Ekeland’s variational principle for interval-valued functions

Abstract

In this paper, we attempt to propose Ekeland’s variational principle for interval-valued functions (IVFs). To develop the variational principle, we study a concept of sequence of intervals. In the sequel, the idea of gH-semicontinuity for IVFs is explored. A necessary and sufficient condition for an IVF to be gH-continuous in terms of gH-lower and upper semicontinuity is given. Moreover, we prove a characterization for gH-lower semicontinuity by the level sets of the IVF. With the help of this characterization result, we ensure the existence of a minimum for an extended gH-lower semicontinuous, level-bounded and proper IVF. To find an approximate minima of a gH-lower semicontinuous and gH-Gâteaux differentiable IVF, the proposed Ekeland’s variational principle is used.

Appendices

A proof of Lemma 2.1

Proof of (i)

Let ${\textbf {A}}=[\underline{a},\overline{a}]$ and ${\textbf {B}}=[\underline{b},\overline{b}].$ Then,

$$\begin{aligned} \Vert {\textbf {A}}\oplus {\textbf {B}}\Vert _{I(\mathbb {R})}=\Vert [\underline{a},\overline{a}]\oplus [\underline{b},\overline{b}]\Vert _{I(\mathbb {R})}=\Vert [\underline{a}+\underline{b},\overline{a}+\overline{b}]\Vert _{I(\mathbb {R})}=\max \{|\underline{a}+\underline{b}|,|\overline{a}+\overline{b}|\}. \end{aligned}$$

We now have the following two possible cases.

• Case 1. $\Vert {\textbf {A}}\oplus {\textbf {B}}\Vert _{I(\mathbb {R})}=|\underline{a}+\underline{b}|.$ Since $|\underline{a}+\underline{b}|\le |\underline{a}|+|\underline{b}|\le \max \{|\underline{a}|,|\overline{a}|\}+\max \{|\underline{b}|,|\overline{b}|\}=\Vert {\textbf {A}}\Vert _{I(\mathbb {R})}+\Vert {\textbf {B}}\Vert _{I(\mathbb {R})},$ we get $\Vert {\textbf {A}}\oplus {\textbf {B}}\Vert _{I(\mathbb {R})}\le \Vert {\textbf {A}}\Vert _{I(\mathbb {R})}+\Vert {\textbf {B}}\Vert _{I(\mathbb {R})}.$

• Case 2. $\Vert {\textbf {A}}\oplus {\textbf {B}}\Vert _{I(\mathbb {R})}=|\overline{a}+\overline{b}|.$ Since $|\overline{a}+\overline{b}|\le |\overline{a}|+|\overline{b}|\le \max \{|\underline{a}|,|\overline{a}|\}+\max \{|\underline{b}|,|\overline{b}|\}=\Vert {\textbf {A}}\Vert _{I(\mathbb {R})}+\Vert {\textbf {B}}\Vert _{I(\mathbb {R})},$ therefore, $\Vert {\textbf {A}}\oplus {\textbf {B}}\Vert _{I(\mathbb {R})}\le \Vert {\textbf {A}}\Vert _{I(\mathbb {R})}+\Vert {\textbf {B}}\Vert _{I(\mathbb {R})}.$

Hence, $\Vert {\textbf {A}}\oplus {\textbf {B}}\Vert _{I(\mathbb {R})}\le \Vert {\textbf {A}}\Vert _{I(\mathbb {R})}+\Vert {\textbf {B}}\Vert _{I(\mathbb {R})}~\text {for all}~{\textbf {A}},~{\textbf {B}}\in I(\mathbb {R}).$ $\square$

Proof of (ii)

Let ${\textbf {A}}=[\underline{a},\overline{a}],~{\textbf {B}}=[\underline{b},\overline{b}],~{\textbf {C}}=[\underline{c},\overline{c}]$ and ${\textbf {D}}=[\underline{d},\overline{d}]$. We note that

$$\begin{aligned} {\textbf {A}}\preceq {\textbf {C}}\implies [\underline{a},\overline{a}]\preceq [\underline{c},\overline{c}]\implies \underline{a}\le \underline{c}~\text {and}~ \overline{a}\le \overline{c}. \end{aligned}$$

(5)

Also,

$$\begin{aligned} {\textbf {B}}\preceq {\textbf {D}}\implies [\underline{b},\overline{b}]\preceq [\underline{d},\overline{d}]\implies \underline{b}\le \underline{d} ~\text {and}~\overline{b}\le \overline{d}. \end{aligned}$$

(6)

From (5) and (6), we have

$$\begin{aligned}{} & {} \underline{a}+\underline{b}\le \underline{c}+\underline{d} ~\text {and}~ \overline{a}+\overline{b}\le \overline{c}+\overline{d}\\{} & {} \quad \implies [\underline{a}+\underline{b},\overline{a}+\overline{b}]\preceq [\underline{c}+\underline{d},\overline{c}+\overline{d}]. \end{aligned}$$

Thus, ${\textbf {A}}\oplus {\textbf {B}}\preceq {\textbf {C}}\oplus {\textbf {D}}$. $\square$

Proof of Lemma 2.2

Proof of (i)

Let ${\textbf {A}}=[\underline{a},\overline{a}],~{\textbf {B}}=[\underline{b},\overline{b}]$ and $\epsilon >0.$ ${\textbf {A}}\ominus _{gH}{} {\textbf {B}}=[\underline{a}-\underline{b},\overline{a}-\overline{b}]$ or $[\overline{a}-\overline{b}, \underline{a}-\underline{b}].$ Let us now consider the following four possible cases.

• Case 1. ${\textbf {A}}\ominus _{gH}{} {\textbf {B}}=[\underline{a}-\underline{b},\overline{a}-\overline{b}]$ and $\Vert {\textbf {A}}\ominus _{gH} {\textbf {B}}\Vert _{I(\mathbb {R})}=|\underline{a}-\underline{b}|.$ So, we have

  $$\begin{aligned} \underline{a}-\underline{b}\le \overline{a}-\overline{b}~\text {and}~|\overline{a}-\overline{b}|\le |\underline{a}-\underline{b}|. \end{aligned}$$

  (7)

  Let $\Vert {\textbf {A}}\ominus _{gH} {\textbf {B}}\Vert _{I(\mathbb {R})}<\epsilon$. Then,

  $$\begin{aligned} |\underline{a}-\underline{b}|<\epsilon . \end{aligned}$$

  (8)

  By eq. (8), we have $-\epsilon<\underline{a}-\underline{b}<\epsilon$, and hence $\underline{b}-\epsilon <\underline{a}.$ By eqs. (7) and (8), we have $|\overline{a}-\overline{b}|<\epsilon$. This implies $\overline{b}-\epsilon <\overline{a}.$ Therefore, ${\textbf {B}}\ominus _{gH}[\epsilon ,\epsilon ]=[\underline{b}-\epsilon ,\overline{b}-\epsilon ]\prec [\underline{a},\overline{a}]={\textbf {A}}.$ Note that by eq. (8), $\underline{a}<\underline{b}+\epsilon$. Also, by eqs. (7) and (8), we have $|\overline{a}-\overline{b}|<\epsilon$. This implies $\overline{a}<\overline{b}+\epsilon .$ Therefore, ${\textbf {A}}=[\underline{a},\overline{a}]\prec [\underline{b}+\epsilon ,\overline{b}+\epsilon ]={\textbf {B}}\oplus [\epsilon ,\epsilon ].$

• Case 2. ${\textbf {A}}\ominus _{gH}{} {\textbf {B}}=[\underline{a}-\underline{b},\overline{a}-\overline{b}]$ and $\Vert {\textbf {A}}\ominus _{gH} {\textbf {B}}\Vert _{I(\mathbb {R})}=|\overline{a}-\overline{b}|.$ So, we have

  $$\begin{aligned} \underline{a}-\underline{b}\le \overline{a}-\overline{b}~\text {and}~|\underline{a}-\underline{b}|\le |\overline{a}-\overline{b}|. \end{aligned}$$

  (9)

  Consider

  $$\begin{aligned}{} & {} \Vert {\textbf {A}}\ominus _{gH} {\textbf {B}}\Vert _{I(\mathbb {R})}<\epsilon \nonumber \\{} & {} \quad \implies |\overline{a}-\overline{b}|<\epsilon \end{aligned}$$

  (10)

  By eq. (10), we have

  $$\begin{aligned} \overline{b}-\epsilon <\overline{a}. \end{aligned}$$

  By eqs. (9) and 10), we have $|\underline{a}-\underline{b}|<\epsilon$. This implies $\underline{b}-\epsilon <\underline{a}.$ Therefore, ${\textbf {B}}\ominus _{gH}[\epsilon ,\epsilon ]=[\underline{b}-\epsilon ,\overline{b}-\epsilon ]\prec [\underline{a},\overline{a}].$ Note that by eq. (10), $\overline{a}<\overline{b}+\epsilon$. Also, by eqs. (9) and (10), we have $|\underline{a}-\underline{b}|<\epsilon$. This implies $\underline{a}<\underline{b}+\epsilon .$ Therefore, ${\textbf {A}}=[\underline{a},\overline{a}]\prec [\underline{b}+\epsilon ,\overline{b}+\epsilon ]={\textbf {B}}\oplus [\epsilon ,\epsilon ].$

• Case 3. ${\textbf {A}}\ominus _{gH}{} {\textbf {B}}=[\overline{a}-\overline{b},\underline{a}-\underline{b}]$ and $\Vert {\textbf {A}}\ominus _{gH}{} {\textbf {B}}\Vert _{I(\mathbb {R})}=|\underline{a}-\underline{b}|.$ This case can be proved by following the steps similar to Case 1.

• Case 4. ${\textbf {A}}\ominus _{gH}{} {\textbf {B}}=[\overline{a}-\overline{b},\underline{a}-\underline{b}]$ and $\Vert {\textbf {A}}\ominus _{gH}{} {\textbf {B}}\Vert _{I(\mathbb {R})}=|\overline{a}-\overline{b}|.$ This case can be proved by following the steps similar to Case 2.

Conversely, let ${\textbf {B}}\ominus _{gH}[\epsilon ,\epsilon ]\prec {\textbf {A}}\prec {\textbf {B}}\oplus [\epsilon ,\epsilon ].$ Note that

$$\begin{aligned} {\textbf {B}}\ominus _{gH}[\epsilon ,\epsilon ]\prec {\textbf {A}}&\implies&[\underline{b}-\epsilon ,\overline{b}-\epsilon ]\prec [\underline{a},\overline{a}]\nonumber \\&\implies&\underline{b}-\epsilon<\underline{a}~\text {and}~\overline{b}-\epsilon <\overline{a}. \end{aligned}$$

(11)

Also,

$$\begin{aligned} {\textbf {A}}\prec {\textbf {B}}\oplus [\epsilon ,\epsilon ]&\implies&[\underline{a},\overline{a}]\prec [\underline{b}+\epsilon ,\overline{b}+\epsilon ]\nonumber \\&\implies&\underline{a}<\underline{b}+\epsilon ~\text {and}~\overline{a}<\overline{b}+\epsilon . \end{aligned}$$

(12)

From eqs. (11) and (12), we have

$$\begin{aligned}{} & {} \underline{b}-\epsilon<\underline{a}<\underline{b}+\epsilon ~\text {and}~\overline{b}-\epsilon<\overline{a}<\overline{b}+\epsilon \\{} & {} \quad \implies |\underline{a}-\underline{b}|<\epsilon ~\text {and}~|\overline{a}-\overline{b}|<\epsilon \\{} & {} \quad \implies \max \{|\underline{a}-\underline{b}|,|\overline{a}-\overline{b}|\}<\epsilon \\{} & {} \quad \text { i.e.,} \Vert {\textbf {A}}\ominus _{gH}{} {\textbf {B}}\Vert _{I(\mathbb {R})}<\epsilon . \end{aligned}$$

This completes the proof of (i). $\square$

Proof of (ii)

Let ${\textbf {A}}=[\underline{a},\overline{a}],~{\textbf {B}}=[\underline{b},\overline{b}]~\text {and}~\epsilon >0.$

Consider ${\textbf {A}}\ominus _{gH}[\epsilon ,\epsilon ]\nprec {\textbf {B}}$. This implies $[\underline{a}-\epsilon ,\overline{a}-\epsilon ]\nprec [\underline{b},\overline{b}].$ Thus, $`\underline{b}\le \underline{a}-\epsilon ~\text { and}~\overline{b}\le \overline{a}-\epsilon$’ or $`\underline{b}<\underline{a}-\epsilon ~\text {and}~\overline{b}>\overline{a}-\epsilon$’ or $`\underline{b}>\underline{a}-\epsilon ~\text {and}~\overline{b}<\overline{a}-\epsilon$’. Let us consider all these three possibilities in the following three cases.

• Case 1. $\underline{b}\le \underline{a}-\epsilon ~\text {and}~\overline{b}\le \overline{a}-\epsilon$. So, we have

  $$\begin{aligned}{} & {} \underline{a}>\underline{b}~\text {and}~\overline{a}>\overline{b},~\text {because}~\epsilon >0\\{} & {} \quad \implies {\textbf {B}}\prec {\textbf {A}}\implies {\textbf {A}}\npreceq {\textbf {B}}. \end{aligned}$$

• Case 2. $\underline{b}<\underline{a}-\epsilon ~\text {and}~\overline{b}>\overline{a}-\epsilon$. Since $\underline{b}<\underline{a}-\epsilon$, so $\underline{a}>\underline{b}$, and thus ${\textbf {A}}\npreceq {\textbf {B}}.$

• Case 3. $\underline{b}>\underline{a}-\epsilon ~\text {and}~\overline{b}<\overline{a}-\epsilon .$ Since $\overline{b}<\overline{a}-\epsilon$, so $\overline{a}>\overline{b},~\text {and thus}~{\textbf {A}}\npreceq {\textbf {B}}.$

Hence, proof of (ii) is complete. $\square$