Well-Posedness and Porosity for Symmetric Optimization Problems

Abstract

In the present work, we investigate a collection of symmetric minimization problems, which is identified with a complete metric space of lower semi-continuous and bounded from below functions. In our recent paper, we showed that for a generic objective function, the corresponding symmetric optimization problem possesses two solutions. In this paper, we strengthen this result using a porosity notion. We investigate the collection of all functions such that the corresponding optimization problem is well-posed and prove that its complement is a $\sigma$-porous set.

1. Introduction

In this paper, we study a class of symmetric minimization problems, which was studied recently in our paper [1]. The results of [1] and of the present paper have prototypes in [2,3], where some minimization problems arising in crystallography were considered. It was shown in [2,3] that a typical symmetric minimization problem possesses exactly two minimizers, and every minimizing sequence converges to them in some natural sense. In [1], we extend the results of [2,3] for a sufficiently large class of symmetric minimization problems by showing that for a generic objective function, the corresponding symmetric optimization problem possesses two solutions. In this paper, we strengthen this result using a porosity notion. We investigate the collection of all functions such that the corresponding optimization problem is well-posed and prove that its complement is a $\sigma$-porous set.

More precisely, we study an optimization problem

$$g\left(\xi \right)\to min,\xi \in X,$$

where X is a complete metric space and g is a lower semi-continuous and bounded from below function.

It is well-known that the above problem possesses a minimizer when the space X is compact or when the objective function f possesses a growth property and all bounded subsets of the space X satisfy certain compactness assumptions. Without such assumptions, the existence problem becomes more difficult. This difficulty is overcome by applying the Baire category approach, which was used for many mathematical problems [4,5,6,7,8,9].

Namely, it is known that the minimization problem stated above can be solved for a generic objective function [8,9,10]. More precisely, there is a collection $\mathcal{F}$ in a complete metric space of objective functions, which is a countable intersection of open and everywhere dense sets such that for every objective function $f\in \mathcal{F}$, the corresponding minimization problem has a unique solution, which is a limit of every minimizing sequence. See [9], which contains this result and its several extensions and modifications. Note that the generic approach in nonlinear analysis is used in [11,12,13,14,15], generic solvability of best approximation problems are discussed in [4,11,13], while generic existence of fixed points for nonlinear operators is established in [7,12,13].

In our recent paper [1] the goal was to establish a generic solvability of optimization problems with symmetry. These results have applications in crystallography [2,3]. In this paper, we strengthen this result using a porosity notion. We investigate the set of all functions for which the corresponding minimization problem is well-posed and show that its complement is a $\sigma$-porous set.

2. The Main Result

We begin this section recalling the following notion of porosity [3,4,7,9,12,13].

Suppose that $(Y,d)$ is a complete metric space and define

$$B_d(y,r)=\{\xi \in X:d(y,\xi )\le r\}.$$

We say that a set $E\subset Y$ is porous with respect to d (or just porous if the metric is understood) if there are a real number $\alpha \in (0,1]$ and a positive number $r_0$ such that for every positive number $r\le r_0$ and every point $y\in Y$ there is a point $z\in Y$ such that

$$B_d(z,\alpha r)\subset B_d(y,r)\setminus E.$$

We say that a set in the complete metric space Y is $\sigma$-porous with respect to d (or just $\sigma$-porous if the metric is understood) if this set is a countable union of porous (with respect to d) subsets of Y.

For every function $h:Y\to (-\infty ,\infty ]$, where the set Y is nonempty, put

$$inf\left(h\right)=inf\left\{h\right(\xi ):\xi \in Y\}$$

and

$$dom\left(h\right)=\{y\in Y:h(y)<\infty \}.$$

Suppose that $(X,\rho )$ is a complete metric space. For every $z\in X$ and every positive $\Delta$ put

$$B(z,\Delta )=\{\xi \in X:\rho (z,\xi )\le \Delta \}.$$

For every $z\in X$ and every subset $D\ne \varnothing$ of the space X, define

$$\rho (z,C)=inf\left\{\rho \right(z,\xi ):\xi \in C\}.$$

Denote by ${\mathcal{M}}_l$ the collection of all functions $f:X\to R^1\cup \{\infty \}$, which are bounded from below, lower semi-continuous, and which are not identical infinity. For each $h_1,d_2\in {\mathcal{M}}_l$, define

$$\tilde{d}(h_1,h_2)=sup\left\{\right|h_1\left(z\right)-h_2\left(z\right)|:z\in X\},$$

(1)

$$d(h_1,h_2)=\tilde{d}(h_1,h_2){(1+\tilde{d}(h_1,h_2))}^{-1}.$$

(2)

Note that by convention, $d(h_1,h_2)=1$ when $\tilde{d}(h_1,h_2)=\infty$.

It is clear that $d:{\mathcal{M}}_l\times {\mathcal{M}}_l\to [0,\infty )$ is a complete metric. We denote by ${\mathcal{M}}_c$ the collection of all continuous finite-valued functions $f:X\to R^1$ which are bounded from below. Clearly, ${\mathcal{M}}_c$ is a closed set in the complete metric space $({\mathcal{M}}_l,d)$. We endow the space ${\mathcal{M}}_c$ with the metric d too.

Suppose that $T:X\to X$ is a continuous operator such that

$$T^2\left(z\right)=z \mathrm{for} \mathrm{every} z\in X.$$

We denote by ${\mathcal{M}}_{l,T}$ the collection of all functions $f\in {\mathcal{M}}_l$ for which

$$f\left(T\right(x\left)\right)=f\left(x\right) \mathrm{for} \mathrm{every} \mathrm{point} x\in X$$

and define

$${\mathcal{M}}_{c,T}=\{f\in {\mathcal{M}}_c:f\circ T=f\}.$$

Evidently, ${\mathcal{M}}_{l,T}$ and ${\mathcal{M}}_{c,T}$ are closed subsets of the complete metric space ${\mathcal{M}}_l$. We endow them with the same metric d too.

We investigate the optimization problem

$$f\left(x\right)\to min,x\in X,$$

where the objective function $f\in {\mathcal{M}}_{l,T}$.

Given $f\in {\mathcal{M}}_{l,T}$, we say that the problem of minimization for f on X is well-posed with respect to $({\mathcal{M}}_l,d)$ if the following properties are true:

There exists $x_f\in X$, which satisfies

$$\{x\in X:f\left(x\right)=inf\left(f\right)\}=\{x_f,T\left(x_f\right)\}$$

and for every $ϵ>0$ there are an open neighborhood $\mathcal{U}$ of f in ${\mathcal{M}}_l$ and a positive number $\delta$ such that if a function $g\in \mathcal{U}$ and if a point $z\in X$ satisfies $g\left(z\right)\le inf\left(g\right)+\delta$, then

$$|g\left(z\right)-f\left(x_f\right)|\le ϵ$$

and

$$min\{\rho (z,\{x_f,T\left(x_f\right)\}),\rho (T\left(z\right),\{x_f,T\left(x_f\right)\})\}\le ϵ.$$

This notion has an analog in the optimization theory [9], where the set of minimizers is a singleton. Here, since the problem is symmetric, the set of minimizers contains two points in general.

The next theorem is our sole main result.

Theorem 1.

Suppose that $\mathcal{A}$ is either ${\mathcal{M}}_{l,T}$ or ${\mathcal{M}}_{c,T}$. Then, there is a set $\mathcal{B}\subset \mathcal{A}$ such that its complement $\mathcal{A}\setminus \mathcal{B}$ is σ-porous in the metric space $(\mathcal{A},d)$ and that for every function $f\in \mathcal{B}$ the minimization problem for f on the space X is well-posed with respect to $({\mathcal{M}}_l,d)$.

3. Auxiliary Results

Lemma 1.

For every positive number $r\le 1$, each $f,g\in {\mathcal{M}}_l$, which satisfy $d(f,g)\le 4^{-1}r$ and each $x\in X$,

$$\left|g\right(x)-f(x\left)\right|\le r.$$

Proof. 

Let $r\in (0,1]$, $f,g\in {\mathcal{M}}_l$ satisfy

$$d(f,g)\le 4^{-1}r$$

(3)

and let $x\in X$ be given. By (2) and (3),

$$d(f,g)\le 4^{-1},$$

$$\tilde{d}(f,g)=d(f,g){(1-d(f,g))}^{-1}\le 2d(f,g)\le 2^{-1}r.$$

In view of (1) and the equation above,

$$|g\left(x\right)-f\left(x\right)|\le 2^{-1}r.$$

 □

Lemma 2.

Suppose that $f\in {\mathcal{M}}_{l,T}$, $ϵ\in (0,1)$, $r\in (0,1]$. Then there are $\overline{f}\in {\mathcal{M}}_{l,T}$ and $\overline{x}\in X$ such that $\overline{f}\in {\mathcal{M}}_{c,T}$ if $f\in {\mathcal{M}}_{c,T}$,

$$f\left(x\right)\le \overline{f}\left(x\right)\le f\left(x\right)+r/2,x\in X$$

(4)

and that for each $y\in X$, which satisfies

$$\overline{f}\left(y\right)\le inf\left(\overline{f}\right)+ϵr/4$$

(5)

the equation

$$min\{\rho (y,\overline{x}),\rho (T\left(y\right),\overline{x})\}\le ϵ$$

is true.

Proof. 

There exists $\overline{x}\in X$ satisfying

$$f\left(\overline{x}\right)\le inf\left(f\right)+ϵr/4.$$

Define a function $\overline{f}\in {\mathcal{M}}_l$ as follows:

$$\overline{f}\left(x\right)=f\left(x\right)+2^{-1}rmin\{\rho (x,\overline{x}),\rho (T\left(x\right),\overline{x}),1\},x\in X.$$

(6)

Clearly, $\overline{f}\in {\mathcal{M}}_{l,T}$, and $\overline{f}\in {\mathcal{M}}_{c,T}$ if $f\in {\mathcal{M}}_{c,T}$ and (4) is true. Let $y\in X$ and (5) hold. By (5) and (6),

$$f\left(y\right)+2^{-1}rmin\{\rho (y,\overline{x}),\rho (T\left(y\right),\overline{x}),1\}=\overline{f}\left(y\right)\le inf\left(\overline{f}\right)+ϵr/4$$

$$\le \overline{f}\left(\overline{x}\right)+ϵr/4=f\left(\overline{x}\right)+ϵr/4\le f\left(y\right)+ϵr/2.$$

Therefore,

$$min\{\rho (y,\overline{x}),\rho (T\left(y\right),\overline{x})\}\le ϵ.$$

 □

4. Proof of Theorem 1

For every integer $n\ge 1$ let ${\mathcal{A}}_n$ be the collection of all functions $f\in \mathcal{A}$ such that:

(i) there are a point $\overline{x}\in X$ and $\delta >0$ such that if $z\in X$ and $f\left(z\right)\le inf\left(f\right)+\delta$, then the inequality $\rho (\overline{x},\{z,T\left(z\right)\})\le 1/n$ is valid.

Let a natural number n be given. We claim that the set $\mathcal{A}\setminus {\mathcal{A}}_n$ is porous.

By Lemma 1, for every positive number $r\le 1$, each $f,g\in {\mathcal{M}}_l$ satisfying $d(f,g)\le 4^{-1}r$ and each $x\in X$,

$$\left|g\right(x)-f(x\left)\right|\le r.$$

(7)

By Lemma 2 applied with $ϵ={\left(2n\right)}^{-1}$, the following property is valid:

(ii) for each function $f\in \mathcal{A}$ and every positive number $r\le 1$, there exist $\overline{f}\in \mathcal{A}$ and $\overline{x}\in X$ such that

$$\tilde{d}(f,\overline{f})\le r/4$$

(8)

and that for each $y\in X$ satisfying

$$\overline{f}\left(y\right)\le inf\left(\overline{f}\right)+{16}^{-1}r{n}^{-1}$$

(9)

the equation

$$min\{\rho (y,\overline{x}),\rho (T\left(y\right),\overline{x})\}\le {\left(2n\right)}^{-1}.$$

(10)

is valid.

Fix

$$\overline{r}=4^{-1},\alpha ={80}^{-1}{n}^{-1}.$$

(11)

Let $f\in \mathcal{A}$ and a positive number $r\le \overline{r}$ be given. By property (ii), there exist $\overline{f}\in \mathcal{A}$ and $\overline{x}\in X$ such that

$$\tilde{d}(f,\overline{f})\le r/4$$

(12)

and that the next property is true:

(iii) for every point $y\in X$ satisfying (9), Equation (10) is true.

Let a function $g\in \mathcal{A}$ satisfy

$$d(g,\overline{f})\le \alpha r.$$

(13)

By (2) and (11)–(13),

$$d(g,f)\le \alpha r+r/4\le r/2.$$

(14)

By (2), (11) and (13),

$$\tilde{d}(g,\overline{f})\le d(g,\overline{f}){(1-d(g,\overline{f}))}^{-1}$$

$$\le \alpha r{(1-\alpha r)}^{-1}\le 2\alpha r$$

(15)

and

$$|inf\left(\overline{f}\right)-inf\left(g\right)|\le 2\alpha r.$$

(16)

Let a point $z\in X$ satisfy the inequality

$$g\left(z\right)\le inf\left(g\right)+\alpha r.$$

(17)

By (15),

$$|g\left(z\right)-\overline{f}\left(z\right)|\le 2\alpha r.$$

(18)

By (11) and (16)–(18),

$$\overline{f}\left(z\right)\le g\left(z\right)+2\alpha r\le inf\left(g\right)+3\alpha r\le inf\left(\overline{f}\right)+5\alpha r$$

$$\le inf\left(\overline{f}\right)+{16}^{-1}r{n}^{-1}.$$

(19)

Property (iii), (9), (10) and (19) imply that

$$min\{\rho (z,\overline{x}),\rho (T\left(z\right),\overline{x})\}\le {\left(2n\right)}^{-1}.$$

Thus

$$g\in {\mathcal{A}}_n$$

by definition. Together with (14), this implies that

$$\{g\in \mathcal{A}:d(g,\overline{f})\le \alpha r\}\subset \{g\in \mathcal{A}:d(g,f)\le r\}\cap {\mathcal{A}}_n.$$

Thus, the set $\mathcal{A}\setminus {\mathcal{A}}_n$ is $\sigma$-porous. Then the set

$$\mathcal{A}\setminus {\cap }_{n=1}^{\infty }{\mathcal{A}}_n={\cup }_{n=1}(\mathcal{A}\setminus {\mathcal{A}}_n)$$

is $\sigma$-porous.

Let

$$f\in {\cap }_{n=1}^{\infty }{\mathcal{A}}_n.$$

(20)

By (20), for every integer $n\ge 1$, there are $x_n\in X$ and ${\delta }_n>0$ such that the following property is valid:

(iv) if a point $z\in X$ satisfies the inequality $f\left(z\right)\le inf\left(f\right)+{\delta }_n$, then the equation $\rho (x_n,\{z,T\left(z\right)\})\le 1/n$ holds.

Suppose that a sequence ${\left\{z_i\right\}}_{i=1}^{\infty }\subset X$ satisfies

$$\underset{i\to \infty }{lim}f\left(z_i\right)=inf\left(f\right).$$

(21)

Let a natural number n be given. By (21) and property (iv), for every large enough positive integer i,

$$\{\rho (x_n,\{z_i,T\left(z_i\right)\})\le n^{-1}.$$

Since n is an arbitrary positive integer, there is a sub-sequence ${\left\{z_{i_p}\right\}}_{p=1}^{\infty }$ such that at least one of the sequences ${\left\{z_{i_p}\right\}}_{p=1}^{\infty }$ and ${\left\{T\left(z_{i_p}\right)\right\}}_{p=1}^{\infty }$ converges. Since T is continuous and $T^2$ is the identity operator, they both converge and

$$T\left(\underset{p\to \infty }{lim}{z}_{i_p}\right)=\underset{p\to \infty }{lim}T\left(z_{i_p}\right).$$

(22)

Set

$$x_f=\underset{p\to \infty }{lim}{z}_{i_p}.$$

(23)

By (21), (23) and the lower semi-continuity of f,

$$f\left(x_f\right)=f\left(T\left(x_f\right)\right)=inf\left(f\right).$$

(24)

Applying property (iv) with $z_i=x_f$ for every natural number i, we obtain that

$$\rho (x_n,\{x_f,T\left(x_f\right)\})\le n^{-1} \mathrm{for} \mathrm{every} \mathrm{natural} \mathrm{number} n\ge 1.$$

(25)

Let $\xi \in X$ be such that

$$f\left(\xi \right)=inf\left(f\right).$$

(26)

By (26) and property (iv) applied with $z_i=\xi$ for every integer $i\ge 1$ we obtain that

$$\rho (x_n,\{\xi ,T\left(\xi \right)\})\le n^{-1} \mathrm{for} \mathrm{every} \mathrm{natural} \mathrm{number} n.$$

(27)

Equations (25) and (27) imply that

$$min\{\rho (\xi ,x_f),\rho (T\left(\xi \right),x_f),\rho (\xi ,T\left(x_f\right)),\rho (T\left(\xi \right),T\left(x_f\right))\}\le 2n^{-1}.$$

Since n is an arbitrary positive integer, we conclude that and at least one of the following equalities is true:

$$\xi =x_f,\xi =T\left(x_f\right).$$

Thus

$$\{x\in X:f\left(x\right)=inf\left(f\right)\}=\{x_f,T\left(x_f\right)\}.$$

(28)

Let $ϵ>0$. Fix a natural number n such that

$$4n^{-1}<ϵ.$$

(29)

Property (iv) and (25) imply that for every $z\in X$ which satisfies the inequality

$$f\left(z\right)\le inf\left(f\right)+{\delta }_n,$$

we have

$$\rho (x_n,\{z,T\left(z\right)\})\le 1/n$$

(30)

and

$$min\{\rho (z,x_f),\rho (T\left(z\right),x_f),\rho (z,T\left(x_f\right)),\rho (T\left(z\right),T\left(x_f\right))\}\le 2n^{-1}.$$

(31)

Fix a positive number

$$\delta <min\{3^{-1}{\delta }_n,8^{-1}ϵ\}.$$

(32)

Let a function $g\in {\mathcal{M}}_l$ satisfy

$$\tilde{d}(g,f)\le \delta$$

(33)

and let a point $z\in X$ be such that

$$g\left(z\right)\le inf\left(g\right)+\delta .$$

(34)

By Equations (32)–(34), we have

$$f\left(z\right)\le g\left(z\right)+\delta \le inf\left(g\right)+2\delta \le inf\left(f\right)+3\delta \le inf\left(f\right)+{\delta }_n.$$

(35)

It follows from (29), (31) and (35) that

$$min\{\rho (z,x_f),\rho (T\left(z\right),x_f),\rho (z,T\left(x_f\right)),\rho (T\left(z\right),T\left(x_f\right))\}\le 2n^{-1}<ϵ.$$

(36)

Thus, (28) holds and for each function $g\in {\mathcal{M}}_l$ which satisfies (33) and every point $z\in X$ satisfying (34) Equation (36) holds. By Equations (33) and (34), we have

$$\left|g\right(z)-inf(f\left)\right|\le 2\delta <ϵ.$$

Thus, the minimization problem for f on X is well-posed with respect to $({\mathcal{M}}_l,d)$ for all $f\in {\cap }_{n=1}^{\infty }{\mathcal{A}}_n$. Theorem 1 is proved.