A Perturbation Approach to Vector Optimization Problems: Lagrange and Fenchel–Lagrange Duality

Abstract

In this paper, we study a general minimization vector problem which is expressed in terms of a perturbation mapping defined on a product of locally convex Hausdorff topological vector spaces with values in another locally convex topological vector space. Several representations of the epigraph of the conjugate of the perturbation mapping are given, and then, variants vector Farkas lemmas associated with the system defined by this mapping are established. A dual problem and another so-called loose dual problem of the mentioned problem are defined and stable strong duality results between these pairs of primal–dual problems are established. The results just obtained are then applied to a general class of composed constrained vector optimization problems. For this class of problems, two concrete perturbation mappings are proposed. These perturbation mappings give rise to variants of dual problems including the Lagrange dual problem and several kinds of Fenchel–Lagrange dual problems of the problem under consideration. Stable strong duality results for these pairs of primal–dual problems are derived. Several classes of concrete vector (and scalar) optimization problems are also considered at the end of the paper to illustrate the significance of our approach.

Appendices

Appendix A Proof of Theorem 2.1 (Extended open mapping theorem)

Consider the set-valued mapping $\mathcal {G}:X\times Y \rightrightarrows Z_0$ defined by

$$\begin{aligned} \mathcal {G}(x,y):= \{z\in Z: \Phi (x,z)\leqq _K y\}\subset Z_0. \end{aligned}$$

Then, (6) simply means that $0_Z\in {{\,\mathrm{int}\,}}_{Z_0} \mathcal {G}(U_0\times V_0)$. The proof of (6) proceeds with three steps as follows:

($\alpha$) Let U and V be the convex neighborhoods of $x_0$ and $\Phi (x_0, 0_Z)$, respectively. It is easy to check that $\mathcal {G}(U\times V)$ is a convex set (using the K-convexity of $\Phi$ and of the sets U, V). We will show that $\mathcal {G}(U\times V)$ is absorbing in $Z_0$. Take $z\in Z_0$, we will show that there exists $\lambda >0$ such that $\lambda z\in \mathcal {G}(U\times V)$. Firstly, one has $0_Z\in \pi ({{\,\mathrm{dom}\,}}\Phi )$ (as $(x_0,0_Z)\in {{\,\mathrm{dom}\,}}\Phi$), so $\hbox {aff} (\pi ({{\,\mathrm{dom}\,}}\Phi ))={{\,\mathrm{lin}\,}}(\pi ({{\,\mathrm{dom}\,}}\Phi ))=Z_0$. Now, as $0_Z\in {{\,\mathrm{icr}\,}}(\pi ({{\,\mathrm{dom}\,}}\Phi ))$, there exists $\delta >0$ such that $\delta z \in \pi ({{\,\mathrm{dom}\,}}\Phi )$. Then, there is $x\in X$ such that $(x,\delta z)\in {{\,\mathrm{dom}\,}}\Phi$, or equivalently, $\Phi (x,\delta z)\in Y$. On the one hand, as U, V are, respectively, the neighborhoods of $x_0$ and $\Phi (x_0,0_Z)$, there exists $\mu \in ]0;1[$ such that $x_0+\mu (x-x_0)\in U$ and $\Phi (x_0,0_Z)+\mu (\Phi (x,\delta z)-\Phi (x_0,0_Z))\in V$. On the other hand, one also has

$$\begin{aligned} \Phi (x_0+\mu (x-x_0), \mu \delta z)&=\Phi ((1-\mu )(x_0,0_Z)+\mu (x,\delta z))\\&\in (1-\mu ) \Phi (x_0,0_Z) + \mu \Phi (x,\delta z)-K (\text {as }\Phi \text { is }K\text {-convex)}\\&=\Phi (x_0,0_Z)+\mu (\Phi (x,\delta z)-\Phi (x_0,0_Z))-K. \end{aligned}$$

Thus, $\lambda z \in \mathcal {G}(U\times V)$, where $\lambda = \mu \delta >0$, meaning that $\mathcal {G}(U\times V)$ is absorbing in $Z_0$.

($\beta$) Now, take an arbitrary neighborhood $U\times V$ of $(x_0, \Phi (x_0, 0_Z))$ and we will prove that $0_Z\in {{\,\mathrm{int}\,}}_{Z_0}{{\,\mathrm{cl}\,}}( \mathcal {G}(U\times V))$. As $X\times Y$ is locally convex, replace U, V by their subsets if necessary, we can suppose that U and V are convex. From ($\alpha$), $\mathcal {G}(U\times V)$ is convex and absorbing in $Z_0$. Consequently, ${{\,\mathrm{cl}\,}}\mathcal {G}(U\times V)$ is a convex, closed and absorbing subset of $Z_0$, and hence, a neighborhood of $0_Z$ (as $Z_0$ is a barreled space). So, $0_Z\in {{\,\mathrm{int}\,}}_{Z_0}{{\,\mathrm{cl}\,}}( \mathcal {G}(U\times V))$.

($\gamma$) We now show that (6) follows from Lemma 2.1. From ($\beta$), $0_Z$ belongs to the intersection of all sets of the form ${{\,\mathrm{int}\,}}_{Z_0}({{\,\mathrm{cl}\,}}( \mathcal {G}(U\times V))$ where $U\times V$ running from the collection of all neighborhoods of $(x_0,\Phi (x_0,0_Z))$. Moreover, by assumption, $X\times Y$ is a complete and first countable space, and $\mathcal {G}$ is a closed convex multifunction, since ${{\,\mathrm{gr}\,}}\mathcal {G}={{\,\mathrm{epi}\,}}\Phi$ and $\Phi$ is K-convex and K-epi closed. Then, Lemma 2.1, applying to the multifunction $\mathcal {G}$ with $X\times Y$, $Z_0$, and $(x_0,\Phi (x_0,0_Z))$ playing the roles of ${\tilde{X}}$, ${\tilde{Y}}$ and ${\tilde{x}}_0$, respectively, gives

$$\begin{aligned} 0_Z \ \in \ \bigcap \limits _{U\times V \in \mathcal {N}(x_0,\Phi (x_0,0_Z))} \ {{\,\mathrm{int}\,}}_{Z_0}(\mathcal {G}(U\times V)), \end{aligned}$$

showing that $0_Z \in {{\,\mathrm{int}\,}}_{Z_0} \mathcal {G}(U_0\times V_0)$ and (6) follows. $\square$

Appendix B Proof of Theorem 3.2

We will show that if one of the conditions $(C_2)$, $(C_3)$, $(C_4)$, or $(C_5)$ holds, then the condition $(C_1)$ in Theorem 3.1 holds, and hence, the conclusion now follows from Theorem 3.1.

$(\alpha )$ Assume that $(C_2)$ holds. Then, for any $L\in \mathcal {L}(X,Y)$, take $y_L:={\widehat{y}}-L({\widehat{x}})$ and $V_L:={\widehat{V}}$, one gets $\Phi ({\widehat{x}},z)-L({\widehat{x}})\leqq _K y_L$ for all $z\in V_L\cap Z_0$. So, $(C_1)$ holds (with ${\widehat{x}}$ playing the role of x).

$(\beta )$ Assume that $(C_3)$ holds. Pick $\bar{k}\in {{\,\mathrm{int}\,}}K$. Then, $\Phi ({\widehat{x}},0_Z)+\bar{k} -{{\,\mathrm{int}\,}}K$ is a neighborhood of $\Phi ({\widehat{x}}, 0_Z)$. So, by the continuity of $\Phi ({\widehat{x}},.)_{| Z_0}$ at $0_Z$, there is a neighborhood ${\widehat{V}}$ of $0_Z$ such that $\Phi ({\widehat{x}}, {\widehat{V}}\cap Z_0) \subset \Phi ({\widehat{x}},0_Z)+\bar{k}-{{\,\mathrm{int}\,}}K$, or equivalently,

$$\begin{aligned} \Phi ({\widehat{x}}, z) \leqq _K \Phi ({\widehat{x}},0_Z)+\bar{k}:={\widehat{y}},\quad \forall z\in {\widehat{V}}\cap Z_0, \end{aligned}$$

showing that $(C_2)$ holds, and consequently, $(C_1)$ holds as well.

$(\gamma )$ Assume that $(C_4)$ holds. It is worth noting that $\hbox {aff} (\pi ({{\,\mathrm{dom}\,}}\Phi ))={{\,\mathrm{lin}\,}}(\pi ({{\,\mathrm{dom}\,}}\Phi ))=Z_0$ (as $0_Z\in \pi ({{\,\mathrm{dom}\,}}\Phi )$). So, ${{\,\mathrm{ri}\,}}(\pi ({{\,\mathrm{dom}\,}}\Phi ))={{\,\mathrm{int}\,}}_{Z_0} (\pi ({{\,\mathrm{dom}\,}}\Phi ))$.

Without loss the generality, assume that $Z_0=\mathbb {R}^n$ and take $\{e_i\}_{i=1}^n$ the standard basis of $\mathbb {R}^n$ (i.e., the $i^{th}$ coordinate of $e_i$ is 1 and the others are 0). As $0_Z\in {{\,\mathrm{int}\,}}_{Z_0} (\pi ({{\,\mathrm{dom}\,}}\Phi ))$, there exists $\{\varepsilon _i\}_{i=1}^n\subset \mathbb {R}_+\setminus \{0\}$ such that $\{\pm \varepsilon _i e_i\}_{i=1}^n\subset \pi ({{\,\mathrm{dom}\,}}\Phi )$. For each $i= 1,2,\ldots , n$, as $\pm \varepsilon _i e_i \in \pi ({{\,\mathrm{dom}\,}}\Phi )$, there exists $x_i,x'_i\in X$ such that $(x_i,\varepsilon _ie_i)$ and $(x'_i, -\varepsilon _i e_i)$ belong to ${{\,\mathrm{dom}\,}}\Phi$. Next, for any $L\in \mathcal {L}(X,Y)$, take $y_L\in Y$ such that (the existence of $y_L$ is guaranteed by [11, Lemma 2.1 (ii)])

$$\begin{aligned} \Phi (x_i, \varepsilon _i e_i)-L(x_i)\!<\!_K y_L \ \ \mathrm{and} \ \, \,\Phi (x'_i, -\varepsilon _ie_i)-L(x_i')\!<\!_K y_L,\quad \forall i\!\in \!\{1,2,\ldots , n\}.\nonumber \\ \end{aligned}$$

(34)

It is easy to see that ${{\,\mathrm{co}\,}}\big (\{\pm \varepsilon _i e_i\}_{i=1}^n\big )$ is a neighborhood of $0_Z$ in $Z_0$, and hence, there exists the neighborhood ${\widehat{V}}$ of $0_Z$ such that ${\widehat{V}}\cap Z_0={{\,\mathrm{co}\,}}\big (\{\pm \varepsilon _i e_i\}_{i=1}^n\big )$. Now, for each $z\in {\widehat{V}}\cap Z_0$, there exists $\{(\lambda _i,\lambda '_i)\}_{i=1}^n\subset \mathbb {R}^2_+$ such that $\sum _{i=1}^{n}(\lambda _i+\lambda '_i)=1$ and $z=\sum _{i=1}^{n}(\lambda _i \varepsilon _i - \lambda '_i \varepsilon ')e_i$. Take ${\tilde{x}}= \sum _{i=1}^{n} (\lambda _i x_i + \lambda '_ix'_i)$, by the convexity of $\Phi$,

$$\begin{aligned} \Phi ({\tilde{x}}, z) \leqq _K \sum _{i=1}^{n} \big (\lambda _i \Phi ( x_i, \varepsilon _ie_i)+ \lambda '_i \Phi ( x'_i, -\varepsilon _ie_i)\big ). \end{aligned}$$

This, together with (34), entails $\Phi ({\tilde{x}}, z)-L({\tilde{x}} )\leqq _K y_L$ which means that $(C_1)$ holds.

$(\delta )$ Assume that $(C_5)$ holds. Take $L\in \mathcal {L}(X,Y)$. As $0_Z\in \pi ({{\,\mathrm{dom}\,}}\Phi )$, there exists $x_0\in X$ such that $(x_0,0_Z)\in {{\,\mathrm{dom}\,}}\Phi$. It is worth noting that as $\Phi$ is K-convex and K-epi closed, $\Phi - L$ is K-convex and K-epi closed as well. Pick $k_0\in {{\,\mathrm{int}\,}}K$. Apply the generalized open mapping theorem, Theorem 2.1, with $\Phi$ being replaced by $\Phi - L$, $U_0=X$ and $V_0=\Phi (x_0,0_Z)-L(x_0)+k_0-{{\,\mathrm{int}\,}}K$, one gets that $0_Z \in {{\,\mathrm{int}\,}}_{Z_0} V_1$, where $V_1$ is the set defined by

$$\begin{aligned} V_1:= \left\{ z\in Z: \begin{array}{l} \Phi (x,z)-L(x) \leqq _K y \text { for some } x\in X \\ \text {and } y\in \Phi (x_0,0_Z)-L(x_0)+k_0-{{\,\mathrm{int}\,}}K \end{array}\right\} . \end{aligned}$$

Take $y_L=\Phi (x_0,0_Z)-L(x_0)+k_0$ and $V_L$ the neighborhood of $0_Z$ such that $V_L\cap Z_0=V_1$. Then, for all $z\in V_L\cap Z_0$, by the definition of the set $V_1$, there exist $x\in X$ and $y\in \Phi (x_0,0_Z)-L(x_0)+k_0-{{\,\mathrm{int}\,}}K$ such that $\Phi (x,z)-L(x)\in y-K .$ Then, it follows that $\Phi (x,z)-L(x)\in \Phi (x_0,0_Z)-L(x_0)+k_0-{{\,\mathrm{int}\,}}K-K = y_L-{{\,\mathrm{int}\,}}K\subset y_L-K,$ yielding $\Phi (x,z)-L(x)\leqq _K y_L$. Consequently, $(C_1)$ holds.

$(\epsilon )$ Assume that $(C_6)$ holds. By Proposition 3.1, to prove (10), it is sufficient to show that ${{\,\mathrm{epi}\,}}{\Phi (\cdot ,T)}^*\subset \mathcal {M}.$ The proof of the last inclusion goes along the line as that of Theorem 3.1.

Firstly, take $(\bar{L},\bar{y})\in {{\,\mathrm{epi}\,}}{\Phi (\cdot ,T)}^*$. Then, (13) holds by (2). Let us set

$$\begin{aligned} \Delta ' _{\bar{L}}:=\bigcup _{(x,z)\in {{\,\mathrm{dom}\,}}\Phi } \Bigg ( \Big (\bar{L}(x)-\Phi (x,z)-K\Big )\times \Big (z-\pi ({{\,\mathrm{dom}\,}}\Phi )\Big ) \Bigg ). \end{aligned}$$

As $\Phi$ is K-convex, it is easy to check that $\Delta '_{\bar{L}} \subset Y\times Z$ is convex. Moreover, as the sets ${{\,\mathrm{dom}\,}}\Phi$, ${{\,\mathrm{int}\,}}K$ and ${{\,\mathrm{int}\,}}\pi ({{\,\mathrm{dom}\,}}\Phi )$ are nonempty, ${{\,\mathrm{int}\,}}\Delta '_{\bar{L}}$ is nonempty as well.

We show that $(\bar{y},0_{Z})\notin {{\,\mathrm{int}\,}}\Delta ' _{\bar{L}}$. Indeed, if $(\bar{y},0_{Z})\in {{\,\mathrm{int}\,}}\Delta '_{\bar{L}}$ then, by the same argument as in the proof of Theorem 3.1, there exists $\bar{k}\in {{\,\mathrm{int}\,}}K$ such that $(\bar{y}+\bar{k},0_{Z})\in \Delta ' _{\bar{L}}$. Hence, there is $(\bar{x},\bar{z}) \in {{\,\mathrm{dom}\,}}\Phi$ such that $\bar{y}+\bar{k}\in {\bar{L}} (\bar{x})-\Phi (\bar{x}, \bar{z})-K$ and $0_Z\in \bar{z}-\pi ({{\,\mathrm{dom}\,}}\Phi )$. Taking $(C_6)$ into account, one gets $\Phi (\bar{x},0_Z) \leqq _K \Phi (\bar{x},\bar{z})$. Consequently, $\bar{y}+\bar{k} \in {\bar{L}}(\bar{x})-\Phi (\bar{x},0_Z)-K,$ yielding $\bar{y} \in {\bar{L}} (\bar{x})-\Phi (\bar{x},0_Z)-{{\,\mathrm{int}\,}}K$, which contradicts (13).

By the separation theorem ([30, Theorem 3.4]) applying to the point $(v,0_{Z})$ and the convex set ${{\,\mathrm{int}\,}}\Delta ' _{\bar{L}}$ in $Y \times Z$, there is $(y^*_0, z^*_0)\in Y^{*}\times Z^* \setminus \{ (0_{Y^*}, 0_{Z^*})\}$ such that

$$\begin{aligned} y^*_0(v) < y^*_0(y) + z^*_0(z),\quad \forall (y,z)\in {{\,\mathrm{int}\,}}\Delta '_{\bar{L}}. \end{aligned}$$

(35)

We now show that

$$\begin{aligned} y^*_0(k')<0, \quad \forall k' \in {{\,\mathrm{int}\,}}K. \end{aligned}$$

(36)

Take $k'\in {{\,\mathrm{int}\,}}K$. On the one hand, it follows from the last part of $(C_6)$ that there exists $({\widehat{x}},{\widehat{z}})\in {{\,\mathrm{dom}\,}}\Phi$ satisfying $0_Z\in {\widehat{z}}- {{\,\mathrm{int}\,}}\pi ({{\,\mathrm{dom}\,}}\Phi )$. On the other hand, according to [11, Lemma 2.1 (i)], there exists $\mu >0$ such that $v-\mu k'\in \bar{L}({\widehat{x}})-\Phi ({\widehat{x}}, {\widehat{z}})-{{\,\mathrm{int}\,}}K$. So, the set

$$\begin{aligned} W:=\Big (\bar{L}({\widehat{x}}) - \Phi ({\widehat{x}},{\widehat{z}}) -{{\,\mathrm{int}\,}}K\Big )\times \Big ({\widehat{z}} -{{\,\mathrm{int}\,}}\pi ({{\,\mathrm{dom}\,}}\Phi )\Big ) \end{aligned}$$

is a neighborhood of $(v-\mu k',0_Z)$. It is easy to see that $W\subset \Delta '_{\bar{L}}$, yielding $(v-\mu k',0_Z)\in {{\,\mathrm{int}\,}}\Delta '_{\bar{L}}$. Applying (35), one has $y^*_0(v)<y^*_0(v-\mu k')+z^*_0(0_Z)$, which yields (36).

Now, take $k_0 \in {{\,\mathrm{int}\,}}K$ such that $y^*_0(k_0)=-1$ and define $T\in \mathcal {L}(Z,Y)$ by $T(z)=-z^{*}_0(z) k_{0}$ for any $z \in Z$. Using the same argument as in the proof of Theorem 3.1 (see (18)), we can show that

$$\begin{aligned} \bar{y}\notin y+T(z) - {{\,\mathrm{int}\,}}K, \,\quad \forall (y,z)\in \Delta '_{\bar{L}} \end{aligned}$$

which, together the fact that $(\bar{L}(x)-\Phi (x,z),z)\in \Delta '_{\bar{L}}$ for all $(x,z)\in {{\,\mathrm{dom}\,}}\Phi$, yields

$$\begin{aligned} \bar{y}\notin \bar{L}(x)-\Phi (x,z) +T (z)-{{\,\mathrm{int}\,}}K,\quad \forall (x,z)\in {{\,\mathrm{dom}\,}}\Phi . \end{aligned}$$

Again, (2) (applying to the mapping $\Phi$) yields $(\bar{L},T,\bar{y})\in {{\,\mathrm{epi}\,}}\Phi ^*$, or equivalently, $(\bar{L},\bar{y})\in {{\,\mathrm{epi}\,}}{\Phi ^*(\cdot ,T)}$. The inclusion ${{\,\mathrm{epi}\,}}{\Phi (\cdot ,T)}^*\subset \mathcal {M}$ has been proved, and so (10) holds.

The proof of (11) (under extra assumption $(C_0)$) goes similarly as the last part the proof of Theorem 3.1. $\square$

Appendix C Proof of Theorem 3.3

Take $k_0 \in {{\,\mathrm{int}\,}}K$, $(\bar{L},\bar{y})\in {{\,\mathrm{epi}\,}}{\Phi (\cdot ,T)}^*$ and consider the set

$$\begin{aligned} \Delta '' _{\bar{L}}:=\bigcup _{(x,z)\in {{\,\mathrm{dom}\,}}\Phi } \Big [ \Big (L(x)-\Phi (x,z)-K\Big )\times \Big (z-S\Big ) \Big ]. \end{aligned}$$

By a similar argument as in “Appendix B” $(\epsilon )$, using $\Delta '' _{\bar{L}}$ instead of $\Delta ^\prime _{\bar{L}}$, we can establish $T\in \mathcal {L}_\Phi$ such that $(\bar{L}, \bar{y}) \in {{\,\mathrm{epi}\,}}\Phi ^*(\cdot , T)$, and we get (10).

For the proof of (11), take $({\tilde{x}},0_Z)\in {{\,\mathrm{dom}\,}}\Phi$. For any $s\in S$ and $\nu >0$, one has $-\nu s\in -S=0_Z-S$, and hence, $(\bar{L}({\tilde{x}})-\Phi ({\tilde{x}}), -\nu z)\in \Delta ''_{\bar{L}}$. The same argument as the last part of the proof of Theorem 3.1 leads to $T \in \mathcal {L}_+(S, K)$. $\square$

Appendix D Proof of Lemma 5.1

• (i) Take $L\in \mathcal {L}(X,Y)$ and $T:=(T_1,T_2)\in \mathcal {L}(W,Y)\times \mathcal {L}(Z,Y)$. By a detailed calculation, one has

  $$\begin{aligned} \Phi _1^*(L,T)&={{\,\mathrm{WSup}\,}}\big [(F+T_1\circ H+T_2\circ G+I_{C})^*(L)+\kappa ^*(T_1)+I^*_{-S}(T_2)\big ]\\&=(F+T_1\circ H+T_2\circ G+I_{C})^*(L)\uplus \kappa ^*(T_1)\uplus I^*_{-S}(T_2). \end{aligned}$$

  Assume further that $T_2\in \mathcal {L}_+(S,K)$. Then, $I^*_{-S}(T_2)={{\,\mathrm{WSup}\,}}[T_2(-S)]={{\,\mathrm{WSup}\,}}\{0_Y\}=-{{\,\mathrm{bd}\,}}K$ (applying Proposition 2.1 (vii) to $N=T_2(-S)$ and $M=\{0_Y\}$), and hence,

  $$\begin{aligned} \Phi _1^*(L,T)&=(F+T_1\circ H_1+T_2\circ H_2+I_{C})^*(L)\uplus \kappa ^*(T_1)\uplus (-{{\,\mathrm{bd}\,}}K)\\&=(F+T_1\circ H_1+T_2\circ H_2+I_{C})^*(L)\uplus \kappa ^*(T_1) \ \text {(by Proposition 2.4(i))}. \end{aligned}$$

• (ii) We prove the first equality, the proof of the second one is similar. Take $T:=(T_1,T_2)\in \mathcal {L}(W,Y)\times \mathcal {L}(Z,Y)$, then

  $$\begin{aligned}&(L,y)\in {{\,\mathrm{epi}\,}}\Phi ^*(.,T)\\&\Longleftrightarrow y\in (F+T_1\circ H+T_2\circ G+I_{C})^*(L)\uplus \kappa ^*(T_1)\uplus I^*_{-S}(T_2)+K\quad \text {(by 32)}\\&\Longleftrightarrow \exists U\in \mathcal {P}_p(Y): y\in U\uplus \kappa ^*(T_1)\uplus I^*_{-S}(T_2), \text { and } (L,U\uplus \kappa ^*(T_1)\uplus I^*_{-S}(T_2))\\&\qquad =(L,U)\boxplus (0_\mathcal {L},\kappa ^*(T_1)\uplus I^*_{-S}(T_2))\in {{\,\mathrm{\mathfrak {E} \mathrm{{pi}}}\,}}(F+T_1\circ H+T_2\circ G+I_{C})^*\\&\qquad \quad \boxplus (0_\mathcal {L},\kappa ^*(T_1)\uplus I^*_{-S}(T_2))\\&\Longleftrightarrow (L,y)\in \Psi \Big ({{\,\mathrm{\mathfrak {E} \mathrm{{pi}}}\,}}(F+T_1\!\circ \! H_1+T_2\!\circ \! H_2+I_{C})^*\boxplus (0_\mathcal {L}, \kappa ^*(T_1)\uplus I_{-S}^*(T_2))\Big ), \end{aligned}$$

  which means,

  $$\begin{aligned} {{\,\mathrm{epi}\,}}\Phi ^*(.,T)\! =\! \Psi \Big ({{\,\mathrm{\mathfrak {E} \mathrm{{pi}}}\,}}(F\!+\!T_1\!\circ \! H_1\!+\!T_2\!\circ \! H_2\!+\!I_{C})^*\boxplus (0_\mathcal {L}, \kappa ^*(T_1)\uplus I_{-S}^*(T_2))\Big ).\nonumber \\ \end{aligned}$$

  (37)

  If $T\notin {{\,\mathrm{dom}\,}}\kappa ^*\times \mathcal {L}_+^w(S,K)$, then $\kappa ^*(T_1)=\{+\infty _Y\}$ or $I^*_{-S}(T_2)=\{+\infty _Y\}$ (see (3)), with (32), yields $\Phi _1^*(L,T)=\{+\infty _Y\}$, $L\in \mathcal {L}(X,Y)$. So, $\mathcal {L}_{\Phi _1}\subset {{\,\mathrm{dom}\,}}\kappa ^*\times \mathcal {L}_+^w(S,K),$ and $(\mathrm {ii})$ follows (see (37), ${{\,\mathrm{epi}\,}}\Phi ^*_1(.,T)=\emptyset$ if $T\notin \mathcal {L}_{\Phi _1}$). $\square$