Borwein–Preiss vector variational principle

Abstract

This article extends to the vector setting the results of our previous work Kruger et al. (J Math Anal Appl 435(2):1183–1193, 2016) which refined and slightly strengthened the metric space version of the Borwein–Preiss variational principle due to Li and Shi (J Math Anal Appl 246(1):308–319, 2000. doi:10.1006/jmaa.2000.6813). We introduce and characterize two seemingly new natural concepts of $\varepsilon$-minimality, one of them dependent on the chosen element in the ordering cone and the fixed “gauge-type” function.

Appendix: Proof of Theorem 11

(i) and (ii). We define sequences $\{x_{i}\}$ and $\{S_{i}\}$ inductively. Set

$$\begin{aligned} S_{0} := \{x \in X\mid f(x_{0})-f(x)-\delta _0\rho (x,x_{0})\bar{c} \in C\}. \end{aligned}$$

(20)

Obviously, $x_{0}\in S_{0}$. Since f is C-lower semicontinuous with respect to $\bar{c}$, by Proposition 8, subset $S_{0}$ is closed: it is sufficient to take $y:=f(x_{0})$ and $g(x):=\delta _0\rho (x,x_{0})$. For any $x \in S_{0}$, we have

$$\begin{aligned} \langle y^{*},f(x_{0}) - f(x)\rangle \ge \delta _0\rho (x,x_{0})\langle y^{*},\bar{c}\rangle = \lambda _0\delta _0\rho (x,x_{0}). \end{aligned}$$

At the same time, by Definition 2, $f(x)-f(x_{0})\in C+\varepsilon \mathbb B$. Hence,

$$\begin{aligned} \langle y^{*},f(x) - f(x_{0})\rangle \ge -\varepsilon . \end{aligned}$$

It follows from the last two inequalities that

$$\begin{aligned} \rho (x,x_{0}) \le \frac{\varepsilon }{\lambda _0\delta _0}. \end{aligned}$$

(21)

For $i=0,1,\ldots$, denote $j_i:=\min \{i,N-1\}$, i.e., $j_i$ is the largest integer $j\le i$ such that $\delta _j>0$. Let $i\in \mathbb N$ and suppose $x_{0}, \ldots ,x_{i-1}$ and $S_{0},\ldots ,S_{i-1}$ have been defined. We choose $x_{i} \in S_{i-1}$ such that

$$\begin{aligned}&\langle y^{*},f(x_{i})\rangle +\lambda _0\sum _{k=0}^{j_i-1}\delta _{k}\rho (x_{i},x_{k}) \nonumber \\&\quad < \inf _{x\in S_{i-1}}\left( \langle y^{*},f(x)\rangle + \lambda _0\sum _{k=0}^{j_i-1}\delta _{k}\rho (x,x_{k})\right) + \lambda _0\delta _{j_i}\varepsilon _i \end{aligned}$$

(22)

and define

$$\begin{aligned} S_{i}:= & {} \Biggl \{x \in S_{i-1}\mid f(x_{i})- f(x) \nonumber \\&-\left( \sum _{k=0}^{j_i-1}\delta _{k} (\rho (x,x_{k})-\rho (x_{i},x_{k}))+ \delta _{j_i}\rho (x,x_{i})\right) \bar{c}\in C\Biggr \}. \end{aligned}$$

(23)

Obviously, $x_{i}\in S_{i}$. Since f is C-lower semicontinuous with respect to $\bar{c}$, by Proposition 8, subset $S_{i}$ is closed: it is sufficient to take $y:=f(x_{i})$ and

$$\begin{aligned} g(x):=\sum _{k=0}^{j_i-1}\delta _{k} (\rho (x,x_{k})-\rho (x_i,x_{k})) + \delta _{j_i}\rho (x,x_{i}). \end{aligned}$$

For any $x \in S_{i}$, we have

$$\begin{aligned} \langle y^{*},f(x_{i}) - f(x)\rangle + \lambda _0\sum _{k=0}^{j_i-1}\delta _{k} (\rho (x_{i},x_{k})-\rho (x,x_{k})) -\lambda _0\delta _{j_i}\rho (x,x_{i})\ge 0, \end{aligned}$$

and consequently, making use of (22),

$$\begin{aligned}&\rho (x,x_{i})\le \frac{1}{\lambda _0\delta _{j_i}} \Biggl (\langle y^{*},f(x_{i})\rangle + \lambda _0\sum _{k=0}^{j_i-1}\delta _{k}\rho (x_{i},x_{k}) \nonumber \\&\quad -\Bigl (\langle y^{*},f(x)\rangle +\lambda _0\sum _{k=0}^{j_i-1} \delta _{k}\rho (x,x_{k})\Bigr )\Biggr )<\varepsilon _i. \end{aligned}$$

(24)

We can see that, for all $i\in \mathbb N$, subsets $S_{i}$ are nonempty and closed, $S_{i}\subset S_{i-1}$, and $\sup _{x\in S_i}\rho (x,x_{i}) \rightarrow 0$ as $i\rightarrow \infty$. Since $\rho$ is a gauge-type function, we also have $\sup _{x\in S_i}d(x,x_{i})\le \varepsilon _i\rightarrow 0$ and consequently, $\mathrm{diam} (S_{i}) \rightarrow 0.$ Since X is complete, $\cap _{i=0}^{\infty }S_{i}$ contains exactly one point; let it be $\bar{x}$. Hence, $\rho (\bar{x},x_{i}) \rightarrow 0$ and $x_{i} \rightarrow \bar{x}$ as $i \rightarrow \infty$. Thanks to (21) and (24), $\bar{x}$ satisfies (i) and (ii).

Before proceeding to the proof of claim (iii), we prepare several building blocks which are going to be used when proving claims (iii) and (iv).

Let integers m, n and i satisfy $0\le m\le i<n$. Since $x_{i+1}\in S_i$ and $\bar{x}\in S_n$, it follows from (20) (when $i=0$) and (23) that

$$\begin{aligned}&f(x_{i})- f(x_{i+1}) -\left( \sum _{k=0}^{j_i-1}\delta _{k} (\rho (x_{i+1},x_{k})-\rho (x_{i},x_{k}))+ \delta _{j_i}\rho (x_{i+1},x_{i})\right) \bar{c}\in C, \end{aligned}$$

(25)

$$\begin{aligned}&f(x_{n})- f(\bar{x}) - \left( \sum _{k=0}^{j_n-1}\delta _{k} (\rho (\bar{x},x_{k})-\rho (x_{n},x_{k}))+ \delta _{j_n}\rho (\bar{x},x_{n})\right) \bar{c}\in C. \end{aligned}$$

(26)

We are going to add together inclusions (25) from $i=m$ to $i=n-1$ and inclusion (26). Depending on the value of N, three cases are possible.

$\underline{\hbox {If}\, N>n}$, then $j_i=i$ and $j_n=n$. Adding inclusions (25) from $i=m$ to $i=n-1$, we obtain

$$\begin{aligned} f(x_{m})- f(x_{n}) -\left( \sum _{k=0}^{n-1}\delta _{k} \rho (x_{n},x_{k})-\sum _{k=0}^{m-1}\delta _{k} \rho (x_{m},x_{k})\right) \bar{c}\in C. \end{aligned}$$

Adding together the last inclusion and inclusion (26), we arrive at

$$\begin{aligned} f(x_{m})- f(\bar{x}) -\left( \sum _{k=0}^{n}\delta _{k} \rho (\bar{x},x_{k})-\sum _{k=0}^{m-1}\delta _{k} \rho (x_{m},x_{k})\right) \bar{c}\in C. \end{aligned}$$

(27)

$\underline{\hbox {If}\, N\le m}$, then $j_i=N-1$ and $j_n=N-1$. Adding inclusions (25) from $i=m$ to $i=n-1$, we obtain

$$\begin{aligned} f(x_{m})- f(x_{n}) -\left( \sum _{k=0}^{N-2}\delta _{k} (\rho (x_{n},x_{k})-\rho (x_{m},x_{k}))+ \delta _{N-1}\sum _{k=m}^{n-1}\rho (x_{k+1},x_{k})\right) \bar{c}\in C. \end{aligned}$$

Adding together the last inclusion and inclusion (26), we arrive at

$$\begin{aligned}&f(x_{m})- f(\bar{x}) - \Biggl (\sum _{k=0}^{N-2}\delta _{k} (\rho (\bar{x},x_{k})-\rho (x_{m},x_{k})) \nonumber \\&\quad +\,\delta _{N-1}\left( \sum _{k=m}^{n-1}\rho (x_{k+1},x_{k})+ \rho (\bar{x},x_{n})\right) \Biggr )\bar{c}\in C. \end{aligned}$$

(28)

$\underline{\hbox {If}\, m<N\le n}$, we add inclusions (25) separately from $i=m$ to $i=N-1$ and from $i=N$ to $i=n-1$ and obtain, respectively,

$$\begin{aligned}&f(x_{m})- f(x_{N}) -\left( \sum _{k=0}^{N-1}\delta _{k} \rho (x_{N},x_{k})-\sum _{k=0}^{m-1}\delta _{k} \rho (x_{m},x_{k})\right) \bar{c}\in C,\\&f(x_{N})- f(x_{n}) -\Biggl (\sum _{k=0}^{N-2}\delta _{k} (\rho (x_{n},x_{k})-\rho (x_{N},x_{k})) \nonumber \\&\quad +\,\delta _{N-1} \sum _{k=N}^{n-1}\rho (x_{k+1},x_{k})\Biggr )\bar{c}\in C. \end{aligned}$$

Adding together the last two inclusions and inclusion (26), we obtain

$$\begin{aligned}&f(x_{m})- f(\bar{x}) - \Biggl (\sum _{k=0}^{N-2}\delta _{k} \rho (\bar{x},x_{k})-\sum _{k=0}^{m-1}\delta _{k} \rho (x_{m},x_{k}) \nonumber \\&\quad +\,\delta _{N-1}\left( \sum _{k=N-1}^{n-1}\rho (x_{k+1},x_{k})+ \rho (\bar{x},x_{n})\right) \Biggr )\bar{c}\in C. \end{aligned}$$

(29)

(iii) When $N=+\infty$, we set $m=0$ in the inclusion (27):

$$\begin{aligned} f(x_{0})- f(\bar{x})- \left( \sum _{k=0}^{n}\delta _{k}\rho (\bar{x},x_{k})\right) \bar{c}\in C. \end{aligned}$$

(30)

Since $\bar{c}\in C\setminus \{0\}$ and C is a pointed cone, we have $-\bar{c}\notin C$. Since C is closed, it holds $(-\bar{c}+r\mathbb B)\cap C=\emptyset$ for some $r>0$, and consequently, $(-s_n\bar{c}+s_nr\mathbb B)\cap C=\emptyset$, where $s_n:=\sum _{k=0}^{n}\delta _{k}\rho (\bar{x},x_{k})$. It follows from (30) that $s_n r\le \Vert f(x_{0})- f(\bar{x})\Vert$ for all $n\in \mathbb N$. This implies that the series $\sum _{k=0}^{\infty } \delta _k\rho (\bar{x},x_k)$ is convergent and, thanks to (30), condition (6) holds true.

When $N<+\infty$, we set $m=0$ and take $n=N-1$ in the inclusion (27) and any $n\ge N$ in the inclusion (29):

$$\begin{aligned}&f(x_{0})- f(\bar{x}) -\left( \sum _{k=0}^{N-1}\delta _{k} \rho (\bar{x},x_{k})\right) \bar{c}\in C,\end{aligned}$$

(31)

$$\begin{aligned}&f(x_{0})-f(\bar{x}) - \Biggl (\sum _{k=0}^{N-2}\delta _{k} \rho (\bar{x},x_{k})\nonumber \\&\quad + \delta _{N-1} \left( \sum _{i=N-1}^{n-1}\rho (x_{i+1},x_{i})+ \rho (\bar{x},x_{n})\right) \Biggr )\bar{c}\in C. \end{aligned}$$

(32)

As above, for some $r>0$ and any $n>N$, it holds $(-\delta _{N-1}s_n\bar{c}+\delta _{N-1}s_nr\mathbb B)\cap C=\emptyset$, where $s_n:=\sum _{i=N-1}^{n-1}\rho (x_{i+1},x_{i})$. It follows from (32) that

$$\begin{aligned} \delta _{N-1}s_nr\le \Vert f(x_{0})-f(\bar{x})\Vert + \sum _{k=0}^{N-2}\delta _{k} \rho (\bar{x},x_{k})+\delta _{N-1}\rho (\bar{x},x_{n})\Vert \bar{c}\Vert . \end{aligned}$$

Since $\rho (\bar{x},x_{n})\rightarrow 0$ as $n\rightarrow \infty$, this implies that the series $\sum _{i=N-1}^{\infty } \rho (x_{i+1},x_{i})$ is convergent. Combining the two inclusions (31) and (32) produces estimate (7).

(iv) For any $x \ne \bar{x},$ there exists an $m_0\in \mathbb N$ such that $x\notin S_{m}$ for all $m\ge m_0$. By (23), this means that

$$\begin{aligned} f(x_{m})- f(x) - \left( \sum _{k=0}^{j_m-1}\delta _{k} (\rho (x,x_{k})-\rho (x_{m},x_{k}))+ \delta _{j_m}\rho (x,x_{m})\right) \bar{c}\notin C. \end{aligned}$$

(33)

Depending on the value of N, we consider two cases.

$\underline{\hbox {If}\, N=+\infty }$, then $j_m=m$. Since the series $\sum _{k=0}^{\infty } \delta _{k}\rho (\bar{x},x_{k})$ is convergent and C is closed, we can pass in (27) to the limit as $n\rightarrow \infty$ to obtain

$$\begin{aligned} f(x_{m})+ \left( \sum _{k=0}^{m-1}\delta _{k}\rho (x_{m},x_{k})\right) \bar{c} - f(\bar{x})-\left( \sum _{k=0}^{\infty } \delta _{k}\rho (\bar{x},x_{k})\right) \bar{c} \in C. \end{aligned}$$

Comparing the last inclusion with (33), we arrive at condition (8).

$\underline{\hbox {If}\, N<\infty }$, we can take $m_0\ge N$. Then $j_m=N-1$ and it follows from (28) that

$$\begin{aligned}&f(x_{m})- f(\bar{x}) - \Biggl (\sum _{k=0}^{N-2}\delta _{k} (\rho (\bar{x},x_{k})-\rho (x_{m},x_{k})) \nonumber \\&\quad +\,\delta _{N-1}\sup _{n\ge m} \left( \sum _{k=m}^{n-1}\rho (x_{k+1},x_{k})+ \rho (\bar{x},x_{n})\right) \Biggr )\bar{c}\in C. \end{aligned}$$

Comparing the last inclusion with (33), we arrive at (9). This completes the proof.