Stability of a convex feasibility problem

Abstract

The 2-sets convex feasibility problem aims at finding a point in the intersection of two closed convex sets A and B in a normed space X. More generally, we can consider the problem of finding (if possible) two points in A and B, respectively, which minimize the distance between the sets. In the present paper, we study some stability properties for the convex feasibility problem: we consider two sequences of sets, each of them converging, with respect to a suitable notion of set convergence, respectively, to A and B. Under appropriate assumptions on the original problem, we ensure that the solutions of the perturbed problems converge to a solution of the original problem. We consider both the finite-dimensional and the infinite-dimensional case. Moreover, we provide several examples that point out the role of our assumptions in the obtained results.

1 Introduction

The convex feasibility problem is the classical problem of finding a point in the intersection of a finite collection of closed and convex sets (see [5, Sect. 4.5] for the main results on this subject). Many concrete problems in applications can be formulated as a convex feasibility problem. As typical examples, we mention solution of convex inequalities, partial differential equations, minimization of convex nonsmooth functions, medical imaging, computerized tomography and image reconstruction. For some details and other applications see, e.g. [2, 7] and the references therein. Moreover, it is worth to mention the recent annotated bibliography [6], about projection methods, containing several references to the convex feasibility problem and its applications.

Many efforts have been devoted to the study of algorithmic procedures to solve convex feasibility problems, both from a theoretical and from a computational point of view (see, e.g. [2,3,4, 9] and the references therein).

Often in concrete applications data are affected by some uncertainties. Hence stability of solutions with respect to data perturbations is a desirable property, also in view of the development of a computational approach to solve the convex feasibility problem. Our paper is devoted to investigate some stability properties of the 2-sets convex feasibility problem by using set convergence notions. We will also consider the case of a pair of closed and convex sets with empty intersection: in this case a solution of the problem is a pair of minimal distance elements of the two sets.

In this paper, we investigate a sequence of perturbed convex feasibility problems whose data are obtained by considering two sequences of closed and convex sets $\left\{ A_n \right\}$ and $\left\{ B_n \right\}$ converging respectively to the sets A and B. If the intersection of $A_n$ and $B_n$ is empty, we consider, as a solution of the n-th perturbed problem, the pair of elements $a_n\in A_n$ and $b_n\in B_n$ such that the distance between $A_n$ and $B_n$ is $\left\| a_{n}-b_{n}\right\|$.

Our aim is to find some conditions that guarantee the convergence of the solutions of the perturbed convex feasibility problems to a solution of the original convex feasibility problem.

We obtain some stability results both in the finite-dimensional and in the infinite-dimensional framework, using the Kuratowski–Painlevé convergence notion in the finite-dimensional case and the Attouch–Wets convergence in the infinite-dimensional setting. Moreover, we give some examples showing that the assumptions that we use to guarantee the stability features of a given convex feasibility problem cannot be avoided, both in the finite and in the infinite-dimensional case.

The paper is organized as follows. Section 2 is devoted to definitions and preliminary results, mainly concerning the various notions of set-convergence. Section 3 presents a stability result for the convex feasibility problem when A and B are contained in a finite-dimensional normed vector space and the sequences of closed and convex sets $\left\{ A_n \right\}$ and $\left\{ B_n \right\}$ converge in the Kuratowski–Painlevé sense respectively to A and B. Section 4 is devoted to study the stability properties of a convex feasibility problem in an infinite-dimensional setting. Here, we use the Attouch–Wets convergence, that is stronger than the Kuratowski–Painlevé convergence, even if they coincide in the finite-dimensional setting. Moreover, it is worth to be noticed that we obtain results concerning both weak and norm convergence of the solutions of perturbed problems to a solution of the original problem. In order to obtain the norm convergence of a sequence of solutions of perturbed problems, we strengthen the convexity assumptions by assuming that A has nonempty interior and it is locally uniformly rotund (LUR) at a given solution a. Finally, in Sect. 5, we provide some rather involved examples in $\ell _2$ that point out the role of our assumptions even in a Hilbert space framework.

2 Notations and preliminaries

Throughout all this paper, X denotes a real normed space with the topological dual $X^*$. We denote by $B_X$ and $S_X$ the closed unit ball and the unit sphere of X, respectively. For $x,y\in X$, [x, y] denotes the closed segment in X with endpoints x and y, and $(x,y)=[x,y]{\setminus }\{x,y\}$ is the corresponding “open” segment. For a subset K of X, $\alpha >0$, and a functional $x^*\in S_{X^{*}}$ bounded on K, let

$$\begin{aligned} S(x^*,\alpha ,K)=\{x\in K;\, x^*(x)\ge \sup x^*(K)-\alpha \} \end{aligned}$$

be the closed slice of K given by $\alpha$ and $x^*$.

For a subset A of X, we denote by ${\mathrm {int}}\,(A)$, ${\mathrm {conv}}\,(A)$ and $\overline{{\mathrm {conv}}}\,(A)$ the interior, the convex hull and the closed convex hull of A, respectively. Moreover,

$$\begin{aligned} \overline{\mathrm {cone}}(A)=\overline{{\mathrm {conv}}}\,{\bigl ([0,\infty )\cdot A\bigr )} \end{aligned}$$

is the closed convex cone generated by the set A. We denote by

$$\begin{aligned} \textstyle \mathrm {diam}(A)=\sup _{x,y\in A}\Vert x-y\Vert , \end{aligned}$$

the (possibly infinite) diameter of A. For $x\in X$, let

$$\begin{aligned} {\mathrm {dist}}(x,A) =\inf _{a\in A} \Vert a-x\Vert . \end{aligned}$$

Moreover, given A, B nonempty subsets of X, we denote by ${\mathrm {dist}}(A,B)$ the usual “distance” between A and B, that is,

$$\begin{aligned} {\mathrm {dist}}(A,B)=\inf _{a\in A} {\mathrm {dist}}(a,B). \end{aligned}$$

2.1 Convergence of sets

By ${\mathrm {c}}(X)$ we denote the family of all nonempty closed subsets of X. Let $\{A_n\}$ be a sequence in ${\mathrm {c}}(X)$ and let us consider the following sets:

$$\begin{aligned} \textstyle \mathrm {Li}\ A_n=\{x\in X;\, x=\lim _n x_n, x_n\in A_n\} \end{aligned}$$

and

$$\begin{aligned} \textstyle \mathrm {Ls}\ A_n=\{x=\lim _k x_k\in X;\, x_k\in A_{n_k}, \{n_k\}\, \hbox {is a subsequence of the integers} \}. \end{aligned}$$

Definition 2.1

Let $\{A_n\}$ be a sequence in ${\mathrm {c}}(X)$ and $A\in {\mathrm {c}}(X)$.

• (i) $\{A_n\}$ converges to A for the lower Kuratowski–Painlevé convergence if $A\subset \mathrm {Li} A_n$.

• (ii) $\{A_n\}$ converges to A for the upper Kuratowski–Painlevé convergence if $A\supset \mathrm {Ls} A_n$.

Moreover, we say that $\{A_n\}$ converges to A for the Kuratowski–Painlevé convergence ($A_{n}{\mathop {\rightarrow }\limits ^{K}}A$) if $\{A_n\}$ converges to A for the upper and the lower Kuratowski–Painlevé convergence.

Now, let us introduce the (extended) Hausdorff metric h on ${\mathrm {c}}(X)$. For $A,B\in {\mathrm {c}}(X)$, we define the excess of A over B as

$$\begin{aligned} e(A,B) = \sup _{a\in A} d(a,B). \end{aligned}$$

Moreover, if $A\ne \emptyset$ and $B=\emptyset$ we put $e(A,B)=\infty$, if $A=\emptyset$ we put $e(A,B)=0$. We define

$$\begin{aligned} h(A,B)=\max \bigl \{ e(A,B),e(B,A) \bigr \}. \end{aligned}$$

Definition 2.2

A sequence $\{A_j\}$ in ${\mathrm {c}}(X)$ is said to Hausdorff converge to $A\in {\mathrm {c}}(X)$ if

$$\begin{aligned} \textstyle \lim _j h(A_j,A) = 0. \end{aligned}$$

Finally, we introduce the so called Attouch–Wets convergence (see, e.g. [11, Definition 8.2.13]), which can be seen as a localization of the Hausdorff convergence. If $N\in {\mathbb {N}}$ and $A,B\in {\mathrm {c}}(X)$, define

$$\begin{aligned} e_N(A,B)= & {} e(A\cap N B_X, B)\in [0,\infty ),\\ h_N(A,B)= & {} \max \{e_N(A,B), e_N(B,A)\}. \end{aligned}$$

Definition 2.3

A sequence $\{A_j\}$ in ${\mathrm {c}}(X)$ is said to Attouch–Wets converge to $A\in {\mathrm {c}}(X)$ if, for each $N\in {\mathbb {N}}$,

$$\begin{aligned} \textstyle \lim _j h_N(A_j,A)= 0. \end{aligned}$$

Several times without mentioning it, we shall use the following theorem in proving the results contained in Sects. 3 and 4.

Theorem 2.4

(See, e.g. [11, Theorem 8.2.14]) The sequence of sets $\{A_n\}$ Attouch–Wets converges to A iff

$$\begin{aligned} \textstyle \sup _{\Vert x\Vert \le N}{|d(x,A_n)-d(x,A)|}\rightarrow 0 \ \ \ (n\rightarrow \infty ), \end{aligned}$$

whenever $N\in {\mathbb {N}}$.

We recall that in the finite-dimensional case the Attouch–Wets convergence and the Kuratowski–Painlevé convergence coincide (see, e.g. [11, Sect. 8.2]).

In the sequel, we shall use the following easy-to-prove fact. For the convenience of the reader we provide a proof.

Fact 2.5

Let A and B be two closed and convex subsets of a normed space X. Let $\{A_n\}$ and $\{B_n\}$ be two sequences of closed convex sets such that $A_n\rightarrow A$ and $B_n\rightarrow B$ for the lower Kuratowski–Painlevé convergence. Then

$$\begin{aligned} \textstyle \limsup _n {\mathrm {dist}}(A_n,B_n)\le {\mathrm {dist}}(A,B). \end{aligned}$$

In particular, if $A\cap B\ne \emptyset$ we have $\lim _n {\mathrm {dist}}(A_n,B_n)=0$.

Proof

Let $\epsilon >0$, then there exist $x\in A$ and $y\in B$ such that $\Vert x-y\Vert \le {\mathrm {dist}}(A,B)+\epsilon$. Since $A_n\rightarrow A$ and $B_n\rightarrow B$ for the lower Kuratowski–Painlevé convergence, there exist two sequences $\{x_n\}$ and $\{y_n\}$ such that $x_n\rightarrow x$, $y_n\rightarrow y$ and, for each $n\in {\mathbb {N}}$, $x_n\in A_n$, $y_n\in B_n$. In particular, it eventually holds $\Vert x_n-x\Vert \le \epsilon$ and $\Vert y_n-y\Vert \le \epsilon$. Hence, the following inequalities eventually hold:

$$\begin{aligned} {\mathrm {dist}}(A_n,B_n)\le \Vert x_n-y_n\Vert \le {\mathrm {dist}}(A,B)+3\epsilon . \end{aligned}$$

By the arbitrariness of $\epsilon >0$, we have the thesis. $\square$

3 Convergence of minimal distance points of a pair of convex sets: the finite-dimensional case

In this section, we denote by X a finite-dimensional normed space.

Definition 3.1

Let A, B be nonempty closed convex set in X. Let

$$\begin{aligned} {\mathrm {m}}(A,B)=\{a\in A;\, {\mathrm {dist}}(a,B)={\mathrm {dist}}(A,B)\}. \end{aligned}$$

It is easy to see that ${\mathrm {m}}(A,B)$ is a closed convex set.

Definition 3.2

Let C be a non empty closed convex subset of X and $x\in C$. Let us define

$$\begin{aligned} D(x)=\{d\in X;\, x+td\in C,\ \forall t>0\}. \end{aligned}$$

Remark 3.3

By [1, Proposition 2.1.5], if $x,y\in C$ then $D(x)=D(y)$. That is, the set D(x) does not depend on $x\in C$. We denote this set, called the asymptotic cone of C, by $C_\infty$.

We prove the following lemma that will be useful in the sequel (it can be seen as a slight generalization of [1, Proposition 2.1.9]).

Lemma 3.4

Let A and B be nonempty closed convex sets in X such that ${\mathrm {m}}(A,B)$ is nonempty. Then

$$\begin{aligned} A_\infty \cap B_\infty =[{\mathrm {m}}(A,B)]_\infty . \end{aligned}$$

Proof

Let $a\in A$ and $b\in B$ be such that $\Vert b-a\Vert ={\mathrm {dist}}(A,B)$.

Let us prove that $A_\infty \cap B_\infty \subset [{\mathrm {m}}(A,B)]_\infty$. Let $d\in A_\infty \cap B_\infty =D(a)\cap D(b)$. Since, for each $t>0$,

$$\begin{aligned} \Vert a-b\Vert =\Vert a+td-(b+td)\Vert ={\mathrm {dist}}(A,B), \end{aligned}$$

we have that $a+td\in {\mathrm {m}}(A,B)$, whenever $t>0$. Hence $d\in [{\mathrm {m}}(A,B)]_\infty$.

For the reverse inclusion, suppose that $a+td\in {\mathrm {m}}(A,B)$, whenever $t>0$. Clearly, $d\in A_\infty$. Now, we prove that $d\in B_\infty$. Let us fix $t>0$ and $n\in {\mathbb {N}}$, and let us observe that

$$\begin{aligned} \Vert b+ntd-(a+ntd)\Vert =\Vert b-a\Vert ={\mathrm {dist}}(A,B). \end{aligned}$$

Hence, there exists $d_n\in B$ such that

$$\begin{aligned} \Vert b+ntd-d_n\Vert \le 2\,{\mathrm {dist}}(A,B). \end{aligned}$$

Then,

$$\begin{aligned} \textstyle \Vert b+td-(b+\frac{d_n-b}{n})\Vert \le \frac{2}{n}\,{\mathrm {dist}}(A,B). \end{aligned}$$

By the arbitrariness of $n\in {\mathbb {N}}$, since $b+\frac{d_n-b}{n}\in B$, and since B is closed, it holds that $b+td\in B$. By the arbitrariness of $t>0$, the thesis is proved. $\square$

The following theorem is the main result of this section. It proves that, under mild assumption, the 2-sets convex feasibility problem has a considerable degree of stability.

Theorem 3.5

Let $\{ A_{n}\}$ and $\{ B_{n}\}$ be two sequences of nonempty closed convex sets in X, A and B two nonempty closed convex subsets of X such that

$$\begin{aligned} A_{n}\rightarrow A\quad \text {and}\quad B_{n}\rightarrow B, \end{aligned}$$

for the Kuratowski–Painlevé convergence. Suppose that ${\mathrm {m}}(A,B)$ is a nonempty bounded set. Let $\left\{ a_{n}\right\}$ and $\left\{ b_{n}\right\}$ be sequences such that $a_{n}\in A_{n}$, $b_{n}\in B_{n}$ ($n\in {\mathbb {N}}$) and

$$\begin{aligned} \text {dist}(A_{n},B_{n})=\left\| a_{n}-b_{n}\right\| . \end{aligned}$$

Then there exists a subsequence $\left\{ a_{n_{k}}\right\}$ such that

$$\begin{aligned} \lim _{k\rightarrow \infty }a_{n_{k}}=c\in {\mathrm {m}}(A\,,B). \end{aligned}$$

Moreover, if ${\mathrm {m}}(A,B)=\{a\}$ then $a_n\rightarrow a$.

Proof

Let us prove the first part of the theorem. By Fact 2.5, it holds

$$\begin{aligned} \limsup _{n} \Vert a_n-b_n\Vert \le {\mathrm {dist}}(A,B). \end{aligned}$$

(1)

We claim that $\{a_n\}$ and $\{b_n\}$ are bounded.

Suppose that this is not the case and let $a\in A$ and $b\in B$ be such that $\Vert a-b\Vert ={\mathrm {dist}}(A,B)$. Without loss of generality, we can suppose that $\Vert a_n\Vert ,\Vert b_n\Vert \rightarrow \infty$. By the lower part of the convergence of $\left\{ A_{n}\right\}$ there exists a sequence $\left\{ a_{n}'\right\}$ such that $a_{n}'\in A_{n}$ and $a_{n}'\rightarrow a$. Since $A_{n}$ is a convex set, for any $\alpha \in \left[ 0,1\right]$ it holds:

$$\begin{aligned} \alpha a_{n}+(1-\alpha )a_{n}'=a_{n}'+\alpha (a_{n}-a_{n}')\in A_{n}. \end{aligned}$$

The sequence

$$\begin{aligned} \left\{ \frac{a_{n}-a_{n}'}{\left\| a_{n}-a_{n}'\right\| }\right\} \end{aligned}$$

has a subsequence converging to $d\ne 0.$ There is no loss of generality in assuming

$$\begin{aligned} \lim _{n\rightarrow +\infty }\frac{a_{n}-a_{n}'}{\left\| a_{n}-a_{n}'\right\| }=d. \end{aligned}$$

Therefore, it holds

$$\begin{aligned} a+\beta d=\lim _{n\rightarrow +\infty }\left( a_{n}'+\frac{\beta }{\left\| a_{n}-a_{n}'\right\| }(a_{n}-a_{n}')\right) . \end{aligned}$$

Since for every $\beta >0$ there exists $n_{2}(\beta )\in {\mathbb {N}}$ such that $\frac{\beta }{\left\| a_{n}-a_{n}'\right\| }\in \left[ 0,1\right] ,$ whenever $n>n_{2}(\beta )$, it holds

$$\begin{aligned} a+\beta d\in A, \end{aligned}$$

for every $\beta >0.$ Hence, $d\in A_{\infty }.$

Analogously, we may prove that

$$\begin{aligned} \lim _{n\rightarrow +\infty }\frac{b_{n}-b_{n}'}{\left\| b_{n}-b_{n}'\right\| }=d'\in B_{\infty }, \end{aligned}$$

where $\left\{ b_{n}'\right\}$ is a sequence such that $b_{n}'\in B_{n}$ and $b_{n}'\rightarrow b$.

Let us observe that $\{a'_n\}$, $\{b'_n\}$ and $\{a_n-b_n\}$ are bounded sequences in X. Since $\Vert a_n\Vert ,\Vert b_n\Vert \rightarrow \infty$, we have $\Vert a_n-a'_n\Vert \sim \Vert b_n-b'_n\Vert$ and hence

$$\begin{aligned} d=\lim _{n\rightarrow +\infty }\frac{a_{n}-a_{n}'}{\left\| a_{n}-a_{n}'\right\| }=\lim _{n\rightarrow +\infty }\frac{b_{n}-b_{n}'}{\left\| b_{n}-b_{n}'\right\| }=d', \end{aligned}$$

Therefore we have

$$\begin{aligned} 0\ne d\in A_{\infty }\cap B_{\infty }. \end{aligned}$$

By Lemma 3.4, we have

$$\begin{aligned} A_{\infty }\cap B_{\infty }=[{\mathrm {m}}(A,B)]_{\infty }. \end{aligned}$$

Then ${\mathrm {m}}(A,B)$ is not a bounded set, a contradiction.

By the claim and compactness, there exist two subsequences $\left\{ a_{n_{k}}\right\}$ and $\left\{ b_{n_{k}}\right\}$, respectively of $\left\{ a_{n}\right\}$ and of $\left\{ b_{n}\right\}$, such that

$$\begin{aligned} \lim _{k\rightarrow +\infty }a_{n_{k}}=u,\quad \lim _{k\rightarrow +\infty }b_{n_{k}}=v, \end{aligned}$$

where $u\in A$ and $v\in B$. By Fact 2.5, $\Vert u-v\Vert ={\mathrm {dist}}(A,B)$ and the thesis is proved.

The second part of the theorem follows easily by the first part. $\square$

Remark 3.6

• (i) The above theorem can be proved in an alternative way, by using known results concerning stability theory for convex optimization problem. However, we preferred to present a direct and more geometrical proof. We give a sketch of the alternative proof below. (See, e.g. [11] for definitions and main results about convergence of functions and well-posed problems).

  Let $f,f_n:X\times X\rightarrow (\infty ,\infty ]$ ($n\in {\mathbb {N}}$) be the convex lower semicontinuous functions defined as follows. For each $(x_1,x_2)\in X\times X$ and $n\in {\mathbb {N}}$, put

  $$\begin{aligned} f(x_1,x_2)={\left\{ \begin{array}{ll} \Vert x_1-x_2\Vert &{}\quad \text {if }x_1\in A\text { and }x_2\in B;\\ \infty &{}\quad \text {otherwise;} \end{array}\right. } \end{aligned}$$

  and

  $$\begin{aligned} f_n(x_1,x_2)={\left\{ \begin{array}{ll} \Vert x_1-x_2\Vert &{}\quad \text {if }x_1\in A_n\text { and }x_2\in B_n;\\ \infty &{}\quad \text {otherwise.} \end{array}\right. } \end{aligned}$$

  Since $A_{n}\rightarrow A$ and $B_{n}\rightarrow B$ for the Kuratowski–Painlevé convergence (equivalently, for the Attouch–Wets convergence), we have that $f_n\rightarrow f$ for the Kuratowski–Painlevé convergence. Moreover, proceeding as in the proof of Theorem 3.5, we may prove that f is Tykhonov well-posed in the generalized sense. Hence, we can apply [11, Theorem 10.2.24] to obtain the thesis.

• (ii) It is interesting to observe that, under the hypothesis of the above theorem, we have $\Vert a_n-b_n\Vert \rightarrow \mathrm {dist}(A,B)$ ($n\rightarrow \infty$). Indeed, if $\{n_k\}$ is a subsequence of the integers, by the proof of Theorem 3.5 there exists $\{n_{k_h}\}$, a subsequence of $\{n_k\}$, such that $\lim _h a_{n_{k_h}}=u\in A$, $\lim _h b_{n_{k_h}}=v\in B$, and $\Vert u-v\Vert =\mathrm {dist}(A,B)$. Then $\lim _h \Vert a_{n_{k_h}}-b_{n_{k_h}}\Vert =\mathrm {dist}(A,B)$ and the proof is complete.

Whenever the two limit sets are such that $A\cap B\ne \emptyset$, we have the following corollary.

Corollary 3.7

Let $\{ A_{n}\}$ and $\{ B_{n}\}$ be two sequences of nonempty closed convex sets in X, A and B two nonempty closed convex subsets of X such that

$$\begin{aligned} A_{n}\rightarrow A\quad \text {and}\quad B_{n}\rightarrow B, \end{aligned}$$

for the Kuratowski–Painlevé convergence. Suppose that $A\cap B$ is a nonempty bounded set. Let $\left\{ a_{n}\right\}$ and $\left\{ b_{n}\right\}$ be sequences such that $a_{n}\in A_{n}$, $b_{n}\in B_{n}$ ($n\in {\mathbb {N}}$) and

$$\begin{aligned} \text {dist}(A_{n},B_{n})=\left\| a_{n}-b_{n}\right\| . \end{aligned}$$

Then there exist two subsequences $\left\{ a_{n_{k}}\right\}$ and $\left\{ b_{n_{k}}\right\}$ such that

$$\begin{aligned} \lim _{k\rightarrow \infty }a_{n_{k}}=\lim _{k\rightarrow \infty }b_{n_{k}}=c\in A\cap B. \end{aligned}$$

Moreover, if $A\cap B=\{c\}$ then $a_n,b_n\rightarrow c$.

The following examples show that both the assumptions in Theorem 3.5 play an independent role and each of them cannot be deleted. The first one focuses on the role of convexity assumptions.

Example 3.8

Let us consider the sets $(n\ge 2)$:

$$\begin{aligned} \begin{aligned} \textstyle A_{n}&=\left\{ \left( x_{1},x_{2}\right) \in \mathbb {R}^{2}:-1\le x_{1}\le -\frac{1}{n}, \,{n}\frac{x_{1}+1}{n-1}\le x_{2}\le 2+{n}\frac{x_{1}+1}{1-n}\right\} \\&\cup \textstyle \left\{ \left( x_{1},x_{2}\right) \in \mathbb {R}^{2}:-1\le x_{1}\le -\frac{1}{2n},\,\frac{2}{x_{1}^{2}}\le x_{2}\le 8n^{2}\right\} \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} B_{n}&=\left\{ \left( x_{1},x_{2}\right) \in \mathbb {R}^{2}:\frac{1}{n}\le x_{1}\le 1,\,{n}\frac{x_{1}-1}{1-n}\le x_{2}\le 2+{n}\frac{x_{1}-1}{n-1}\right\} \\&\cup \left\{ \left( x_{1},x_{2}\right) \in \mathbb {R}^{2}:\frac{1}{2n}\le x_{1}\le 1,\,\frac{2}{x_{1}^{2}}\le x_{2}\le 8n^{2}\right\} . \end{aligned} \end{aligned}$$

The sequences $\left\{ A_{n}\right\}$ and $\left\{ B_{n}\right\}$ converge respectively to

$$\begin{aligned} \begin{aligned} A&=\left\{ \left( x_{1},x_{2}\right) \in \mathbb {R}^{2}:-1\le x_{1}\le 0,\, x_{1}+1\le x_{2}\le 1-x_{1}\right\} \\&\cup \left\{ \left( x_{1},x_{2}\right) \in \mathbb {R}^{2}:-1\le x_{1}<0,\, x_{2}\ge \frac{2}{x_{1}^{2}}\right\} \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} B&=\left\{ \left( x_{1},x_{2}\right) \in \mathbb {R}^{2}:0\le x_{1}\le 1,\,1-x_{1}\le x_{2}\le 1+x_{1}\right\} \\&\cup \left\{ \left( x_{1},x_{2}\right) \in \mathbb {R}^{2}:0\le x_{1}\le 1,\, x_{2}\ge \frac{2}{x_{1}^{2}}\right\} . \end{aligned} \end{aligned}$$

It is easy to see that $A\cap B=\left\{ (0,1)\right\}$ and

$$\begin{aligned} A_{n}{\mathop {\rightarrow }\limits ^{K}}A,\quad B_{n}{\mathop {\rightarrow }\limits ^{K}} B. \end{aligned}$$

All the assumptions of Theorem 3.5 are satisfied except for the convexity of $A_{n}$ and $B_{n}$. The minimal distance between the sets $A_{n}$ and $B_{n}$ is achieved only at the pair of points

$$\begin{aligned} a_{n}=\left( -\frac{1}{2n},8n^{2}\right) \in A_{n}\,\mathrm {and}\, b_{n}=\left( \frac{1}{2n},8n^{2}\right) \in B_{n}. \end{aligned}$$

It is clear that the sequences $\left\{ a_{n}\right\}$ and $\left\{ b_{n}\right\}$ are not bounded. Hence, the thesis of Theorem 3.5 does not hold.

The second example proves that the boundedness assumption on the set ${\mathrm {m}}(A,B)$ cannot be dropped.

Example 3.9

Let $A_{n}$ and $B_{n}$ be defined as in Example 3.8. Let us consider the sets

$$\begin{aligned} C_{n}=\mathrm {conv}(A_{n})\quad \mathrm {and}\quad D_{n}=\mathrm {conv}(B_{n}). \end{aligned}$$

It is easy to see that

$$\begin{aligned} C_{n}{\mathop {\rightarrow }\limits ^{K}}C,\quad D_{n}{\mathop {\rightarrow }\limits ^{K}}D, \end{aligned}$$

where

$$\begin{aligned} C=\left\{ \left( x_{1},x_{2}\right) \in \mathbb {R}^{2}:-1\le x_{1}\le 0,\, x_{1}+1\le x_{2}\right\} \end{aligned}$$

and

$$\begin{aligned} D=\left\{ \left( x_{1},x_{2}\right) \in \mathbb {R}^{2}:0\le x_{1}\le 1,\,1-x_{1}\le x_{2}\right\} . \end{aligned}$$

Moreover, we have $C\cap D=\left\{ \left( x_{1},x_{2}\right) \in \mathbb {R}^{2}:x_{1}=0,\, x_{2}\ge 1\right\}$.

All the assumptions of Theorem 3.5 are satisfied except for the boundedness of the set $C\cap D$. The minimal distance between the sets $C_{n}$ and $D_{n}$ is achieved only at the same pair of points $a_{n}\in C_{n}$ and $b_{n}\in D_{n}$ as in Example 3.8. Of course, as in the previous example both the sequences $\left\{ a_{n}\right\}$ and $\left\{ b_{n}\right\}$ have no convergent subsequences. Therefore the thesis of Theorem 3.5 does not hold.

4 Convergence of minimal distance points of a pair of convex sets: the infinite-dimensional case

In an infinite-dimensional setting, we need some strengthenings of the assumptions to obtain stability results for our problems. Indeed, Example 5.2, in Sect. 5, shows that an analogue of Theorem 3.5 does not hold, even if we assume that the sequences of sets converge for the Hausdorff convergence and that the space X is a Hilbert space. In this section, we prove that an additional geometric condition on the limit sets ensures the stability result (see Theorem 4.5 below). Moreover, we use the Attouch–Wets convergence of sets instead of the Kuratowski–Painlevé convergence (cf. Example 5.6).

We start with some definitions and preliminary results. Let us recall that a body in X is a closed convex set in X with nonempty interior.

Definition 4.1

(See, e.g. [8, Definition 7.10]) Let A be a nonempty subset of a normed space X. A point $a\in A$ is called a strongly exposed point of A if there exists a support functional $f\in X^*{\setminus }\{0\}$ for A in a (i.e., $f (a) = \sup f(A)$), such that $x_n\rightarrow a$ for all sequences $\{x_n\}$ in A such that $\lim f(x_n) = \sup f(A)$. In this case, we say that f strongly exposes A at a.

Let us observe that $f\in S_{X^*}$ strongly exposes A at a iff $f(a)=\sup f(A)$ and

$$\begin{aligned} \mathrm {diam}\bigl (S(f,\alpha ,A)\bigr )\rightarrow 0 \text { as } \alpha \rightarrow 0. \end{aligned}$$

Definition 4.2

(See, e.g. [10, Definition 1.3]) Let $A\subset X$ be a body. We say that $x\in \partial A$ is an LUR (locally uniformly rotund) point of A if for each $\epsilon >0$ there exists $\delta >0$ such that if $y \in A$ and ${\mathrm {dist}}(\partial A,(x+y)/2)<\delta$ then $\Vert x-y\Vert <\epsilon$.

If $A=B_X$, the previous definition coincides with the standard definition of local uniform rotundity of the norm at x. We say that A is an LUR body if each point in $\partial A$ is an LUR point of A.

Lemma 4.3

Let A be a body in X and suppose that $a\in \partial A$ is an LUR point of A. Then, if $f\in S_{X^*}$ is a support functional for A in a, f strongly exposes A at a. Moreover, every slice S of the form $S=S(f,\alpha ,A)$ is a bounded set.

The first part of the lemma is well-known in the case the body is a ball (see, e.g. [8, Exercise 8.27]) and in the general case the proof is similar. However, for the convenience of the reader we include a proof.

Proof

Without loss of generality, we can suppose that $a=0$. Fix $w\in {\mathrm {int}}\,A$ and observe that $f(w)<0$.

Let us prove the first part of the lemma. Let $\alpha >0$, $z\in S=S(f,\alpha ,A)$ and

$$\begin{aligned} \textstyle z'=z-\frac{f(z)}{f(w)}w. \end{aligned}$$

Since $\frac{z}{2}\in A$ and $f(z')=0$, we have that $[\frac{z}{2},\frac{z'}{2}]\cap \partial A\ne \emptyset$. Hence

$$\begin{aligned} \textstyle {\mathrm {dist}}(\partial A,\frac{z}{2})\le \frac{1}{2}\Vert z'-z\Vert \le \frac{1}{2}\frac{\Vert w\Vert }{|f(w)|}\alpha . \end{aligned}$$

Since $a=0$ is an LUR point of A, if $\alpha \rightarrow 0$ then $\mathrm {diam}(S)\rightarrow 0$ and the proof is concluded.

Now, the second part of the lemma follows easily. Suppose on the contrary that there exists $\alpha >0$ such that $S=S(f,\alpha ,A)$ is unbounded. Then there exists a sequence $\{y_n\}$ in $S{\setminus }\{0\}$ such that $\Vert y_n\Vert \rightarrow \infty$. Put $z_n=\frac{y_n}{\Vert y_n\Vert }$ and observe that $\Vert z_n\Vert =1$ and $z_n\in S(f,\alpha /\Vert y_n\Vert ,A)$, a contradiction by the first part of the lemma. $\square$

Lemma 4.4

Let X be a normed space. There exists a constant $\Gamma >0$ such that if $R>1$, if $x,y,a,b\in X$ are such that $\Vert x\Vert ,\Vert y\Vert < R$ and $\Vert a\Vert ,\Vert b\Vert >2 R$, then, if $[x,a]\cap RS_X=\{a'\}$ and $[y,b]\cap RS_X=\{b'\}$, it holds

$$\begin{aligned} \Vert b'-a'\Vert \le \Gamma \max \{\Vert x\Vert ,\Vert y\Vert ,\Vert a-b\Vert \}. \end{aligned}$$

Proof

Let $\lambda ,\mu \in (0,1)$ be such that $a'=\lambda a+(1-\lambda )x$ and $b'=\mu b+(1-\mu )y$. By the triangle inequality, it follows easily that

$$\begin{aligned} \textstyle \frac{R-\Vert y\Vert }{\Vert b\Vert -\Vert y\Vert }\le \mu . \end{aligned}$$

Moreover, since

$$\begin{aligned} \textstyle R=\Vert \lambda a+(1-\lambda )x\Vert \ge \lambda \Vert a\Vert -(1-\lambda )\Vert x\Vert , \end{aligned}$$

we have

$$\begin{aligned} \textstyle \lambda \le \frac{R+\Vert x\Vert }{\Vert a\Vert +\Vert x\Vert }. \end{aligned}$$

Without loss of generality, we can assume that $\lambda \ge \mu$. If we denote

$$\begin{aligned} d=\max \{\Vert x\Vert ,\Vert y\Vert ,\Vert a-b\Vert \}, \end{aligned}$$

we have

$$\begin{aligned} \textstyle \Vert b'-a'\Vert\le & {} \textstyle \lambda \Vert a-b\Vert +(1-\lambda )\Vert x-y\Vert +|\lambda -\mu |(\Vert y\Vert +\Vert b\Vert )\\\le & {} \textstyle 3d+\bigl (\frac{R+\Vert x\Vert }{\Vert a\Vert +\Vert x\Vert }-\frac{R-\Vert y\Vert }{\Vert b\Vert -\Vert y\Vert }\bigr )(\Vert y\Vert +\Vert b\Vert )\\\le & {} \textstyle 3d+\frac{R(\Vert b\Vert -\Vert a\Vert -\Vert y\Vert -\Vert x\Vert )+\Vert a\Vert \,\Vert y\Vert +\Vert b\Vert \,\Vert x\Vert }{(\Vert a\Vert +\Vert x\Vert )(\Vert b\Vert -\Vert y\Vert )}(\Vert y\Vert +\Vert b\Vert )\\\le & {} \textstyle 3d+\frac{R(\Vert b\Vert -\Vert a\Vert -\Vert y\Vert -\Vert x\Vert )+\Vert a\Vert \,\Vert y\Vert +\Vert b\Vert \,\Vert x\Vert }{(\Vert a\Vert +\Vert x\Vert )(\Vert b\Vert /2)}(2\Vert b\Vert )\\\le & {} \textstyle 3d+4\frac{R(\Vert b\Vert -\Vert a\Vert -\Vert y\Vert -\Vert x\Vert )}{2R}+4\frac{\Vert a\Vert \,\Vert y\Vert +\Vert b\Vert \,\Vert x\Vert }{\Vert a\Vert +\Vert x\Vert }\\\le & {} \textstyle 3d+2\bigl (\bigl |\Vert b\Vert -\Vert a\Vert \bigr |\bigr )+\frac{4\Vert a\Vert \,\Vert y\Vert }{\Vert a\Vert +\Vert x\Vert }+\frac{4\Vert b\Vert \,\Vert x\Vert }{\Vert a\Vert +\Vert x\Vert }\\\le & {} \textstyle 5d+\frac{4\Vert a\Vert }{\Vert a\Vert +\Vert x\Vert }d+\frac{4\Vert b-a+a\Vert \,\Vert x\Vert }{\Vert a\Vert +\Vert x\Vert }\\\le & {} \textstyle 5d+4d+\frac{4\Vert x\Vert }{\Vert a\Vert +\Vert x\Vert }\Vert b-a\Vert +\frac{4\Vert a\Vert }{\Vert a\Vert +\Vert x\Vert }\Vert x\Vert \le \textstyle 17d\\ \end{aligned}$$

The proof is concluded if we set $\Gamma =17$$\square$

The following theorem is the main result of this section.

Theorem 4.5

Let X be a normed space, B a nonempty closed convex subset of X, A a body in X and $a\in \partial A$ an LUR point of A. Let $\{A_n\}$ and $\{B_n\}$ be two sequences of closed convex sets such that $A_n\rightarrow A$ and $B_n\rightarrow B$ for the Attouch–Wets convergence. Suppose that $\{a_n\}$ and $\{b_n\}$ are sequences in X such that $a_n\in A_n,\ b_n\in B_n$ ($n\in {\mathbb {N}}$) and

$$\begin{aligned} {\mathrm {dist}}(A_n,B_n)=\Vert a_n-b_n\Vert . \end{aligned}$$

Suppose that at least one of the following conditions holds.

• 1. $A\cap B=\{a\}$.

• 2. $A\cap B=\emptyset$ and there exists $b\in B$ such that ${\mathrm {dist}}(A,B)=\Vert a-b\Vert$.

Then $a_n\rightarrow a$ in the $\Vert {\cdot }\Vert$-topology.

Proof

There is no loss of generality in assuming $a=0$. Let us assume that (1) holds.

Since ${\mathrm {int}}\,(A)\cap B=\emptyset$, by the Hahn–Banach theorem there exists $f\in S_{X^{*}}$ such that

$$\begin{aligned} \sup f(A)=0=\inf f(B). \end{aligned}$$

In particular, f is a support functional for A in 0. Let $\alpha >0$ and observe that, by Lemma 4.3, there exists $r>1$ such that $S=S(f,3\alpha ,A)\subset rB_X$. Put $R=r+\alpha$.

We claim that $\{a_n\}$ and $\{b_n\}$ are eventually contained in $2R B_X$. Suppose that this is not the case and let $\{a_{n_k}\}$ and $\{b_{n_k}\}$ be two subsequences such that $\Vert a_{n_k}\Vert >2R$ and $\Vert b_{n_k}\Vert >2R$ whenever $k\in {\mathbb {N}}$. Now, let $x_{n_k}\in A_{n_k}$ and $y_{n_k}\in B_{n_k}$ be such that $\Vert x_{n_k}\Vert \rightarrow 0$ and $\Vert y_{n_k}\Vert \rightarrow 0$ as $k\rightarrow \infty$. Let $[x_{n_k},a_{n_k}]\cap RS_X=\{a'_{n_k}\}$ and $[y_{n_k},b_{n_k}]\cap RS_X=\{b'_{n_k}\}$, and observe that, by Lemma 4.4, it holds $\Vert b'_{n_k}-a'_{n_k}\Vert \rightarrow 0$ as $n\rightarrow \infty$.

Since $A_n\rightarrow A$ for the Attouch–Wets convergence, $a'_{n_k}\in A_{n_k}\cap RS_X$ and

$$\begin{aligned} A= S(f,3\alpha ,A)\cup [A\cap \{x\in X;\,f(x)\le -3\alpha \}]\subset rB_X\cup \{x\in X;\,f(x)\le -3\alpha \}, \end{aligned}$$

it eventually holds $a'_{n_k}\in \{x\in X;\,f(x)\le -2\alpha \}$.

Analogously, since $B_n\rightarrow B$ for the Attouch–Wets convergence, $b'_{n_k}\in B_{n_k}\cap RS_X$ and

$$\begin{aligned} B\subset \{x\in X;\,f(x)\ge 0\}, \end{aligned}$$

it eventually holds $b'_{n_k}\in \{x\in X;\,f(x)\ge -\alpha \}$.

In particular, it eventually holds $\Vert b'_{n_k}-a'_{n_k}\Vert \ge f(b'_{n_k}-a'_{n_k})\ge \alpha$, a contradiction. Therefore our claim is proved.

Now, since $\{a_n\}$ and $\{b_n\}$ are bounded, there exist sequences $\{w_n\}\subset A$ and $\{z_n\}\subset B$ such that $\Vert w_n-a_n\Vert \rightarrow 0$ and $\Vert z_n-b_n\Vert \rightarrow 0$. Since clearly $\lim _n\Vert z_n-w_n\Vert =0$, it holds

$$\begin{aligned} 0\le \liminf _n[f(z_n)-\Vert w_n-z_n\Vert ]\le \liminf _n f(w_n)\le \limsup _n f(w_n)\le 0, \end{aligned}$$

and hence that $f(w_n)\rightarrow 0$ as $n\rightarrow \infty$. Since, by Lemma 4.3, f strongly exposes 0, we have that $w_n\rightarrow 0$ and hence that $a_n\rightarrow 0$ in the $\Vert {\cdot }\Vert$-topology. This concludes the proof in case (1).

If assumption (2) holds, the proof is similar, but some additional efforts are needed. Let $d={\mathrm {dist}}(A,B)$ and observe that:

• (i) ${\mathrm {int}}\,(A)\cap (B+dB_X)=\emptyset$;

• (ii) $0\in B+dB_X$;

• (iii) $\limsup _n \Vert a_n-b_n\Vert \le d$

Then there exists $f\in S_{X^{*}}$ such that

$$\begin{aligned} \sup f(A)=0=\inf f(B+dB_X). \end{aligned}$$

In particular, f is a support functional for A in 0 and $\inf f(B)=d$. Let $\Gamma$ be the constant given by Lemma 4.4 and let us consider $S=S(f,(\Gamma +2)d,A)$ and observe that, by Lemma 4.3, there exists $r>1$ such that $S\subset rB_X$. Let $R=r+d$.

We claim that $\{a_n\}$ and $\{b_n\}$ are eventually contained in $2R B_X$. Suppose that this is not the case and let $\{a_{n_k}\}$ and $\{b_{n_k}\}$ be two subsequences such that $\Vert a_{n_k}\Vert >2R$ and $\Vert b_{n_k}\Vert >2R$ whenever $k\in {\mathbb {N}}$. Now, let $x_{n_k}\in A_{n_k}$$y_{n_k}\in B_{n_k}$ be such that $x_{n_k}\rightarrow a$ and $y_{n_k}\rightarrow b$ as $k\rightarrow \infty$. Let $[x_{n_k},a_{n_k}]\cap RS_X=\{a'_{n_k}\}$ and $[y_{n_k},b_{n_k}]\cap RS_X=\{b'_{n_k}\}$, and observe that, by Lemma 4.4, it eventually holds $\Vert b'_{n_k}-a'_{n_k}\Vert < (\Gamma +1)d$.

Since $A_n\rightarrow A$ for the Attouch–Wets convergence, $a'_{n_k}\in A_n\cap RS_X$ and

$$\begin{aligned} A= & {} S(f,(\Gamma +2)d,A)\cup [A\cap \{x\in X;\,f(x)\le -(\Gamma +2)d\}]\\\subset & {} rB_X\cup \{x\in X;\,f(x)\le -(\Gamma +2)d\}, \end{aligned}$$

it eventually holds $a'_{n_k}\in \{x\in X;\,f(x)\le -(\Gamma +1)d\}$.

Analogously, since $B_n\rightarrow B$ for the Attouch–Wets convergence, $b'_{n_k}\in B_n\cap RS_X$ and

$$\begin{aligned} B\subset \{x\in X;\,f(x)\ge d\}, \end{aligned}$$

it eventually holds $b'_{n_k}\in \{x\in X;\,f(x)\ge 0\}$.

In particular, it eventually holds $\Vert b'_{n_k}-a'_{n_k}\Vert \ge f(b'_{n_k}-a'_{n_k})\ge (\Gamma +1)d$, a contradiction and our claim is proved.

Now, since $\{a_n\}$ and $\{b_n\}$ are bounded, there exist sequences $\{w_n\}\subset A$ and $\{z_n\}\subset B$ such that $\Vert w_n-a_n\Vert \rightarrow 0$ and $\Vert z_n-b_n\Vert \rightarrow 0$. Let us observe that, by Fact 2.5,

$$\begin{aligned} d\le \liminf \Vert z_n-w_n\Vert \le \limsup _n\Vert z_n-w_n\Vert =\limsup _n\Vert a_n-b_n\Vert \le d \end{aligned}$$

and

$$\begin{aligned} 0\le \liminf _n[f(z_n)-\Vert w_n-z_n\Vert ]\le \liminf _n f(w_n)\le \limsup _n f(w_n)\le 0. \end{aligned}$$

Hence, we obtain $f(w_n)\rightarrow 0$ as $n\rightarrow \infty$. Since, by Lemma 4.3, f strongly exposes 0, we have that $w_n\rightarrow 0$ and hence that $a_n\rightarrow 0$ in the $\Vert {\cdot }\Vert$-topology. $\square$

Remark 4.6

• (i) As in the finite-dimensional case (see Remark 3.6), the theorem above can be proved in an alternative way, by using known results concerning stability theory for convex optimization problem. However, the well-posedness of the involved problems requires a proof with techniques similar to those used in Theorem 4.5. As in the finite-dimensional case, we preferred to present a direct and more geometrical proof.

• (ii) Similarly to the finite-dimensional case, under the hypothesis of the above theorem, we have $\Vert a_n-b_n\Vert \rightarrow \mathrm {dist}(A,B)$. Indeed, by the proof of Theorem 4.5, $\{a_n\}$ and $\{b_n\}$ are bounded, and hence there exist sequences $\{w_n\}\subset A$ and $\{z_n\}\subset B$ such that $\Vert w_n-a_n\Vert \rightarrow 0$ and $\Vert z_n-b_n\Vert \rightarrow 0$. In particular, $\liminf _n \Vert a_{n}-b_{n}\Vert =\liminf _n \Vert w_{n}-z_{n}\Vert \ge \mathrm {dist}(A,B)$. Since, by Fact 2.5, $\mathrm {dist}(A,B)\ge \limsup _n \Vert a_{n}-b_{n}\Vert$, the proof is complete.

• (iii) It is not difficult to see that, in general, condition (2) in Theorem 4.5 does not ensure that the sequence $\{b_n\}$ is convergent. However, under the additional requirement that $\frac{b-a}{\Vert b-a\Vert }$ is an LUR point of the unit ball of X, it is easy to prove (proceeding as in the proof of the above theorem) that $b_n\rightarrow b$.

If the limit sets A and B satisfy a strong condition about non-separation, we obtain a result similar to Corollary 3.7.

Proposition 4.7

Let A and B be two closed convex subsets of a reflexive Banach space X such that $A\cap B$ is bounded and such that $({\mathrm {int}}\,A)\cap B\ne \emptyset$. Let $\{A_n\}$ and $\{B_n\}$ be two sequences of closed convex sets such that $A_n\rightarrow A$ and $B_n\rightarrow B$ for the Attouch–Wets convergence. Suppose that $\{a_n\}$ and $\{b_n\}$ are sequences in X such that $a_n\in A_n,\ b_n\in B_n$ ($n\in {\mathbb {N}}$) and

$$\begin{aligned} {\mathrm {dist}}(A_n,B_n)=\Vert a_n-b_n\Vert . \end{aligned}$$

Then there exist two subsequences $\{a_{n_k}\}$ and $\{b_{n_k}\}$ that weakly converge to a point of $A\cap B$.

Proof

By [11, Corollary 9.2.8], the sequence $\{A_n\cap B_n\}$ converges to $A\cap B$ for the Attouch–Wets convergence. In particular, the sets $A_n\cap B_n$ ($n\in {\mathbb {N}}$) are eventually nonempty and hence $a_n$ and $b_n$ eventually coincide. Since $A\cap B$ is bounded and X is reflexive, the thesis holds. $\square$

By combining the above proposition with Theorem 4.5, we obtain the following corollary.

Corollary 4.8

Let X be a reflexive Banach space. Let A be an LUR body of X and B a closed convex subset of X such that $A\cap B$ is nonempty and bounded. Let $\{A_n\}$ and $\{B_n\}$ be two sequences of closed convex sets such that $A_n\rightarrow A$ and $B_n\rightarrow B$ for the Attouch–Wets convergence. Suppose that $\{a_n\}$ and $\{b_n\}$ are sequences in X such that $a_n\in A_n,\ b_n\in B_n$ ($n\in {\mathbb {N}}$) and

$$\begin{aligned} {\mathrm {dist}}(A_n,B_n)=\Vert a_n-b_n\Vert . \end{aligned}$$

Then there exist subsequences $\{a_{n_k}\}$ and $\{b_{n_k}\}$ that weakly converge to a point $c\in A\cap B$. Moreover, if $({\mathrm {int}}\,A)\cap B=\emptyset$ then $a_n, b_n\rightarrow c$ with respect to the norm convergence.

5 Examples and final remarks

In this section we provide three examples to illustrate the role of the assumptions in the infinite-dimensional case. We point out that all of them are in $\ell _2$, therefore the assumptions used in Sect. 4 cannot be avoided even in the “simplest” infinite-dimensional space.

The following example shows that an analogous of Theorem 3.5 does not hold in the infinite-dimensional setting.

Example 5.1

Let $X=\ell _2$ and let $\{e_n\}_n$ be its standard basis. We denote by $\{e_n^*\}_n$ the dual basis. Let $A,B,A_n,B_n\subset X$ ($n\in {\mathbb {N}}$, $n\ge 2$) be defined as follows:

$$\begin{aligned} A= & {} \textstyle \overline{{\mathrm {cone}}}\,(\{ e_k+\frac{1}{k} e_1;\, k\in {\mathbb {N}}\});\\ B= & {} \{x\in X;\, e_1^*(x)= 0\};\\ A_n= & {} \textstyle \overline{{\mathrm {conv}}}\,\bigl (\{\ln n\, e_n+\frac{1}{n} e_1\}\cup (\frac{1}{n} e_1+A)\bigr );\\ B_n= & {} B. \end{aligned}$$

Let $a_n=\ln n\, e_n+\frac{1}{n} e_1\in A_n$ and $b_n=\ln n\, e_n\in B_n$. Then:

• (i) $A\cap B=\{0\}$;

• (ii) $A_n\rightarrow A$ and $B_n\rightarrow B$ for the Hausdorff convergence (and, hence, for the Attouch–Wets convergence);

• (iii) ${\mathrm {dist}}(A_n,B_n)=\Vert a_n-b_n\Vert$;

• (iv) $\Vert a_n\Vert ,\Vert b_n\Vert \rightarrow \infty$.

Proof

We just have to prove (i) and (ii), since the proofs of (iii) and (iv) are straightforward.

• (i) For $n\in {\mathbb {N}}{\setminus }\{1\}$, let $f_n=n e_1^*-e_n^*$ and $g_n=e_n^*$ and observe that

  $$\begin{aligned} \textstyle \{ e_k+\frac{1}{k} e_1;\, k\in {\mathbb {N}}\}\subset \{x\in X;\, f_n(x)\ge 0,\, g_n(x)\ge 0\}. \end{aligned}$$

  Then $A\subset \bigcap _{n=1}^\infty \{x\in X;\, f_n(x)\ge 0,\, g_n(x)\ge 0\}$. Now, if $x\in A\cap B$, it holds $e_1^*(x)=0$, $f_n(x)=-e^*_n(x)\ge 0$ and $g_n(x)=e^*_n(x)\ge 0$. Then $x=0$.

• (ii) We just have to prove that $A_n\rightarrow A$ for the Hausdorff convergence. Let us observe that

  $$\begin{aligned} \textstyle {\mathrm {dist}}(a_n,\frac{1}{n} e_1+A)\le \Vert \frac{1}{n} e_1+\ln n(e_n+\frac{1}{n} e_1)-a_n\Vert =\frac{\ln n}{n}. \end{aligned}$$

  Hence, it holds

  $$\begin{aligned} \textstyle h(A_n,A)\le h(A_n,\frac{1}{n} e_1+A)+h(\frac{1}{n} e_1+A,A)\le \frac{\ln n}{n}+\frac{1}{n}, \end{aligned}$$

  and the proof is concluded.

$\square$

Given two sets $A,B\subset X$, we say that A and B are separated iff there exists $x^*\in X^*{\setminus }\{0\}$ such that

$$\begin{aligned} \textstyle \sup x^*(A)\le \inf x^*(B). \end{aligned}$$

The following example shows that, in Proposition 4.7, the condition

$$\begin{aligned} ({\mathrm {int}}\,A)\cap B\ne \emptyset \end{aligned}$$

cannot be replaced with the weaker condition “A and B are not separated”.

Example 5.2

Let $X=\ell _2$ and for $n\in {\mathbb {N}}$ let us consider the following subsets of X:

$$\begin{aligned} C_n= & {} \textstyle \overline{{\mathrm {cone}}}\,\bigl (\{e_{2n-1},e_{2n}+\frac{1}{n} e_{2n-1}\}\bigr )-\frac{1}{n} e_{2n-1};\\ D_n= & {} \textstyle \overline{{\mathrm {cone}}}\,\bigl (\{-e_{2n-1},e_{2n}-\frac{1}{n} e_{2n-1}\}\bigr )+\frac{1}{n} e_{2n-1};\\ C'_n= & {} \textstyle C_n-\frac{1}{\ln n}{e_{2n-1}};\\ D'_n= & {} \textstyle D_n-\frac{1}{\ln n}{e_{2n-1}};\\ A= & {} \textstyle \overline{{\mathrm {conv}}}\,\bigl (\bigcup _{n\in {\mathbb {N}}}C_n\bigr );\\ B= & {} \textstyle \overline{{\mathrm {conv}}}\,\bigl (\bigcup _{n\in {\mathbb {N}}}D_n\bigr );\\ A_n= & {} \textstyle \overline{{\mathrm {conv}}}\,\bigl (\bigcup _{k\in {\mathbb {N}}{\setminus }\{n\}}C_k\cup C'_n\bigr );\\ B_n= & {} \textstyle \overline{{\mathrm {conv}}}\,\bigl (\bigcup _{k\in {\mathbb {N}}{\setminus }\{n\}}D_k\cup D'_n\bigr ). \end{aligned}$$

Then:

• (i) A and B are not separated;

• (ii) $A\cap B$ is bounded;

• (iii) $A_n\rightarrow A$ and $B_n\rightarrow B$ for the Hausdorff convergence (and, hence, for the Attouch–Wets convergence);

• (iv) let $a_n=b_n=(1+\frac{n}{\ln n})e_{2n}\in A_n\cap B_n$ then ${\mathrm {dist}}(A_n,B_n)=\Vert a_n-b_n\Vert =0$ and $\Vert a_n\Vert =\Vert b_n\Vert = 1+\frac{n}{\ln {n}}\rightarrow \infty$.

In order to prove Example 5.2 we need some preliminary work. Let us define $X_n={\mathrm {span}}\,\{e_{2n-1},e_{2n}\}$, $B_{X_n}=B_X\cap X_n$ and $Y_N={\mathrm {span}}\,(\bigcup _{n=1}^N X_n)$. Observe that

$$\begin{aligned} Y_N=X_1\oplus _2\cdots \oplus _2 X_N, \end{aligned}$$

where we denote by $X_1\oplus _2\cdots \oplus _2 X_N$ the direct sum $X_1\oplus \cdots \oplus X_N$ endowed with the norm

$$\begin{aligned} \textstyle \Vert (x_1,\ldots ,x_N)\Vert =(\Vert x_1\Vert ^2+\cdots +\Vert x_N\Vert ^2)^{\frac{1}{2}}. \end{aligned}$$

The easy proof of the following lemma is left to the reader.

Lemma 5.3

Let $C_n, D_n\subset X_n$ be defined as above, then the following inclusion holds:

$$\begin{aligned} \textstyle (C_n+\frac{1}{\sqrt{n^2+1}} B_{X_n})\cap (D_n+\frac{1}{\sqrt{n^2+1}} B_{X_n})\subset 2B_{X_n}. \end{aligned}$$

Lemma 5.4

Let $W_n$ be convex subsets of $X_n$ containing the origin ($n=1,\ldots ,N$) and let $\epsilon >0$, then the following inclusion holds:

$$\begin{aligned} \textstyle {\mathrm {conv}}\,\left( \bigcup _{n=1}^N W_n+\epsilon B_{Y_N}\right) \subset 2\,{\mathrm {conv}}\,\left( \bigcup _{n=1}^N [W_n+\sqrt{N}\epsilon B_{X_n}]\right) . \end{aligned}$$

Proof

Since $Y_N=X_1\oplus _2\cdots \oplus _2 X_N$, it is not difficult to prove that

$$\begin{aligned} B_{Y_N}\subset \sqrt{N}{\mathrm {conv}}\,\left( \bigcup _{n=1}^N B_{X_n}\right) , \end{aligned}$$

hence the following inclusions hold:

$$\begin{aligned} \textstyle {\mathrm {conv}}\,\left( \bigcup _{n=1}^N W_n+\epsilon B_{Y_N}\right)\subset & {} \textstyle {\mathrm {conv}}\,\left( \bigcup _{n=1}^N W_n+\epsilon \sqrt{N}{\mathrm {conv}}\,\left( \bigcup _{n=1}^N B_{X_n}\right) \right) \\\subset & {} 2\,{\mathrm {conv}}\,\left( \bigcup _{n=1}^N [W_n+\sqrt{N}\epsilon B_{X_n}]\right) . \end{aligned}$$

$\square$

Lemma 5.5

For $n=1,\ldots ,N$, let $W_n$ and $Z_n$ be convex subsets of $X_n$ containing the origin. Then the following inclusion holds:

$$\begin{aligned} \textstyle {\mathrm {conv}}\,\bigl (\bigcup _{n=1}^N W_n\bigr )\cap {\mathrm {conv}}\,\bigl (\bigcup _{n=1}^N Z_n\bigr )\subset 2{\mathrm {conv}}\,\bigl (\bigcup _{n=1}^N W_n\cap Z_n\bigr ) \end{aligned}$$

Proof

Let $x\in {\mathrm {conv}}\,\bigl (\bigcup _{n=1}^N W_n\bigr )\cap {\mathrm {conv}}\,\bigl (\bigcup _{n=1}^N Z_n\bigr )$, then there exist $\alpha _n, \beta _n\in [0,1]$, $w_n\in W_n$ and $z_n\in Z_n$ ($n=1,\ldots ,N$) such that

$$\begin{aligned} \textstyle x=\sum _{i=1}^N \alpha _n w_n=\sum _{i=1}^N \beta _n z_n. \end{aligned}$$

Since $Y_N=X_1\oplus \cdots \oplus X_N$, it holds $\alpha _n w_n=\beta _n z_n$ ($n=1,\ldots ,N$). Now suppose that $\alpha _n\ge \beta _n>0$, then $w_n=\frac{\beta _n}{\alpha _n}z_n\in W_n\cap Z_n$. Analogously, if $0<\alpha _n\le \beta _n$, then $z_n=\frac{\alpha _n}{\beta _n}w_n\in W_n\cap Z_n$. Hence

$$\begin{aligned} x\in (\alpha _1+\beta _1)(W_1\cap Z_1)+\cdots +(\alpha _N+\beta _N)(W_N\cap Z_N)\subset 2{\mathrm {conv}}\,\bigl [\bigcup _{n=1}^N (W_n\cap Z_n)\bigr ]. \end{aligned}$$

$\square$

Proof of Example 5.2

Let us prove assertions (i), (ii) and (iii); the proof of (iv) is obvious.

• (i) Let us observe that, for each $n\in {\mathbb {N}}$, the segments $[-\frac{1}{n} e_{2n-1},\frac{1}{n} e_{2n-1}]$ and $[0, e_{2n}]$ are contained in $A\cap B$. Now, suppose that there exists $f\in X^*$ such that $\sup f(A)\le \inf f(B)$, then f is constant on $A\cap B$ and, by the above remark, it holds $f(e_n)=0$ whenever $n\in {\mathbb {N}}$. Hence $f=0$.

• (ii) Let us prove that $A\cap B$ is bounded. For $k\in {\mathbb {N}}$, let us denote by $P_k$ the canonical projection on the first k coordinates. Let $x\in A\cap B$ and let $N\in {\mathbb {N}}$ be such that $\Vert x-P_{2N}x\Vert \le 1$. To conclude the proof it suffices to show that $\Vert P_{2N}x\Vert \le 8$ We claim that $P_{2N} x\in \overline{{\mathrm {conv}}}\,\bigl (\bigcup _{n=1}^N C_n\bigr )$. Indeed, since $x\in A$, there exists a sequence $\{y_k\}$, converging in norm to x, such that $y_k\in {\mathrm {conv}}\,\bigl (\bigcup _{n=1}^k C_n\bigr )$. Then the sequence $\{P_{2N}y_k\}\subset \overline{{\mathrm {conv}}}\,\bigl (\bigcup _{n=1}^N C_n\bigr )$ converges in norm to $P_{2N}x$ and the claim is proved.

  Analogously, it holds $P_{2N} x\in \overline{{\mathrm {conv}}}\,\bigl (\bigcup _{n=1}^N D_n\bigr )$ and hence,

  $$\begin{aligned} P_{2N} x\in & {} \textstyle [{\mathrm {conv}}\,\bigl (\bigcup _{n=1}^N C_n\bigr )+\frac{1}{\sqrt{N^3+N}}B_{Y_N}]\cap [{\mathrm {conv}}\,\bigl (\bigcup _{n=1}^N D_n\bigr )+\frac{1}{\sqrt{N^3+N)}}B_{Y_N}]\\\subset & {} \textstyle 2\, {\mathrm {conv}}\,\bigl (\bigcup _{n=1}^N [C_n+\frac{1}{\sqrt{N^2+1}}B_{X_n}]\bigr )\cap 2\, {\mathrm {conv}}\,\bigl (\bigcup _{n=1}^N [D_n+\frac{1}{\sqrt{N^2+1}}B_{X_n}]\bigr )\\\subset & {} \textstyle 4\, {\mathrm {conv}}\,\bigl (\bigcup _{n=1}^N [C_n+\frac{1}{\sqrt{N^2+1}}B_{X_n}]\cap [D_n+\frac{1}{\sqrt{N^2+1}}B_{X_n}]\bigr )\\\subset & {} \textstyle 4\, {\mathrm {conv}}\,\bigl (\bigcup _{n=1}^N 2B_{X_n}\bigr )\subset 8B_X, \end{aligned}$$

  where the above inclusions hold by Lemmas 5.3, 5.4 and 5.5, respectively.

• (iii) Let us prove that $A_n\rightarrow A$ for the Hausdorff convergence, the proof that $B_n\rightarrow B$ for the Hausdorff convergence is similar. Let us observe that $h(C_n,C'_n)=\frac{1}{\ln n}$, hence we have:

  $$\begin{aligned} \textstyle h(A,A_n)=h(\overline{{\mathrm {conv}}}\,\bigl (\bigcup _{k\in {\mathbb {N}}{\setminus }\{n\}}C_k\cup C'_n\bigr ), \overline{{\mathrm {conv}}}\,\bigl (\bigcup _{k\in {\mathbb {N}}{\setminus }\{n\}}C_k\cup C_n\bigr ))\le \frac{1}{\ln n}, \end{aligned}$$

  and the proof is concluded.

$\square$

The following example shows that in Theorem 4.5 it is not possible to replace the Attouch–Wets convergence with the Kuratowski–Painlevé convergence.

Example 5.6

Let $X=\ell _2$ and for $n\in {\mathbb {N}}$ let us consider the following closed convex subsets of X:

$$\begin{aligned} A= & {} \textstyle e_1+B_X;\\ B= & {} \textstyle \{x\in X;\, e_1^*(x)=0\};\\ A_n= & {} \textstyle {\mathrm {conv}}\,\bigl (A\cup \{e_n\});\\ B_n= & {} B. \end{aligned}$$

Then:

• (i) $A\cap B=\{0\}$ and 0 is an LUR point of A;

• (ii) $A_n\rightarrow A$ and $B_n\rightarrow B$ for the Kuratowski–Painlevé convergence;

• (iii) $e_n\in A_n\cap B_n$;

• (iv) $\{e_n\}$ does not converge in norm.

Proof

(iii) and (iv) are trivial. (i) follows by the well-known fact that the unit ball of $\ell _2$ is uniformly rotund and hence LUR. It remains to prove that $A_n\rightarrow A$ for the Kuratowski–Painlevé convergence. Since $A\subset A_n$ ($n\in {\mathbb {N}}$), it is clear that $A\subset \mathrm {Li} A_n$. On the other hand, if $\{n_k\}$ is a subsequence of the integers and $x=\lim _k x_k\in X$ with $x_k\in A_{n_k}$, it is easy to see that $x\in A$ (observe that $x_k=\lambda _k a_k+(1-\lambda _k)e_{n_k}$ for some $\lambda _k\in [-1,1]$ and $a_k\in A$, since $e_{n_k}\rightarrow 0$ in the weak topology and A is weakly closed we have $x\in A$). Then $A\supset \mathrm {Ls} A_n$ and the proof is concluded. $\square$