Generalized Hukuhara Hadamard derivative of interval-valued functions and its applications to interval optimization

Abstract

In this article, we study the notion of gH-Hadamard derivative for interval-valued functions (IVFs) and apply it to solve interval optimization problems (IOPs). It is shown that the existence of gH-Hadamard derivative implies the existence of gH-Fréchet derivative and vise-versa. Further, it is proved that the existence of gH-Hadamard derivative implies the existence of gH-continuity of IVFs. We found that the composition of a Hadamard differentiable real-valued function and a gH-Hadamard differentiable IVF is gH-Hadamard differentiable. Further, for finite comparable IVF, we prove that the gH-Hadamard derivative of the maximum of all finite comparable IVFs is the maximum of their gH-Hadamard derivative. The proposed derivative is observed to be useful to check the convexity of an IVF and to characterize efficient points of an optimization problem with IVF. For a convex IVF, we prove that if at a point the gH-Hadamard derivative does not dominate to zero, then the point is an efficient point. Further, it is proved that at an efficient point, the gH-Hadamard derivative does not dominate zero and also contains zero. For constraint IOPs, we prove an extended Karush–Kuhn–Tucker condition using the proposed derivative. The entire study is supported by suitable examples.

Appendices

Appendix A: Proof of Lemma 1

Proof

Let ${\textbf {P}} = [\underline{p}, \overline{p}]$ and ${\textbf {Q}} = [\underline{q}, \overline{q}]$.

• (i) Since ${\textbf {Q}}\nprec {\textbf {0}}$ and ${\textbf {Q}}\preceq {\textbf {P}}$, then

  $$\begin{aligned} \overline{q}\ge 0~\text {and}~\overline{q}\ge \overline{p} \implies \overline{p}\ge 0 \implies {\textbf {P}}\nprec {\textbf {0}}. \end{aligned}$$

• (ii) Since ${\textbf {P}} \oplus {\textbf {Q}} \nprec {\textbf {0}}$ and ${\textbf {Q}} \preceq {\textbf {0}}$, then

  $$\begin{aligned} \overline{p}+\overline{q}\ge 0 \text { and } \overline{q}\le 0 \implies \overline{p} \ge 0 \implies {\textbf {P}} \nprec {\textbf {0}}. \end{aligned}$$

$\square$

Appendix B: Proof of Lemma 4

Proof

• (i) If

  $$\begin{aligned} \varvec{\Phi }(t) \nprec {\textbf {0}} \text { for all } t \in \mathscr {S}, \end{aligned}$$

  (7.1)

  then due to linearity of $\varvec{\Phi }$, we have

  $$\begin{aligned} \varvec{\Phi }(t)=(-1) \odot \varvec{\Phi }(-t) \nsucc {\textbf {0}} \text { for all } t \in \mathscr {S} \end{aligned}$$

  (7.2)

  since $\varvec{\Phi }(-t) \nprec {\textbf {0}}$ by (7.1). From (7.1) and (7.2), it is clear that ${\textbf {0}}$ and $\varvec{\Phi }(t)$ are not comparable.

• (ii) If $\varvec{\Phi }(t) \preceq {\textbf {0}} \text { for all } x \in \mathscr {S},$ then due to linearity of $\varvec{\Phi }$, we have $\varvec{\Phi }(t)= (-1) \odot \varvec{\Phi }(-t) \succeq {\textbf {0}} \text { for all } t \in \mathscr {S}.$ Hence, $\varvec{\Phi }(t)= {\textbf {0}}$.

$\square$