\{x\} \cup P^u_M(x)) \quad \forall\, x \in S(0,R).
\end{equation*}
\notag
Let V \subset \mathbb{R}^d be a bounded starlike domain, and let \partial V be a C^1-smooth closed (d-1)-dimensional manifold. Given a point y \in \partial V, we denote the unit outward normal vector to V at y by N_{\partial V}(y). We also consider the half-space
\begin{equation}
\Pi_{\partial V}(y, t):= \bigl\{ x \in \mathbb{R}^d\colon \langle x, N_{\partial V}(y)\rangle \leqslant\langle y, N_{\partial V}(y)\rangle-t \bigr\}.
\end{equation}
\tag{7}
The diameter of M is defined by
\begin{equation*}
\operatorname{diam}(M)=\sup\{ \| y-y' \|\colon y, y' \in M \}.
\end{equation*}
\notag
Since \partial U is C^1-smooth and compact, there exists \lambda_0 > 0 such that, for all \lambda \geqslant \lambda_0 and y \in \partial U,
\begin{equation}
\lambda \overline{U} \cap \overline{B}(\lambda y, \operatorname{diam}(M) ) \subset \Pi_{\lambda \partial U}(\lambda y, -\delta) \cap \overline{B}(\lambda y, \operatorname{diam}(M) )
\end{equation}
\tag{8}
and
\begin{equation}
\lambda U \cap \overline{B}(\lambda y, \operatorname{diam}(M) ) \supset \Pi_{\lambda \partial U}(\lambda y, \delta) \cap \overline{B}(\lambda y, \operatorname{diam}(M) ).
\end{equation}
\tag{9}
Next, there exists R > 0 such that
\begin{equation}
d_u(x,M) \geqslant \lambda_0 \quad \forall\, x \in S(0,R).
\end{equation}
\tag{10}
We prove that
\begin{equation*}
B(a,\delta) \cap \operatorname{conv}(\{x\} \cup P^u_M(x))=\varnothing \quad \forall\, x \in S(0,R).
\end{equation*}
\notag
Assume the contrary. Consider an arbitrary b \in B(a,\delta) \cap \operatorname{conv}(\{x\} \cup P^u_M(x)) and take y_1, \dots, y_m \in P^u_M(x) such that b \in \operatorname{conv}(\{x, y_1, \dots, y_m \}). We also set
\begin{equation*}
U' := U(x, d_u(x,M)).
\end{equation*}
\notag
It is clear that \partial U' is homothetic to \partial U and y_1, \dots, y_m \in \partial U'. Consider
\begin{equation*}
N(y_1) := N_{\partial U'}(y_1)\quad\text{and} \quad \Pi(y_1,t) := \Pi_{\partial U'}(y_1, t).
\end{equation*}
\notag
Since U' is a starlike domain with respect to the point x, we have x \in \Pi(y_1, 0), and therefore x \in \Pi(y_1, -\delta). Substituting \lambda U=U'-x and \lambda y=y_1-x into (8) and (9) and adding x to the resulting expressions we obtain
\begin{equation}
\overline{U'} \cap \overline{B}(y_1, \operatorname{diam}(M) ) \subset \Pi(y_1, -\delta) \cap \overline{B}(y_1, \operatorname{diam}(M) )
\end{equation}
\tag{11}
and
\begin{equation}
U' \cap \overline{B}(y_1, \operatorname{diam}(M) ) \supset \Pi(y_1, \delta) \cap \overline{B}(y_1, \operatorname{diam}(M) ).
\end{equation}
\tag{12}
It is clear that \| y_1-y_i \| \leqslant \operatorname{diam}(M) for all i=1, \dots, m. Hence (11) implies that y_i \in \Pi(y_1,-\delta). As a result, \operatorname{conv}\{x, y_1, \dots, y_m\} \subset \Pi(y_1,-\delta), and therefore {b \in \Pi(y_1,-\delta)}. Consequently, c := b-2\delta N(y_1) \in \Pi(y_1, \delta). It is clear that {c\in B(a, 3\delta)}, and now it follows from (6) that some of the points a_1, \dots, a_\nu also lie in \Pi(y_1,\delta). Without loss of generality we can assume below that a_1 \in \Pi(y_1,\delta). On the other hand, since M \subset \overline{B}(y_1, \operatorname{diam}(M)), using (12) we obtain
\begin{equation*}
U' \cap M \supset \Pi(y_1,\delta) \cap M,
\end{equation*}
\notag
which, however, is impossible, because U' \cap M=\varnothing and a_1 \in \Pi(y_1,\delta) \cap M.
Now consider the case when \operatorname{aff} M \neq \mathbb{R}^d. We can assume without loss of generality that 0 \in M, that is, \operatorname{aff} M= \operatorname{span} M. We set l:= d-\dim (\operatorname{span} M) > 0. It is clear that there exists a set W=\{w_1, \dots, w_l\}, where w_1, \dots, w_l \in \mathbb{R}^d, such that
\begin{equation}
\operatorname{aff} (M \cup W)=\operatorname{span} (M \cup W)=\mathbb{R}^d.
\end{equation}
\tag{13}
By what has already been proved, for each point a \in \operatorname{ri}(\operatorname{conv} M) and
\begin{equation}
b := \frac{a}{2}+\frac{1}{2l}(w_1+\dots+w_l) \in \operatorname{int}(\operatorname{conv}(M \cup W))
\end{equation}
\tag{14}
there exists x_* \in \mathbb{R}^d such that
\begin{equation*}
b \in \operatorname{conv}(P^u_{M \cup W}(x_*)).
\end{equation*}
\notag
Therefore, we have
\begin{equation*}
b=\lambda y+(1-\lambda)w,
\end{equation*}
\notag
where
\begin{equation*}
\lambda \in [0,1], \qquad y \in \operatorname{conv}(P^u_{M \cup W}(x_*) \cap M )\quad\text{and} \quad w \in \operatorname{conv}(P^u_{M \cup W}(x_*) \cap W).
\end{equation*}
\notag
Since \dim{(\operatorname{span}{M})}+\dim{(\operatorname{span}{W})}=n, we have \operatorname{span}{M} \cap \operatorname{span}{W}=\{ 0\}. Now (14) implies that
\begin{equation*}
\lambda=\frac{1}{2}, \qquad y=a\quad\text{and} \quad w=\frac{w_1+\dots+w_l}{l}.
\end{equation*}
\notag
As a result,
\begin{equation*}
a \in \operatorname{conv}(P^u_{M \cup W}(x_*) \cap M)=\operatorname{conv}(P^u_{M}(x_*)),
\end{equation*}
\notag
as claimed.
2) Let us now prove the equality \operatorname{conv} M=\overline{\bigcup_{x \in \mathbb{R}^d} \operatorname{conv}(P^u_M(x))}. Since M is compact, we have \operatorname{conv} M=\overline{\operatorname{ri}(\operatorname{conv} M)} \subset \overline{\bigcup_{x \in \mathbb{R}^d} \operatorname{conv}(P^u_M(x))}. It is clear that the reverse inclusion also holds because \operatorname{conv}(P^u_M(x) ) \subset \operatorname{conv} M for all x \in \mathbb{R}^d.
Theorem 2 is proved. Remark 1. Under the hypotheses of Theorem 2, for each d \geqslant 2 there exists a compact set M \subset \mathbb{R}^d such that \operatorname{conv} M \neq \bigcup_{x \in \mathbb{R}^d} \operatorname{conv}(P^u_M(x)). Proof. As U we take the Euclidean ball B(0,1) in \mathbb{R}^d. Let e_1, \dots, e_d be an orthonormal basis for \mathbb{R}^d, and let M=\overline{B}(e_2,1) \cup \overline{B}(-e_2,1). Then P^u_M is the metric projection P_M in the Euclidean space \mathbb{R}^d. It is clear that (e_1 \pm e_2) \in M, since e_1 \in \operatorname{conv} M. It is easily seen that if e_1 \in \operatorname{conv}(P_M(x)), then P_M(x)=\{e_1+e_2, e_1-e_2 \}. Hence the ball \overline{B}(x, \| x-e_2\|-1) touches the balls \overline{B}(e_2,1) and \overline{B}(-e_2,1) at the points e_1+e_2 and e_1-e_2, respectively. Now one can draw a straight line through each triple of points \{e_2, e_1+e_2, x\} and \{-e_2, e_1-e_2, x\} (here it is important that the space \mathbb{R}^d is Euclidean). Clearly, these two straight lines intersect at x. On the other hand, the straight lines through the pairs of points \{e_2, e_1+e_2 \} and \{-e_2, e_1 -e_2 \}, are parallel. This contradiction completes the proof. Corollary 1. Let U \subset \mathbb{R}^d be a bounded starlike domain with respect to 0, let \partial U be a C^1-smooth closed (d-1)-dimensional manifold, and let M \subset \mathbb{R}^d be a compact set. Assume that |P^u_M(x)| \leqslant k for all x \in \mathbb{R}^d. Then \operatorname{conv} M=\operatorname{conv}_k M. Proof. By Theorem 2
\begin{equation*}
\operatorname{ri} (\operatorname{conv} M) \subset \bigcup_{x \in \mathbb{R}^d} \operatorname{conv} ( P^u_M(x) ),
\end{equation*}
\notag
and therefore, since |P^u_M(x)| \leqslant k, we have
\begin{equation*}
\overline{\operatorname{conv} M}=\overline{\operatorname{ri} (\operatorname{conv} M)} \subset \overline{\bigcup_{x \in \mathbb{R}^d} \operatorname{conv} ( P^u_M(x) )} \subset \overline{\operatorname{conv}_k M}.
\end{equation*}
\notag
The reverse inclusion is clear. So we have shown that \overline{\operatorname{conv} M}=\overline{\operatorname{conv}_k M}. The set M is compact, hence the sets \operatorname{conv} M and \operatorname{conv}_k M are closed. Therefore, \operatorname{conv} M=\operatorname{conv}_k M.
This proves Corollary 1.
§ 4. Results for convex domain Let U be a bounded convex domain in a Banach space X, let 0 \in U, and let p_u be the Minkowski functional of U. It is clear that there exist c_1, c_2 > 0 such that
\begin{equation*}
c_1 \| x \| \leqslant p_u(x) \leqslant c_2 \| x \| \quad \forall x \in X.
\end{equation*}
\notag
We recall (see [26], § 2.4.7) that a convex domain U is smooth if the Minkowski functional p_u is Gâteaux differentiable on X \setminus \{ 0 \}, that is, for each x \in X \setminus \{ 0 \} there exists a continuous linear functional p'_u(x;\cdot) \in X^* such that for any h \in X the limit
\begin{equation}
\lim_{t \to 0}\frac{p_u(x+th)-p_u(x)}{t}=p'_u(x;h)
\end{equation}
\tag{15}
exists. The modulus of smoothness of a convex body U is defined by
\begin{equation*}
\omega_u(t)=\sup\biggl\{ \frac{1}{2}(p_u(x+y)+p_u(x-y)-2)\colon x \in \partial U, \, y \in X, \, p_u(y)=t \biggr\}.
\end{equation*}
\notag
A convex body U is said to be uniformly smooth if its Minkowski functional p_u is uniformly Fréchet differentiable on U, that is, (15) holds uniformly for x \in \partial U, h \in U. It is known (see [26], § 2.4.7) that the modulus of smoothness of a uniformly smooth body U satisfies \omega_u(t)=o(t) as t \searrow 0. Let x \in \partial U and let \Pi_x be the support hyperplane to U at x. Then p_u(x+y)-1 \geqslant 0 and p_u(x-y)-1 \geqslant 0 for all y \in \Pi_x-x. Hence
\begin{equation}
\sigma(t) := \sup\bigl\{ p_u(x+y)-1\colon x \in \partial U, \, y \in \Pi_x-x, \, p_u(y) \leqslant t \bigr\}=o(t), \qquad t \to 0.
\end{equation}
\tag{16}
For x \in \partial U let f_x \in S_{X^*} be the support functional to U at x. We claim that, for each \varepsilon \in (0,c_1), there exists a positive \delta=o(\varepsilon), \varepsilon \searrow 0, such that
\begin{equation}
\bigl\{ z \in X\colon f_x(z) \leqslant f_x(x)-\delta \bigr\} \cap \overline{B}(x, \varepsilon) \subset U \cap \overline{B}(x, \varepsilon) \quad \forall\, x \in \partial U.
\end{equation}
\tag{17}
Let \varepsilon and \delta be some positive numbers. We represent each point z \in B(x, \varepsilon) such that f_x(z) \leqslant f_x(x)-\delta as follows:
\begin{equation*}
z=(1-\tau)x+y, \quad\text{where}\quad \tau \geqslant \frac{\delta}{f_x(x)}\quad\text{and} \quad y \in (\Pi_x-x).
\end{equation*}
\notag
Now by (16) we have
\begin{equation*}
p_u(z)=p_u((1-\tau)x+y)=(1-\tau) p_u\biggl(x+\frac{y}{1-\tau}\biggr) \leqslant 1-\tau+ \sigma(p_u(y))
\end{equation*}
\notag
and
\begin{equation*}
\tau=\tau p_u(x) \leqslant \tau p_u\biggl(x-\frac{y}{\tau}\biggr)=p_u(\tau x-y)=p_u(x-z) \leqslant c_2 \| x-z \| \leqslant c_2 \varepsilon.
\end{equation*}
\notag
Since z \in B(x, \varepsilon), we have \| y \| \leqslant \| z-x\|+\tau \| x \| \leqslant \varepsilon+\tau/ c_1, and therefore
\begin{equation}
p_u(y) \leqslant c_2 \| y \| \leqslant c_2 \varepsilon+\tau \frac{c_2}{c_1} \leqslant \biggl(c_2+\frac{c_2^2}{c_1}\biggr) \varepsilon =: C \varepsilon.
\end{equation}
\tag{18}
It is clear that for each \varepsilon > 0 one can choose \delta=\delta(\varepsilon) > 0 so that c_1 \delta > \sigma(C \varepsilon) and \delta=o(\varepsilon) as \varepsilon \searrow 0. By (18) and since f_x(x) \leqslant \| x \| \leqslant 1/c_1, for \delta= \delta(\varepsilon) we have
\begin{equation*}
p_u(z) \leqslant 1-\tau+\sigma(p_u(y)) \leqslant 1-\frac{\delta}{f_x(x)}+\sigma(C\varepsilon) \leqslant 1-c_1 \delta+\sigma(C \varepsilon) < 1.
\end{equation*}
\notag
Therefore, z \in U. This proves (17). Note that in each smooth convex body a finite-dimensional space is uniformly smooth. Theorem 3. Let U \subset \mathbb{R}^d be a bounded smooth convex domain such that 0 \in U, and let M \subset \mathbb{R}^d be closed. Then
\begin{equation*}
\operatorname{ri}(\operatorname{conv} M) \subset \bigcup_{x \in \mathbb{R}^d} \operatorname{conv}(P^u_M(x)) \quad\textit{and}\quad \overline{\operatorname{conv} M}=\overline{\bigcup_{x \in \mathbb{R}^d} \operatorname{conv}(P^u_M(x))}.
\end{equation*}
\notag
Proof. 1) First let us show that \operatorname{ri}(\operatorname{conv} M)\,{\subset} \bigcup_{x \in \mathbb{R}^d} \operatorname{conv}(P_M(x)). Consider the case when \operatorname{aff} M= \mathbb{R}^d. Then
\begin{equation*}
\operatorname{ri} (\operatorname{conv} M)=\operatorname{int} (\operatorname{conv} M).
\end{equation*}
\notag
Let a \in \operatorname{ri} (\operatorname{conv} M) \setminus M. Then there exist points a_1, \dots, a_\nu \in M and \delta \in (0,1) such that
\begin{equation}
B(a, 2\delta) \subset \operatorname{conv}(\{a_1, \dots a_\nu\})\quad\text{and} \quad B(a, 2\delta) \cap M=\varnothing.
\end{equation}
\tag{19}
We claim that there exists a point x_* \,{\in}\, \mathbb{R}^d such that a \!\in\! \operatorname{conv}(P_M(x_*)). By Lemma 3 the map P_M^u is upper semicontinuous. Now, in view of Theorem 1, in order to show that the required point x_* exists it suffices to find R > 0 such that
\begin{equation*}
a \notin \operatorname{conv}(\{x, P_M^u(x)\}) \quad \forall\, x \in S(0,R).
\end{equation*}
\notag
We set R_a :=\max_{i=1, \dots, \nu}\| a-a_i \|. Now we recall the definitions of the unit outward normal vector N_{\partial V}(y) and the half-space \Pi_Y(y,t) introduced in the proof of Theorem 2 (see (7)). Since U is uniformly smooth, we have (17). Hence there exists \lambda_0 > 0 such that, for all \lambda \geqslant \lambda_0 and all y \in \lambda \partial U,
\begin{equation}
\Pi_{\lambda \partial U}(y, \delta) \cap \overline{B}(y, R_a +1 ) \subset U(0,\lambda) \cap \overline{B}(y, R_a +1 ).
\end{equation}
\tag{20}
Let R > 0 be such that
\begin{equation*}
p_u( a-x) > \lambda_0+\sup_{z \in B(a,\delta)} p_u(a-z) \quad \forall\, x \in S(0,R).
\end{equation*}
\notag
We show that
\begin{equation}
B(a, \delta) \cap \overline{U}(x, d_u(x,M))=\varnothing \quad \forall\, x \in S(0,R).
\end{equation}
\tag{21}
Assume the contrary. Let b \in B(a,\delta) \cap \overline{U}(x, d_u(x,M)). It is clear that in this case U(x, p_u(b-x)) \cap M=\varnothing. Set U' := U(x, p_u(b-x)). Since
\begin{equation*}
p_u(b-x) \geqslant p_u( a-x)-p_u(a-b) > \lambda_0,
\end{equation*}
\notag
from (20) we obtain
\begin{equation*}
\Pi_{\partial U'} (b, \delta) \cap \overline{B}(b,R_a +1) \subset U' \cap \overline{B}(b, R_a +1).
\end{equation*}
\notag
The inclusion \overline{B}(a,R_a) \subset \overline{B}(b, R_a+1) is straightforward, hence
\begin{equation}
\Pi_{\partial U'} (b, \delta) \cap \overline{B}(a,R_a) \subset U' \cap \overline{B}(a, R_a).
\end{equation}
\tag{22}
We set c := b-\delta N_{\partial U'}(b) \in \Pi_{\partial U'}(b,\delta). Since b \in B(a,\delta), we find that
\begin{equation*}
c \in B(a,2\delta) \subset \operatorname{conv}( \{a_1, \dots, a_\nu\}).
\end{equation*}
\notag
Consequently, the half-space \Pi_{\partial U'} (b, \delta) contains the point c and one of the points a_1, \dots, a_\nu. We can assume without loss of generality that a_1 \in \Pi_{\partial U'} (b, \delta). It is clear from the definition of R_a that a_1 \in \overline{B}(a,R_a). Hence
\begin{equation}
a_1 \in \Pi_{\partial U'} (b, \delta) \cap \overline{B}(a,R_a).
\end{equation}
\tag{23}
On the other hand a_1 \in M, and so a_1 \notin U', which implies that
\begin{equation}
a_1 \notin U' \cap \overline{B}(a, R_a).
\end{equation}
\tag{24}
It is clear that (23) and (24) are in contradiction to (22). This therefore proves (21). As a result, B(a,\delta) \cap \operatorname{conv}( \{ x\} \cup P_M^u(x) )=\varnothing for all x \in S(0,R). This completes the case when \operatorname{aff} M=\mathbb{R}^d.
The case when \operatorname{aff} M \neq \mathbb{R}^d can be reduced to \operatorname{aff} M=\mathbb{R}^d as in Theorem 2.
2) We claim that \overline{\operatorname{conv} M}=\overline{\bigcup_{x \in \mathbb{R}^d} \operatorname{conv}(P_M^u(x))}. By 1) we have \overline{\operatorname{conv} M}=\overline{\operatorname{ri}(\operatorname{conv} M)} \subset \overline{\bigcup_{x \in \mathbb{R}^d} \operatorname{conv}(P^u_M(x))}. It is clear that the reverse inclusion also holds, since \operatorname{conv}(P^u_M(x) ) \subset \operatorname{conv} M for all x \in \mathbb{R}^d.
Theorem 3 is proved. Remark 2. For a closed unbounded set M, in general \operatorname{conv} M \neq \overline{\operatorname{conv} M} and therefore \operatorname{conv} M \neq \overline{\bigcup_{x \in \mathbb{R}^d} \operatorname{conv}(P^u_M(x))}, in contrast to the conclusion of Theorem 2. Corollary 2. Let U \subset \mathbb{R}^d be a bounded smooth convex domain and M \subset \mathbb{R}^d be a closed set with at most k-valued metric projection P^u_M. Then \overline{\operatorname{conv} M}=\overline{\operatorname{conv}_k M}. The proof is similar to the proof of Corollary 1.
§ 5. Infinite-dimensional results Theorem 4. Let X be a Banach space, U \subset X be a bounded uniformly smooth convex domain, and let M \subset X be boundedly compact. Then
\begin{equation*}
\overline{\operatorname{conv} M}=\overline{\bigcup_{x \in X}\operatorname{conv}(P^u_M(x))}.
\end{equation*}
\notag
Proof. Let p_u be the Minkowski functional of U, and let c_1 and c_2, c_1 < c_2, be positive numbers such that
\begin{equation}
c_1 \| x \| \leqslant p_u(x) \leqslant c_2 \| x \| \quad \forall x \in X.
\end{equation}
\tag{25}
We assume without loss of generality that c_2=1 (for otherwise we can replace the norm \| \cdot \| by c_1 \| \cdot \|). Let a_1, \dots, a_\nu \in M, A := \{a_1, \dots, a_\nu\}, and let {a \in \operatorname{ri}(\operatorname{conv} A)}. We claim that a can be approximated arbitrarily well by points in \bigcup_{x \in X}\operatorname{conv}(P^u_M(x)). Given Y \subset X and t > 0, we set
\begin{equation}
Y_t := \bigl\{x \in X\colon d(x, Y) \leqslant t \bigr\}\quad\text{and} \quad Y^u_t := \bigl\{x \in X\colon d_u(x, Y) \leqslant t \bigr\}.
\end{equation}
\tag{26}
Fix \varepsilon \in (0,1/2). Since U is uniformly smooth, (17) holds, and therefore here exists \lambda_0 > 0 such that for all \lambda \geqslant \lambda_0, all y \in \lambda \partial U and any support functional f_y \in S_{X^*} to \lambda U at y we have
\begin{equation}
\bigl\{z\,{\in}\, X\colon f_y(z) \,{\leqslant}\, f_y(y)-\varepsilon \bigr\} \cap \overline{B}(y, \operatorname{diam}(A)+1) \,{\subset}\, U(0,\lambda) \cap \overline{B}(y, \operatorname{diam}(A)+1).
\end{equation}
\tag{27}
Let r > 0 be such that
\begin{equation*}
d_u(x, A_{2\varepsilon} \cup \{ a \}) \geqslant \lambda_0 \quad \forall\, x \in X \setminus B(0, r).
\end{equation*}
\notag
We claim that
\begin{equation}
a \notin \overline{U}(x, d_u(x,A_{2\varepsilon})) \quad \forall\, x \in X \setminus B(0,r).
\end{equation}
\tag{28}
Assume the contrary. Then
\begin{equation}
U(x, p_u(a-x)) \subset U(x, d_u(x, A_{2\varepsilon})).
\end{equation}
\tag{29}
We denote the norm-one support functional to the domain \overline{U}(x, p_u(a-x)) at the point a by f_a. For t \in \mathbb{R} we set
\begin{equation*}
\Pi(a, t)=\bigl\{z \in X\colon f_a(z) \leqslant f_a(a)-t \bigr\}.
\end{equation*}
\notag
Hence by (27) we have
\begin{equation}
\Pi(a, \varepsilon) \cap \overline{B}(a, \operatorname{diam}(A)+1) \subset U(x, p_u(a-x)) \cap \overline{B}(a, \operatorname{diam}(A)+ 1).
\end{equation}
\tag{30}
It is clear that some points of A also lie in the half-space \Pi(a, 0). We can assume without loss of generality that a_1 \in \Pi(a, 0), that is, f_a(a_1) \leqslant f_a(a). Hence there exists b_1 \in B(a_1, 2\varepsilon) such that f_a(b_1) \leqslant f_a(a)-\varepsilon. Since
\begin{equation*}
\| b_1-a\| \leqslant \| b_1-a_1\|+\| a_1-a \| \leqslant 2\varepsilon+\operatorname{diam}(A) < 1+\operatorname{diam}(A),
\end{equation*}
\notag
we have b_1 \in \Pi(a, \varepsilon) \cap \overline{B}(a, \operatorname{diam}(A)+1). On the other hand b_1 \notin U(x, p_u(a-x)), since A_{2\varepsilon} \cap U(x, p_u(a-x))= \varnothing in view of inclusion (29). But the existence of such b_1 contradicts (30). This therefore proves (28). Since c_2=1, from the definition (26) we obtain A_{2\varepsilon} \subset A_{2c_2\varepsilon}^u= A_{2\varepsilon}^u. Now (28) implies that
\begin{equation}
a \notin \overline{U}(x, d_u(x,A^u_{2\varepsilon})) \quad \forall\, x \in X \setminus B(0,r).
\end{equation}
\tag{31}
Next, from the definition of P_M^u it is clear that the supremum
\begin{equation}
\sup\bigl\{ \| y \|\colon y \in P_M^u(x), \,\| x \| \leqslant r \bigr\} =: R
\end{equation}
\tag{32}
is finite. Since M is boundedly compact, there exists an \varepsilon-net x_1, \dots, x_m \in X for {\overline{B}(0, R) \cap M}. Let L := \operatorname{span}\{x_1, \dots, x_m\}. For x \in L we set
\begin{equation*}
D(x) := \overline{U}(x, d_u(x,M^u_{2\varepsilon})) \cap L
\end{equation*}
\notag
and consider the set-valued mapping
\begin{equation*}
\Phi\colon L \to 2^L \setminus \{\varnothing\}, \qquad x \mapsto P^u_{D(x)} \circ P_M^u(x).
\end{equation*}
\notag
We claim that \Phi is upper semicontinuous. To prove this it suffices to show that if x_n \to x as n\to\infty, z_n \in \Phi(x_n), and z_n \to z as n\to\infty, then z \in \Phi(x). By the definition of \Phi there exist y_n \in M such that
\begin{equation*}
y_n \in P_M^u(x_n)\quad\text{and} \quad z_n \in P^u_{D(x_n)}(y_n).
\end{equation*}
\notag
Passing to a subsequence we can assume that y_n \to y as n\to\infty and, in addition, y \in P^u_M(x), since the map P_M^u is upper semicontinuous. The sets D(x_n) tend to D(x) in the Hausdorff metric, because the function d_u(\, \cdot \,, M^u_{2\varepsilon}) is continuous on X. Therefore,
\begin{equation*}
z \in D(x)\quad\text{and} \quad p_u(z_n-y_n)=d_u(y_n, D(x_n)) \to d_u(y,D(x)),\qquad n\to\infty.
\end{equation*}
\notag
On the other hand p_u(z_n-y_n) \to p_u(z-y) as n\to\infty. Hence z \in P^u_{D(x)}(y), which implies that z \in \Phi(x), as required.
Now consider the map
\begin{equation*}
F\colon L \to \operatorname{Conv}(L), \qquad x \mapsto \operatorname{conv} \Phi(x).
\end{equation*}
\notag
It is upper semicontinuous because \Phi is. Note that for all x \in L
\begin{equation*}
\operatorname{conv}( \{ x\} \cup F(x)) \subset D(x)=\overline{U}(x, d_u(x, M^u_{2\varepsilon})) \cap L \subset \overline{U}(x, d_u(x, A^u_{2\varepsilon})),
\end{equation*}
\notag
so that by (31)
\begin{equation*}
a \notin \operatorname{conv}( \{ x\} \cup F(x)) \qquad \forall\, x \in L \setminus B(0,r).
\end{equation*}
\notag
Now by Lemma 2 there exists a point x_* \in L \cap \overline{B}(0,r) such that
\begin{equation*}
a \in F(x_*)=\operatorname{conv} \Phi(x_*).
\end{equation*}
\notag
Let z_1, \dots, z_n \in \Phi(x_*) be points such that a=\sum_{i=1}^n \lambda_i z_i for \lambda_i \geqslant 0, i=1,\dots,n, \sum_{i=1}^n{\lambda_i}=1. Since \Phi(x_*)=P^u_{D(x_*)} \circ P_M^u(x_*), there exist y_i \in P_M^u(x_*) such that z_i \in P^u_{D(x_*)}(y_i).
For each i \in 1, \dots, n and all z \in D(x_*), \widetilde{z}_i \in P^u_L(y_i), we have
\begin{equation}
p_u(z_i-y_i) \leqslant p_u(z-y_i) \leqslant p_u(z-\widetilde{z}_i)+p_u(\widetilde{z}_i-y_i).
\end{equation}
\tag{33}
It is clear that
\begin{equation}
p_u(\widetilde{z}_i-y_i)=d_u(y_i, L) \leqslant c_2 d(y_i, L)=d(y_i, L) \leqslant \varepsilon,
\end{equation}
\tag{34}
because y_i \in M \cap B(0,R) in view of (32) and since L contains an \varepsilon-net for {M \cap B(0,R)}. Assume that \widetilde{z}_i \notin D(x_*), that is, p_u( \widetilde{z}_i-x_*) > d_u(x_*, M_{2\varepsilon}^u). Hence, for
\begin{equation*}
z=x_*+d_u(x_*,M^u_{2\varepsilon})\frac{\widetilde{z}_i-x_*}{p_u(\widetilde{z}_i-x_*)}
\end{equation*}
\notag
we have the estimate
\begin{equation*}
\begin{aligned} \, p_u(\widetilde{z}_i-z) &=p_u(\widetilde{z}_i-x_*)-p_u(z-x_*)= p_u(\widetilde{z}_i-x_*)-d_u(x_*, M^u_{2\varepsilon}) \\ &\leqslant p_u(\widetilde{z}_i-y_i)+p_u( y_i-x_*)-d_u(x_*,M^u_{2\varepsilon}) \\ &\leqslant \varepsilon+d_u(x_*, M)-d_u(x_*, M^u_{2\varepsilon}) \leqslant 3\varepsilon. \end{aligned}
\end{equation*}
\notag
Therefore,
\begin{equation}
p_u(z-\widetilde{z}_i) \leqslant c_2\| z- \widetilde{z}_i\|=c_2\| \widetilde{z}_i-z \| \leqslant \frac{c_2}{c_1}p_u(\widetilde{z}_i- z) \leqslant \frac{3 \varepsilon}{c_1}.
\end{equation}
\tag{35}
Now from (33)– (35) we obtain
\begin{equation*}
\| z_i-y_i\| \leqslant \frac{p_u(z_i-y_i)}{c_1} \leqslant \frac{p_u(z- \widetilde{z}_i)}{c_1}+\frac{p_u(\widetilde{z}_i-y_i)}{c_1} \leqslant \frac{3\varepsilon}{c_1^2}+\frac{\varepsilon}{c_1} \leqslant \frac{4\varepsilon}{c_1^2}.
\end{equation*}
\notag
If \widetilde{z}_i \in D(x_*), then
\begin{equation*}
p_u(z_i-y_i) =p_u(\widetilde{z}_i-y_i) \leqslant \varepsilon \quad \Longrightarrow \quad \| z_i-y_i \| \leqslant p_u(z_i-y_i) \leqslant \frac{\varepsilon}{c_1} < \frac{4\varepsilon}{c_1^2}.
\end{equation*}
\notag
So, for b=\sum_{i=1}^n \lambda_i y_i \in \operatorname{conv} P_M^u(x_*) we have
\begin{equation*}
\| a-b\|=\biggl\| \sum_{i=1}^n \lambda_i (z_i-y_i)\biggr\| < \frac{4\varepsilon}{c_1^2} \sum_{i=1}^n \lambda_i=\frac{4\varepsilon}{c_1^2}.
\end{equation*}
\notag
Since \varepsilon can be chosen arbitrarily small, a \in \overline{\bigcup_{x \in X}\operatorname{conv}(P_M^u(x))}, and therefore \operatorname{conv} M \subset \overline{\bigcup_{x \in X}\operatorname{conv}(P_M^u(x))}. The reverse inclusion \overline{\operatorname{conv} M} \supset \overline{\bigcup_{x \in X}\operatorname{conv}(P_M^u(x))} is clear.
Theorem 4 is proved. Corollary 3. Let X be a Banach space, U \subset X be a bounded uniformly smooth convex domain and M \subset X be a compact set with at most k-valued metric projection P_M^u. Then \overline{\operatorname{conv} M}=\overline{\operatorname{conv}_k M}. Proof. By Theorem 4, \overline{\operatorname{conv} M}=\overline{\bigcup_{x \in X}\operatorname{conv}(P^u_M(x))} \subset \overline{\operatorname{conv}_k M}. The reverse inclusion \overline{\operatorname{conv}_k M} \subset \overline{\operatorname{conv} M} is clear.
Corollary 3 is proved. Theorem 5. Let X be a Banach space, U \subset X be a bounded smooth convex domain and M \subset X be a compact set. Then \overline{\operatorname{conv} M}=\overline{\bigcup_{x \in X}\operatorname{conv}(P_M^u(x))}. Proof. Let p_u be the Minkowski functional of U. Let c_1 and c_2, c_1 < c_2, be the fixed positive numbers from (25). We can assume without loss of generality that c_2 < 1. Let a_1, \dots, a_\nu \in M, A := \{a_1, \dots, a_\nu\}, and let a\,{\in} \operatorname{ri}(\operatorname{conv} A). We claim that a can be approximated arbitrarily well by \bigcup_{x \in X}\operatorname{conv}(P_M^u(x)). Since M is compact, there exists an \varepsilon-net x_1, \dots, x_m \in X for M. Set
\begin{equation*}
L := \operatorname{span}( A \cup \{x_1, \dots, x_m\}).
\end{equation*}
\notag
It is clear that V := U \cap L is a uniformly smooth convex domain in L, and it follows from the proof of Theorem 4 that there exits r > 0 such that
\begin{equation*}
a \notin \overline{V}(x, d_u(x,A_{2\varepsilon}^u)) \quad \forall\, x \in L \setminus B(0,r)
\end{equation*}
\notag
(see (31) and above). For x \in L we set
\begin{equation*}
D(x) := \overline{B}(x, d_u(x,M_{2\varepsilon}^u)) \cap L
\end{equation*}
\notag
and consider the set-valued mapping
\begin{equation*}
\Phi\colon L \to 2^L \setminus \{ \varnothing \}, \qquad x \mapsto P^u_{D(x)} \circ P_M^u(x).
\end{equation*}
\notag
Arguing as in Theorem 4, it can be shown that there exist x_* \in B(0,r) \cap L and b \in \operatorname{conv} P_M^u(x_*) such that \| a- b\| < 4\varepsilon/c_1^2. It follows that \overline{\operatorname{conv} M} \subset \overline{\bigcup_{x \in X}\operatorname{conv}(P_M^u(x))}. The reverse inclusion \overline{\bigcup_{x \in X}\operatorname{conv}(P_M^u(x))} \subset \overline{\operatorname{conv} M} is clear.
The theorem is proved. Corollary 4. Let X be a Banach space, U \subset X be a smooth convex domain and M \subset X be a compact set. 1) If the metric projection P^u_M is at most k-valued, then \overline{\operatorname{conv} M}=\operatorname{conv} M=\operatorname{conv}_k M. 2) If \operatorname{conv} M is not closed, then for each k there exists a point x_k \in X such that |P_M^u(x_k)| \geqslant k. Proof. 1) By Theorem 4
\begin{equation*}
\overline{\operatorname{conv} M}=\overline{\bigcup_{x \in X}\operatorname{conv}(P^u_M(x))} \subset \overline{\operatorname{conv}_k M}.
\end{equation*}
\notag
We claim that the set \operatorname{conv}_k M is closed. Indeed, if x_n\,{\in} \operatorname{conv}_k M and x_n \to x as {n\to\infty}, then for any natural n there exist y_{n}^i \in M (some of which can repeat) and \lambda_n^{i} \geqslant 0, i=1, \dots, k, \sum_{i=1}^k \lambda_n^{i}=1, such that
\begin{equation*}
x_n=\sum_{i=1}^k \lambda_n^{i} y_n^{i}.
\end{equation*}
\notag
Since M is compact, there exists an increasing subsequence of indices \{n_j\}_{j=1}^\infty \subset \mathbb{N} such that y_{n_j}^{i} \to y^{i} and \lambda_{n_j}^{i} \to \lambda^i as j \to \infty, for all i=1, \dots, k. It is clear that x=\sum_{i=1}^k \lambda^i y^i \in \operatorname{conv}_k M, and thus the set \operatorname{conv}_k M is closed. Hence {\overline{\operatorname{conv} M} \subset \operatorname{conv}_k M}. The inclusion \operatorname{conv}_k M \subset \operatorname{conv} M is clear. The last two inclusions imply that \overline{\operatorname{conv} M}=\operatorname{conv} M= \operatorname{conv}_k M.
Assertion 1) is immediate from 2).
Corollary 4 is proved.
§ 6. Results for balls in Banach spaces If as U we take the unit ball B(0,1) in a Banach space X, then P_M^u is the ordinary metric projection P_M. Now the following result is immediate from Theorems 3–5. Theorem 6. Let X be a Banach space, and let M \subset X. Then the following assertions hold. 1) If X is a smooth finite-dimensional space and M is boundedly compact, then \operatorname{ri}(\operatorname{conv} M) \subset \bigcup_{x \in \mathbb{R}^d} \operatorname{conv}(P_M(x)). 2) If X is a uniformly smooth space and M is boundedly compact, then \overline{\operatorname{conv} M}=\overline{\bigcup_{x \in \mathbb{R}^d} \operatorname{conv}(P_M(x))}. 3) If X is smooth and M is compact, then \overline{\operatorname{conv} M}=\overline{\bigcup_{x \in \mathbb{R}^d} \operatorname{conv}(P_M(x))}. Of course, similar results can also be derived from other theorems and corollaries presented in § 4 and § 5. The author is greatly indebted to P. A. Borodin for stating the problem and making valuable comments, and also to A. R. Alimov and I. G. Tsar’kov for helpful discussions.
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Citation:
K. S. Shklyaev, “The convex hull and the Carathéodory number of a set in terms of the metric projection operator”, Sb. Math., 213:10 (2022), 1470–1486
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https://www.mathnet.ru/eng/sm9742https://doi.org/10.4213/sm9742e https://www.mathnet.ru/eng/sm/v213/i10/p167
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