Let F be a setvalued map from a finitely dimensional normed space X to {\mathbb{R}}^{m}. We recall that the epigraph of F with respect to C is defined as the set
epiF:=\{(x,y)\in X\times {\mathbb{R}}^{m}:y\in F(x)+C\}.
The effective domain of F is the set
domF:=\{x\in X:F(x)\ne \mathrm{\varnothing}\}.
F is called convex (resp., closed) with respect to C if epiF is convex (resp., closed) in X\times {\mathbb{R}}^{m}. Sometimes a vector function f:D\subseteq {\mathbb{R}}^{n}\to {\mathbb{R}}^{m} is identified with the setvalued map
F(x):=\{\begin{array}{cc}\{f(x)\},\hfill & x\in D,\hfill \\ \mathrm{\varnothing},\hfill & x\notin D.\hfill \end{array}
Definition 3.1 [[16], Definition 3.1]
Assume that domF\ne \mathrm{\varnothing}. The conjugate map of F, denoted by {F}^{\ast}, is a setvalued map from \mathcal{L}(X,{\mathbb{R}}^{m}) to {\mathbb{R}}^{m} defined as follows.
{F}^{\ast}(A):=Sup\bigcup _{x\in X}[A(x)F(x)],\phantom{\rule{1em}{0ex}}\mathrm{\forall}A\in \mathcal{L}(X,{\mathbb{R}}^{m}),
where \mathcal{L}(X,{\mathbb{R}}^{m}) denotes the space of continuous linear maps from X to {\mathbb{R}}^{m}.
Definition 3.2 [[16], Definition 3.2]
Let F be a setvalued map from {\mathbb{R}}^{n} to {\mathbb{R}}^{m}. Assume that dom{F}^{\ast}\ne \mathrm{\varnothing}. The biconjugate map of F, denoted by {F}^{\ast \ast}, is a setvalued map from {\mathbb{R}}^{n} to {\mathbb{R}}^{m} defined as follows.
{F}^{\ast \ast}(x):=Sup\bigcup _{A\in \mathcal{L}({\mathbb{R}}^{n},{\mathbb{R}}^{m})}[A(x){F}^{\ast}(A)],\phantom{\rule{1em}{0ex}}\mathrm{\forall}x\in {\mathbb{R}}^{n}.
Remark 3.3 Let F be a setvalued map from {\mathbb{R}}^{n} to {\mathbb{R}}^{m} with dom{F}^{\ast}\ne \mathrm{\varnothing}. By identifying x\in {\mathbb{R}}^{n} with the linear map \overline{x}:\mathcal{L}({\mathbb{R}}^{n},{\mathbb{R}}^{m})\to {\mathbb{R}}^{m} defined as follows:
\overline{x}(A):=A(x),\phantom{\rule{1em}{0ex}}\mathrm{\forall}A\in \mathcal{L}({\mathbb{R}}^{n},{\mathbb{R}}^{m}),
we see that {F}^{\ast \ast} is the restriction of {({F}^{\ast})}^{\ast} on {\mathbb{R}}^{n}, i.e.,
{F}^{\ast \ast}={\left({F}^{\ast}\right)}^{\ast}{}_{{\mathbb{R}}^{n}}.
In the rest of this section, we assume that the ordering cone C\subseteq {\mathbb{R}}^{m} is closed, convex, pointed and intC\ne \mathrm{\varnothing}.
Lemma 3.4 [[16], Proposition 3.5]
Let F be a setvalued map from {\mathbb{R}}^{n} to {\mathbb{R}}^{m} with domF\ne \mathrm{\varnothing}. Then

(i)
{F}^{\ast} is closed and convex.

(ii)
If dom{F}^{\ast}\ne \mathrm{\varnothing}, then F(x)\subseteq {F}^{\ast \ast}(x)+C, \mathrm{\forall}x\in {\mathbb{R}}^{n}.
Lemma 3.5 Let F be a setvalued map from {\mathbb{R}}^{n} to {\mathbb{R}}^{m} with dom{F}^{\ast}\ne \mathrm{\varnothing}. Then {F}^{\ast \ast} is closed and convex.
Proof It is immediate from Remark 3.3 and Lemma 3.4. □
Lemma 3.6 [[16], Proposition 3.6]
Let f be a convex vector function from a nonempty convex set D\subseteq {\mathbb{R}}^{n} to {\mathbb{R}}^{m}, and let x\in D, A\in \mathcal{L}({\mathbb{R}}^{n},{\mathbb{R}}^{m}). Then A\in \partial f(x) if and only if
Lemma 3.7 Let f be a convex vector function from a nonempty convex set D\subseteq {\mathbb{R}}^{n} to {\mathbb{R}}^{m}. Then
D\subseteq dom{f}^{\ast \ast}\subseteq \overline{D}.
Proof Let x\in riD be arbitrary. By Lemma 2.7, \partial f(x)\ne \mathrm{\varnothing}. Then, by Lemma 3.6, \partial f(x)\subseteq dom{f}^{\ast}. Consequently, dom{f}^{\ast}\ne \mathrm{\varnothing}. Then, by Lemma 3.4, D\subseteq dom{f}^{\ast \ast}. Now, suppose on the contrary that dom{f}^{\ast \ast}\u2288\overline{D}. Then there is {x}_{0}\in dom{f}^{\ast \ast} such that {x}_{0}\notin \overline{D}. Using the strong separation theorem, one can find \xi \in \mathcal{L}({\mathbb{R}}^{n},\mathbb{R})\setminus \{0\} so that
\xi ({x}_{0})>\underset{x\in \overline{D}}{sup}\xi (x).
(1)
Pick any {y}_{0}\in riD and {A}_{0}\in \partial f({y}_{0}). By Lemma 3.6, {f}^{\ast}({A}_{0}) is a singleton. For each c\in C, we define a linear map {\beta}_{c}:\mathbb{R}\to {\mathbb{R}}^{m} as follows:
{\beta}_{c}(t)=tc\phantom{\rule{1em}{0ex}}(\mathrm{\forall}t\in \mathbb{R}).
By (1) and by Lemma 2.6(ii),
ISup\bigcup _{x\in D}\{({\beta}_{c}\xi )(x)\}=(\underset{x\in D}{sup}\xi (x))c.
Then we have
\begin{array}{rcl}{f}^{\ast}({A}_{0})+(\underset{x\in D}{sup}\xi (x))c& =& {f}^{\ast}({A}_{0})+ISup\bigcup _{x\in D}\{({\beta}_{c}\xi )(x)\}\\ =& Sup\bigcup _{x\in D}\{{A}_{0}(x)f(x)\}+ISup\bigcup _{x\in D}({\beta}_{c}\xi )(x)\\ =& Sup(\bigcup _{x\in D}\{{A}_{0}(x)f(x)\}+\bigcup _{x\in D}\{({\beta}_{c}\xi )(x)\})\\ \text{(by Lemma 2.6(iv))}\\ \subseteq & Sup\bigcup _{x\in D}\{{A}_{0}(x)f(x)+({\beta}_{c}\xi )(x)\}+C\\ \text{(by Lemma 2.6(iv))}\\ =& {f}^{\ast}({A}_{0}+{\beta}_{c}\xi )+C.\end{array}
Then there exists {y}_{c}\in {f}^{\ast}({A}_{0}+{\beta}_{c}\xi ) such that
{f}^{\ast}({A}_{0})+(\underset{x\in D}{sup}\xi (x))c\u2ab0{y}_{c}.
Let z\in {f}^{\ast \ast}({x}_{0}) be arbitrary. From the definition of {f}^{\ast \ast}, one has
\begin{array}{rl}z& \u2ab0({A}_{0}+{\beta}_{c}\xi )({x}_{0}){y}_{c}\\ \u2ab0[{A}_{0}({x}_{0}){f}^{\ast}({A}_{0})]+[\xi ({x}_{0})\underset{x\in D}{sup}\xi (x)].c\phantom{\rule{1em}{0ex}}(\mathrm{\forall}c\in C).\end{array}
By (1), this is impossible since C\ne \{0\} and pointed. Thus, dom{f}^{\ast \ast}\subseteq \overline{D}. The proof is complete. □
Let {x}_{0},x\in {\mathbb{R}}^{n}, {\{{x}_{k}\}}_{k}\subseteq [{x}_{0},x]. Then we write ‘{x}_{k}\uparrow x’ if
\{\begin{array}{c}{x}_{k}\to x,\hfill \\ \parallel {x}_{k+1}{x}_{0}\parallel \ge \parallel {x}_{k}{x}_{0}\parallel \phantom{\rule{1em}{0ex}}(\mathrm{\forall}k).\hfill \end{array}
Lemma 3.8 [[16], Lemma 3.16]
Let f be a convex function from a nonempty convex set D\subseteq {\mathbb{R}}^{n} to {\mathbb{R}}^{m}, and let x\in D. If there exists {x}_{0}\in riD such that
f(x)=\underset{t\uparrow 1}{lim}f(tx+(1t){x}_{0}),
then for every sequence {\{({A}_{k},{x}_{k})\}}_{k}\subseteq \mathcal{L}({\mathbb{R}}^{n},{\mathbb{R}}^{m})\times [{x}_{0},x] such that {x}_{k}\uparrow x and {A}_{k}\in \partial f({x}_{k}), we have
\underset{k\to \mathrm{\infty}}{lim}{A}_{k}(x{x}_{k})=0.
Although biconjugate maps of vector functions have a setvalued structure, under certain conditions, they reduce to singlevalued maps. Such conditions are the convexity and closedness of the functions. Moreover, we have the following theorem.
Theorem 3.9 (Generalized FenchelMoreau theorem) Let f be a vector function from a nonempty convex set D\subseteq {\mathbb{R}}^{n} to {\mathbb{R}}^{m}. Then f is closed and convex if and only if
Proof \underline{\Rightarrow}: Let x\in D be arbitrary. Pick a point {x}_{0}\in riD. By Lemma 2.8, f is continuous relative to [{x}_{0},x]. Hence
f(x)=\underset{t\uparrow 1}{lim}f(tx+(1t){x}_{0}).
(2)
Let {\{{\lambda}_{k}\}}_{k}\subseteq (0,1) be an increasing sequence that converges to 1. Put {x}_{k}={\lambda}_{k}x+(1{\lambda}_{k}){x}_{0}. Then {\{{x}_{k}\}}_{k}\subseteq riD\cap [{x}_{0},x] and {x}_{k}\uparrow x. By Lemma 2.7, \partial f({x}_{k})\ne \mathrm{\varnothing}. For each k, pick {A}_{k}\in \partial f({x}_{k}). By Lemma 3.6, f({x}_{k})={A}_{k}({x}_{k}){f}^{\ast}({A}_{k}). Hence,
\begin{array}{rcl}\parallel f(x)[{A}_{k}(x){f}^{\ast}({A}_{k})]\parallel & =& \parallel f(x)[{A}_{k}({x}_{k}){f}^{\ast}({A}_{k})]+[{A}_{k}({x}_{k}){A}_{k}(x)]\parallel \\ \le & \parallel f(x)f({x}_{k})\parallel +\parallel {A}_{k}({x}_{k}x)\parallel .\end{array}
(3)
Take k\to \mathrm{\infty} in (3), by (2) and by Lemma 3.8, we have
\parallel f(x)[{A}_{k}(x){f}^{\ast}({A}_{k})]\parallel \to 0,
which together with Lemma 3.4(ii) implies
f(x)\in Ub\left(\bigcup _{A\in \mathcal{L}({\mathbb{R}}^{n},{\mathbb{R}}^{m})}[A(x){f}^{\ast}(A)]\right)\cap cl(co\left(\bigcup _{A\in \mathcal{L}({\mathbb{R}}^{n},{\mathbb{R}}^{m})}[A(x){f}^{\ast}(A)]\right)).
Hence, by Lemma 2.6(i), Remark 2.5 and by the definition of biconjugate maps, we have
f(x)=ISup\bigcup _{A\in \mathcal{L}({\mathbb{R}}^{n},{\mathbb{R}}^{m})}[A(x){f}^{\ast}(A)]={f}^{\ast \ast}(x).
(4)
Finally, we shall show that
Indeed, by the proof above, we have dom{f}^{\ast \ast}\supseteq D. Let {x}_{0}\in dom{f}^{\ast \ast} be arbitrary. By Lemma 3.7, {x}_{0}\in \overline{D}. Let {y}_{0}\in {f}^{\ast \ast}({x}_{0}) and x\in riD. Then (x,f(x)),({x}_{0},{y}_{0})\in epi{f}^{\ast \ast}. For every natural number k\ge 1, put
\begin{array}{c}{x}_{k}=\frac{1}{k}x+(1\frac{1}{k}){x}_{0},\hfill \\ {y}_{k}=\frac{1}{k}f(x)+(1\frac{1}{k}){y}_{0}.\hfill \end{array}
Obviously, ({x}_{k},{y}_{k})\to ({x}_{0},{y}_{0}) and ({x}_{k},{y}_{k})\in epi{f}^{\ast \ast}, ∀k, since {f}^{\ast \ast} is convex. By (4), f({x}_{k})={f}^{\ast \ast}({x}_{k}) since {x}_{k}\in D. Hence, ({x}_{k},{y}_{k})\in epif (∀k). This fact together with closedness of f implies
({x}_{0},{y}_{0})\in epif.
Hence {x}_{0}\in D. Thus, dom{f}^{\ast \ast}=D and then f={f}^{\ast \ast}.
\underline{\Leftarrow}: It is immediate from Lemma 3.5. The theorem is proved. □
When m=1 and C={\mathbb{R}}_{+}, Theorem 3.9 is the famous FenchelMoreau theorem in convex analysis.